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Solution manual for introduction to electric-circuits

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• Solution Manual to accompany Introduction to Electric Circuits, 6e By R. C. Dorf and J. A. Svoboda 1
• Table of Contents Chapter 1 Electric Circuit Variables Chapter 2 Circuit Elements Chapter 3 Resistive Circuits Chapter 4 Methods of Analysis of Resistive Circuits Chapter 5 Circuit Theorems Chapter 6 The Operational Amplifier Chapter 7 Energy Storage Elements Chapter 8 The Complete Response of RL and RC Circuits Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements Chapter 10 Sinusoidal Steady-State Analysis Chapter 11 AC Steady-State Power Chapter 12 Three-Phase Circuits Chapter 13 Frequency Response Chapter 14 The Laplace Transform Chapter 15 Fourier Series and Fourier Transform Chapter 16 Filter Circuits Chapter 17 Two-Port and Three-Port Networks 2
• Errata for Introduction to Electric Circuits, 6th Edition Errata for Introduction to Electric Circuits, 6th Edition Page 18, voltage reference direction should be + on the right in part B: Page 28, caption for Figure 2.3-1: "current" instead of "cuurent" Page 41, line 2: "voltage or current" instead of "voltage or circuit" Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit. Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..." Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources, then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read: "Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab. Then Rt = Vab/1." Page 340, Problem P8.3-5: The answer should be Page 340, Problem P8.3-6: The answer should be Page 341, Problem P.8.4-1: The answer should be Page 546, line 4: The angle is instead of . Page 554, Problem 12.4.1 Missing parenthesis: Page 687, Equation 15.5-2: Partial t in exponent: http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM . .
• Errata for Introduction to Electric Circuits, 6th Edition Page 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2 (s) and Hc(s) = V1(s) / Vs(s). http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM
• Chapter 1 – Electric Circuit Variables Exercises Ex. 1.3-1 i (t ) = 8 t 2 − 4 t A q(t ) = ∫ t 0 i dτ + q(0) = ∫ t 0 t 8 8 (8τ 2 − 4τ ) dτ + 0 = τ 3 −2τ 2 = t 3 − 2 t 2 C 0 3 3 Ex. 1.3-3 t t 0 0 q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 3τ dτ + 0 = − 4 4 4 t cos 3τ 0 = − cos 3 t + C 3 3 3 Ex. 1.3-4 dq ( t ) i (t ) = dt 0  i (t ) = 2  −2( t − 2 ) −2e t <0 0< t < 2 t >2 Ex. 1.4-1 i1 = 45 µA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A Ex. 1.4-2 ∆ q = i∆ t = ( 4000 A )( 0.001 s ) = 4 C Ex. 1.4-3 i= ∆ q 45 × 10−9 = = 9 × 10−6 = 9 µA −3 ∆t 5 × 10 Ex. 1.4-4 electron   C   −19 i = 10 billion  1.602 ×10 electron  = s    C   9 electron   −19 10×10  1.602 × 10 electron  s    electron C = 1010 × 1.602 ×10−19 electron s C = 1.602 × 10−9 = 1.602 nA s 1-1
• Ex. 1.6-1 (a) The element voltage and current do not adhere to the passive convention in Figures 1.6-1B and 1.6-1C so the product of the element voltage and current is the power supplied by these elements. (b) The element voltage and current adhere to the passive convention in Figures 1.6-1A and 1.6-1D so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) The element voltage and current do not adhere to the passive convention in Figure 1.6-1B, so the product of the element voltage and current is the power delivered by this element: (2 V)(6 A) = 12 W. The power received by the element is the negative of the power delivered by the element, -12 W. (d) The element voltage and current do not adhere to the passive convention in Figure 1.6-1B, so the product of the element voltage and current is the power supplied by this element: (2 V)(6 A) = 12 W. (e) The element voltage and current adhere to the passive convention in Figure 1.6-1D, so the product of the element voltage and current is the power delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the element is the negative of the power delivered to the element, -12 W. Problems Section 1-3 Electric Circuits and Current Flow P1.3-1 i (t ) = d 4 1 − e −5t = 20 e −5t A dt ( ) P1.3-2 t t t t 4 4 q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 1 − e −5τ dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C 0 0 0 0 5 5 ( ) P1.3-3 t t −∞ t −∞ q ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0. t q ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2 τ 2 = 2 t − 4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C. 2 t 2 t 4 t 4 t 8 t 8 q ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8 − t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C. t q ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t . 1-2
• P1.3-4 i = 600 A = 600 C s C s mg Silver deposited = 600 ×20 min×60 = 8.05×105 mg=805 g ×1.118 s min C Section 1-6 Power and Energy P1.6-1 a.) q = ∫ i dt = i∆t = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10 4 C b.) P = v i = (110 V )(10 A ) = 1100 W c.) Cost = P1.6-2 0.06\$ × 1.1kW × 2 hrs = 0.132 \$ kWhr P = ( 6 V )(10 mA ) = 0.06 W ∆t = 200 W⋅s ∆w = = 3.33×103 s 0.06 W P P1.6-3 for 0 ≤ t ≤ 10 s: v = 30 V and i = 30 t = 2t A ∴ P = 30(2t ) = 60t W 15 25 t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V 5 v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W for 10 ≤ t ≤ 15 s: v ( t ) = − 30 t +b A 10 ⇒ b = 75 ⇒ i (t ) = −3t + 75 A for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = − i (25) = 0 ∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W 1-3
• Energy = ∫ P dt = = 30t 2 10 0 ∫ 10 0 60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt 15 15 25 25 + 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J 3 10 2 15 P1.6-4 a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). 5( 3600 ) t 5 ( 3600 )  0.5 τ  0.5 2 2 11 + w = ∫ Pdt = ∫0 vi dτ = ∫0  dτ = 22 t + 3600 τ 3600   0 = 441× 103 J = 441 kJ b.) Cost = 441kJ × 1 hr 10¢ × = 1.23¢ 3600s kWhr P1.6-5 1 1 ( cos 3 t )( sin 3 t ) = sin 6 t 3 6 1 p ( 0.5 ) = sin 3 = 0.0235 W 6 1 p (1) = sin 6 = −0.0466 W 6 p (t ) = 1-4
• Here is a MATLAB program to plot p(t): clear t0=0; tf=2; dt=0.02; t=t0:dt:tf; % % % % initial time final time time increment time v=4*cos(3*t); i=(1/12)*sin(3*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power plot(t,p) xlabel('time, s'); ylabel('power, W') P1.6-6 p ( t ) = 16 ( sin 3 t )( sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t W Here is a MATLAB program to plot p(t): clear t0=0; tf=2; dt=0.02; t=t0:dt:tf; % % % % initial time final time time increment time v=8*sin(3*t); i=2*sin(3*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power plot(t,p) xlabel('time, s'); ylabel('power, W') 1-5
• P1.6-7 ( ) ( ) p ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e −2t Here is a MATLAB program to plot p(t): clear t0=0; tf=2; dt=0.02; t=t0:dt:tf; % % % % initial time final time time increment time v=4*(1-exp(-2*t)); i=2*exp(-2*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power plot(t,p) xlabel('time, s'); ylabel('power, W') P1.6-8 P = V I =3 × 0.2=0.6 W w = P ⋅ t = 0.6 × 5 × 60=180 J 1-6
• Verification Problems VP 1-1 Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are: (-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W =0W The element voltages and currents satisfy conservation of energy and may be correct. VP 1-2 Notice that the element voltage and current of some branches do not adhere to the passive convention. The sum of the powers absorbed by each branch are: -(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A) = -9 W + 6 W + 6 W + 12 W + 9 W -12 W ≠0W The element voltages and currents do not satisfy conservation of energy and cannot be correct. Design Problems DP 1-1 The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25) = 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust the estimates of the maximum voltage and current and a Grade A device otherwise. 1-7
• DP1-2 ( ) ( ) p ( t ) = 20 1 − e −8 t × 0.03 e −8 t = 0.6 1 − e−8t e−8t Here is a MATLAB program to plot p(t): clear t0=0; tf=1; dt=0.02; t=t0:dt:tf; v=20*(1-exp(-8*t)); i=.030*exp(-8*t); for k=1:length(t) p(k)=v(k)*i(k); end % % % % initial time final time time increment time % device voltage % device current % power plot(t,p) xlabel('time, s'); ylabel('power, W') Here is the plot: The circuit element must be able to absorb 0.15 W. 1-8
• Chapter 2 - Circuit Elements Exercises Ex. 2.3-1 m ( i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied Therefore the element is linear. Ex. 2.3-2 m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied Therefore the element is not linear. Ex. 2.5-1 v 2 (10 ) P= = =1 W R 100 2 Ex. 2.5-2 P= v 2 (10 cos t ) 2 = = 10 cos 2 t W R 10 Ex. 2.8-1 ic = − 1.2 A, v d = 24 id = 4 ( − 1.2) = − 4.8 V A id and vd adhere to the passive convention so P = vd id = (24) (−4.8) = −115.2 W is the power received by the dependent source 2-1
• Ex. 2.8-2 vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V id and vd adhere to the passive convention so P = vd id = (2.2) (−8) = −17.6 W is the power received by the dependent source. The power supplied by the dependent source is 17.6 W. Ex. 2.8-3 ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A id and vd adhere to the passive convention so P = vd id = (2.5) (1.75) = 4.375 W is the power received by the dependent source. 2-2
• Ex. 2.9-1 θ = 45° , I = 2 mA, R p = 20 kΩ a= θ 45 (20 kΩ) = 2.5 kΩ ⇒ aR = p 360 360 vm = (2 ×10−3 )(2.5 ×103 ) = 5 V Ex. 2.9-2 v = 10 V, i = 280 µA, k = 1 µA °K for AD590  °K  i  = 280° K i = kT ⇒ T = = (280µA)1  µA  k   Ex. 2.10-1 At t = 4 s both switches are open, so i = 0 A. Ex. 2.10.2 At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V . At t = 6 s the switch is in the down position, so v = 0 V. Problems Section 2-3 Engineering and Linear Models P2.3-1 The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P2.3-2 (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed linear. (b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV 4 (c) When v = 4 V, i = = 33 A = 33 A. 0.12 2-3
• P2.3-3 (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is v = 256.5i . The element is indeed linear. (b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V 12 (c) When v = 12 V, i = = 0.04678 A = 46.78 mA. 256.5 P2.3-4 Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence, the property of homogeneity is not satisfied. The element is not linear. Section 2-5 Resistors P2.5-1 i = is = 3 A and v = Ri = 7 × 3 = 21 V v and i adhere to the passive convention ∴ P = v i = 21 × 3 = 63 W is the power absorbed by the resistor. P2.5-2 i = is = 3 mA and v = 24 V v 24 = = 8000 = 8 k Ω i .003 P = (3×10 −3 )× 24 = 72×10 −3 = 72 mW R = P2.5-3 v = vs =10 V and R = 5 Ω v 10 = =2 A R 5 v and i adhere to the passive convention ∴ p = v i = 2⋅10 = 20 W is the power absorbed by the resistor i = 2-4
• P2.5-4 v = vs = 24 V and i = 2 A v 24 = = 12 Ω i 2 p = vi = 24⋅2 = 48 W R= P2.5-5 v1 = v 2 = vs = 150 V; R1 = 50 Ω; R2 = 25 Ω v 1 and i1 adhere to the passive convention so v 1 150 = =3 A R 1 50 v 150 v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = − = −6 A R2 25 i1 = The power absorbed by R1 is P = v1 i1 = 150 ⋅ 3 = 450 W 1 The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W P2.5-6 i1 = i 2 = is = 2 A ; R1 =4 Ω and R2 = 8 Ω v 1 and i 1 do not adhere to the passive convention so v 1 =− R 1 i 1 =−4⋅2=−8 V. The power absorbed by R 1 is P1 =−v 1i 1 =−(−8)(2) = 16 W. v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V . The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W. P2.5-7 Model the heater as a resistor, then (250) 2 v2 v2 ⇒ R = = = 62.5 Ω 1000 R P v 2 (210) 2 = 705.6 W with a 210 V source: P = = R 62.5 with a 250 V source: P = 2-5
• P2.5-8 The current required by the mine lights is: i = P 5000 125 A = = 3 v 120 Power loss in the wire is : i 2 R Thus the maximum resistance of the copper wire allowed is 0.05P 0.05×5000 = = 0.144 Ω (125/3) 2 i2 now since the length of the wire is L = 2×100 = 200 m = 20,000 cm R= thus R = ρ L / A with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1 A= ρL R = 1.7×10−6 ×20,000 = 0.236 cm 2 0.144 Section 2-6 Independent Sources P2.6-1 v s 15 2 = = 3 A and P = R i 2 = 5 ( 3 ) = 45 W R 5 (b) i and P do not depend on is . (a) i = The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A. P2.6-2 v 2 102 = 20 W (a) v = R i s = 5 ⋅ 2 = 10 V and P = = R 5 (b) v and P do not depend on v s . The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V 2-6
• P2.6-3 Consider the current source: i s and v s do not adhere to the passive convention, so Pcs =i s v s =3⋅12 = 36 W is the power supplied by the current source. Consider the voltage source: i s and v s do adhere to the passive convention, so Pvs = i s vs =3 ⋅12 = 36 W is the power absorbed by the voltage source. ∴ The voltage source supplies −36 W. P2.6-4 Consider the current source: i s and vs adhere to the passive convention so Pcs = i s vs =3 ⋅12 = 36 W is the power absorbed by the current source. Current source supplies − 36 W. Consider the voltage source: i s and vs do not adhere to the passive convention so Pvs = i s vs = 3 ⋅12 =36 W is the power supplied by the voltage source. P2.6-5 (a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW 1 1 1  (b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t  = 10 + 5 sin 2 mJ 2 4 0 1 1 2 2-7
• Section 2-7 Voltmeters and Ammeters P2.7-1 (a) R = v 5 = = 10 Ω i 0.5 (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source adhere to the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers -6 W. P2.7-2 The voltmeter current is zero so the ammeter current is equal to the current source current except for the reference direction: i = -2 A The voltage v is the voltage of the current source. The power supplied by the current source is 40 W so 40 = 2 v ⇒ v = 20 V 2-8
• Section 2-8 Dependent Sources P2.8-1 r = vb 8 = =4 Ω ia 2 P2.8-2 vb = 8 V ; g v b = i a = 2 A ; g = ia 2 A = = 0.25 vb 8 V i b = 8 A ; d i b = i a = 32A ; d = i a 32 A = =4 ib 8 A va = 2 V ; b va = vb = 8 V ; b = vb 8 V = =4 va 2 V P2.8-3 P2.8-4 Section 2-9 Transducers P2.9-1 a= θ = θ 360 , θ = 360 vm Rp I (360)(23V) = 75.27° (100 kΩ)(1.1 mA) P2.9-2 AD590 : k =1 µA , K v =20 V (voltage condition satisfied) ° 4 µ A < i < 13 µ A    i T =  k  ⇒ 4 ° K< T <13° K 2-9
• Section 2-10 Switches P2.10-1 At t = 1 s the left switch is open and the right switch is closed so the voltage across the resistor is 10 V. i= v 10 = = 2 mA R 5×103 At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 15 V. i= v 15 = = 3 mA R 5×103 P2.10-2 At t = 1 s the current in the resistor is 3 mA so v = 15 V. At t = 4 s the current in the resistor is 0 A so v = 0 V. Verification Problems VP2-1 vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .) So vo 40 V = = 20 2 A is Your lab partner is wrong. VP2-2 vs 12 = 0.48 A. The power absorbed by = R 25 this resistor will be P = i vs = (0.48) (12) = 5.76 W. We expect the resistor current to be i = A half watt resistor can't absorb this much power. You should not try another resistor. 2-10
• Design Problems DP2-1 1.) 10 10 > 0.04 ⇒ R < = 250 Ω R 0.04 2.) 102 1 < ⇒ R > 200 Ω R 2 Therefore 200 < R < 250 Ω. For example, R = 225 Ω. DP2-2 1.) 2 R > 40 ⇒ R > 20 Ω 15 2.) 2 2 R < 15 ⇒ R < = 3.75 Ω 4 Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously. DP2-3 P = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W 1 2 2 P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W 2 2 P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W 2 2 2-11
• Chapter 3 – Resistive Circuits Exercises Ex 3.3-1 Apply KCL at node a to get 2 + 1 + i3 = 0 ⇒ i3 = -3 A Apply KCL at node c to get 2 + 1 = i4 ⇒ i4 = 3 A Apply KCL at node b to get i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A Apply KVL to the loop consisting of elements A and B to get -v2 – 3 = 0 ⇒ v2 = -3 V Apply KVL to the loop consisting of elements C, E, D, and A to get 3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V Apply KVL to the loop consisting of elements E and F to get v6 – 6 = 0 ⇒ v6 = 6 V Check: The sum of the power supplied by all branches is -(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0 3-1
• Ex 3.3-2 Apply KCL at node a to determine the current in the horizontal resistor as shown. Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A Ex 3.3-3 −18 + 0 − 12 − va = 0 ⇒ va = −30 V and im = Ex 3.3-4 −va − 10 + 4va − 8 = 0 ⇒ va = 2 va + 3 ⇒ im = 9 A 5 18 = 6 V and vm = 4 va = 24 V 3 Ex 3.4-1 From voltage division  3   = 3V  3+9  v3 = 12  then v i = 3 = 1A 3 The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W The power supplied by the source is (12)(1) = 12 W. 3-2
• Ex 3.4-2 P = 6 W and R1 = 6 Ω i2 = P 6 = = 1 or i =1 A R1 6 v0 = i R1 =(1) (6)=6V from KVL: − v + i (2 + 4 + 6 + 2) = 0 s ⇒ v = 14 i = 14 V s Ex 3.4-3 25 From voltage division ⇒ v = (8) = 2 V m 25+75 Ex 3.4-4 25 From voltage division ⇒ v = ( −8 ) = −2 V m 25+75 Ex. 3.5-1 1 R eq = 1 1 1 1 4 + 3+ 3+ 3= 3 3 10 10 10 10 10 ⇒ R eq = By current division, the current in each resistor = 103 1 = kΩ 4 4 1 -3 1 (10 ) = mA 4 4 Ex 3.5-2 10 From current division ⇒ i = ( −5 ) = − 1 A m 10+40 3-3
• Problems Section 3-3 Kirchoff’s Laws P3.3-1 Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 A The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is power received by element B. The power supplied by element B is 12 W. Apply KVL to the loop consisting of elements D, F, E, and C to get 4 + v + (-5) – 12 = 0 ⇒ v = 13 V The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W is the power supplied by element F. Check: The sum of the power supplied by all branches is -(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0 3-4
• P3.3-2 Apply KCL at node a to get 2 = i2 + 6 = 0 ⇒ i2 = -4 A Apply KCL at node b to get 3 = i4 + 6 ⇒ i4 = -3 A Apply KVL to the loop consisting of elements A and B to get -v2 – 6 = 0 ⇒ v2 = -6 V Apply KVL to the loop consisting of elements C, D, and A to get -v3 – (-2) – 6 = 0 ⇒ v4 = -4 V Apply KVL to the loop consisting of elements E, F and D to get 4 – v6 + (-2) = 0 ⇒ v6 = 2 V Check: The sum of the power supplied by all branches is -(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0 3-5
• P3.3-3 KVL : −12 − R 2 (3) + v = 0 (outside loop) v = 12 + 3R 2 or R 2 = KCL i+ 12 − 3 = 0 (top node) R1 i = 3− (a) v − 12 3 12 12 or R1 = 3−i R1 v = 12 + 3 ( 3) = 21 V i = 3− (b) R2 = 12 =1 A 6 2 − 12 10 12 = − Ω ; R1 = =8Ω 3 3 3 − 1.5 (checked using LNAP 8/16/02) (c) 24 = − 12 i, because 12 and i adhere to the passive convention. 12 = 2.4 Ω 3+ 2 because 3 and v do not adhere to the passive convention ∴ i = − 2 A and R1 = 9 = 3v, ∴ v = 3V and R 2 = 3 − 12 = −3 Ω 3 The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative. 3-6
• P3.3-4 12 =2A i = 1 6 20 = 5A i = 2 4 i = 3−i = − 2 A 3 2 i = i +i = 3A 4 2 3 Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W 2 Power absorbed by the 6 Ω resistor = 6 ⋅ i 2 = 24 W 1 Power absorbed by the 8 Ω resistor = 8 ⋅ i 2 = 72 W 4 P3.3-5 (checked using LNAP 8/16/02) v1 = 8 V v2 = −8 + 8 + 12 = 12 V (checked using LNAP 8/16/02) v3 = 2⋅ 4 = 8 V v2 4Ω : P = 3 = 16 W 4 2 v2 = 24 W 6Ω : P = 6 v2 8Ω : P = 1 = 8 W 8 P3.3-6 P2 mA = − 3 × ( 2 ×10−3 )  = −6 × 10−3 = −6 mW   P1 mA = −  −7 × (1× 10−3 )  = 7 × 10−3 = 7 mW   (checked using LNAP 8/16/02) 3-7
• P3.3-7 P2 V = +  2 × (1× 10−3 )  = 2 × 10−3 = 2 mW   −3 P3 V = + 3 × ( −2 × 10 )  = −6 × 10−3 = −6 mW   (checked using LNAP 8/16/02) P3.3-8 KCL: iR = 2 + 1 ⇒ iR = 3 A KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V ∴ R= vR 12 = =4Ω iR 3 (checked using LNAP 8/16/02) P3.3-9 KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V KCL: iR + 8 = 0 ⇒ iR = −8 A ∴ R= vR −80 = = 10 Ω iR −8 (checked using LNAP 8/16/02) 3-8
• P3.3-10 KCL at node b: 5.61 3.71 − 5.61 12 − 5.61 −1.9 = + ⇒ 0.801 = + 1.278 7 R1 5 R1 ⇒ R1 = KCL at node a: 1.9 = 3.983 ≈ 4 Ω 1.278 − 0.801 3.71 3.71 − 5.61 3.71 − 12 −8.29 + + = 0 ⇒ 1.855 + ( −0.475 ) + =0 2 4 R2 R2 ⇒ R2 = 8.29 = 6.007 ≈ 6 Ω 1.855 − 0.475 (checked using LNAP 8/16/02) 3-9
• Section 3-4 A Single-Loop Circuit – The Voltage Divider P3.4-1 6 6 12 = 12 = 4 V v = 1 6+3+5+ 4 18 3 5 10 12 = 2 V ; v = 12 = V v = 2 18 3 18 3 4 8 v = 12 = V 4 18 3 (checked using LNAP 8/16/02) P3.4-2 (a) R = 6 + 3 + 2 + 4 = 15 Ω 28 28 = = 1.867 A (b) i = R 15 ( c ) p = 28 ⋅ i =28(1.867)=52.27 W (28 V and i do not adhere to the passive convention.) (checked using LNAP 8/16/02) 3-10
• P3.4-3 i R2 = v = 8 V 12 = i R1 + v = i R1 + 8 ⇒ 4 = i R1 8 8 4 4 ⋅ 100 ; R1 = = = = 50 Ω 8 R 2 100 i 4 4 8 8 ⋅ 100 ; R2 = = = 200 Ω (b) i = = 4 R1 100 i 4 8 ( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω i i (a) i= (checked using LNAP 8/16/02) P3.4-4 Voltage division 16 v1 = 12 = 8 V 16 + 8 4 v3 = 12 = 4 V 4+8 KVL: v3 − v − v1 = 0 v = −4 V (checked using LNAP 8/16/02) P3.4-5 v  ⇒ R = 50  s − 1 v   o  with v = 20 V and v > 9 V, R < 61.1 Ω  s 0   R = 60 Ω with v = 28 V and v < 13 V, R > 57.7 Ω  s 0   100  v using voltage divider: v =  0  100 + 2 R  s  3-11
• P3.4-6  240  a.)   18 = 12 V  120 + 240  18   b.) 18   = 0.9 W  120 + 240   R  c.)   18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω  R + 120  R d.) 0.2 = ⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω R + 120 (checked using LNAP 8/16/02) 3-12
• Section 3-5 Parallel Resistors and Current Division P3.5-1 i = 1 i = 2 i = 3 i = 4 1 1 1 6 4= 4= A 1 + 1 + 1 +1 1+ 2 + 3 + 6 3 6 3 2 1 1 2 3 4 = A; 1 + 1 + 1 +1 3 6 3 2 1 1 2 4 =1 A 1 + 1 + 1 +1 6 3 2 1 1 4=2 A 1 + 1 + 1 +1 6 3 2 P3.5-2 (a) (b) (c) 1 1 1 1 1 = + + = ⇒ R = 2Ω R 6 12 4 2 v = 6 ⋅ 2 = 12 V p = 6 ⋅12 = 72 W P3.5-3 i= 8 8 or R1 = R1 i 8 = R 2 (2 − i ) ⇒ i = 2 − (a) (b) 8 8 or R 2 = 2−i R2 8 4 8 = A ; R1 = =6Ω 4 12 3 3 8 2 8 i = = A ; R2 = =6Ω 2 12 3 2− 3 i = 2− 3-13
• ( c ) R1 = R 2 2 ⋅ R1 R 2 R1 + R 2 1 will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A. 2 1 = 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω 2 P3.5-4 Current division: 8 i = −6 = −2 A 1 16 + 8 ( ) 8 i = −6 = −3 A 2 8+8( ) i = i −i 1 2 = +1 A P3.5-5   R 1  i and current division: i =  2 R + R  s  1 2 Ohm's Law: v = i R yields o 2 2  v  R + R  2 i =  o  1 s  R  R   2  1  plugging in R = 4Ω, v > 9 V o 1 and R = 6Ω, v < 13 V gives o 1 So any 3.15 A < i < 3.47 A s gives i > 3.15 A s i < 3.47 A s keeps 9 V < v < 13 V. o 3-14
• P3.5-6  24  a)   1.8 = 1.2 A  12 + 24   R  b)   2 = 1.6 ⇒ 2 R = 1.6 R + 1.6 (12 ) ⇒ R = 48 Ω  R + 12  R c) 0.4 = ⇒ ( 0.4 )(12 ) = 0.6 R ⇒ R = 8 Ω R + 12 Section 3-7 Circuit Analysis P3.7-1 (a) (b) (c) 48 ⋅ 24 = 32 Ω 48 + 24 32 ⋅ 32 v = 32 + 32 24 = 16 V ; 32 ⋅ 32 8+ 32 + 32 16 1 = A i= 32 2 48 1 1 ⋅ = A i2 = 48 + 24 2 3 R = 16 + 3-15
• P3.7-2 3⋅ 6 =6Ω 3+ 6 1 1 1 = + + ⇒ R p = 2.4 Ω then 12 6 6 (a) R1 = 4 + (b) 1 Rp R 2 = 8 + R p = 10.4 Ω (c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0 ⇒ −24+6 (i1 −2)+10.4i1 = 0 36 =2.2 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.88 V ⇒ i1 = 16.4 1 6 (d ) i2 = ( 2.2 ) = 0.878 A, 1 1 1 + + 6 6 12 v2 = ( 0.878 ) (6) = 5.3 V (e) i3 = 6 2 i2 = 0.585 A ⇒ P = 3 i3 = 1.03 W 3+ 6 3-16
• P3.7-3 Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with a 2 Ω resistor by the equivalent 1 Ω resistor This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors: i1 = 1+1 (1.5 ) = 0.75 A 2 + (1 + 1) 3-17
• P3.7-4 (a) 1 1 1 1 = + + ⇒ R2 24 12 8 R2 = 4 Ω and R1 = (10 + 8) ⋅ 9 = 6Ω 10 + 8 + 9 b g (b) First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next, apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A . (c) 1 8 2.25 = 1125 A . i2 = 1 1 1 + + 24 8 12 b gLM b10 +98g + 9 3OP = −10 V Q N and v1 = − 10 3-18
• P3.7-5 30 v1 = 6 ⇒ v1 = 8 V 10 + 30 R2 12 = 8 ⇒ R2 = 20 Ω R2 + 10 20 = b g R1 10 + 30 R1 + 10 + 30 b g ⇒ R1 = 40 Ω Alternate values that can be used to change the numbers in this problem: meter reading, V 6 4 4 4.8 Right-most resistor, Ω 30 30 20 20 R1, Ω 40 10 15 30 3-19
• P3.7-6 P3.7-7 1× 10−3 = 24 12 ×103 + R p ⇒ R p = 12 ×103 = 12 kΩ ( 21×10 ) R = ( 21×10 ) + R 3 12 × 10 = R p 3 3 ⇒ R = 28 kΩ P3.7-8   130 500 = 15.963 V Voltage division ⇒ v = 50   130 500 + 200 + 20      100   10  ∴v = v   = (15.963)   = 12.279 V h  100 + 30   13  v ∴ i = h = .12279 A h 100 3-20
• P3.7-9 3-21
• P3.7-10 Req = 15 ( 20 + 10 ) = 10 Ω 15 + ( 20 + 10 ) ia = − 60  30   60   20  = −6 A, ib =  = 4 A, vc =    ( −60 ) = −40 V R   Req  30 + 15   eq   20 + 10  P3.7-11 a) Req = 24 12 = b) (24)(12) =8Ω 24 + 12 from voltage division: 100 5  20  100 v = 40  V∴ i = 3 = A = x x 20 3 3  20 + 4  from current division: i = i 5  8  A = x 8+8 6   3-22
• P3.7-12 9 + 10 + 17 = 36 Ω a.) b.) 36 (18 ) = 12 Ω 36+18 36 R = 18 ⇒ 18 R = (18 )( 36 ) ⇒ R = 36 Ω 36+R P3.7-13 2 R( R ) 2 = R 2R + R 3 v 2 240 Pdeliv. = = =1920 W Req 2 R to ckt 3 Thus R =45 Ω Req = P3.7-14 R = 2 + 1 + ( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω eq ∴i = 40 40 = =5 A Req 8 ( )  6  5 1 i1 = i   = ( 5) 3 = 3 A  6 + 12   2  5 1 i2 = i   = ( 5) 2 = 2 A  2+2 from current division ( ) 3-23
• Verification Problems VP3-1 KCL at node a: i = i + i 3 1 2 − 1.167 = − 0.833 + ( −0.333) − 1.167= − 1.166 OK KVL loop consisting of the vertical 6 Ω resistor, the 3 Ω and4Ω resistors, and the voltage source: 6i + 3i + v + 12 = 0 3 2 yields v = −4.0 V not v = −2.0 V VP3-2 reduce circuit: 5+5=10 in parallel with 20 Ω gives 6.67Ω  6.67  by current division: i =   5 = 1.25 A  20 + 6.67  ∴Reported value was correct. VP3-3 320   v = ( 24 ) = 6.4 V o  320 + 650 + 230   ∴Reported value was incorrect. 3-24
• VP3-4 KVL bottom loop: − 14 + 0.1iA + 1.2iH = 0 KVL right loop: − 12 + 0.05iB + 1.2iH = 0 KCL at left node: iA + iB = iH This alone shows the reported results were incorrect. Solving the three above equations yields: iA = 16.8 A iH = 10.3 A iB = −6.49 A ∴ Reported values were incorrect. VP3-5 1   Top mesh: 0 = 4 i a + 4 i a + 2  i a + − i b  = 10 ( −0.5 ) + 1 − 2 ( −2 ) 2   Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V The KVL equations are satisfied so the analysis is correct. 3-25
• VP3-6 Apply KCL at nodes b and c to get: KCL equations: Node e: −1 + 6 = 0.5 + 4.5 Node a: 0.5 + i c = −1 ⇒ i c = −1.5 mA Node d: i c + 4 = 4.5 ⇒ i c = 0.5 mA That's a contradiction. The given values of ia and ib are not correct. Design Problems DP3-1 Using voltage division: vm = R 2 + aR p R1 + (1 − a ) R p + R 2 + aR p 24 = R 2 + aR p R1 + R 2 + R p 24 vm = 8 V when a = 0 ⇒ R2 R1 + R 2 + R p vm = 12 V when a = 1 ⇒ R2 + R p R1 + R 2 + R p The specification on the power of the voltage source indicates 242 1 ≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω R1 + R 2 + R p 2 = 1 3 = 1 2 Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives 3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives R1 = 6000 Ω and R 2 = 4000 Ω . With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate 24 mW, 16 mW and 8 mW respectively. Therefore the design is complete. 3-26
• DP3-2 Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage 200 division, 12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2 R 2 + 200 82 1 = 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the is 400 8 series combination of two 200 Ω resistors. The power required to be dissipated by each of these 42 1 = 0.08 W < W . resistors is 200 8 Now let’s check the voltage: 190 210 11.88 < v < 12.12 0 190 + 420 210 + 380 3.700 < v0 < 4.314 4 − 7.5% < v0 < 4 + 7.85% Hence, vo = 4 V ± 8% and the design is complete. DP3-3 Vab ≅ 200 mV 10 10 120 Vab = (120) (0.2) 10 + R 10 + R 240 ⇒ R=5Ω let v = 16 = 10 + R 162 = 25.6W ∴P= 10 v= DP3-4 N 1 N 1 = N  i = G v = v where G = ∑ T T R  R n = 1 Rn ∴N= iR ( 9 )(12 ) = = 18 bulbs v 6 3-27
• 28
• Chapter 4 – Methods of Analysis of Resistive Circuits Exercises Ex. 4.3-1 v −v a a b + + 3 = 0 ⇒ 5 v − 3 v = −18 a b 3 2 v KCL at a: KCL at b: v −v b a − 3 −1 = 0 ⇒ v − v = 8 b a 2 Solving these equations gives: va = 3 V and vb = 11 V Ex. 4.3-2 KCL at a: v −v a a b + + 3 = 0 ⇒ 3 v − 2 v = −12 a b 4 2 v v −v b a b − −4=0 3 2 ⇒ − 3 v + 5 v = 24 a b v KCL at a: Solving: va = −4/3 V and vb = 4 V Ex. 4.4-1 Apply KCL to the supernode to get v + 10 v 2+ b + b =5 20 30 Solving: v = 30 V and v = v + 10 = 40 V b a b 4-1
• Ex. 4.4-2 ( vb + 8) − ( −12) + vb = 3 10 40 ⇒ v = 8 V and v = 16 V b a Ex. 4.5-1 Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into vb = 4 ia and solve for vb . 6 vb + =i 8 12 a  9 + vb  ⇒ v = 4i = 4   ⇒ v = 4.5 V b a b  12    Ex. 4.5-2 The controlling voltage of the dependent source is a node voltage so it is already expressed as a function of the node voltages. Apply KCL at node a. v −6 v −4v a a = 0 ⇒ v = −2 V + a a 20 15 Ex. 4.6-1 Mesh equations: −12 + 6 i + 3  i − i  − 8 = 0 ⇒ 9 i − 3 i = 20  1 2 1 1 2   8 − 3  i − i  + 6 i = 0 ⇒ − 3 i + 9 i = −8  1 2 2 1 2   Solving these equations gives: 13 1 i = A and i = − A 1 6 2 6 The voltage measured by the meter is 6 i2 = −1 V. 4-2
• Ex. 4.7-1  3 Mesh equation: 9 + 3 i + 2 i + 4  i +  = 0 ⇒  4 The voltmeter measures 3 i = −4 V ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i= −12 A 9 Ex. 4.7-2 Mesh equation: 15 + 3 i + 6 ( i + 3) = 0 ⇒ ( 3 + 6 ) i = −15 − 6 ( 3) ⇒ i= −33 2 = −3 A 9 3 Ex. 4.7-3 3 3 = i1 − i 2 ⇒ i1 = + i 2 . 4 4 3  Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4  + i 2  + 5 i 2 = 9 ⇒ 9 i 2 = 6 4  2 4 so i 2 = A and the voltmeter reading is 2 i 2 = V 3 3 Express the current source current in terms of the mesh currents: 4-3
• Ex. 4.7-4 Express the current source current in terms of the mesh currents: 3 = i1 − i 2 ⇒ i1 = 3 + i 2 . Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3 Finally, i 2 = − 1 A is the current measured by the ammeter. 3 Problems Section 4-3 Node Voltage Analysis of Circuits with Current Sources P4.3-1 KCL at node 1: 0= v −v −4 − 4 − 2 1 1 2 + +i = + + i = −1.5 + i ⇒ i = 1.5 A 8 6 8 6 v (checked using LNAP 8/13/02) 4-4
• P4.3-2 KCL at node 1: v −v v 1 2 1 + + 1 = 0 ⇒ 5 v − v = −20 1 2 20 5 KCL at node 2: v −v v −v 1 2 2 3 +2= ⇒ − v + 3 v − 2 v = 40 1 2 3 20 10 KCL at node 3: v −v v 2 3 3 +1 = ⇒ − 3 v + 5 v = 30 2 3 10 15 Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V. (checked using LNAP 8/13/02) P4.3-3 KCL at node 1: v −v v 4 − 15 4 1 2 1 + =i ⇒ i = + = −2 A 1 5 20 1 5 20 KCL at node 2: v −v v −v 1 2 2 3 +i = 2 5 15  4 − 15  15 − 18 ⇒ i = − =2A + 2 15  5  (checked using LNAP 8/13/02) 4-5
• P4.3-4 Node equations: −.003 + − v1 v1 − v2 + =0 R1 500 v1 − v2 v2 + − .005 = 0 500 R2 When v1 = 1 V, v2 = 2 V 1 1 −1 + = 0 ⇒ R1 = = 200 Ω 1 R1 500 .003 + 500 2 −1 2 − + − .005 = 0 ⇒ R2 = = 667 Ω 1 500 R2 .005 − 500 −.003 + (checked using LNAP 8/13/02) P4.3-5 Node equations: v1 v − v 2 v1 − v3 + 1 + =0 500 125 250 v − v3 v − v2 − 1 − .001 + 2 =0 125 250 v − v3 v1 − v3 v3 − 2 − + =0 250 250 500 Solving gives: v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V Finally, v = v1 − v3 = 0.022 V (checked using LNAP 8/13/02) 4-6
• Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage Sources P4.4-1 Express the branch voltage of the voltage source in terms of its node voltages: 0 − va = 6 ⇒ va = −6 V KCL at node b: va − vb v −v +2= b c 6 10 KCL at node c: Finally: ⇒ −6 − vb v −v +2= b c 6 10 vb − vc vc = 10 8 ⇒ −1− ⇒ 4 vb − 4 vc = 5 vc 9  30 = 8  vc  − 3 vc 4  vb v −v +2= b c 6 10 ⇒ vb = ⇒ 30 = 8 vb − 3 vc 9 vc 4 ⇒ vc = 2 V (checked using LNAP 8/13/02) P4.4-2 Express the branch voltage of each voltage source in terms of its node voltages to get: va = −12 V, vb = vc = vd + 8 4-7
• KCL at node b: vb − va = 0.002 + i ⇒ 4000 vb − ( −12 ) = 0.002 + i ⇒ vb + 12 = 8 + 4000 i 4000 KCL at the supernode corresponding to the 8 V source: v 0.001 = d + i ⇒ 4 = vd + 4000 i 4000 vb + 4 = 4 − vd ⇒ ( vd + 8 ) + 4 = 4 − vd ⇒ vd = −4 V so Consequently vb = vc = vd + 8 = 4 V and i = 4 − vd = 2 mA 4000 (checked using LNAP 8/13/02) P4.4-3 Apply KCL to the supernode: va − 10 va va − 8 + + − .03 = 0 ⇒ va = 7 V 100 100 100 (checked using LNAP 8/13/02) P4.4-4 Apply KCL to the supernode: va + 8 ( va + 8 ) − 12 va − 12 va + + + =0 500 125 250 500 Solving yields va = 4 V (checked using LNAP 8/13/02) 4-8
• P4.4-5 The power supplied by the voltage source is v −v v −v   12 − 9.882 12 − 5.294  va ( i1 + i 2 ) = va  a b + a c  = 12  +  6  4 6    4 = 12(0.5295 + 1.118) = 12(1.648) = 19.76 W (checked using LNAP 8/13/02) P4.4-6 Label the voltage measured by the meter. Notice that this is a node voltage. Write a node equation at the node at which the node voltage is measured.  12 − v m  v m v −8 − + 0.002 + m =0 + 3000  6000  R That is 6000  6000  3 +  v m = 16 ⇒ R = 16 R   −3 vm (a) The voltage measured by the meter will be 4 volts when R = 6 kΩ. (b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ. 4-9
• Section 4-5 Node Voltage Analysis with Dependent Sources P4.5-1 Express the resistor currents in terms of the node voltages: va − vc = 8.667 − 10 = −1.333 A and 1 v −v 2 − 10 i 2= b c = = −4 A 2 2 i 1= Apply KCL at node c: i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333) ⇒ A= −5.333 =4 −1.333 (checked using LNAP 8/13/02) P4.5-2 Write and solve a node equation: va − 6 v v − 4va + a + a = 0 ⇒ va = 12 V 1000 2000 3000 ib = va − 4va = −12 mA 3000 (checked using LNAP 8/13/02) P4.5-3 First express the controlling current in terms of the node voltages: 2 − vb i = a 4000 Write and solve a node equation: − 2 − vb v  2 − vb  + b − 5  = 0 ⇒ vb = 1.5 V 4000 2000  4000  (checked using LNAP 8/14/02) 4-10
• P4.5-4 Apply KCL to the supernode of the CCVS to get 12 − 10 14 − 10 1 + − + i b = 0 ⇒ i b = −2 A 4 2 2 Next 10 − 12 1 =−  −2 V =4 4 2 ⇒ r = 1 A − r i a = 12 − 14   2 ia = (checked using LNAP 8/14/02) P4.5-5 First, express the controlling current of the CCVS in v2 terms of the node voltages: i x = 2 Next, express the controlled voltage in terms of the node voltages: v2 24 ⇒ v2 = 12 − v 2 = 3 i x = 3 V 2 5 so ix = 12/5 A = 2.4 A. (checked using ELab 9/5/02) 4-11
• Section 4-6 Mesh Current Analysis with Independent Voltage Sources P 4.6-1 2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0 15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0 −6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0 or 14 i1 − 3 i 2 − 9 i 3 = 0 −3 i1 + 9 i 2 − 6 i 3 = −15 −9 i1 − 6 i 2 + 15 i 3 = 21 so i1 = 3 A, i2 = 2 A and i3 = 4 A. (checked using LNAP 8/14/02) P 4.6-2 Top mesh: 4 (2 − 3) + R(2) + 10 (2 − 4) = 0 so R = 12 Ω. Bottom, right mesh: 8 (4 − 3) + 10 (4 − 2) + v 2 = 0 so v2 = −28 V. Bottom left mesh −v1 + 4 (3 − 2) + 8 (3 − 4) = 0 so v1 = −4 V. (checked using LNAP 8/14/02) 4-12
• P 4.6-3 Ohm’s Law: i 2 = −6 = −0.75 A 8 KVL for loop 1: R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0 KVL for loop 2 + (−6) − 3 − 4 ( i1 − i 2 ) = 0 ⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0 ⇒ i 1 = −3 A R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω (checked using LNAP 8/14/02) P4.6-4 KVL loop 1: 25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0 450 ia −100 ib = −2 KVL loop 2: −100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0 −100 ia + 500 ib = − 4 ⇒ ia = − 6.5 mA , ib = − 9.3 mA (checked using LNAP 8/14/02) P4.6-5 Mesh Equations: mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0 Solving: 5 i = i2 ⇒ i = − = −0.294 A 17 (checked using LNAP 8/14/02) 4-13
• Section 4-7 Mesh Current Analysis with Voltage and Current Sources P4.7-1 1 A 2 mesh 2: 75 i2 + 10 + 25 i2 = 0 mesh 1: i1 = ⇒ i2 = − 0.1 A ib = i1 − i2 = 0.6 A (checked using LNAP 8/14/02) P4.7-2 mesh a: ia = − 0.25 A mesh b: ib = − 0.4 A vc = 100(ia − ib ) = 100(0.15) =15 V (checked using LNAP 8/14/02) P4.7-3 Express the current source current as a function of the mesh currents: i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5 Apply KVL to the supermesh: 30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10 50 i2 − 15 = − 10 ⇒ i2 = 5 = .1 A 50 i1 =−.4 A and v2 = 20 i2 = 2 V (checked using LNAP 8/14/02) 4-14
• P4.7-4 Express the current source current in terms of the mesh currents: ib = ia − 0.02 Apply KVL to the supermesh: 250 ia + 100 (ia − 0.02) + 9 = 0 ∴ ia = − .02 A = − 20 mA vc = 100(ia − 0.02) = −4 V (checked using LNAP 8/14/02) P4.7-5 Express the current source current in terms of the mesh currents: i 3 − i 1 = 2 ⇒ i1 = i 3 − 2 Supermesh: 6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8 Lower, left mesh: −12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3 Eliminating i1 and i2 from the supermesh equation: 6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24  24  The voltage measured by the meter is: 3 i 3 = 3   = 8 V  9  (checked using LNAP 8/14/02) 4-15
• P4.7-6 Mesh equation for right mesh: 4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = − 10 5 A=− A 12 6 (checked using LNAP 8/14/02) P 4.7-7 i2 = −3 A i1 − i2 = 5 ⇒ i1 − ( −3) = 5 ⇒ i1 = 2 A 2 ( i3 − i1 ) + 4 i3 + R ( i3 − i2 ) = 0 ⇒ 2 ( −1 − 2 ) + 4 ( −1) + R ( −1 − ( −3) ) = 0 ⇒ R=5 Ω (checked using LNAP 8/14/02) 4-16
• P 4.7-8 Express the controlling voltage of the dependent source as a function of the mesh current v2 = 50 i1 Apply KVL to the right mesh: −100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A v2 = 50 i1 = 10 V (checked using LNAP 8/14/02) P 4.7-9 ib = 4ib − ia ⇒ ib = 1 ia 3 1  −100  ia  + 200ia + 8 = 0 3  ⇒ ia = − 0.048 A (checked using LNAP 8/14/02) P4.7-10 Express the controlling current of the dependent source as a function of the mesh current: ib = .06 − ia Apply KVL to the right mesh: −100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒ ia = 10 mA Finally: vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V (checked using LNAP 8/14/02) 4-17
• P4.7-11 Express the controlling voltage of the dependent source as a function of the mesh current: vb = 100 (.006 − ia ) Apply KVL to the right mesh: −100 (.006 − ia ) + 3[100(.006 − ia )] + 250 ia = 0 ⇒ ia = −24 mA (checked using LNAP 8/14/02) P4.7-12 apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3 apply KVL to right mesh : 5 ×103 i1 + 100 × 103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2 Solving (1) & ( 2 ) simultaneously Power delevered to cathode = ⇒ i1 = ( 2) 6 3 mA, i2 = mA 55 220 ( 5 i1 ) ( i2 ) + 100 ( i2 )2 ( 55)( 3 220) + 100 ( 3 220) = 5 6 ∴ Energy in 24 hr. = Pt = (1) 2 = 0.026 mW ( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr ) = 2.25 J 4-18
• P4.7-13 (a) (b) vo = − g R L v and v = ∴ vo = −g vi R2 R1 + R 2 vi ⇒ (5 ×103 )(103 ) = −170 1.1×103 RL R2 vo = −g vi R1 + R 2 ⇒ g = 0.0374 S PSpice Problems SP 4-1 4-19
• SP 4-2 From the PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E+00 -2.250E+00 V_V3 -7.500E-01 The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current in the voltage source labeled V3). SP 4-3 The PSpice schematic after running the simulation: The PSpice output file: **** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2 V_V4 0 N01588 12Vdc 4-20
• R_R4 V_V5 R_R5 V_V6 I_I1 I_I2 N01588 N01565 4k N01542 N01565 0Vdc 0 N01516 4k N01542 N01516 8Vdc 0 N01565 DC 2mAdc 0 N01542 DC 1mAdc VOLTAGE SOURCE CURRENTS NAME CURRENT V_V4 V_V5 V_V6 -4.000E-03 2.000E-03 -1.000E-03 From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA. SP 4-4 The PSpice schematic after running the simulation: The PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V7 V_V8 -5.613E-01 -6.008E-01 The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of the circuit. However this current is directed from right to left in the 2 Ω resistor while the current i is directed from left to right. Consequently, i = +5.613 A. 4-21
• Verification Problems VP 4-1 Apply KCL at node b: vb − va v −v 1 − + b c = 0 4 2 5 −4.8 − 5.2 1 − 4.8 − 3.0 − + ≠0 4 2 5 The given voltages do not satisfy the KCL equation at node b. They are not correct. VP 4-2 Apply KCL at node a: v v −v  − b a  − 2 + a = 0 2  4  4  20 − 4  − = −4≠ 0 −2+ 2  4  The given voltages do not satisfy the KCL equation at node a. They are not correct. 4-22
• VP 4-3 Writing a node equation:  12 − 7.5  7.5 7.5 − 6 − + =0 + R2  R1  R3 so 4.5 7.5 1.5 + + =0 R1 R3 R2 There are only three cases to consider. Suppose R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then − − 4.5 7.5 1.5 −0.9 + 0.75 + 0.15 + + = = 0 R1 R3 R2 1000 This choice of resistance values corresponds to branch currents that satisfy KCL. Therefore, it is indeed possible that two of the resistances are 10 kW and the other resistance is 5 kW. The 5 kW is R1. VP 4-4 KCL at node 1: 0= v1 − v 2 20 + v1 5 +1 ⇒ −8 − ( −20 ) −8 + +1 = 0 20 5 KCL at node 2: v1 − v 2 20 = 2+ v 2 − v3 KCL at node 3: 10 −8 − ( −20 ) −20 − ( −6 ) = 2+ 20 10 12 6 ⇒ = 20 10 ⇒ v2 − v3 10 +1 = v3 15 ⇒ −20 − ( −6 ) −6 −4 −6 +1 = ⇒ = 10 15 10 15 KCL is satisfied at all of the nodes so the computer analysis is correct. 4-23
• VP 4-5 Top mesh: 10 (2 − 4) + 12(2) + 4 (2 − 3) = 0 Bottom right mesh 8 (3 − 4) + 4 (3 − 2) + 4 = 0 Bottom, left mesh: 28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0 (Perhaps the polarity of the 28 V source was entered incorrectly.) KVL is not satified for the bottom, left mesh so the computer analysis is not correct. 4-24
• Design Problems DP 4-1 Model the circuit as: a) We need to keep v2 across R2 as 4.8 ≤ v2 ≤ 5.4 display is active 0.3 A For I =   0.1 A display is not active KCL at a: v2 − 15 v2 + +I =0 R1 R2 Assumed that maximum I results in minimum v2 and visa-versa. Then 4.8 V v2 =  5.4 V when I = 0.3 A when I = 0.1 A Substitute these corresponding values of v2 and I into the KCL equation and solve for the resistances 4.8 − 15 4.8 + + 0.3 = 0 R1 R2 5.4 − 15 5.4 + + 0.1 = 0 R1 R2 ⇒ R1 = 7.89 Ω, R2 = 4.83 Ω b) 15 − 4.8 = 1.292 A ⇒ PR = (1.292)2 (7.89) = 13.17 W 1max 7.89 ( 5.4 )2 = 6.03 W 5.4 IR = = 1.118 A ⇒ PR = 2max 2 max 4.83 4.83 maximum supply current = I R = 1.292 A IR 1max = 1max c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop below 4.8V or rise above 5.4V. The power dissipated in the resistors is excessive. Most of the power from the supply is dissipated in the resistors, not the display. 4-25
• DP 4-2 Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL to the supernode corresponding to the 8 V source: va − v1 R + va vb vb − ( −v 2 ) + + = 0 ⇒ 2 va − v1 + 2 vb + v 2 = 0 R R R ⇒ 2 va − v1 + 2 ( va + 8 ) + v 2 = 0 ⇒ 4 va − v1 + v 2 + 16 = 0 ⇒ va = Next set va = 0 to get 0= v1 − v 2 4 v1 − v 2 4 −4 − 4 ⇒ v1 − v 2 = 16 V For example, v1 = 18 V and v2 = 2 V. 4-26
• DP 4-3 a) pply KCL to left mesh: −5 + 50 i1 + 300 (i1 − I ) = 0 Apply KCL to right mesh: ( R + 2) I + 300 ( I − i1 ) = 0 150 1570 + 35 R We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA ﬁ l amp so the lamp will not light. Solving for I: I= b) From the equation for I, we see that decreasing R increases I: try R = 50 Ω ⇒ I = 45 mA (won't light) try R = 25Ω ⇒ I = 61 mA ⇒ will light Now check R±10% to see if the lamp will light and not burn out: −10% → 22.5Ω → I = 63.63 mA  lamp will  +10% → 27.5Ω → I = 59.23 mA  stay on DP 4-4 Equivalent resistance: R = R1 || R 2 || ( R 3 + R 4 ) R ( 25 ) 10 + R We require vab = 10 V. Apply the voltage division principle in the left circuit to get: Voltage division in the equivalent circuit: v1 = 4-27
• 10 = ( ) R1 R 2 ( R3 + R4 ) R4 R4 × × 25 v1 = R3 + R4 R3 + R4 10 + R1 R 2 ( R3 + R4 ) ( ) This equation does not have a unique solution. Here’s one solution: choose R1 = R2 = 25 Ω and R3 + R4 = 20 Ω then 10 = (12.5 20 ) × 25 ⇒ R = 18.4Ω R4 × 4 20 10 + (12.5 20 ) and R3 + R4 = 20 ⇒ R3 = 1.6 Ω DP 4-5 Apply KCL to the left mesh: (R 1 + R 3 ) i1 − R 3 i2 − v1 = 0 Apply KCL to the left mesh: (R − R 3 i1 + 2 + R 3 ) i2 + v2 = 0 Solving for the mesh currents using Cramer’s rule: − R3  v1  ( R1 + R 3 )     −v2 ( R 2 + R 3 )    and i =  − R 3 i1 = 2 ∆ ∆ 2 where ∆ = ( R1 + R 3 ) ( R 2 + R 3 ) − R 3 v1   − v2  Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then ∆ = 3 MΩ. The mesh currents will be given by i = [ 2v1 − v2 ] 1000 and i 2 = [ −2v2 + v1 ] 1000 3 × 10 3 × 10 Now check the extreme values of the source voltages: 1 6 6 if v1 = v2 = 1 V ⇒ i = 2 mA 3 if v1 = v2 = 2 V ⇒ i = 4 mA 3 ⇒ i = i1 − i2 = v1 + v2 3000 okay okay 4-28
• Chapter 5 Circuit Theorems Exercises Ex 5.3-1 R = 10 Ω and is = 1.2 A. Ex 5.3-2 R = 10 Ω and is = −1.2 A. Ex 5.3-3 R = 8 Ω and vs = 24 V. Ex 5.3-4 R = 8 Ω and vs = −24 V. Ex 5.4-1 vm =   20 10 2 (15) + 20  − ( 2 ) = 6 + 20(− ) = −2 V 10 + 20 + 20 5  10 + (20 + 20)  Ex 5.4-2 im = 25 3 − ( 5) = 5 − 3 = 2 A 3+ 2 2+3 Ex 5.4-3   3 3 vm = 3  ( 5) − (18 ) = 5 − 6 = −1 A  3 + (3 + 3)  3 + (3 + 3) 5-1
• Ex 5.5-1 5-2
• Ex 5.5-2 ia = 2 i a − 12 ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a 3 i sc = 2 i a Rt = ⇒ i a = −3 A ⇒ i sc = 2 ( −3 ) = − 2 A 3 −6 =3Ω −2 Ex 5.6-1 5-3
• Ex 5.6-2 ia = 2 i a − 12 ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a 3 i sc = 2 i a Rt = ⇒ i a = −3 A ⇒ i sc = 2 ( −3 ) = − 2 A 3 −6 =3Ω −2 5-4
• Ex 5.6-3 12 × 24 12 × 24 = = 8Ω 12 + 24 36 24 voc = ( 30 ) = 20 V 12 + 24 Rt = i= 20 8+ R Ex 5.7-1 voc = 6 (18) = 12 V 6+3 Rt = 2 + ( 3)( 6 ) = 4 Ω 3+ 6 For maximum power, we require R L = Rt = 4 Ω Then 2 pmax = voc 122 = =9 W 4 Rt 4 ( 4 ) 5-5
• Ex 5.7-2 1 50 3 i sc = ( 5.6 ) = ( 5.6 ) = 5 A 1 1 1 50 + 1 + 5 + + 3 150 30 150 ( 30 ) Rt = 3 + = 3 + 25 = 28 Ω 150 + 30 2 R t i sc ( 28 ) 52 = 175 W pmax = = 4 4 Ex 5.7-3  10   R L  100 R L p=iv= (10 ) =  2     Rt + R L   Rt + R L  ( Rt + R L ) The power increases as Rt decreases so choose Rt = 1 Ω. Then pmax = i v = 100 ( 5 ) (1 + 5) 2 = 13.9 W Ex 5.7-4 From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: Rt = 20 Ω and 2 pmax v = oc 4 Rt ⇒ voc = pmax 4 Rt = 5 ( 4 ) 20 = 20 V 5-6
• Problems Section 5-3: Source Transformations P5.3-1 (a) 5-7
• ∴ Rt = 2 Ω (b) (c) −9 − 4i − 2i + (−0.5) = 0 −9 + (−0.5) i = = −1.58 A 4+2 v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V ia = i = − 1.58 A vt = − 0.5 V (checked using LNAP 8/15/02) P5.3-2 Finally, apply KVL: −10 + 3 ia + 4 ia − 16 =0 3 ∴ ia = 2.19 A (checked using LNAP 8/15/02) 5-8
• P5.3-3 Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left: 5-9
• Finally, apply KVL to loop − 6 + i (9 + 19) − 36 − vo = 0 i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V (checked using LNAP 8/15/02) P5.3-4 − 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0 ∴ ia = 375 µ A (checked using LNAP 8/15/02) 5-10
• P5.3-5 −12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A (checked using LNAP 8/15/02) P5.3-6 A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor: Source transformations on both the right side and the left side of the circuit: 5-11
• Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source: Finally, va = 50 (100 ) 100 ( 0.21) = ( 0.21) = 7 V 50 + 100 3 (checked using LNAP 8/15/02) 5-12
• Section 5-4 Superposition P5.4–1 Consider 6 A source only (open 9 A source) Use current division: v1  15  = 6  ⇒ v1 = 40 V 20 15 + 30   Consider 9 A source only (open 6 A source) Use current division: v2  10  = 9  ⇒ v2 = 40 V 20 10 + 35   ∴ v = v1 + v2 = 40 + 40 = 80 V (checked using LNAP 8/15/02) P5.4-2 Consider 12 V source only (open both current sources) KVL: 20 i1 + 12 + 4 i1 + 12 i1 = 0 ⇒ i1 = −1/ 3 mA Consider 3 mA source only (short 12 V and open 9 mA sources) Current Division: 4  16  i2 = 3   = 3 mA 16 + 20  5-13
• Consider 9 mA source only (short 12 V and open 3 mA sources) Current Division:  12  i3 = −9  = −3 mA  24 + 12   ∴ i = i1 + i2 + i3 = − 1/ 3 + 4 / 3 − 3 = − 2 mA (checked using LNAP 8/15/02) P5.4–3 Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30 mA current source.  2  ia = 30   = 6 mA ⇒  2+8  6  i1 = ia   = 2 mA  6 + 12  Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the 15 mA current source.  4  ib = 15   = 6 mA ⇒  4+6  6  i2 = ib   = 2 mA  6 + 12  5-14
• Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V voltage source.  6 || 6   3  i3 = − 2.5   ( 6 || 6 ) + 12  = − 10  3 + 12  = −0.5 mA      Finally, i = i1 + i2 + i3 = 2 + 2 − 0.5 = 3.5 mA (checked using LNAP 8/15/02) P5.4–4 Consider 10 V source only (open 30 mA source and short the 8 V source) Let v1 be the part of va due to the 10 V voltage source. 100 ||100 (10 ) (100 ||100 ) + 100 v1 = = Consider 8 V source only (open 30 mA source and short the 10 V source) 50 10 (10 ) = V 150 3 Let v2 be the part of va due to the 8 V voltage source. v1 = = 100 ||100 (8) (100 ||100 ) + 100 50 8 (8) = V 150 3 5-15
• Consider 30 mA source only (short both the 10 V source and the 8 V source) Let v2 be the part of va due to the 30 mA current source. v3 = (100 ||100 ||100)(0.03) = va = v1 + v2 + v3 = Finally, 100 (0.03) = 1 V 3 10 8 + +1 = 7 V 3 3 (checked using LNAP 8/15/02) P5.4-5 Consider 8 V source only (open the 2 A source) Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh: 6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0 i1 = Consider 2 A source only (short the 8 V source) 8 2 = A 12 3 Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh: 6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0 i2 = Finally, i x = i1 + i 2 = −6 1 =− A 12 2 2 1 1 − = A 3 2 6 5-16
• Section 5-5: Thèvenin’s Theorem P5.5-1 (checked using LNAP 8/15/02) 5-17
• P5.5-2 The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps: (a) (b) (c) (d) Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V. 5-18
• P5.5-3 The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps: (a) (b) (c) (d) (e) Comparing (e) to Figure P5.5-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02) 5-19
• P5.5-4 Find Rt: Rt = 12 (10 + 2 ) =6Ω 12 + (10 + 2 ) Write mesh equations to find voc: Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i 1 = 18 36 i1 = 18 ⇒ i1 = i2 = Finally, 1 A 2 14  1  7  = A 3 2 3 7 1 voc = 3 i 2 + 10 i1 = 3   + 10   = 12 V  3  2 (checked using LNAP 8/15/02) 5-20
• P5.5-5 Find voc: Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −voc Apply KCL at node a:  6 − voc  voc  3  − +  − voc  = 0 +  8  4  4  −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces va = 0 Apply KCL at node a:  6−0 0  3  −  + +  − 0  + i sc = 0  8  4  4  i sc = Rt = 6 3 = A 8 4 voc −2 8 = =− Ω i sc 3 4 3 (checked using LNAP 8/15/02) 5-21
• P5.5-6 Find voc: 2 va − va va = + 3 + 0 ⇒ va = 18 V 3 6 The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Apply KCL at the top, middle node: Find isc: 2 va − va va v = + 3 + a ⇒ va = −18 V 3 6 3 va −18 i sc = = = −6 V Apply Ohm’s law to the right-hand 3 Ω resistor : 3 3 v 18 Finally: R t = oc = = −3 Ω i sc −6 Apply KCL at the top, middle node: (checked using LNAP 8/15/02) 5-22
• P5.5-7 (a) −vs + R1 ia + ( d + 1) R 2 ia = 0 ia = vs R1 + ( d + 1) R 2 v oc = ( d + 1) R 2vs R1 + ( d + 1) R 2 vs R1 ia = i sc = ( d + 1) ia = −ia − d ia + ( d + 1) vs R1 vT − iT = 0 R2 R1 ia = −vT iT = ( d + 1) vT vT R 2 ( d + 1) + R1 + = × vT R1 R 2 R1 R 2 Rt = R1 R 2 vT = iT R1 + ( d + 1) R 2 (b) Let R1 = R2 = 1 kΩ. Then 625 Ω = R t = and 5 = voc = 1000 1000 ⇒ d= − 2 = −0.4 A/A 625 d +2 ( d + 1) vs d +2 ⇒ vs = −0.4 + 2 5 = 13.33 V −0.4 + 1 (checked using LNAP 8/15/02) 5-23
• P5.5-8 From the given data: 2000  voc  R t + 2000   voc = 1.2 V  ⇒  4000  R t = −1600 Ω voc  2= R t + 4000   6= When R = 8000 Ω, v= R voc Rt + R v= 8000 (1.2 ) = 1.5 V −1600 + 8000 P5.5-9 From the given data: voc  R t + 2000   voc = 24 V   ⇒  voc  R t = 4000 Ω  0.003 = R t + 4000   0.004 = i= voc Rt + R (a) When i = 0.002 A: 24 0.002 = ⇒ R = 8000 Ω 4000 + R (b) Maximum i occurs when R = 0: 24 = 0.006 = 6 mA ⇒ i ≤ 6 mA 4000 P5.5-10 The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA. The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V. The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so 1 0 − 0.002 − = ⇒ R t = −150 Ω Rt −3 − 0 5-24
• P5.5-11 −12 + 6000 ia + 2000 ia + 1000 ia = 0 ia = 4 3000 A voc = 1000 ia = 4 V 3 ia = 0 due to the short circuit −12 + 6000 isc = 0 ⇒ isc = 2 mA 4 voc Rt = = 3 = 667 Ω isc .002 4 3 ib = 667 + R ib = 0.002 A requires 4 3 − 667 = 0 R = 0.002 (checked using LNAP 8/15/02) 5-25
• P5.5-12 10 = i + 0 ⇒ i = 10 A voc + 4 i − 2 i = 0 ⇒ voc = −2 i = −20 V i + i sc = 10 ⇒ i = 10 − i sc 4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A Rt = −2 = iL = voc −20 = = −2 Ω isc 10 −20 ⇒ RL = 12 Ω RL − 2 (checked using LNAP 8/15/02) 5-26
• Section 5-6: Norton’s Theorem P5.6-1 When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off. P5.6-2 (checked using LNAP 8/16/02) 5-27
• P5.6-3 P5.6-4 To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − isc Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10 (1) Apply KVL to mesh 2 to get 5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 = 11 i2 6 Substituting into equation 1 gives  11  7  i 2  − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A 6  5-28
• Figure (a) Calculating the short circuit current, isc, using mesh equations. To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v Rt = T iT In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 + i T Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i1 − 4 i 2 = 0 ⇒ i1 = 4 i2 7 (2) Apply KVL to mesh 2 to get 5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT Substituting for i1 using equation 2 gives 4  −6  i 2  + 11 i 2 = −vT 7  Finally, Rt = ⇒ 7.57 i 2 = −vT vT −vT −vT = = = 7.57 Ω −iT iT i2 5-29
• Figure (b) Calculating the Thevenin resistance, R t = vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − 0 = i1 Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0 ⇒ i1 = 10 = 1.43 A 7 Apply KVL to mesh 2 to get 5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V Figure (c) Calculating the open circuit voltage, voc, using mesh equations. As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 ≈ voc (checked using LNAP 8/16/02) 5-30
• P5.6-5 To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 . Apply KCL at node 2 to get v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 = 3 v a ⇒ v a = −16 V Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 = isc 3 ⇒ 9 v a = isc 6 ⇒ isc = 9 ( −16 ) = −24 A 6 Figure (a) Calculating the short circuit current, Isc, using mesh equations. To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v R th = T iT Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. 5-31
• In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT . Apply KCL at node 2 to get v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ vT = 3 v a Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0 3 ⇒ 9 v a − vT + 6 iT = 0 ⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT Finally, Rt = vT = −3 Ω iT Figure (b) Calculating the Thevenin resistance, R th = vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get 5-32
• v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 + v oc = 3 v a Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc 3 Combining these equations gives 3 ( −48 + voc ) = 9 v a = voc ⇒ voc = 72 V Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that R th I sc = ( −3)( −24 ) = 72 = Voc (checked using LNAP 8/16/02) Section 5-7: Maximum Power Transfer P5.7-1 a) For maximum power transfer, set RL equal to the Thevenin resistance: R L = R t = 100 + 1 = 101 Ω b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit: 5-33
• The voltage across RL is vL = Then pmax 101 (100 ) = 50 V 101 + 101 2 v 502 = L = = 24.75 W R L 101 P5.7-2 Reduce the circuit using source transformations: Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is P = i 2 ( R) = (0.03)2 (60) = 54 mW max R 5-34
• P5.7-3  RL  v L = vS    RS + R L    2 2 vL vS RL ∴ pL = = R L ( RS + R L )2 By inspection, pL is max when you reduce RS to get the smallest denominator. ∴ set RS = 0 P5.7-4 Find Rt by finding isc and voc: The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix. Apply KCL at the top-left node to get ix + 0.9 = 10 ix ⇒ ix = 0.9 = 0.1 A 9 so Next isc = 10 ix = 1A Apply KCL at the top-left node to get 5-35
• ix + 0.9 = 10 ix ⇒ ix = 0.9 = 0.1 A 9 Apply Ohm’s law to the 3 Ω resistor to get voc = 3 (10 ix ) = 30 ( 0.1) = 3 V For maximum power transfer to RL: R L = Rt = voc 3 = =3Ω isc 1 The maximum power delivered to RL is given by 2 pmax = voc 32 3 = = W 4 R t 4 ( 3) 4 P5.7-5 The required value of R is R = Rt = 8 + ( 20 + 120 ) (10 + 50 ) = 50 Ω ( 20 + 120 ) + (10 + 50 ) 30  170  voc =  ( 20 )  10 −   170 + 30 ( 20 )  50 170 + 30    170(20)(10) − 30(20)(50) 4000 = = = 20 V 200 200 The maximum power is given by 2 v 202 pmax = oc = =2W 4 R t 4 ( 50 ) 5-36
• PSpice Problems SP5-1 a = 0.3333 b = 0.3333 c =33.33 V/A (a) (b) vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3 7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3 18 3 = 3 = 30 mA ⇒ i3 = 100 100 3 7− 5-37
• SP5-2 Before the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E-02 -4.000E-02 After the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V2 -4.000E-02 5-38
• SP5-3 voc = −2 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V3 V_V4 -7.500E-01 7.500E-01 isc = 0.75 A Rt = −2.66 Ω 5-39
• SP5-4 voc = 8.571 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V5 -2.075E+00 V_V6 1.132E+00 X_H1.VH_H1 9.434E-01 isc = 1.132 A Rt = 7.571 Ω 5-40
• Verification Problems VP5-1 Use the data in the first two lines of the table to determine voc and Rt: voc   Rt + 0  voc = 39.9 V  ⇒  voc   R t = 410 Ω 0.0438 = R t + 500   0.0972 = Now check the third line of the table. When R= 5000 Ω: v 39.9 i = oc = = 7.37 mA R t + R 410 + 5000 which disagree with the data in the table. The data is not consistent. VP5-2 Use the data in the table to determine voc and isc: voc = 12 V (line 1 of the table) isc = 3 mA so Rt = (line 3 of the table) voc = 4 kΩ isc Next, check line 2 of the table. When R = 10 kΩ: v 12 i = oc = = 0.857 mA 3 R t + R 10 (10 ) + 5 (103 ) To cause i = 1 mA requires which agrees with the data in the table. v 12 0.001 = i = oc = ⇒ R = 8000 Ω R t + R 10 (103 ) + R I agree with my lab partner’s claim that R = 8000 causes i = 1 mA. 5-41
• VP5-3 60 60 voc 11 i= = = 11 = 54.55 mA R t + R 6 110 + 40 60 + 40 ( ) 11 The measurement supports the prelab calculation. Design Problems DP5-1 The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0−5 open circuit voltage. Therefore: R t = − = 625 Ω and voc = 5 V. 0.008 − 0 Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then 5= R2 1 vs = vs 2 R1 + R 2 ⇒ vs = 10 V and R1 R2 = R3 + 500 ⇒ R3 = 125 Ω R1 + R2 Now vs, R1, R2 and R3 have all been specified so the design is complete. 625 = R 3 + DP5-2 The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0 − ( −3 ) = 500 Ω and voc = −3 V. open circuit voltage. Therefore: R t = − −0.006 − 0 From the circuit we calculate R 3 ( R1 + R 2 ) R1 R 3 and voc = − is Rt = R1 + R 2 + R 3 R1 + R 2 + R 3 so R 3 ( R1 + R 2 ) R1 R 3 500 Ω = is and −3 V = − R1 + R 2 + R 3 R1 + R 2 + R 3 5-42
• Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and −3 = − 1000 R1 is = − R1 i s ⇒ 6 = R1 i s 2000 2 This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω = 400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete. DP5-3 The slope of the graph is positive so the Thevenin resistance is negative. This would require R1 R 2 R3 + < 0 , which is not possible since R1, R2 and R3 will all be non-negative. R1 + R 2 Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b. DP5-4 The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, the slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open −5 − 0 circuit voltage. Therefore: R t = − = −625 Ω and voc = −5 V. 0 − 0.008 The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt, of this circuit are given by R 2 ( d + 1) voc = vs R1 + ( d + 1) R 2 , ( d + 1) v isc = s R1 and R1 R 2 Rt = R1 + ( d + 1) R 2 Let R1 = R2 = 1 kΩ. Then −625 Ω = R t = and −5 = ( d + 1) vs d +2 1000 1000 ⇒ d= − 2 = −3.6 A/A d +2 −625 ⇒ vs = −3.6 + 2 ( − 5 ) = −3.077 V −3.6 + 1 Now vs, R1, R2 and d have all been specified so the design is complete. 5-43
• Chapter 6: The Operational Amplifier Exercises Ex. 6.4-1 vs vs − vo + +0 = 0 R1 R2 vo R = 1+ 2 vs R1 Ex. 6.4-2 a) va = R2 vs R1 + R2 va va − v0 + +0 =0 R3 R4 vo R = 1+ 4 va R3 b) When R2 >> R1 then ⇒ vo  R2   R4  =  1 +  vs  R1 + R2  R3  R2 R vo R − 2 = 1 and − 1+ 4 R1 + R2 R2 vs R3 6-1
• Ex. 6.5-1 vo vo − vs + +0= 0 R2 R1 vo R2 = vs R1 + R2 Ex. 6.6–1  R  vin − vout vin + + 0 = 0 ⇒ vout = 1 + f  vin  Rf R1 R1    when R f = 100 kΩ and R1 = 25 kΩ then  100 ⋅103  vout = 1 +  = 5 vin 25 ⋅103   6-2
• Ex. 6.7-1  R2 R1   10 × 103    10 ×103   v3 =  − v  v2 +  1 +   1 + 3 3 3 3  1  10 × 10   R 2 + 10 × 10   10 × 10   10 × 10      R2 R1   = − v + 2 1 + v 3  2 3  1    10 × 10   R 2 + 10 × 10  1 We require v3 = ( 4 ) v1 −   v2 , so 5 R1   3 4 = 2 1 +  ⇒ R1 = 10 × 10 = 10 kΩ 10 ×103   and R2 1 = 5 R 2 + 10 × 103 ⇒ R 2 + 10 × 103 = 5 R 2 ⇒ R 2 = 2.5 kΩ 6-3
• Ex. 6.7-2 As in Ex 6.7-1   R2 R1   v3 = −  v + 2 1 + v 3  2 3  1  R 2 + 10 × 10   10 × 10     4 We require v3 = ( 6 ) v1 −   v2 , so 5 R1   3 6 = 2 1 +  ⇒ R1 = 20 ×10 = 20 kΩ 10 ×103   and R2 4 = 5 R 2 + 10 × 103 ⇒ 4 R 2 + 40 × 103 = 5 R 2 ⇒ R 2 = 40 kΩ Ex. 6.8-1 Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 vos + (50 × 103 ) ib1 For a µ A741 op amp, vos ≤ 1 mV and ib1 ≤ 80 nA so output offset voltage = 6 vos + (50 × 103 )ib1 ≤ 6 (10−3 )+(50.103 )(80×10−9 ) = 10 mV Ex. 6.8-2 vo = −  R  R2 vin + 1 + 2  vos + R2ib1 R1  R1  When R2 = 10 kΩ, R1 = 2 kΩ, vos ≤ 5 mV and ( ) ( output offset voltage ≤ 6 5 × 10−3 + 10 × 103 ib1 ≤ 500 nA then ) ( 500.10 ) ≤ 35×10 −9 −3 = 35 mV 6-4
• Ex. 6.8-3 Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 vos + ( 50 ×103 ) ib1 For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so output offset voltage ≤ 6 0.1×10−3  + ( 50 ×103 ) 0.012 ×10−9      −3 −6 −3 ≤ 0.6 ×10 + 0.6 ×10 − 0.6 ×10 = 0.6 mV Ex. 6.8-4 Writing node equations v− − vs v− − vo v− + + =0 Ra Rb Ri + Rs   R i v  vo −  − A −   R +R i s  + vo − v− = 0  R0 Rb After some algebra Av = R0 ( Ri + Rs ) + ARi R f vo = vs ( R f + R0 ) ( Ri + Rs ) + Ra ( R f + R0 + Ri + Rs ) − ARi Ra For the given values, Av = −2.00006 V/V. 6-5
• Problems Section 6-4: The Ideal Operational Amplifier P6.4-1 (checked using LNAP 8/16/02) P6.4-2 Apply KVL to loop 1: − 12 + 3000 i1 + 0 + 2000 i1 = 0 12 = 2.4 mA 5000 The currents into the inputs of an ideal op amp are zero so io = i1 = 2.4 mA ⇒ i1 = i2 = − i1 = − 2.4 mA va = i2 (1000 ) + 0 = −2.4 V Apply Ohm’s law to the 4 kΩ resistor vo = va − io ( 4000 ) = −2.4 − ( 2.4 ×10−3 ) ( 4000 ) = −12 V (checked using LNAP 8/16/02) 6-6
• P6.4-3 The voltages at the input nodes of an ideal op amp are equal so va = −2 V . Apply KCL at node a: vo − ( −2 ) 12 − ( −2 ) + = 0 ⇒ vo = −30 V 8000 4000 Apply Ohm’s law to the 8 kΩ resistor io = −2 − vo = 3.5 mA 8000 (checked using LNAP 8/16/02) P6.4-4 The voltages at the input nodes of an ideal op amp are equal so v = 5 V . Apply KCL at the inverting input node of the op amp:  v −5  −3 − a  − 0.1× 10 − 0 = 0 ⇒ va = 4 V  10000  Apply Ohm’s law to the 20 kΩ resistor va 1 i = = mA 20000 5 (checked using LNAP 8/16/02) P6.4-5 The voltages at the input nodes of an ideal op amp are equal so va = 0 V . Apply KCL at node a:  v − 0   12 − 0  −3 − o −  − 2 ⋅10 = 0  3000   4000  ⇒ vo = − 15 V Apply KCL at the output node of the op amp: v v io + o + o = 0 ⇒ io = 7.5 mA 6000 3000 (checked using LNAP 8/16/02) 6-7
• P6.4-6 The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal so va = 2.5 V . Apply Ohm’s law to the 4 kΩ resistor: v 2.5 ia = a = = 0.625 mA 4000 4000 Apply KCL at node a: ib = ia = 0.625 mA Apply KVL: vo = 8000 ib + 4000 ia = (12 × 103 )( 0.625 × 10−3 ) = 7.5 V (checked using LNAP 8/16/02) P6.4-7 R2  v − 0   va − 0  vs − s  + 0 = 0 ⇒ va = − − R1  R1   R 2    io =  R 2 + R3  R 2 + R3 0 − va 0 − va + =− va =  v  R1 R 3  s  R2 R3 R 2 R3    v − 0   va − 0  R4 R2 R4 − o va = vs −  + 0 = 0 ⇒ vo = −  R 4   R3  R3 R1 R 3     6-8
• P6.4-8 The node voltages have been labeled using: 1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal. 2. KCL 3. Ohm’s law Then v0 = 11.8 − 1.8 = 10 V and io = 10 = 2.5 mA 4000 (checked using LNAP 8/16/02) P6.4-9 KCL at node a: va − ( −18 ) v + a + 0 = 0 ⇒ va = −12 V 4000 8000 The node voltages at the input nodes of ideal op amps are equal so vb = va . Voltage division: vo = 8000 vb = −8 V 4000 + 8000 (check using LNAP 8/16/02) 6-9
• Section 6-5: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers P6.5-1 KCL at node b: vb − 2 vb v +5 1 + + b = 0 ⇒ vb = − V 20000 40000 40000 4 The node voltages at the input nodes of an ideal op amp are equal so ve = vb = − KCL at node e: ve v −v 10 + e d = 0 ⇒ vd = 10 ve = − V 1000 9000 4 1 V. 4 (checked using LNAP 8/16/02) P6.5-2 Apply KCL at node a: 0= va − 12 v v −0 + a + a ⇒ va = 4 V 6000 6000 6000 Apply KCL at the inverting input of the op amp: v −0  0 − vo  − a +0+  = 0  6000   6000  ⇒ vo = −va = −4 V Apply KCL at the output of the op amp: vo  0 − vo  = 0 io −  +  6000  6000 v ⇒ io = − o = 1.33 mA 3000 (checked using LNAP 8/16/02) 6-10
• P6.5-3 Apply KCL at the inverting input of the op amp:  v − 0   vs − 0  − a −  = 0  R2   R1  R ⇒ va = − 2 vs R1 Apply KCL at node a:  1 1 1  va −v0 va va − 0 R R +R R +R R + + = 0 ⇒ v0 = R4  + +  va = 2 3 2 4 3 4 va R4 R3 R2 R2 R3  R4 R3 R2  =− Plug in values ⇒ yields R2 R3 + R2 R4 + R3 R4 vs R1 R3 vo 30 + 900 + 30 =− = −200 V/V 4.8 vs P6.5-4 Ohm’s law: i= v1 − v2 R2 KVL: v0 = ( R1 + R2 + R3 ) i = R1 + R2 + R3 ( v1 − v2 ) R2 6-11
• P6.5-5  R  v1 −va v1 −v2 R + + 0 = 0 ⇒ va = 1+ 1  v1 − 1 v2 R1 R7 R7  R7   R  v2 − vb v −v R − 1 2 + 0 = 0 ⇒ vb =  1+ 2  v2 − 2 v1 R2 R7 R7  R7   v −v  v −0 R6 −  b c  + c + 0 = 0 ⇒ vc = vb R4 + R6  R4  R6  v −v   v −v  R R − a c  + c 0  + 0 = 0 ⇒ v0 = − 5 va + (1+ 5 )vc R3 R3  R3   R5   R R R (R +R ) R R (R +R ) R  R  R v0 =  5 1 + 6 3 5 (1+ 2 )  v2 −  5 (1+ 1 ) + 6 3 5 2  v1 R7  R7 R3 ( R4 + R6 ) R7   R3 R7 R3 ( R4 + R6 )  R3 v −v i0 = c 0 = R5 6-12
• P6.5-6 KCL at node b: KCL at node a: So va vc 5 + = 0 ⇒ vc = − va 3 3 20 × 10 25 × 10 4  5  va −  − va  va − ( −12 ) va va + 0  4  = 0 ⇒ v = − 12 V + + + a 3 3 3 40 × 10 10 × 10 20 ×10 10 ×103 13 5 15 vc = − va = − . 4 13 6-13
• P6.5-7 Apply KCL at the inverting input node of the op amp  ( va + 6 ) − 0   v −0  − a  = 0 +0−  10000   30000  ⇒va = −1.5 V Apply KCL to the super node corresponding the voltage source: va −0 va + 6− 0 + 10000 30000 ( v + 6 )−vb = 0 v −v + a b + a 30000 10000 ⇒ 3va + va + 6 + va − vb + 3 ( va + 6 )− vb  = 0   ⇒ vb = 2va + 6 = 3 V Apply KCL at node b: vb v − v  v − v   ( v + 6 ) − vb  + b 0 − a b − a  = 0 10000 30000  30000   10000  ⇒ 3vb +( vb − v0 )−( va − vb )−3( va + 6 ) − vb  = 0   ⇒ v0 = 8vb − 4va −18 = 12 V Apply KCL at the output node of the op amp: i0 + v0 v −v + 0 b = 0 ⇒ i0 = − 0.7 mA 30000 30000 6-14
• P6.5-8 Apply KVL to the bottom mesh: −i0 (10000) − i0 (20000) + 5 = 0 ⇒ i0 = 1 mA 6 The node voltages at the input nodes of an ideal op amp are equal. Consequently va = 10000 i0 = 10 V 6 Apply KCL at node a: va v −v + a 0 = 0 ⇒ 10000 20000 v0 = 3va = 5 V P6.5-9 vb + 12 vb + = 0 ⇒ vb = −4 V 40000 20000 The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V . KCL at node b: The node voltages at the input nodes of an ideal op amp are equal, so vd = vc + 0 × 10 4 = −4 V . KCL at node g: vg  v f − vg  2 − + = 0 ⇒ vg = v f 3  3 3  20 × 10  40 ×10 6-15
• The node voltages at the input nodes of an ideal op amp are equal, so ve = vg = 2 vd − v f vd − 3 v f vd − ve KCL at node d: 0 = + = + 20 ×103 20 ×103 20 ×103 20 ×103 vd − v f Finally, ve = vg = 2 vf . 3 6 24 V ⇒ v f = vd = − 5 5 2 16 vf = − V. 3 5 P6.5-10 By voltage division (or by applying KCL at node a) va = R0 vs R1 + R0 Applying KCL at node b: vb − vs vb −v0 + = 0 R1 R0 +∆R ⇒ R0 +∆R ( vb −vs )+ vb = v0 R1 The node voltages at the input nodes of an ideal op amp are equal so vb = va .  R +∆R  R0  R +∆R  R0  ∆R ∆R v0 =  0 +1 − 0 vs =  −vs  vs = −  R1  R1 + R0 R1 + R0  R0  R1 + R0   R1 6-16
• Section 6-6: Design Using Operational Amplifier P6.6-1 Use the current-to-voltage converter, entry (g) in Figure 6.6-1. P6.6-2 Use the voltage –controlled current source, entry (i) in Figure 6.6-1. P6.6-3 Use the noninverting summing amplifier, entry (e) in Figure 6.6-1. 6-17
• P6.6-4 Use the difference amplifier, entry (f) in Figure 6.6-1. P6.6-5 Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1. P6.6-6 Use the negative resistance converter, entry (h) in Figure 6.6-1. 6-18
• P6.6-7 Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the current iin to be zero. P6.6-8 Summing Amplifier: va = − ( 6 v1 + 2 v2 )   ⇒ vo = 6 v1 + 2 v2 Inverting Amplifier: vo = −va  P6.6-9 6-19
• Using superposition, vo = v1 + v2 + v3 = −9 − 16 + 32 = 7 V P6.6-10 R1 6 12 24 6||12 6||24 R2 -vo/vs 12||12||24 0.8 6||12||24 0.286 6||12||12 0.125 12||24 2 12||12 1.25 12||24 6||12||12 6||12||24 12||12||24 6||12 0.5 24 8 12 3.5 6 1.25 R1 R2 -vo/vs 12||12 6||24 0.8 6-20
• Section 6-7: Operational Amplifier Circuits and Linear Algebraic Equations P6.7-1 6-21
• P6.7-2 6-22
• Section 6-8: Characteristics of the Practical Operational Amplifier P6.8-1 The node equation at node a is: vout − vos vos = + ib1 3 100×10 10×103 Solving for vout:  100×103  3 3 vout = 1+  vos + (100 × 10 ) ib1 = 11vos + (100 × 10 ) ib1 10×103   ( ) = 11 ( 0.03×10−3 )+(100×103 ) 1.2×10−9 = 0.45 mV P6.8-2 The node equation at node a is: vos v −v + ib1 = 0 os 10000 90000 Solving for vo:  90×103  vo = 1+ v + ( 90 × 103 ) ib1 = 10 vos + ( 90 × 103 ) i b1 3  os  10×10  = 10(5 ×10−3 )+ ( 90 ×103 ) (.05 × 10−9 ) = 50.0045 × 10−3 − 50 mV 6-23
• P6.8-3 v1 −vin v1 v1 −v0  + + = 0 R1 Rin R2 v0 Rin ( R0 − AR2 )  =  ⇒ v0 + Av1 v0 − v1 vin ( R1 + Rin )( R0 + R2 ) + R1 Rin (1+ A) + =0   R0 R2  P6.8-4 a) v0 R 49×103 = − 2 = − = −9.6078 vin R1 5.1×103 ( ) b) ( 2×106 ) 75−( 200,000)( 50×103 ) v0 = = −9.9957 vin (5×103 + 2×106 )(75+50×103 ) + (5×103 )(2×106 )(1+ 200,000) c) v0 2×106 (75− (200,000)(49×103 )) = = −9.6037 vin (5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000) 6-24
• P6.8-5 Apply KCL at node b: R3 vb = (vcm − v p ) R2 + R3 Apply KCL at node a: va −v0 va −(vcm + vn ) + = 0 R4 R1 The voltages at the input nodes of an ideal op amp are equal so va = vb . R R +R v0 = − 4 (vcm + vn ) + 4 1 va R1 R1 R v0 = − 4 (vcm + vn ) + R1 ( R4 + R1 ) R3 (vcm − v p ) R1 ( R2 + R3 ) when R4 R1 R4 +1 ( R4 + R1 ) R3 R3 R R R1 then = = × 3 = 4 R3 R2 R1 ( R2 + R3 ) R1 +1 R2 R2 so v0 = − R4 R R (vcm + vn ) + 4 (vcm − v p ) = − 4 (vn + v p ) R1 R1 R1 6-25
• PSpice Problems SP6-1: (a) v z = a vw + b v x + c v y The following three PSpice simulations show 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V -5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V Therefore v z = vw + 4 v x − 5 v y (b) When vw= 2 V: vz = 4 vx − 5 v y + 2 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V: 6-26
• 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V: -5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V: 6-27
• SP6-2 a) Using superposition and recognizing the inverting and noninverting amplifiers: vo = − 80  80  vs +  1 +  ( −2 ) = −3.2 vs − 8.4 25  25  b) Using the DC Sweep feature of PSpice produces the plot show below. Two points have been labeled in anticipation of c). c) Notice that the equation predicts ( −3.2 ) ( −5 ) − 8.4 = 7.6 and ( −3.2 ) ( 0 ) − 8.4 = −8.4 Both agree with labeled points on the plot. 6-28
• SP6-3 VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E-04 -7.000E-04 v34 = −1.5 − −12 × 10−6 ≅ −1.5 V v23 = 4.5 − ( −1.5 ) = 6 V v50 = 12 − 0 = 12 V io = −7 ×10−4 = −0.7 mA 6-29
• SP6-4 V4 is a short circuit used to measure io. The input of the VCCS is the voltage of the left-hand voltage source. (The nominal value of the input is 20 mV.) The output of the VCCS is io. A plot of the output of the VCCS versus the input is shown below. The gain of the VCVS is gain = 50 × 10−6 − ( −50 × 10−6 ) 1 A = ×10−3 V 100 × 10 − ( −100 × 10 ) 2 −3 −3 (The op amp saturates for the inputs larger than 0.1 V, limiting the operating range of this VCCS.) 6-30
• Verification Problems VP6-1 Apply KCL at the output node of the op amp io = v − ( −5 ) vo + o =0 10000 4000 Try the given values: io =−1 mA and vo = 7 V −1×10−3 ≠ 3.7 × 10−3 = 7 − ( −5 ) 7 + 10000 4000 KCL is not satisfied. These cannot be the correct values of io and vo. VP6-2 va = ( 4 ×103 )( 2 ×10−3 ) = 8 V 12 ×103 va = −1.2 ( 8 ) = −9.6 V 10 ×103 So vo = −9.6 V instead of 9.6 V. vo = − 6-31
• VP6-3 First, redraw the circuit as: Then using superposition, and recognizing of the inverting and noninverting amplifiers:  6  4   4 vo =  −  −  ( −3) + 1 +  ( 2 ) = −18 + 6 = −12 V  2  2   2 The given answer is correct. VP6-4 First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f, hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op amp are equal.) The given voltages satisfy all the node equations at nodes b, c and d: node b: 0− (−5) 0 0− 2 + + =0 10000 40000 4000 node c: 0− 2 2 −5 = +0 4000 6000 node d: 2 −5 5 5−11 = + 6000 5000 4000 Therefore, the analysis is correct. 6-32
• VP6-5 The given voltages satisfy the node equations at nodes b and e: node b: −.25− 2 −.25 −.25−( −5 ) + + =0 20000 40000 40000 node e: −2.5−( −0.25 ) −0.25 ≠ +0 9000 1000 Therefore, the analysis is not correct. Notice that −2.5−( +0.25 ) +0.25 = +0 9000 1000 So it appears that ve = +0.25 V instead of ve = −0.25 V. Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2 = 1/ 4 and K4 = 9. The given node voltages satisfy the equation −2.5 = vd = K 4 ( K v + K v ) = 10  1 ( 2 )+ 1 ( −5)    4 2  1 a 2 c None-the-less, the analysis is not correct. 6-33
• Design Problems DP6-1 From Figure 6.6-1g, this circuit is described by vo = R f i in . Since i out = Notice that i oa = i in + i in < (1250 ) i in 5000 = Rf vo 1 i out , we require = = , or Rf = 1250 Ω 5000 4 i in 5000 5 5 i in . To avoid current saturation requires i in < i sat or 4 4 4 i sat . For example, if isat = 2 mA, then iin < 1.6 mA is required to avoid current saturation. 5 DP6-2 3 3 3  3   12   3   12  vo = − vi + 3 = − vi + 1 +    5 − vi +  1 +   5 4 4 4  4   35   4   12 + 23  6-34
• DP6-3 1 1 1 1  (a) 12 v i + 6 = 24  v i + ( 5 )  ⇒ K 4 = 24, K 1 = , and K 2 = . Take Ra = 18 kΩ and 20 2 20 2  Rb = 1 kΩ to get (b) (c) (d) 6-35
• DP6-4 6-36
• DP6-5 We require a gain of 4 = 200 . Using an inverting amplifier: 20 ×10−3 Here we have 200 = − R2 10 ×103 + R1 . For example, let R1 = 0 and R2 = 1 MΩ. Next, using the noninverting amplifier: Here we have 200 = 1 + R2 R1 . For example, let R1 = 1 kΩ and R2 = 199 kΩ. The gain of the inverting amplifier circuit does not depend on the resistance of the microphone. Consequently, the gain does not change when the microphone resistance changes. 6-37
• Chapter 7 Energy Storage Elements Exercises Ex. 7.3-1 2 2<t <4 d  and i C ( t ) = 1 v s ( t ) = −1 4 < t < 8 dt  0 otherwise  2 t − 2 2 < t < 4  so i (t ) = i C ( t ) + i R ( t ) =  7 − t 4<t <8  0 otherwise  2 t − 4 2 < t < 4  i R (t ) = 1 v s (t ) =  8 − t 4<t <8  0 otherwise  Ex. 7.3-2 1 t 1 t v(t ) = ∫ i s (τ ) dτ + v(t 0 ) = ∫ i s (τ ) dτ − 12 1 0 C t0 3 t v(t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4 0 t v(t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10 4 In particular, v(4) = 36 V. In particular, v(10) = 0 V. t v(t ) = 3∫ 0 dτ + 0 = 0 for 10 < t 10 Ex. 7.4-1 Cv 2 1 2 = ( 2×10 −4 ) (100 ) = 1 J 2 2 + − vc ( 0 ) = vc ( 0 ) = 100 V W = 7-1
• Ex. 7.4-2 a) W ( t ) = W ( 0 ) + ∫ 0 vi dt t First, W ( 0 ) = 0 since v ( 0 ) = 0 Next, v( t ) = v( 0 ) + ∴ W (t ) = 1 t 4 t 4 ∫ 0 i dt = 10 ∫ 0 2 dt = 2×10 t C ∫ ( 2×10 ) t ( 2 )dt t 4 0 = 2 × 104 t 2 W (1s ) = 2 ×104 J = 20 kJ b) W (100s ) = 2 × 104 (100 ) 2 = 2 × 108 J = 200 MJ Ex. 7.4-3 We have v (0+ ) = v (0− ) = 3 V vc ( t ) = t 1 t 5t 5t 5t ∫ 0 i(t ) dt + vc (0) = 5 ∫ 0 3 e dt + 3 = 3 ( e −1)+3 = 3 e V, 0<t <1 C a) v(t ) = vR ( t ) + vc ( t ) = 5 i ( t ) + vc ( t ) = 15 e5t + 3 e5t = 18 e5t V, 0 < t < 1 b)  W (t ) t =0.2 s = 6.65 J 2  W ( t ) = 1 Cvc2 ( t ) = 1 × 0.2 ( 3e5t ) = 0.9e10t J ⇒  2 2  W ( t ) t =0.8 s = 2.68 kJ Ex. 7.5-1 7-2
• Ex. 7.5-2 v1 = v2 ⇒ dv1 dv2 i i C = ⇒ 1 = 2 ⇒ i1 = 1 i2 dt dt C1 C2 C2 C  KCL: i = i1 + i2 =  1 + 1 i2  C2  ⇒ i2 = C2 i C1 +C2 Ex. 7.5-3 (a) to (b) : (c) to (d) : 1 1 = mF , 1 1 1 9 + + 1 1 1 3 3 3 1 1 1 1 = + + 2 2 10 Ceq 9 (b) to (c) : 1+ ⇒ Ceq = 1 10 = mF , 9 9 10 mF 19 7-3
• Ex. 7.6-1  2 2<t <4 d  and v L ( t ) = 1 i s ( t ) = −1 4 < t < 8 dt  0 otherwise  2 t − 2 2 < t < 4  so v(t ) = v L ( t ) + v R ( t ) =  7 − t 4<t <8  0 otherwise  2 t − 4 2 < t < 4  4<t <8 v R (t ) = 1 is (t ) =  8 − t  0 otherwise  Ex. 7.6-2 i (t ) = 1 t 1 t ∫t 0 v s (τ ) dτ + i(t 0 ) = 1 ∫0 v s (τ ) dτ − 12 L 3 t i (t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4 0 t i (t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10 4 In particular, i(4) = 36 A. In particular, i(10) = 0 A. t i (t ) = 3∫ 0 dτ + 0 = 0 for 10 < t 10 Ex. 7.7-1 v = L di 1 d =   ( 4 t e−t ) = (1−t ) e−t V dt  4  dt P = vi =  (1−t ) e− t  ( 4 t e− t ) = 4 t (1−t ) e −2t W   2 1 11 W = Li 2 =   ( 4 t e− t ) = 2 t 2 e − 2t J 2 24 7-4
• v (t ) = L  0  2t  and i ( t ) =  −2( t − 2 )  0  di 1 di = dt 2 dt p (t ) = v (t ) i (t ) Ex. 7.7-2 t <0  0  0<t <1  1 ⇒ v( t ) =  1<t < 2  −1  0 t >2  t <0 0<t <1 1<t < 2 t >2 t <0  0 2t 0<t <1  =  2( t − 2 ) 1<t < 2  0 t >2  W ( t ) = W ( t0 ) + ∫ tt p( t ) dt 0 i (t ) = 0 for t < 0 ⇒ p ( t ) = 0 for t < 0 ⇒ W ( t0 ) = 0 0 < t < 1: W ( t ) = t ∫ 2 t dt = t 1<t < 2 : W ( t ) = W (1)+ ∫ 2 ( t − 2 ) dt = t 2 0 t 1 2 − 4t + 4 t >2 : W (t ) = W ( 2) = 0 Ex. 7.8-1 7-5
• Ex. 7.8-2 Ex. 7.8-3 i1 = 1 t ∫ v dt + i1 ( t0 ) , L1 t 0 i2 = 1 t ∫ v dt + i 2 ( t0 ) L 2 t0 but i1 ( t0 ) = 0 and i 2 ( t0 ) = 0 1 1 t 1  t 1 t t ∫ t v dt + ∫ t v dt =  + ∫ v dt  ∫ t v dt = L1 0 LP t 0 0  L1 L2  0 1 t 1 ∫ t v dt i L2 L 0 L1 ∴1 = 1 = = 1 t 1 1 i L1 + L2 + ∫ t v dt LP 0 L1 L2 i = i1 + i2 = 7-6
• Problems Section 7-3: Capacitors P7.3-1 v (t ) = v (0) + 1 t i (τ ) dτ C ∫0 and q = Cv In our case, the current is constant so ∴ Cv ( t ) = Cv ( 0 ) + i t t ∫ i (τ ) dτ . 0 −6 −6 q −Cv( 0 ) 150×10 −(15×10 )( 5 ) ∴ t= = = 3 ms i 25×10−3 P7.3-2 i (t ) = C d 1d 1 v (t ) = 12 cos ( 2t + 30° ) = (12 )( −2 ) sin ( 2t + 30° ) = 3cos ( 2t + 120° ) A dt 8 dt 8 P7.3-3 d ( 3×10 ) cos ( 500t + 45 ) = C dt −3 so ° 12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° ) = C ( 6000 ) cos ( 500t + 45° ) 3×10−3 1 1 = ×10−6 = µ F C= 3 6×10 2 2 7-7
• P7.3-4 v (t ) = 0 < t < 2 ×10−9 1 t 1 ∫0 i (τ ) dτ + v ( 0 ) = 2 ×10−12 C is ( t ) = 0 ⇒ v ( t ) = 2 × 10−9 < t < 3 × 10−9 1 2 × 10−12 t ∫ i (τ ) dτ − 10 −3 0 t ∫ 0 dτ − 10 0 −3 = −10−3 is ( t ) = 4 ×10−6 A t 1 4 ×10−6 ) dτ − 10−3 = −5 × 10−3 + ( 2 × 106 ) t −12 ∫2ns ( 2 × 10 In particular, v ( 3 × 10−9 ) = −5 ×10−3 + ( 2 × 106 ) ( 3 × 10−9 ) = 10−3 ⇒ v (t ) = 3 × 10−9 < t < 5 ×10−9 is ( t ) = −2 × 10−6 A t 1 −2 × 10−6 ) dτ + 10−3 = 4 × 10−3 − (106 ) t −12 ∫3ns ( 2 × 10 In particular, v ( 5 × 10−9 ) = 4 ×10−3 − (106 ) ( 5 × 10−9 ) = −10−3 V ⇒ v (t ) = 5 ×10−9 < t is ( t ) = 0 ⇒ v ( t ) = 1 2 × 10−12 ∫ t 5ns 0 dτ − 10−3 = −10−3 V P7.3-5 (b) (a)  0 0 < t <1  d  4 1< t < 2 i (t ) = C v(t ) =  dt  −4 2 < t < 3 0 3<t  t 1 t v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = ∫ i (τ ) dτ 0 C 0 t v ( t ) = ∫ 0 dτ + 0 = 0 V For 0 < t < 1, i(t) = 0 A so 0 For 1 < t < 2, i(t) = (4t − 4) A so ( t v ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = 2τ 2 − 4τ 1 ( ) t ) 1=2 t 2 − 4t + 2 V v(2) = 2 22 − 4 ( 2 ) + 2 = 2 V . For 2 < t < 3, i(t) = (−4t + 12) A so ( t v ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = −2τ 2 + 12τ 2 ( ) t ) 1 +2= ( −2 t 2 ) + 12 t − 14 V v(3) = −2 32 + 12 ( 3) − 14 = 4 V t For 3 < t, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 4 = 4 V 0 7-8
• P7.3-6 (a) (b)  0 0<t <2 d  i (t ) = C v(t ) = 0.1 2 < t < 6 dt 0 6<t  t 1 t v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ 0 C 0 t For 0 < t < 2, i(t) = 0 A so v ( t ) = 2 ∫ 0 dτ + 0 = 0 V 0 For 2 < t < 6, i(t) = 0.2 t − 0.4 V so ( t v ( t ) = 2 ∫ ( 0.2τ − 0.4 ) dτ + 0 = 0.2τ 2 − 0.8τ 1 ( ) t ) 2 =0.2 t 2 − 0.8 t + 0.8 V v(6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 V . For 6 < t, i(t) = 0.8 A so t v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6 t − 6.4 V 6 P7.3-7 v (t ) = v ( 0) + 1 t 4 t −3 −6τ ∫ 0 i(τ ) dτ = 25 + 2.5 ×10 ∫ 0 ( 6×10 ) e dτ C t = 25 + 150∫ 0 e −6τ dτ t  1  = 2 5 + 150  − e −6τ  = 50 − 25e−6t V  6 0 P7.3-8 v 1 = (1− 2e−2t ) × 10−3 = 25 (1− 2e −2t ) µ A 3 200×10 40 dv iC = C = 10×10−6 ( −2 ) ( −10 e−2t ) = 200 e−2 t µ A dt i = iR + i C = 200 e−2t + 25 − 50 e −2t iR = ( ) = 25 + 150e −2t µA 7-9
• Section 7-4: Energy Storage in a Capacitor P7.4-1 Given 0 t<2   i ( t ) = 0.2 ( t − 2 ) 2 < t < 6  0.8 t >6  The capacitor voltage is given by v (t ) = For t < 2 t 1 t i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ + v ( 0 ) 0 0.5 ∫ 0 t v ( t ) = 2 ∫ 0 dτ + 0 = 0 0 In particular, v ( 2 ) = 0. For 2 < t < 6 v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V t t 2 2 In particular, v ( 6 ) = 3.2 V. For 6 < t t v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V t 6 Now the power and energy are calculated as 0   2 p ( t ) = v ( t ) i ( t ) = 0.04 ( t − 2 )  1.28 ( t − 4 )  t<2 2<t <6 6<t and W (t ) = ∫ t 0 0  t<2  4  2<t<6 p (τ ) dτ =  0.01( t − 2 )  2 6<t 0.8 ( t − 4 ) − 0.64  7-10
• 7-11
• These plots were produced using three MATLAB scripts: capvol.m function v = CapVol(t) if t<2 v = 0; elseif t<6 v = 0.2*t*t - .8*t +.8; else v = 1.6*t - 6.4; end capcur.m function i = CapCur(t) if t<2 i=0; elseif t<6 i=.2*t - .4; else i =.8; end c7s4p1.m t=0:1:8; for k=1:1:length(t) i(k)=CapCur(k-1); v(k)=CapVol(k-1); p(k)=i(k)*v(k); w(k)=0.5*v(k)*v(k); end plot(t,i,t,v,t,p) text(5,3.6,'v(t), V') text(6,1.2,'i(t), A') text(6.9,3.4,'p(t), W') title('Capacitor Current, Voltage and Power') xlabel('time, s') % plot(t,w) % title('Energy Stored in the Capacitor, J') % xlabel('time, s') 7-12
• P7.4-2  ic ( 0 ) = 0.2 A dv  = (10×10−6 ) ( −5 )( −4000 ) e −4000t = 0.2e−4000t A ⇒  −19 dt  ic (10ms ) = 8.5×10 A  1 W ( t ) = Cv 2 ( t ) and v ( 0 ) = 5 − 5e0 = 0 ⇒ W ( 0 ) = 0 2 −3 v (10×10 ) = 5 − 5 e −40 = 5 − 21.2 × 10−18 ≅ 5 ⇒ W (10 ) = 1.25×10−4 J ic = C P7.4–3 dvc so read off slope of vc (t ) to get i (t ) dt p (t ) = vc (t ) i (t ) so multiply vc (t ) & i(t ) curves to get p (t ) i (t ) = C P7.4-4 1 t 1 t π  ∫0 i dτ = vc ( 0 ) + 2 ∫0 50 cos10t + 6 dτ = C   π 5 Now since vc ( t )ave = 0 ⇒ vc ( 0 ) − sin = 0 ⇒ vc ( t ) 2 6 vc ( t ) = vc ( 0 ) + ( 2×10 ) ( 2.5) = 6.25 µ J 1 = C v2 = c max 2 2 −6 ∴ Wmax 5 π 5 π      vc ( 0 ) − 2 sin 6  + 2 s in 10t + 6    π 5  = sin 10t +  V 2 6  2 First non-negative t for max energy occurs when: 10t + π 6 = π 2 ⇒t = π 30 = 0.1047 s 7-13
• P7.4-5 Max. charge on capacitor = C v = (10×10−6 ) ( 6 ) = 60 µ C ∆q 60×10−6 = = 6 sec to charge i 10×10−6 1 1 2 stored energy = W = C v 2 = (10×10−6 ) ( 6 ) =180 µ J 2 2 ∆t = Section 7-5: Series and Parallel capacitors P7.5-1 2µ F 4 µ F = 6µ F 6µ F in series with 3µ F = i (t ) = 2 µ F 6µ F⋅3µ F = 2µ F 6µ F+3µ F d (6 cos100t ) = (2×10−6 ) (6) (100) (− sin100t ) A = −1.2 sin100t mA dt P7.5-2 4 µ F in series with 4 µ F = 4 µ F×4 µ F = 2 µF 4 µ F+4 µ F 2 µF 2 µF = 4 µF 4 µ F in series with 4 µ F = 2 µ F i (t ) =(2×10−6 ) d (5+ 3 e −250t ) = (2×10−6 ) (0+ 3(−250) e −250t ) A = −1.5 e −250t mA dt P7.5-3 C in series with C = C C C ⋅C C = C +C 2 C 5 = C 2 2 C⋅ 5 C 5 5 2 C in series with C = = C 5C 2 7 C+ 2 5  d 5  (25×10−3 ) cos 250t =  C  (14sin 250t ) =  C (14)(250) cos 250t  7  dt 7  so 25×10−3 = 2500 C ⇒ C = 10×10−6 = 10 µF 7-14
• Section 7-6: Inductors P7.6-1 di = 2 0 0 [1 0 0 ( 4 0 0 ) c o s 4 0 0 t ] V dt 8 ×1 0 6 V = 4 ×1 0 6 V ∴ v m ax = 8 ×1 0 6 V th u s h a ve a fie ld o f m m 2 6 w h ic h e x ce e d s d ie le c tric stre n g th in a ir o f 3 ×1 0 V /m ∴ W e g e t a d isc h a rg e a s th e a ir is io n iz e d . F in d m a x . v o ltag e a c ro ss c o il: v (t ) = L P7.6-2 v=L di + R i = (.1) (4e− t − 4te− t ) + 10(4te− t ) = 0.4 e − t + 39.6t e − t V dt P7.6-3 (a) (b)  0 0 < t <1  4 1< t < 2 d  v(t ) = L i (t ) =  dt  −4 2 < t < 3  3<t 0 i (t ) = t 1 t v (τ ) dτ + i ( 0 ) = ∫ v (τ ) dτ 0 L ∫0 For 0 < t < 1, v(t) = 0 V so t i ( t ) = ∫ 0 dτ + 0 = 0 A 0 For 1 < t < 2, v(t) = (4t − 4) V so t t i ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = ( 2τ 2 − 4τ ) =2 t 2 − 4 t + 2 A 0 1 i (2) = 4 ( 22 ) − 4 ( 2 ) + 2 = 2 A For 2 < t < 3, v(t) = −4t + 12 V so t t i ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = ( −2τ 2 + 12τ ) +2= ( −2 t 2 + 12 t − 14 ) A 2 2 i (3) = −2 ( 32 ) + 12 ( 3) − 14 = 4 A t For 3 < t, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 4 = 4 A 3 7-15
• P7.6-4 v (t ) = (250 × 10 −3 ) d (120 × 10 −3 ) sin(500t − 30° ) = (0.25)(0.12)(500) cos(500 t − 30° ) dt = 15 cos(500t − 30° ) P7.6-5 iL (t ) = for 1 5 ×10−3 0< t < 1 µ s iL (t ) = 1 5×10−3 t ∫ v (τ ) dτ 0 s − 2 ×10−6 vs (t ) = 4 mV  4×10−3  −6 −6 4 ×10−3 dτ − 2×10−6 =   t − 2×10 = 0.8− 2×10 A ∫0 5×10−3   t  4×10−3  6 iL (1µs) =  1×10−6 )  − 2×10−6 = − ×10−6 A =−1.2 A −3 ( 5  5×10  for 1µ s <t < 3 µ s vs (t ) = −1 mV t 1 6 1×10−3 6 −1×10−3 ) dτ − ×10−6 =− (t −1×10−6 ) − ×10−6 =( −0.2 t −10−6 ) A −3 ∫1µ s ( −3 5×10 5 5×10 5  1×10−3  iL ( 3µ s ) =  − + 3×10−6  − 1×10−6 = −1.6 µA −3  5×10  iL (t ) = for 3µ s < t vs (t ) = 0 so iL (t ) remains − 1.6 µA 7-16
• P7.6-6 v(t ) = ( 2 ×10 ) i 3 s for 0<t <1 µ s (t ) + d ( 4 ×10 ) dt −3 is (t ) (in general)  1×10−3  d is (t ) = (1)  t = 103 t ⇒ is (t ) = 1×103 −6  dt  1×10  v(t ) = (2×103 )(1×103 ) t + 4×10−3 (1×103 ) = ( 2×106 t + 4 ) V d is (t ) = 0 dt v(t ) = (2×103 )(1×10−3 ) + ( 4×10−3 )×0 = 2 V for 1µ s <t < 3µ s is (t ) = 1 mA ⇒  1×10−3  t ⇒ for 3µ s< t < 5µ s is (t ) = 4×10−3 − −6   1×10  d 1×10−3 =−103 is (t ) = − −6 dt 1×10 v(t ) = ( 2×103 )( 4×10−3 −103 t )+ 4×10−3 ( −103 ) = 4 − ( 2×106 ) t when 5µ s <t< 7µ s is (t ) = −1×10−3 and v(t ) = ( 2×103 )(10−3 ) = − 2 V d is (t ) = 0 dt  1×10−3  d when 7µ s< t < 8µ s is (t ) =  t − 8×10−3 ⇒ is (t ) = 1×103 −6  dt  1×10  v(t ) = ( 2 × 103 )(103 t − 8 × 10−3 ) + ( 4 × 10−3 )(103 ) = −12 + ( 2 × 106 ) t when 8µ s < t , then is (t ) = 0 ⇒ d is (t ) = 0 dt v(t )= 0 P7.6-8 (a) (b)  0 0<t <2 d  v(t ) = L i (t ) = 0.1 2 < t < 6 dt 0 6<t  t 1 t i ( t ) = ∫ v (τ ) dτ + i ( 0 ) = 2 ∫ v (τ ) dτ 0 L 0 t For 0 < t < 2, v(t) = 0 V so i ( t ) = 2 ∫ 0 dτ + 0 = 0 A 0 For 2 < t < 6, v(t) = 0.2 t − 0.4 V so t t i ( t ) = 2∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 A 2 2 ( ) i (6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 A . For 6 < t, v(t) = 0.8 V so 7-17
• t i ( t ) = 2 ∫ 0.8 dτ + 3.2 = (1.6 t − 6.4 ) A 6 Section 7-7: Energy Storage in an Inductor P7.7-1 t<0 0 d  v( t ) =100×10 i ( t ) = 0.4 0≤ t ≤1 dt 0 t>1  t <0 0  p( t ) =v( t ) i( t ) = 1.6t 0≤t ≤1 0 t >1  −3 W (t ) = ∫ t 0  0  p (τ ) dτ = 0.8t 2  0.8  t <0 0<t <1 t >1 7-18
• P7.7-2  d  p (t ) = v (t ) i (t ) = 5 (4sin 2t )  (4sin 2t )  dt  = 5 (8cos 2t ) (4sin 2t ) = 80 [2 cos 2t sin 2t ] = 80 [sin(2t + 2t ) + sin(2t − 2t )] = 80 sin 4t W t t 80 W(t ) = ∫ p(τ ) dτ = 80∫ sin4τ dτ = − [cos 4τ |t0 ] = 20 (1 − cos 4t ) 0 0 4 P7.7-3 t 1 6 cos 100τ dτ + 0 25×10−3 ∫ 0 6 t [sin 100τ | 0 ] = 2.4sin100 t = −3 (25×10 )(100) p(t ) = v(t ) i(t ) = (6 cos100 t )(2.4 sin100t ) i(t ) = = 7.2 [ 2(cos100 t )(sin100 t ) ] = 7.2 [sin 200 t + sin 0] = 7.2 sin 200 t t t W (t ) = ∫0 p (τ ) dτ = 7.2 ∫0 sin 200τ dτ 7.2  cos 200τ |t0   200  = 0.036[1 − cos 200t ] J = 36 [1 − cos 200t ] mJ =− Section 7-8: Series and Parallel Inductors P7.8-1 6×3 = 2 H and 2 H + 2 H = 4 H 6+3 1 t 6 sin100τ | t0  = 0.015sin100 t A = 15sin100 t mA i (t ) = ∫ 0 6 cos100τ dτ =  4 4×100  6H 3H = P7.8-2 (8×10 )×(8×10 ) −3 4 mH + 4 mH = 8 mH , 8mH 8mH = −3 8×10−3 +8×10−3 = 4 mH and 4 mH + 4 mH = 8 mH d v(t ) = (8×10−3 ) (5+ 3e −250t ) = (8×10−3 ) (0+ 3(−250) e −250t ) =−6 e −250t V dt 7-19
• P7.8-3 L⋅ L L L 5 L = and L + L + = L+ L 2 2 2 5  d 25cos 250 t =  L  (14×10−3 ) sin 250 t =  5 L (14×10−3 )(250) cos 250 t    2  dt 2  25 so L = = 2.86 H 5 −3 (14×10 ) (250) 2 L L = ( ) Section 7-9: Initial Conditions of Switched Circuits P7.9-1 Then i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 12 V Next 7-20
• P7.9-2 Then i L ( 0+ ) = i L ( 0− ) = 1 mA and v C ( 0+ ) = v C ( 0− ) = 6 V Next P7.9-3 Then i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 0 V Next 7-21
• P7.9-4 at t = 0− KVL: − vc (0− ) + 32 − 15 = 0 ⇒ vc (0− ) = vc (0+ ) =17 V at t = 0+ Apply KCL to supernode shown above: −15 − 9 15 ic ( 0+ ) + − + 0.003 = 0 ⇒ ic ( 0+ ) = 6 mA 4000 5000 Then ic ( 0+ ) 6 ×10−3 dvc V = = = 3000 −6 2 ×10 s dt t =0+ C 7-22
• Section 7-10: Operational amplifier Circuits and Linear Differential Equations P7.10-1 P7.10-2 7-23
• P7.10-3 P7.10-4 7-24
• Verification Problems VP7-1 We need to check the values of the inductor current at the ends of the intervals. 1 ? + 0.065 = 0.025 ( Yes!) at t = 1 0.025 = − 25 at t = 3 − 3 ? 3 − 0.115 + 0.065 = 25 50 −0.055 = −0.055 ( Yes!) 9 ? − 0.115 = 0.065 50 at t = 9 0.065 = 0.065 ( Yes!) The given equations for the inductor current describe a current that is continuous, as must be the case since the given inductor voltage is bounded. VP7-2 We need to check the values of the inductor current at the ends of the intervals. 1 1 ? + 0.025 = − + 0.03 ( Yes!) at t = 1 − 200 100 4 ? 4 − 0.03 ( No!) + 0.03 = 100 100 The equation for the inductor current indicates that this current changes instantaneously at t = 4s. This equation cannot be correct. at t = 4 − Design Problems DP7-1 i (t ) d = 0.5 F . v ( t ) = −6 e −3t is proportional to i(t) so the element is a capacitor. C = d dt v (t ) dt v (t ) d = 0.5 H . b) i ( t ) = −6 e −3t is proportional to v(t) so the element is an inductor. L = d dt i (t ) dt v (t ) c) v(t) is proportional to i(t) so the element is a resistor. R = = 2 Ω. i (t ) a) 7-25
• DP7-2 1.131cos ( 2t + 45° ) = 1.131  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t )    = 0.8 cos 2 t − 0.8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. The noticing that ∫ t −∞ t v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t −∞ d d v ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with a capacitor to get the minus sign. Then 4 cos 2 t 4 cos 2 t R= = = 5 Ω and i1 (t ) 0.8 cos 2 t −0.8 sin 2 t i2 (t ) = = 0.1 F C= d −8 sin 2 t 4 cos 2 t dt 1.131cos ( 2t − 45° ) = 1.131  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t )    = 0.8 cos 2 t + 0.8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. Then noticing that t t −∞ −∞ ∫ v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t d d v ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with an inductor to get the plus sign. Then 4 cos 2 t 4 cos 2 t = = 5 Ω and i1 (t ) 0.8 cos 2 t R= L= ∫ t −∞ 4 cos 2 t dτ i2 (t ) = 2 sin 2 t = 2.5 H 0.8 sin 2 t 7-26
• DP7-3 a) 11.31cos ( 2t + 45° ) = 11.31  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t )    = 8 cos 2 t − 8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. The noticing that t t −∞ −∞ ∫ i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t d d i ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with an inductor to get the minus sign. Then v (t ) 8 cos 2 t v2 (t ) −8 sin 2 t R= 1 = = 2 Ω and L = = =1H d 4 cos 2 t 4 cos 2 t 4 cos 2 t −8 sin 2 t dt b) 11.31cos ( 2t + 45° ) = 11.31  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t )    = 8 cos 2 t + 8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. The noticing that ∫ t −∞ t i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t −∞ d d i ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with a capacitor to get the minus sign. Then v (t ) 8 cos 2 t R= 1 = = 2 Ω and C = 4 cos 2 t 4 cos 2 t ∫ t −∞ 4 cos 2 t dτ v2 (t ) = 2 sin 2 t = 0.25 F 8 sin 2 t 7-27
• DP7-4 iL ( 0− ) = 0 at t=0− By voltage division: vC ( 0− ) = We require vC ( 0− ) = 3 V so VB 4 VB = 12 V at t=0+ Now we will check First: and dvC dt t = 0+ iL ( 0+ ) = iL ( 0− ) = 0 vC ( 0+ ) = vC ( 0− ) = 3 V Apply KCL at node a: VB − vC ( 0+ ) + + iL ( 0 ) + iC ( 0 ) = 3 0 + iC ( 0+ ) = 12 − 3 ⇒ iC ( 0+ ) = 3 A 3 Finally iC ( 0 dvC = dt t =0+ C as required. + )= 3 V = 24 0.125 s DP7-5 1 1 2 L i L2 = C v C where iL and vC are the steady state inductor current and capacitor 2 2 v voltage. At steady state, i L = C . Then R We require 2 v  L  C  = C vC2 R ⇒ C= L R2 ⇒ R = L = C 10−2 = 10−6 104 = 10 2 Ω so R = 100 Ω . 7-28
• Chapter 8 – The Complete Response of RL and RC Circuit Exercises Ex. 8.3-1 Before the switch closes: After the switch closes: 2 = 8 Ω so τ = 8 ( 0.05 ) = 0.4 s . 0.25 Finally, v (t ) = voc + (v (0) − voc ) e− t τ = 2 + e−2.5 t V for t > 0 Therefore R t = Ex. 8.3-2 Before the switch closes: After the switch closes: 8-1
• 2 6 = 8 Ω so τ = = 0.75 s . 0.25 8 1 1 −1.33 t Finally, i (t ) = isc + (i (0) − isc ) e−t τ = + e A for t > 0 4 12 Therefore R t = Ex. 8.3-3 At steady-state, immediately before t = 0:  12  10   i ( 0) =     = 0.1 A  10 + 40   16+ 40||10  After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be so I sc = 0.3 A, Rt = 40 Ω, τ = Finally: L 20 1 s = = 40 2 Rt i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A 8-2
• Ex. 8.3-4 At steady-state, immediately before t = 0 After t = 0, we have: so Voc = 12 V, Rt = 200 Ω, τ = Rt C = (200)(20 ⋅10 −6 ) = 4 ms Finally: v(t ) = (12 − 12) e −t 4 + 12 = 12 V Ex. 8.3-5 Immediately before t = 0, i (0) = 0. After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the inductor current: I sc = 0.2 A, Rt = 45 Ω, τ = L 25 5 = = Rth 45 9 So i (t ) = 0.2 (1 − e−1.8t ) A Now that we have the inductor current, we can calculate v(t): d i (t ) dt = 8(1 − e −1.8t ) + 5(1.8)e −1.8t v(t ) = 40 i (t ) + 25 = 8 + e −1.8t V for t > 0 8-3
• Ex. 8.3-6 At steady-state, immediately before t = 0 so i(0) = 0.5 A. After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get I sc = 93.75 mA, Rt = 640 Ω, τ = L .1 1 s = = Rt 640 6400 i (t ) = 406.25 e −6400t + 93.75 mA Finally: v (t ) = 400 i (t ) + 0.1 Ex. 8.4-1 d i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t ) dt = 37.5 − 97.5e −6400t V τ = ( 2×103 )(1×10−6 ) = 2 ×10−3 s vc (t ) = 5 + (1.5−5 ) e −500 t V vc (0.001) = 5 − 3.5e − 0.5 = 2.88 V So vc (t ) will be equal to vT at t = 1 ms if v = 2.88 V. T 8-4
• Ex. 8.4-2 iL (0) = 1 mA, I sc = 10 mA    ⇒ iL ( t ) = 10 − 9 e L Rt =500 Ω, τ =  500  vR (t ) = 300 iL ( t ) = 3 − 2.7 e − 500 t L −500 t L mA V We require that vR ( t ) = 1.5 V at t = 10 ms = 0.01 s . That is 1.5 = 3 − 2.7e Ex. 8.6-1 0 < t < t1 − 500 (0.01) L ⇒ L= 5 = 8.5 H 0.588 v(t ) = v(∞)+Ae− t / RC where v(∞) = (1 A)(1 Ω ) = 1 V v(t ) = 1 + A e− t /(1)(.1) = 1 + A e−10 t Now v(0− ) = v(0+ ) = 0 = 1 + A ⇒ A = −1 ∴ v(t ) =1−e−10t V t > t1 t1 = 0.5 s, v(t ) = v(t1 ) e − t −.5 (1)(.1) = v(0.5) e−10(t −.5) Now v(0.5) 1 − e −10(0.5) = 0.993 V ∴ v(t ) = 0.993 e−10(t −0.5) V 8-5
• Ex. 8 6-2 t < 0 no sources ∴ v(0− ) = v(0+ ) = 0 0 < t <t1 − t v (t ) = v (∞ ) + A e − t / RC = v (∞ ) + A e 2 × 1 0 5 (1 0 − 7 ) where for t = ∞ (steady-state) ∴capacitor becomes an open ⇒ v(∞) = 10 V v(t ) = 10 + Ae −50t Now v(0) = 0 = 10 + A ⇒ A = −10 ∴ v(t ) = 10(1− e −50t ) V t > t1 , t1 = .1 s v(t ) = v(.1) e−50(t −.1) where v(.1) = 10(1− e −50(.1) ) = 9.93 V ∴ v(t ) = 9.93e−50( t −.1) V Ex. 8.6-3 For t < 0 i = 0. For 0 < t < 0.2 s KCL: −5 + v / 2 + i = 0   di  + 10i = 50 di also: v = 0.2  dt dt  −10 t ∴ i (t ) = 5+ Ae i (0) = 0 = 5+ A ⇒ A = −5 so we have i (t ) = 5 (1−e−10t ) A For t > 0.2 s i (0.2) = 4.32 A ∴ i(t ) = 4.32e−10(t −.2) A 8-6
• Ex. 8.7-1 vs ( t ) = 10sin 20t V  d v( t )  KVL: −10sin 20t +10 .01  + v( t ) = 0 dt   d v( t ) ⇒ +10 v( t ) = 100sin 20t dt Natural response: vn (t ) = Ae−t τ where τ = Rt C ∴ vn (t ) = Ae−10t Forced response: try v f (t ) = B1 cos 20t + B2 sin 20 t Plugging v f (t ) into the differential equation and equating like terms yields: B1 = − 4 & B2 = 2 Complete response: v(t ) = vn (t ) + v f (t ) v(t ) = Ae −10t − 4 cos 20t + 2 sin 20t Now v(0− ) = v(0+ ) = 0 = A − 4 ∴ A = 4 ∴ v(t ) = 4 e −10t − 4 cos 20t + 2sin 20 t V Ex. 8.7-2 is ( t ) = 10 e −5t for t > 0 KCL at top node: −10e −5t + i ( t ) + v( t ) /10 = 0 Now v ( t ) = 0.1 Natural response: in (t ) = Ae −t τ di ( t ) di ( t ) ⇒ +100 i ( t ) = 1000 e −5t dt dt where τ = L R t ∴ in (t ) = Ae −100t Forced response: try i f (t ) = Be −5t & plug into the differential equation −5 Be −5t + 100 Be −5t = 1000e −5t Complete response: i (t ) = Ae − −100 t + 10.53e ⇒ B = 10.53 −5t + Now i (0 ) = i (0 ) = 0 = A + 10.53 ⇒ A = −10.53 ∴ i (t ) = 10.53 (e −5t − e −100t ) A Ex. 8.7-3 When the switch is closed, the inductor current is iL = vs / R = vs . When the switch opens, the inductor current is forced to change instantaneously. The energy stored in the inductor instantaneously dissipates in the spark. To prevent the spark, add a resistor (say 1 kΩ) across the switch terminals. 8-7
• Problems Section 8.3: The Response of a First Order Circuit to a Constant Input P8.3-1 Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the initial capacitor voltage, v (0). By voltage division v (0) = 6 (12 ) = 4 V 6+6+6 Next, consider the circuit after the switch closes. The closed switch is modeled as a short circuit. We need to find the Thevenin equivalent of the part of the circuit connected to the capacitor. Here’s the circuit used to calculate the open circuit voltage, Voc. Voc = 6 (12 ) = 6 V 6+6 Here is the circuit that is used to determine Rt. A short circuit has replaced the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. ( 6 )( 6 ) = 3 Ω Rt = 6+6 Then τ = R t C = 3 ( 0.25 ) = 0.75 s Finally, v ( t ) = Voc + ( v ( 0 ) − Voc ) e−t / τ = 6 − 2 e−1.33 t V for t > 0 8-8
• P8.3-2 Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. An inductor in a steady-state dc circuit acts like an short circuit, so a short circuit replaces the inductor. The current in that short circuit is the initial inductor current, i(0). i ( 0) = 12 =2 A 6 Next, consider the circuit after the switch opens. The open switch is modeled as an open circuit. We need to find the Norton equivalent of the part of the circuit connected to the inductor. Here’s the circuit used to calculate the short circuit current, Isc. I sc = 12 =1 A 6+6 Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. ( 6 + 6 )( 6 ) = 4 Ω Rt = ( 6 + 6) + 6 Then L 8 τ= = =2 s Rt 4 Finally, i ( t ) = I sc + ( i ( 0 ) − I sc ) e −t / τ = 1 + e −0.5 t A for t > 0 8-9
• P8.3-3 Before the switch closes: After the switch closes: −6 = 3 Ω so τ = 3 ( 0.05 ) = 0.15 s . −2 Finally, v (t ) = voc + (v (0) − voc ) e− t τ = −6 + 18 e−6.67 t V for t > 0 Therefore R t = P8.3-4 Before the switch closes: After the switch closes: 8-10
• −6 6 = 3 Ω so τ = = 2 s . −2 3 t − 10 Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 + e−0.5 t A for t > 0 3 Therefore R t = P8.3-5 Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the circuit connected to the capacitor can be replaced by it's Thevenin equivalent circuit to get: Therefore τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s . Next, v C (t ) = voc + (v (0) − voc ) e − t τ = 10 − 5 e−12.5 t V for t > 0 Finally, v 0 (t ) = vC (t ) = 10 − 5 e −12.5 t V for t > 0 8-11
• P8.3-6 Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the circuit connected to the capacitor can be replaced by it's Norton equivalent circuit to get: Therefore τ = 5 = 0.25 ms . 20 × 103 Next, i L (t ) = isc + (i L (0) − isc ) e Finally, vo ( t ) = 5 − t τ = 0.5 − 0.25 e−4000 t mA for t > 0 d i L ( t ) = 5 e −4000 t V dt for t > 0 P8.3-7 At t = 0− (steady-state) Since the input to this circuit is constant, the inductor will act like a short circuit when the circuit is at steady-state: for t > 0 iL ( t ) = iL ( 0 ) e − ( R L ) t = 6 e−20t A 8-12
• P8.3-8 Before the switch opens, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and voltages, including the capacitor voltage, will have constant values. Opening the switch disturbs the circuit. Eventually the disturbance dies out and the circuit is again at steady state. All the element currents and voltages will again have constant values, but probably different constant values than they had before the switch opened. Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the capacitor voltage, vo(t). Because the circuit is at steady state, the value of the capacitor voltage will be constant. This constant is the value of the capacitor voltage just before the switch opens. In the absence of unbounded currents, the voltage of a capacitor must be continuous. The value of the capacitor voltage immediately after the switch opens is equal to the value immediately before the switch opens. This value is called the initial condition of the capacitor and has been labeled as vo(0). There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equal to the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore vo ( 0 ) = Vs The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives vo ( 0 ) = 2 + 8 e0 = 10 V Consequently, Vs = 10 V 8-13
• Next, consider the circuit after the switch opens. Eventually (certainly as t →∞) the circuit will again be at steady state. Here is the circuit at t = ∞, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the steady-state capacitor voltage, vo(∞). There is no current in the horizontal resistor and vo(∞) is equal to the voltage across the vertical resistor. Using voltage division, 10 vo ( ∞ ) = (10 ) R + 10 The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives vo ( ∞ ) = 2 + 8 e −∞ = 2 V Consequently, 10 (10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω R + 10 −t τ Finally, the exponential part of vo(t) is known to be of the form e where τ = R t C and 2= Rt is the Thevenin resistance of the part of the circuit connected to the capacitor. Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. R t = 10 + so ( 40 )(10 ) = 18 40 + 10 Ω τ = R t C = 18 C From the equation for vo(t) t −0.5 t = − ⇒ τ =2s τ Consequently, 2 = 18 C ⇒ C = 0.111 = 111 mF 8-14
• P8.3-9: Before the switch closes, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and voltages, including the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting out the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All the element currents and voltages will again have constant values, but probably different constant values than they had before the switch closed. The inductor current is equal to the current in the 3 Ω resistor. Consequently, − 0.35 t vo (t ) 6 − 3 e − 0.35 t = = 2− e A when t > 0 i (t ) = 3 3 In the absence of unbounded voltages, the current in any inductor is continuous. Consequently, the value of the inductor current immediately before t = 0 is equal to the value immediately after t = 0. Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(0). Apply KVL to the loop to get R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0 ⇒ i (0) = 24 R1 + R 2 + 3 The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives i ( 0 ) = 2 − e0 = 1 A Consequently, 1= 24 ⇒ R1 + R 2 = 21 R1 + R 2 + 3 Next, consider the circuit after the switch closes. Here is the circuit at t = ∞, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R1. 8-15
• An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to the loop to get 24 R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) = R2 + 3 The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives i ( ∞ ) = 2 − e −∞ = 2 A Consequently 2= 24 ⇒ R2 = 9 Ω R2 + 3 Then R1 = 12 Ω Finally, the exponential part of i(t) is known to be of the form e −t τ where τ = L and Rt Rt is the Thevenin resistance of the part of the circuit that is connected to the inductor. Here is shows the circuit that is used to determine Rt. A short circuit has replaced combination of resistor R1 and the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. R t = R 2 + 3 = 9 + 3 = 12 Ω so τ= L L = R t 12 From the equation for i(t) −0.35 t = − t τ ⇒ τ = 2.857 s Consequently, 2.857 = L ⇒ L = 34.28 H 12 8-16
• P8.3-10 First, use source transformations to obtain the equivalent circuit for t > 0: for t < 0: So iL ( 0 ) = 2 A, I sc 1 L 1 s = 0, Rt = 3 + 9 = 12 Ω, τ = = 2 = 24 Rt 12 and iL ( t ) = 2e−24t t >0 Finally v ( t ) = 9 iL ( t ) = 18 e−24t t >0 8-17
• Section 8-4: Sequential Switching P8.4-1 Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc = 5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore v (t ) = voc + (v (0) − voc ) e−t τ = 5 + 5e−5 t V for 0 < t < 1.5 s At t =1.5 s, v (1.5) = 5 + 5e ( τ = 8 × 0.05 = 0.4 s . Therefore −0.05 1.5) = 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω so v (t ) = voc + (v (1.5) − voc ) e −( t −1.5) τ = 10 − 5 e −2.5 ( t −1.5 ) V for 1.5 s < t Finally −5 t  for 0 < t < 1.5 s  5+5 e V v (t ) =  −2.5 ( t −1.5) for 1.5 s < t V 10 − 5 e  P8.4-2 Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get: Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2 12 A and Rt = 6 Ω so τ = = 2 s . Therefore 6 i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s 8-18
• At t =1.5 s, i (1.5) = 2 + e −0.5 (1.5) = 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ = Therefore i (t ) = isc + (i (1.5) − isc ) e −( t −1.5 ) τ = 3 − 0.53 e −0.667 ( t −1.5 ) 12 = 1.5 s . 8 V for 1.5 s < t Finally  2 + e −0.5 t A for 0 < t < 1.5 s  i (t ) =  −0.667 ( t −1.5 ) for 1.5 s < t A 3 − 0.53 e  P8.4-3 At t = 0−: KVL : − 52 + 18 i + (12 8) i = 0 ⇒ i (0− ) =104 39 A  6  + ∴ iL = i   = 2 A = iL (0 )  6+ 2  For 0 < t < 0.051 s iL (t ) = iL (0) e −t τ where τ = L R t R t = 6 12 + 2 = 6 Ω iL (t ) = 2 e −6t A  6 +12  −6 t ∴ i (t ) = iL (t )   =6e A  6  For t > 0.051 s i L ( t ) = i L (0.051) e − ( R L ) ( t − 0.051) i L (0.051) = 2 e − 6 (.051) = 1.473 A i L ( t ) = 1.473 e − 14 ( t − 0.051) A i ( t ) = i L ( t ) = 1.473 e − 14 ( t − 0.051) A 8-19
• P8.4-4 At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 µF capacitor is 3 V. The charge stored by the capacitor is q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C 0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at t = 0+). The voltage across the parallel capacitors is determined by considering charge conservation: q ( 0+ ) = (100 µ F) v ( 0+ ) + (400 µ F) v ( 0+ ) v (0 + ) = 100 ×10 ( ) q ( 0+ ) −6 + 400 ×10−6 = q ( 0− ) 500 × 10−6 = 300 × 10−6 500 × 10−6 v 0+ = 0.6 V 10 ms < t < l s: Combine 100 µF & 400 µF in parallel to obtain v(t ) = v ( 0+ ) e − (t −.01) RC 3 = 0.6e − (t −.01) (10 ) (5 x10 v(t ) = 0.6 e−2(t −.01) V −4 ) P8.4-5 For t < 0: Find the Thevenin equivalent of the circuit connected to the inductor (after t >0). First, the open circuit voltage: 40 =1 A 20 + 20 voc = 20 i1 − 5 i1 = 15 V i1 = 8-20
• Next, the short circuit current: 20 i 1 = 5 i 1 ⇒ i 1 = 0 i sc + 0 = 40 ⇒ i sc = 2 A 20 Then voc 15 = = 7.5 Ω i sc 2 Replace the circuit connected to the inductor by its Norton equivalent circuit. First L 15 × 10−3 1 τ= = = Rt 7.5 500 Next i ( t ) = 2 − 2 e − 500 t A t >0 Rt = After t = 0, the steady state inductor current is 2 A. 99% of 2 is 1.98. 1.98 = 2 − 2 e − 500 t P8.4-6 ⇒ t = 9.2 ms v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s ∴ v ( t ) = 5 e −10 t V for t > 0 2.5 = 5 e −10 t1 i (t1 ) = v (t1 ) 100 ×10 3 t 1 = 0.0693 s = 2.5 = 25 µ A 100 × 103 8-21
• Section 8-5: Stability of First Order Circuits P8.5-1 This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as 100  v T (400− R) 100 iT  = 100+ 400  ⇒ Rt = iT 100+ 400  vT = 400 i (t ) − R i (t )  i (t ) = The circuit is stable when R < 400 Ω. P8.5-2 The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as Ohm’s law: v ( t ) = 1000 iT ( t ) KVL: A v ( t ) + vT ( t ) − v ( t ) − 4000 iT ( t ) = 0 ∴ vT ( t ) = (1 − A ) 1000 iT ( t ) + 4000 iT ( t ) vT ( t ) = ( 5 − A ) ×1000 iT ( t ) The circuit is stable when A < 5 V/V. Rt = P8.5-3 The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as v (t ) Ohm’s law: i ( t ) = − T 6000 v (t ) KCL: i ( t ) + B i ( t ) + i T ( t ) = T 3000  v ( t )  vT ( t ) ∴ i T ( t ) = − ( B + 1)  − T +  6000  3000 ( B + 3) vT ( t ) = 6000 vT ( t ) 6000 Rt = = iT ( t ) B + 3 The circuit is stable when B > −3 A/A. 8-22
• P8.5-4 The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as (1000 )( 4000 ) i (t ) = 800 i (t ) v(t ) = T T 1000 + 4000 vT (t ) = v(t ) − Av(t ) = (1 − A ) v(t ) vT (t ) = 800 (1 − A) iT (t ) The circuit is stable when A < 1 V/V. Rt = 8-23
• Section 8-6: The Unit Step Response P8.6-1 The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0 where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is −a 0 vo ( 0 ) = A + B e ( ) = A + B The steady-state circuit for t < 0. Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives A+ B = 8 V Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is vo ( ∞ ) = A + B e The steady-state circuit for t > 0. − a (∞) =A Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives A = -7 V Therefore B = 15 V 8-24
• The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit connected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt = 6 Ω Therefore a= 1 1 = 2.5 −3 s ( 6 ) ( 66.7 ×10 ) (The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .) Putting it all together: 8 V for t ≤ 0  vo ( t ) =  − 2.5 t V for t ≥ 0 −7 + 15 e P8.6-2 The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0 where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The steady-state circuit for t < 0. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is −a 0 vo ( 0 ) = A + B e ( ) = A + B Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives 8-25
• A+ B = 6 ( 3) = 2 V 3+ 6 Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is vo ( ∞ ) = A + B e The steady-state circuit for t > 0. − a (∞) =A Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives A= 6 (6) = 4 V 3+ 6 Therefore B = −2 V The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit connected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt = ( 3)( 6 ) = 2 3+ 6 Ω Therefore a= 1 ( 2 )(.5 ) =1 1 s (The time constant is τ = ( 2 )( 0.5 ) = 1 s .) Putting it all together: 2 V for t ≤ 0  vo ( t ) =  − t  4 − 2 e V for t ≥ 0 8-26
• P8.6-3 The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0 where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The value of the inductor current at time t = 0, will be equal to the steady state inductor current before the input changes. At time t = 0 the output current is The steady-state circuit for t < 0. io ( 0 ) = A + B e − a ( 0) = A+ B Consequently, the inductor current is labeled as A + B. Analysis of the circuit gives A+ B = −7 = −1.4 A 5 Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The value of the inductor current at time t = ∞, will be equal to the steady state inductor current after the input changes. At time t = ∞ the output current is The steady-state circuit for t > 0. io ( ∞ ) = A + B e − a (∞) =A Consequently, the inductor current is labeled as A. Analysis of the circuit gives A= 6 = 1.2 A 5 Therefore B = −2.6 V 8-27
• The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit connected to the inductor. 1 L =τ = a Rt Here is the circuit used to calculate Rt. Rt = ( 5 ) ( 4 ) = 2.22 5+ 4 Ω Therefore a= 2.22 1 = 1.85 1.2 s 1.2 = 0.54 s .) 2.22 (The time constant is τ = Putting it all together: −1.4 A for t ≤ 0  io ( t ) =  − 1.85 t A for t ≥ 0 1.2 − 2.6 e P8.6-4 v (t ) = 4u (t ) − u (t − 1) − u (t − 2) + u (t − 4) − u (t − 6) P8.6-5 Assume that the circuit is at steady state at t = 1−. Then 0  vs ( t ) = 4 0  t <1 1<t < 2 t >2 v ( t ) = 4 − 4 e − (t −1) V for 1 ≤ t ≤ 2 τ = R C = ( 5 ×105 )( 2 × 10−6 ) = 1 s so v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V and v ( t ) = 2.53 e − (t − 2) V for t ≥ 2 ∴ 0  v(t ) = 4− 4e − ( t −1) 2.53e − (t − 2)  t ≤1 1≤t ≤ 2 t ≥2 8-28
• P8.6-6 The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0. For 0 < t < 0.5 s the switch is open: v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 × 1 1 = s 6 2 so v ( t ) = 10 e − 2 t V In particular, v ( 0.5 ) = 10 e For t > 0.5 s the switch is closed: − 2 ( 0.5) = 3.679 V v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω, 1 6 τ = 2× = so 1 s 3 v ( t ) = 10 + ( 3.679 − 10 ) e = 10 − 6.321 e − 3 ( t − 0.5) − 3 ( t − 0.5 ) V V P8.6-7 Assume that the circuit is at steady state before t = 0. Then the initial inductor current is i(0−) = 0 A. For 0 < t < 1 ms: The steady state inductor current will be 30 i ( ∞ ) = lim i ( t ) = ( 40 ) = 24 A t →∞ 30 + 20 The time constant will be 50 × 10−3 1 τ= = 10−3 = s 30 + 20 1000 The inductor current is i ( t ) = 24 (1 − e −1000 t ) A In particular, i ( 0.001) = 24 (1 − e −1 ) = 15.2 A For t > 1 ms Now the initial current is i(0.001) = 15.2 A and the steady state current is 0 A. As before, the time constant is 1 ms. The inductor current is i ( t ) = 15.2 e −1000 ( t − 0.001) A 8-29
• The output voltage is 480 (1 − e −1000 t ) V t < 1 ms  v ( t ) = 20 i ( t ) =  −1000 ( t − 0.001) V t > 1 ms 303 e  P8.6-8 For t < 0, the circuit is: After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: vc ( t ) = −15 + ( −6 − ( −15 ) ) e − t / ( 4000×0.00005 ) = −15 + 9 e − 5 t V 8-30
• Section 8-7 The Response of an RL or RC Circuit to a Nonconstant Source P8.7-1 Assume that the circuit is at steady state before t = 0: KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 A Then vc (0− ) = −12 ix = 22 V = vc (0+ ) After t = 0: KVL : 12i (t ) − 8e −5t + v ( t ) = 0 x c dv ( t ) 1 dvc ( t ) KCL : −ix ( t )−2ix ( t ) + (1 36) c = 0 ⇒ ix ( t ) = dt 108 dt  1 dvc ( t )  −5t ∴ 12  + v (t ) = 0  − 8e c 108 dt  dv (t ) c + 9v (t ) = 72e−5t ⇒ v (t ) = Ae −9t c cn dt Try v (t ) = Be −5t & substitute into the differential equation ⇒ B = 18 cf ∴ v (t ) = Ae −9t + 18 e−5t c v (0) = 22 = A + 18 ⇒ A = 4 c ∴ v (t ) = 4e−9t + 18e−5t V c 8-31
• P8.7-2 Assume that the circuit is at steady state before t = 0: iL (0+ ) = iL (0− ) = 12 = 3A 4 After t = 0: v( t ) −12 v( t ) + iL ( t ) + = 6 e −2t 4 2 di ( t ) also : v ( t ) = (2 / 5) L dt di ( t )  3 iL ( t ) + (2 / 5) L  = 3 + 6 e −2t 4 dt  KCL : diL ( t ) 10 + iL ( t ) = 10 + 20 e −2t 3 dt ∴ in (t ) = Ae − (10 / 3)t , try i f (t ) = B + Ce−2t , substitute into the differential equation, and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t ∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15 ∴ iL (t ) = −15e − (10 / 3) t + 3 + 15e−2t Finally, v(t ) =( 2 / 5 ) diL = 20 e − (10 / 3) t −12 e −2t V dt P8.7-3 Assume that the circuit is at steady state before t = 0:  6  Current division: iL (0− ) = −5   = −1 mA  6 + 24  8-32
• After t = 0: KVL: − 25sin 4000 t + 24iL ( t ) + .008 di L ( t ) 25 +3000i L ( t ) = sin4000t dt .008 diL ( t ) =0 dt in (t ) = Ae −3000t , try i f (t ) = B cos 4000t + C sin 4000t , substitute into the differential equation and equate like terms ⇒ B = −1/ 2, C = 3 / 8 ⇒ i f (t ) = −0.5cos 4000 t + 0.375 sin 4000 t iL (t ) = in (t ) + i f (t ) = Ae −3000t − 0.5cos 4000 t + 0.375 sin 4000 t iL (0+ ) = iL (0− ) =−10−3 = A− 0.5 ⇒ A ≅ 0.5 ∴ iL (t ) = 0.5 e −3000t − 0.5cos 4000 t + 0.375 sin 4000 t mA but v(t ) = 24iL (t ) = 12 e −3000t − 12 cos 4000t + 9sin 4000t V P8.7-4 Assume that the circuit is at steady state before t = 0: Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get: dvc ( t )   KVL: − 10 cos 2t + 15  1 +v t =0 ⇒ 30 dt  c ( )   dvc ( t ) + 2vc ( t ) = 20 cos 2t dt vn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to get B = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2t Now vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e−2t + 5cos 2t + 5sin 2t V 8-33
• P8.7-5 Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) = 0 A. After t = 0, we have: di ( t ) KVL : − 10sin100t + i ( t ) + 5 + v (t ) = 0 dt v( t ) Ohm's law : i ( t ) = 8 dv( t ) ∴ +18 v( t ) = 160sin100t dt ∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equation and equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t ∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 t v(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55 so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t V P8.7-6 Assume that the circuit is at steady state before t = 0. vo ( t ) = −vc ( t ) vC (0+ ) = vC (0− ) = −10 V After t = 0, we have v (t ) 8 e−5 t i (t ) = s = = 0.533 e−5 t mA 15000 15000 The circuit is represented by the differential dv ( t ) vC ( t ) equation: i ( t ) = C C + . Then dt R ( 0.533 ×10 ) e −3 −5 t = ( 0.25 ×10−6 ) dvc ( t ) + (10−3 ) vc ( t ) ⇒ dt dvc ( t ) + 4000 vc ( t ) = 4000 e−5t dt Then vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get ( d B e−5t dt ) + 4000 ( B e ) = 4000 e −5t −5t ⇒ B= 4000 = −1.00125 ≅ −1 −3995 8-34
• vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae −4000t vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e−2t − 11 e−4000t V Finally vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0 P8.7-7 From the graph iL (t ) = 1 t mA . Use KVL to get 4 (1) iL (t ) + 0.4 diL (t ) = v1 (t ) ⇒ dt diL (t ) + 2.5 iL (t ) = 2.5 v1 (t ) dt Then di  1  t + 2.5  1 t  = 2.5 v1 (t ) ⇒ 4  dt  4  v1 = 0.1+ 0.25t V P8.7-8 Assume that the circuit is at steady state before t = 0. v (0+ ) = v (0− ) = 2 30 = 10 V 4+ 2 After t = 0 we have KVL :  1 d v( t )  5 d v( t ) + v (t ) + 4  −i  = 30 2 dt  2 dt   1 d v( t )  −3t 2 i ( t ) + 4 i ( t ) −  + 30 = e 2 dt   The circuit is represented by the differential equation d v( t ) 6 6 2 (10 + e −3t ) v (t ) = + 19 19 3 dt Take vn ( t ) = Ae −( 6 /19 ) t . Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get 8-35
• −3Ce −3t + 6 60 4 −3t ( B + Ce −3t ) = + e 19 19 19 Equate coefficients to get B = 10 , C = − 4 4 ⇒ v f ( t ) = e −3t + Ae− (6 /19) t 51 51 Then v ( t ) = vn ( t ) + v f ( t ) = 10 − 4 −3t e + Ae− (6 /19) t 51 Finally vc (0+ ) = 10 V, ⇒ 10 = 10 − ∴ vc (t ) = 10 + 4 +A ⇒ 51 A= 4 51 4 − (6 /19) t −3t (e −e ) V 51 P8.7-9 We are given v(0) = 0. From part b of the figure: 5t 0 ≤ t ≤ 2 s vs ( t ) =  t > 2s 10 Find the Thevenin equivalent of the part of the circuit that is connected to the capacitor: The open circuit voltage: The short circuit current: (ix=0 because of the short across the right 2 Ω resistor) Replace the part of the circuit connected to the capacitor by its Thevenin equivalent: KVL: dv( t ) + v ( t ) − vs ( t ) = 0 dt dv( t ) v ( t ) vs ( t ) + = dt 2 2 2 vn ( t ) = Ae−0.5 t 8-36
• For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation and equating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0, we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) . At t = 2 s, v( 2 ) = 10e−1 = 3.68 . Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation and equating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e − (t −2) / 2 − (t − 2) / 2 . Using v ( 2 ) = 3.68 , we . P8.7-10  d v (t )  KVL: − kt + Rs C C  + vC ( t ) = 0 dt   d vC ( t ) k 1 vC ( t ) = t ⇒ + Rs C Rs C dt vc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t & plug into D.E. ⇒ B1 + 1 k t thus B0 = − kRs C , B1 = k . [ B0 + B1t ] = Rs C Rs C Now we have vc (t ) = Ae − t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C. ∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF get vc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )] v(t) and vC(t) track well on a millisecond time scale. 8-37
• Spice Problems SP 8-1 8-38
• SP 8-2 8-39
• SP 8-3 v(t ) = A + B e −t / τ for t > 0 ⇒ 7.2 = A + B    ⇒ B = −0.8 V −∞ ⇒ A = 8.0 V  8.0 = v(∞) = A + B e  0.05  8 − 7.7728  7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⇒ − = ln   = −1.25878 0.8 τ   0.05 ⇒ τ= = 39.72 ms 1.25878 7.2 = v(0) = A + B e 0 Therefore v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0 8-40
• SP 8-4 i (t ) = A + B e −t / τ 0 = i (0) = A + B e 0 ⇒ 0 = A+ B 4 × 10−3 = i (∞) = A + B e −∞ ⇒ A = 4 × 10−3 2.4514 ×10 = v(5 ×10 ) = ( 4 ×10 −3 ⇒ − Therefore for t > 0 −6 5 × 10−6 τ −3   −3  ⇒ B = −4 × 10 A A  ) − ( 4 ×10 ) e −3 ( ) − 5×10−6 / τ  ( 4 − 2.4514 ) × 10−3  = ln   = −0.94894 4 × 10−3   5 × 10−6 ⇒ τ= = 5.269 µ s 0.94894 i (t ) = 4 − 4 e −t / 5.269×10 −6 mA for t > 0 8-41
• Verification Problems VP 8-1 First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitor voltage is vc = 4 V. We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot. Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of ( 2000 )( 4000 ) = 4 kΩ so the time constant is the circuit connected to the capacitor is R t = 2000 + 4000 3 4 2   τ = R t C =  × 103  ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is 3 3  vc (t ) = 4 e− t 0.67 +4 V where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then vc (t1 ) =  1.33  −  4 e  .67  + 4 = 4.541 ~ 4.5398 V So the plot is correct. VP 8-2 The initial and steady-state inductor currents shown on the plot agree with the values obtained from the circuit. Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of ( 2000 )( 4000 ) = 4 kΩ so the time constant is the circuit connected to the inductor is R t = 2000 + 4000 3 L 5 15 ms . Thus inductor current is τ= = = R t 4 ×103 4 3 iL (t ) − 2 e− t 3.75 + 5 mA where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then iL (t1 ) =  3.75  −  −2 e  3.75  + 5 = 4.264 mA ≠ 4.7294 mA so the plot does not correspond to this circuit. 8-42
• VP 8-3 Notice that the steady-state inductor current does not depend on the inductance, L. The initial and steady-state inductor currents shown on the plot agree with the values obtained from the circuit. After t = 0 So I sc = 5 mA and τ = The inductor current is given by iL (t ) = −2e −1333t has units of Henries. Let t 1 = 3.75 ms, then L L 1333 + 5 mA , where t has units of seconds and L 4.836 = iL (t1 ) = −2 e −(1333)⋅(0.00375) L + 5 = −2e−5 L + 5 so 4.836−5 = e −5 L −2 and L= −5 =2 H  4.836−5  ln    −2  is the required inductance. VP 8-4 First consider the circuit. When t < 0 and the circuit is at steady-state: For t > 0 So Voc = R2 R1 R2 RRC ( A + B) , Rt = and τ = 1 2 R1 + R2 R1 + R2 R1 + R2 8-43
• Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor voltage is (Voc =) 4 V, so vC (t ) = − 6e −t τ + 4 At t 1 = 1.333 ms 3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4 so τ = −0.001333 = 0.67 ms  −4+ 3.1874  ln   −6   Combining the information obtained from the circuit with the information obtained from the plot gives R2 R2 R1 R2C A = −2, ( A + B ) = 4, = 0.67 ms R1 + R2 R1 + R2 R1 + R2 There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is one convenient way. Pick R = 3000 and R = 6000. Then 1 1 2 2 2A = −2 ⇒ A = −3 3 2( A+ B) = 4 ⇒ B −3 = 6 ⇒ B = 9 3 2 1 µF = C 2000 ⋅ C = ms ⇒ 3 3 Design Problems DP 8-1 R3 6= 12 ⇒ R1 + R 2 = R 3 . Steady-state response when the switch is open: R1 + R 2 + R 3 Steady-state response when the switch is open: 10 = 10 ms = 5 τ = ( R 1 || R 3 ) C = R3 6 R3 R1 + R 3 12 ⇒ R1 = R3 5 . C Let C = 1 µF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ. 8-44
• DP 8-2 12 steady state response when the switch is open : 0.001 = R + R 1 2 steady state response when the switch is open: 0.004 = 12 R1 ⇒ R + R = 12 kΩ . 1 2 ⇒ R1 = 3 kΩ . Therefore, R 2 = 9 kΩ.  L  L 10 ms = 5 τ = 5  ⇒ L = 240 H =  R1 + R 2  2400   DP 8-3 Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use Rt = 50 kΩ. 10−6 (a) ∆t = 5 Rt C ⇒ C = = 4 pF 5 50×103 (b) ∆ t = 5 ( 50×103 )( 2×10−6 ( ) ) = 0.5 s DP 8-4 Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use Rt = 50 kΩ. ∆t −∆t τ When the switch is open: 5 e = (1 − k ) 5 ⇒ ln (1− k ) = − ⇒ ∆ t = −τ ln (1− k ) When the switch is open: 5 − 5 e (a) C = −∆t τ τ = k 5 ⇒ ∆ t = −τ ln (1− k ) 10−6 = 6.67 pF − ln (1−.95 ) ( 50×103 ) (b) ∆ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 s DP 8-5 i (0) = R1 20 × 40 R1 R1 + 40 40 + 40 + R1 8-45
• For t > 0: i (t ) = i (0) e −t τ where τ = L 10−2 = R t 40+ R 2 At t < 200 µ s we need i ( t ) > 60 mA and i ( t ) <180 mA First let's find a value of R 2 to cause i (0) < 180 mA. 1 t A = 166.7 mA so i (t ) = 0.1667 e− τ . 6 Next, we find a value of R 2 to cause i (0.0002) > 60 mA. Try R 2 = 40 Ω . Then i (0) = 10−2 1 s. = 0.2 ms = 50 5000 i (0.0002) = 166.7×10−3 e −5000×0.0002 = 166.7×10−3 e −1 = 61.3 mA Try R 2 = 10Ω, then τ = DP 8-6 The current waveform will look like this: We only need to consider the rise time: −t Vs A τ −t τ iL (t ) = (1 − e ) = (1 − e ) R+2 R+2 where L 0.2 1 s τ = = = 3 15 Rt A ∴ iL (t ) = (1 − e −15t ) 3 2 Now find A so that iL R fuse ≥ 10 W during 0.25 ≤ t ≤ 0.75 s 2 ∴ we want [iL (0.25)]R fuse = 10 W ⇒ A2 (1−e −15(.25) ) 2 (1) =10 ⇒ A = 9.715 V 9 8-46
• Chapter 9 - Complete Response of Circuits with Two Energy Storage Elements Exercises Apply KVL to right mesh: Ex. 9.3-1 di ( t ) + v ( t ) + 1( i ( t ) −is ( t ) ) = 0 dt di ( t ) ⇒ v ( t ) = −2 − ( i ( t )−is ( t ) ) dt 2 The capacitor current and voltage are related by i (t ) =  1 dis ( t ) 1 di ( t ) d 2i ( t ) 1 dv ( t ) 1 d  di ( t ) = −2 −i ( t )+ is ( t )  = − −  dt dt 2 2 dt 2 dt  2 dt  2 dt ∴ d 2i ( t ) 1 di ( t ) 1 dis ( t ) + + i( t ) = 2 dt 2 dt 2 dt The inductor voltage is related to the inductor current by di ( t ) v (t ) = 1 dt Ex. 9.3-2 Apply KCL at the top node: is ( t ) = Using the operator s = v (t ) 1 dv ( t ) + i (t ) + 1 2 dt d we have dt v (t ) = s i (t )  v (t ) 1  + sv ( t )  ⇒ is ( t ) = v ( t ) + 1 s 2 is ( t ) = v ( t ) + i ( t ) + sv ( t )   2 Therefore 2s is ( t ) = 2 s v ( t ) + 2 v ( t ) + s 2 v ( t ) ⇒ d 2v ( t ) dv ( t ) d i (t ) +2 + 2 v (t ) = 2 s 2 dt dt dt 9-1
• Ex. 9.3-3 Using the operator s = d , apply KVL to the dt left mesh: i1 ( t ) + s ( i1 ( t ) −i 2 ( t ) ) = vs ( t ) Apply KVL to the right mesh: 1 2 i 2 ( t ) + 2 i 2 ( t ) + s ( i 2 ( t ) −i1 ( t ) ) = 0 s 1 2 i1 ( t ) = 2 i 2 ( t ) + 2 i 2 ( t ) + i 2 ( t ) s s Combining these equations gives: 3s i2 ( t ) + 4si2 ( t ) + 2i2 ( t ) = s vs ( t ) 2 2 or Ex. 9.4-1 d 2i2 ( t ) di2 ( t ) d 2 vs ( t ) 3 +4 + 2i2 ( t ) = dt 2 dt dt 2 Using the operator s = top node: i s (t ) = d , apply KCL at the dt v (t ) 1 + i (t ) + s v (t ) 4 4 Apply KVL to the right-most mesh: v (t ) − ( s i (t ) + 6 i (t )) = 0 Combining these equations gives: s 2 i ( t ) + 7 s i ( t ) + 10 i ( t ) = 4 i s ( t ) The characteristic equation is: s 2 + 7 s + 10 = 0 . The natural frequencies are: s = −2 and s = −5 . Ex. 9.4-2 Assume zero initial conditions . Write mesh d equations using the operator s = : dt 1 s i1 ( t ) − i 2 ( t )  + 7 + 10 i1 ( t ) − 10 = 0  2  and 1   v ( t ) − 7 − s i1 ( t ) − i 2 ( t )  = 0 2 9-2
• Now 0.005 s v ( t ) = i 2 ( t ) ⇒ v ( t ) = 200 200 i2 (t ) s i2 (t ) s so the second mesh equation becomes: 1   − 7 − s  i1 ( t ) − i 2 ( t )  = 0 2 Writing the mesh equation in matrix form: s  10 + 2   −1 s  2  1  s   i1 ( t )   3  2  = 1 200  i 2 ( t )  7    s+  2 s  − Obtain the characteristic equation by calculating a determinant: 10 + − 1 s 2 = s 2 + 20s + 400 = 0 ⇒ s1,2 = −10 ± j 17.3 1 200 s+ 2 s s 2 − 1 s 2 Ex. 9.5-1 After t = 0 , we have a parallel RLC circuit with 1 1 7 1 1 2 = = = =6 α = and ω o = 2 RC 2(6)(1/ 42) 2 LC (7)(1/ 42) 2 7   − 6 = −1, − 6  2 dv ( t ) ∴ vn (t ) = A1e −t + A2 e−6t . We need vn (0) and n dt 7 ∴ s 1 , s 2 = −α ± α −ω = − ± 2 2 2 o to evaluate A1 & A2 . t =0 At t = 0+ we have: iC ( 0+ ) = −10 A ⇒ dv ( t ) 10 = = 420 V s 1 dt t =0+ 42 Then vn (0+ ) = 0 = A1 + A2    A1 = −84 , A2 = 84 dvn =−420= − A1 − 6 A2  dt t =0+  9-3
• Finally ∴ vn (t ) = −84e −t + 84e −6t V 1 1 = 0 ⇒ s 2 + 40s + 100 = 0 s+ RC LC Therefore s 1 , s 2 = −2.68 , −37.3 Ex. 9.5-2 s2 + vn ( t ) = A1 e −2.68t + A2 e−37.3t , v(0) = 0 = A1 + A2 v(0+ ) 1 dv(0+ ) + KCL at t = 0 yields + i (0 ) + = 0 so 1 40 dt dv(0+ ) = − 40 v(0+ ) − 40 i (0+ ) = − 40(0) − 40(1) = − 2.7 A1 − 37.3 A2 dt Therefore: A1 = −1.16 , A2 = 1.16 ⇒ v(t ) = vn (t ) = −1.16e −2.68t + 1.16e−37.3t + Ex. 9.6-1 For parallel RLC circuits: α = 1 1 1 1 2 = = 50, ω o = = = 2500 −3 LC (0.4)(10−3 ) 2 RC 2(10)(10 ) The roots of the characteristic equations are:∴ s1,2 = −50 ± (50) 2 − 2500 = −50, −50 The natural response is vn (t ) = A1 e−50t + A2 t e−50t . At t = 0+ we have: −v(0+ ) ic (0 ) = = − .8V 10Ω dv( t ) i (0+ ) ∴ = c = −800 V / s dt c + t =0 + 9-4
• So vn (0+ ) = 8 = A1 ⇒ vn (t ) = 8e−50t + A2 t e−50t dv(0+ ) = −800 = − 400 + A2 ⇒ A2 = − 400 dt ∴ vn (t ) = 8 e −50t − 400 t e−50t V 1 1 = = 8000 2 RC 2(62.5)(10−6 ) 1 1 ω o2 = = = 1 08 −6 LC (.01)(10 ) α= Ex. 9.7-1 2 ∴ s = − α ± α 2 −ω o = − 8000 ± (8000) 2 −108 = − 8000 ± j 6000 ∴ vn (t ) = e −8000t [ A1 cos 6000 t + A2 sin 6000 t ] at t = 0+ 10 + ic (0+ ) = 0 62.5 ⇒ ic (0+ ) = −.24 A 0.08 + ∴ dv(0+ ) ic (0+ ) = = −2.4 ×10+5 V/s dt C dvn (0+ ) vn (0 ) = 10 = A1 and = −2.4 × 105 = 6000 A2 − 8000 A1 ⇒ A2 = −26.7 dt + ∴ vn (t ) = e−8000t [10 cos 6000 t − 26.7 sin 6000 t ] V Ex. 9.8-1 d 2v ( t ) dv ( t ) +5 + 6 v ( t ) = vs ( t ) so the characteristic equation is 2 dt dt s 2 + 5 s + 6 = 0 . The roots are s 1 , s 2 = −2, − 3 . The differential equation is d 2v ( t ) dv ( t ) (a) +5 + 6 v ( t ) = 8 . Try v f ( t ) = B . Substituting into the differential equation gives 2 dt dt 6 B = 8 ∴ v f (t ) = 8 / 6 V . 9-5
• (b) d 2v ( t ) dt 2 +5 dv ( t ) dt + 6 v ( t ) = 3 e − 4 t . Try v f ( t ) = B e− 4 t . Substituting into the differential 3 3 equation gives (−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = . ∴ v f ( t ) = e−4t . 2 2 2 d v (t ) dv ( t ) (c) +5 + 6 v ( t ) = 2 e − 2 t . Try v f ( t ) = B t e− 2 t because –2 is a natural frequency. 2 dt dt Substituting into the differential equation gives (4t − 4) B + 5 B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2. ∴ v f ( t ) = 2 t e −2t . Ex. 9.8-2 d 2i ( t ) di ( t ) + 20 i ( t ) = 36 + 12 t . Try i f ( t ) = A + B t . Substituting into the differential dt 2 dt equation gives 0 + 9 B + 20( A + Bt ) = 36 + 12t ⇒ B = 0.6 and A = 1.53. +9 ∴ i f ( t ) = 1.53 + 0.6 t A Ex. 9.9-1 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get 9-6
• vs ( t ) = L   v (t )  d  vC ( t ) d d + C vC ( t )  + R1  C + C vC ( t )  + vC ( t )    R2  dt  R 2 dt dt     = LC  L d  R1  d2 + R1 C  vC ( t ) +  1 + vC ( t ) +  v t  R2  dt  R2  C ( )  dt 2     (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 1= V v f = vC ( ∞ ) = 2 R1 + R 2 The characteristic equation is R1    1+   1 R1   R2  2 2 s + + s+ = s + 6 s + 8 = ( s + 2 )( s + 4 ) = 0  R 2 C L   LC           so the natural response is vn = A1 e −2 t + A2 e −4 t V The complete response is vc ( t ) = iL ( t ) = + vC ( t ) 1.309 + 1 + A1 e −2 t + A2 e−4 t V 2 d vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 dt At t = 0 ( ) 0 = vc 0+ = A1 + A2 + 0.5 ( ) = −1.236 A − 3.236 A 0 = iL 0 + 1 2 + 0.3819 Solving these equations gives A1 = −1 and A2 = 0.5, so vc ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 v f = vC ( ∞ ) = 1= V R1 + R 2 4 The characteristic equation is 9-7
• R1    1+   1 R1   R2  2 2 2 s + + s+ = s + 4s + 4 = ( s + 2 ) = 0  R 2 C L   LC           so the natural response is vn = ( A1 + A2 t ) e −2 t V The complete response is vc ( t ) = iL ( t ) = vC ( t ) + At t = 0+ 1 + ( A1 + A2 t ) e −2 t V 4 d 1 vC ( t ) = + dt 4 ( ) (( A 0 = vc 0+ = A1 + ( ) 0 = iL 0+ = 2 ) − A1 ) − A2 t e −2 t 1 4 1 + A2 − A1 4 Solving these equations gives A1 = −0.25 and A2 = −0.5, so vc ( t ) = 1  1 1  −2 t −  + t e V 4 4 2  (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Use the steady state response as the forced response: R2 4 v f = vC ( ∞ ) = 1= V R1 + R 2 5 The characteristic equation is R1    1+   1 R1   R2  2 2 s + + s+ = s + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0  R 2 C L   LC           so the natural response is vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V The complete response is iL ( t ) = vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vC ( t ) 4 + A 2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 9-8
• At t = 0+ ( ) 0 = vc 0+ = 0.8 + A1 ( ) 0 = iL 0+ = 0.2 + A2 2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V Ex 9.9-2 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C d  d   d   L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d = R1 LC 2 iL ( t ) + ( L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt Using iL ( t ) = vo ( t ) gives R2  L d  R1 + R 2  d2 vs ( t ) = LC 2 vo ( t ) +  + R1 C  vo ( t ) +  v t  R2  dt  R2  o ( )  R2 dt     R1 9-9
• (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 1= V v f = vo ( ∞ ) = 2 R1 + R 2 The characteristic equation is R2    1+   1 R2   R1  2 2 s + + s+ = s + 6 s + 8 = ( s + 2 )( s + 4 ) = 0  R1 C L   LC           so the natural response is vn = A1 e −2 t + A2 e −4 t V The complete response is 1 + A1 e −2 t + A2 e −4 t V 2 A1 −2 t A2 −4 t v (t ) 1 iL ( t ) = o e + e V = + 1.309 2.618 1.309 1.309 vo ( t ) = vC ( t ) = 1.309 iL ( t ) + At t = 0+ 1 d 1 iL ( t ) = + 0.6180 A1 e −2 t + 0.2361 A2 e −4 t 4 dt 2 ( ) 0 = iL 0+ = ( ) 0 = vC 0+ = A1 A2 1 + + 2.618 1.309 1.309 1 + 0.6180 A1 + 0.2361 A2 2 Solving these equations gives A1 = −1 and A2 = 0.5, so vo ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 (b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω Use the steady-state response as the forced response: R2 3 v f = vo ( ∞ ) = 1= V R1 + R 2 4 The characteristic equation is R2    1+   1 R2   R1  2 2 s2 +  + s+ = s + 4s + 4 = ( s + 2 ) = 0  R1 C L   LC           9-10
• so the natural response is vn = ( A1 + A2 t ) e −2 t V The complete response is vo ( t ) = iL ( t ) = vo ( t ) 3 vC ( t ) = 3 iL ( t ) + At t = 0+ 3 + ( A1 + A2 t ) e −2 t V 4 = 1  A1 A2  −2 t t e V + + 4  3 3  3   A1 A2  A2  −2 t d iL ( t ) = +   + t e + 4  3 3  3  dt   ( ) 0 = iL 0+ = ( ) 0 = vC 0+ A1 + 1 4 3 3 A1 A2 = + + 4 3 3 Solving these equations gives A1 = -0.75 and A2 = -1.5, so vo ( t ) = 3  3 3  −2 t −  + t e V 4 4 2  (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 v f = vo ( ∞ ) = 1= V R1 + R 2 5 The characteristic equation is R2    1+   1 R2   R1  2 2 s + + s+ = s + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0  R1 C L   LC           so the natural response is vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V The complete response is vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) = vo ( t ) 1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 9-11
• vC ( t ) = iL ( t ) + At t = 0+ 1 d iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t 2 dt ( ) ( 0 ) = 0.2 + 2 A 0 = iL 0+ = 0.2 + A1 0 = vC + 2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V Ex. 9.10-1 At t = 0+ no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0 3 di (0+ ) di (0+ ) +0=0 ⇒ =0 10 dt dt 0 dv1 (0+ ) + ⇒ = 0 KCL at A : + i1 (0 ) + 0 = 0 1 dt 5 dv2 (0+ ) = 10 ⇒ KCL at B : − 0 + i2 (0+ ) − 10 = 0 ⇒ i2 (0+ ) = 6 dt KVL : − 0 + For t > 0: dv2 (0+ ) = 12 V s dt v1 1 d v1 + +i = 0 1 12 dt 5 d v2 = 10 KCL at B : − i + 6 dt 3 di − v1 + + v2 = 0 KVL: 10 dt Eliminating i yields 1 d v1 5 d v2 v1 + + − 10 = 0 12 dt 6 dt 3  5 d 2 v2  −v1 +   + v2 = 0 10  6 dt 2  KCL at A : Next 9-12
• v1 = v2 + 1 d 2 v2 4 dt 2 ⇒ d v1 d 2 v2 1 d 3 v2 = + dt dt 2 4 dt 3 Now, eliminating v1 v2 + 1 d 2 v2 1  d v2 1 d 3 v2  5 d v2 +  + = 10 + 4 dt 2 12  dt 4 dt 3  6 dt Finally, the circuit is represented by the differential equation: d 3 v2 d2 v dv + 12 2 2 + 44 2 + 48v2 = 480 3 dt dt dt The characteristic equation is s 3 + 12s 2 + 44s + 48 = 0 . It’s roots are s1,2,3 = −2, −4, −6 . The natural response is vn = A1e −2t + A2 e −4t + A3e −6t Try v f = B as the forced response. Substitute into the differential equation and equate coefficients to get B = 10. Then v2 (t ) = vn (t ) + v f (t ) = A1e −2t + A2 e −4t + A3e−6t + 10 We have seen that v2 (0+ ) = 0 and dv2 (0+ ) d 2 v2 (0+ ) = 12 V/s . Also = 4[v1 (0+ ) − v2 (0+ )] = 0 . dt dt 2 Then v2 (0+ ) = 0 = A1 + A2 + A3 + 10 dv2 (0+ ) = 12 = −2 A1 − 4 A2 − 6 A3 dt d 2 v2 (0+ ) = 0 = 4 A1 + 16 A2 + 36 A3 dt 2 Solving these equations yields A1 = −15, A2 = 6, A3 = −1 so v2 (t ) = ( −15 e −2t + 6 e −4t − e −6t + 10 ) V Ex. 9.11-1 1 1 s2 + s+ =0 RC LC In our case L = 0.1, C = 0.1 so we have s 2 + 10 s + 100 = 0 R 9-13
• a) R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0 s1,2 = −5, − 20 b) R = 1 Ω ⇒ s 2 + 10 s + 100 = 0 s1,2 = − 5 ± j 5 3 9-14
• Problems Section 9-3: Differential Equations for Circuits with Two Energy Storage Elements P9.3-1 KCL: iL ( t ) = v (t ) dv ( t ) +C R2 dt KVL: vs ( t ) = R1iL ( t ) + L diL ( t ) + v (t ) dt dv ( t )  L dv ( t ) d 2v ( t )  v (t ) vs ( t ) = R1  +C + LC + v (t ) + dt  R2 dt dt 2  R2 d 2v ( t ) R   L  dv ( t ) =  1 + 1 v ( t ) +  R1C +  + [ LC ] vs ( t ) R2  dt dt 2  R2   In this circuit R1 = 2 Ω, R 2 = 100 Ω, L = 1 mH, C = 10 µ F so 2 dv ( t ) −8 d v ( t ) vs ( t ) = 1.02v ( t ) + .00003 + 10 dt dt 2 dv ( t ) d 2 v ( t ) 8 8 10 vs ( t ) = 1.02 × 10 v ( t ) + 3000 + dt dt 2 P9.3-2 Using the operator s = KCL: is ( t ) = d we have dt v (t ) + iL ( t ) + Csv ( t ) R1 KVL: v ( t ) = R2iL ( t ) + LsiL ( t ) Solving by usingCramer's rule for iL ( t ) : iL ( t ) = is ( t ) R2 Ls + + R2Cs + LCs 2 + 1 R1 R1  R2  L  2 1 +  iL ( t ) +  + R2C  siL ( t ) + [ LC ] s iL ( t ) = is ( t ) R1  R1    9-15
• In this circuit R1 = 100 Ω, R 2 = 10 Ω, L = 1 mH, C = 10 µ F so 1.1iL ( t ) + .00011siL ( t ) + 10−8 s 2iL ( t ) = is ( t ) diL ( t ) d 2iL ( t ) + = 108 is ( t ) 1.1× 10 iL ( t ) + 11000 2 dt dt 8 P9.3-3 After the switch closes, a source transformation gives: KCL: iL ( t ) + C dvc ( t ) vs ( t ) + vc ( t ) + =0 dt R2 KVL: diL ( t ) − vc ( t ) − vs ( t ) = 0 dt di ( t ) vc ( t ) = R1is ( t ) + R1iL ( t ) + L L − vs ( t ) dt Differentiating R1is ( t ) + R1iL ( t ) + L d vc ( t ) d is ( t ) d iL ( t ) d 2iL ( t ) d vs ( t ) = R1 + R1 +L − dt dt dt dt 2 dt Then  d is ( t ) d iL ( t ) d 2iL ( t ) d vs ( t )  vs ( t ) + R1 +L − iL ( t ) + C  R1 + dt dt dt 2 dt  R2  di ( t )  1  +  R1is ( t ) + R1iL ( t ) + L L − vs ( t )  = 0 R2  dt  Solving for i L ( t ) : d 2i L ( t )  R1 − R1 R1 di s ( t ) 1 dv s ( t ) 1  di L ( t )  R1 1  + + + + i s (t ) − +   i L (t ) = 2 dt LCR 2 L dt L dt  L R 2C  dt  LR 2C LC      9-16
• Section 9-4: Solution of the Second Order Differential Equation - The Natural Response P9.4-1 From Problem P 9.3-2 the characteristic equation is: 1.1×108 +11000 s + s 2 = 0 ⇒ s1 , s2 = −11000± (11000) 2 − 4(1.1×108 ) = −5500± j8930 2 P9.4-2 KVL: 40 ( is ( t ) − iL ( t ) ) = (100 × 10−3 ) diL ( t ) + vc ( t ) dt The current in the inductor is equal to the current in the capacitor so 1  dv ( t ) iL ( t ) =  ×10−3  c 3  dt 2  40  dv ( t )  100  d vC ( t ) vC ( t ) = 40 is ( t ) −  × 10−3  C −  × 10−6  2  3  dt  3  dt d 2 vC ( t ) dv ( t ) + 400 C + 30000 vC ( t ) = 40 is ( t ) 2 dt dt 2 s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100, s2 = −300 P9.4-3 v ( t ) − vs ( t ) dv ( t ) + iL ( t ) + (10 × 10−6 ) =0 1 dt di ( t ) KVL: v ( t ) = 2iL ( t ) + (1× 10−3 ) L dt KCL: 0 = 2 iL ( t ) + (1× 10−3 ) vs ( t ) = 3iL ( t ) + 0.00102 diL ( t ) dt diL ( t ) dt − vs ( t ) + iL ( t ) + (10 × 10−6 ) ( 2 ) + 10−8 d 2iL ( t ) dt 2 ⇒ d 2iL ( t ) dt diL ( t ) dt + (10 × 10−6 )(10−3 ) + 102000 diL ( t ) dt d 2iL ( t ) dt + 3 × 108 iL ( t ) = 108 vs ( t ) s 2 + 102000s + 3 ×108 = 0 ⇒ s1 = 3031, s2 = −98969 9-17
• Section 9.5: Natural Response of the Unforced Parallel RLC Circuit P9.5-1 dv( 0 ) d = −3000 V/s . Using the operator s = , the node dt dt − vs ( t ) v( t ) v( t ) L   + = 0 or  LCs 2 + s + 1 v ( t ) = vs ( t ) equation is Csv ( t ) + R sL R   1 1 s+ = 0 ⇒ s 2 + 500 s + 40, 000 = 0 The characteristic equation is: s 2 + RC LC The initial conditions are v ( 0 ) = 6 V, The natural frequencies are: s1,2 = −250 ± 2502 − 40, 000 = −100, −400 The natural response is of the form v ( t ) = Ae−100t + Be−400t . We will use the initial conditions to evaluate the constants A and B. v ( 0) = 6 = A + B    ⇒ dv ( 0 ) = −3000 = −100 A − 400 B  dt  A = −2 and B = 8 Therefore, the natural response is v ( t ) = −2e−100t + 8e−400t t >0 P9.5-2 The initial conditions are v ( 0 ) = 2 V, i ( 0 ) = 0 . 1 1 s+ = 0 ⇒ s 2 + 4s + 3 = 0 RC LC The natural frequencies are: s1 , s2 = −1, − 3 The natural response is of the form v ( t ) = Ae − t + Be −3t . We will use the initial conditions to evaluate the constants A and B. dv ( t ) = − Ae − t − 3Be −3t . At t = 0 this becomes Differentiating the natural response gives dt dv ( t ) v ( t ) dv ( t ) v (t ) i (t ) dv ( 0 ) + + i ( t ) = 0 or = − − . = − A − 3B . Applying KCL gives C dt R dt RC C dt dv ( 0 ) v ( 0) i ( 0) = − At t = 0 this becomes − . Consequently dt RC C The characteristic equation is: s 2 + −1A − 3B = − v ( 0) i ( 0) 2 − =− −0 = − 8 RC C 14 9-18
• Also, v ( 0 ) = 2 = A + B . Therefore A = −1 and B = 3. The natural response is v ( t ) = −e − t + 3e −3t V P9.5-3 di1 ( t ) di ( t ) −3 2 =0 dt dt di ( t ) di ( t ) KVL : − 3 1 +3 2 + 2i2 ( t ) = 0 dt dt KVL : i1 + 5 (1) ( 2) d , the KVL equations are dt  (1+5s )i1 + ( −3s )i2 = 0  3s 2 i1 = 0 ⇒ (1 + 5s )( 3s + 2 ) − ( 3s )  i1 = 0  ⇒ (1 + 5s ) i1 − ( 3s )   3s + 2 ( −3s ) i1 + ( 3s + 2 ) i2 = 0  1 The characteristic equation is (1+5s )( 3s + 2 ) − 9 s 2 = 6 s 2 + 13s + 2 = 0 ⇒ s1,2 = − , −2 6 Using the operator s = −t −t The currents are i1 ( t ) = Ae 6 + Be−2t and i2 ( t ) = Ce 6 + De −2t , where the constants A, B, C and D must be evaluated using the initial conditions. Using the given initial values of the currents gives i1 ( 0 ) = 11 = A + B and i 2 ( 0 ) = 11 = C + D Let t = 0 in the KCL equations (1) and (2) to get di1 ( 0 ) dt =− di2 ( 0 ) A C 33 143 = − − 2 B and =− =− −D 2 6 6 6 dt So A = 3, B = 8, C = −1 and D = 12. Finally, i1 (t ) = 3e− t / 6 + 8e−2t A and i 2 (t ) = −e −t / 6 + 12e −2t A 9-19
• Section 9.6: Natural Response of the Critically Damped Unforced Parallel RLC Circuit P9.6-1 After t = 0 Using KVL: 100 ic ( t ) + 0.025 dic ( t ) + vc ( t ) = 0 dt The capacitor current and voltage are related by: ic ( t ) = 10−5 ∴ d 2 vc ( t ) dt 2 + 4000 dvc ( t ) dt dvc ( t ) dt + 4 × 106 vc ( t ) = 0 The characteristic equation is: s 2 + 4000 s + 4 × 106 = 0 The natural frequencies are: s1,2 = − 2000, − 2000 The natural response is of the form: vc ( t ) = A1e−2000t + A2 t e−2000t (The capacitor current is continuous at t = 0 in this circuit because it is equal to the inductor current.) Before t = 0 the circuit is at steady state vc ( 0+ ) = 3 = A1 dvc ( 0+ ) = 0 = −2000 A1 + A2 ⇒ A2 = 6000 dt ∴ vc ( t ) = ( 3 + 6000t ) e−2000t V for t ≥ 0 P9.6-2 After t = 0 Using KCL: ∫ t −∞ 1 dvc ( t ) =0 4 dt d 2 vc ( t ) dv ( t ) ⇒ +4 c + 4vc ( t ) = 0 dt dt vc (τ ) dτ + vc ( t ) + The characteristic equation is: s 2 + 4 s + 4 = 0 9-20
• The natural frequencies are: s1,2 = −2, −2 The natural response is of the form: vc ( t ) = A1e−2t + A2 t e−2 t Before t = 0 the circuit is at steady state i L ( 0+ ) = i L ( 0 − ) and vC ( 0+ ) = vC ( 0− ) i C ( 0 + ) = −i L ( 0 + ) = − 2 A dv ( 0+ ) dt vc ( 0 + )=0= A1 and dvc ( 0+ ) dt = −8 = A2 = i C ( 0+ ) 14 = −8 V ⇒ vc ( t ) = −8 t e−2 t V P9.6-3 Assume that the circuit is at steady state before t = 0. The initial conditions are vc ( 0− ) = 104 V & iL ( 0− ) = 0 A After t = 0 − vc ( t ) + .01 KVL: diL ( t ) + 106 iL ( t ) = 0 dt (1) KCL:  d 2iL ( t ) dvC ( t ) di ( t )  = −C .01 + 106 L  iL ( t ) = −C 2 dt dt dt   ∴ 0.01 C ( 2) d 2iL ( t ) di ( t ) + 106 C L + iL ( t ) = 0 2 dt dt The characteristic equation is: ( 0.01 C ) s 2 + (106 C ) s + 1 = 0 The natural frequencies are: s1,2 = −106 C ± (10 C ) 6 2 − 4 ( 0.01C ) 2 ( 0.01C ) For critically-damped response: 1012 C 2 − .04C = 0 ⇒ C = 0.04 pF so s1,2 = −5 ×107 , −5 × 107 . 9-21
• The natural response is of the form: iL ( t ) = A1e −5×10 t + A2 t e −5×10 7 diL + ( 0 ) = 100 dt di ( 0 ) = 106 So iL ( 0 ) = 0 = A1 and L dt Now from (1) ⇒ 7 t vc ( 0+ ) − 106 iL ( 0+ )  = 106 A   s = A2 ∴ iL ( t ) = 106 t e−5×10 t A 7 Now v ( t ) = 106 iL ( t ) = 1012 t e−5×10 t V 7 P9.6-4 The characteristic equation can be shown to be: s 2 + The natural frequencies are: s1,2 = −250, − 250 1 1 = s 2 + 500 s + 62.5 × 103 = 0 s+ RC LC The natural response is of the form: v ( t ) = Ae −250t + B t e−250t dv ( 0 ) = −3000 = −250 A + B ⇒ B = −1500 dt − 1500 t e−250t V v ( 0 ) = 6 = A and ∴ v ( t ) = 6e−250t P9.6-5 After t=0, using KVL yields: di ( t ) t + Ri ( t ) + 2 + 4 ∫0 i (τ ) dτ = 6 (1) dt v( t ) Take the derivative with respect to t: d 2i ( t ) di ( t ) +R + 4i ( t ) = 0 dt 2 dt The characteristic equation is s 2 + Rs + 4 = 0 Let R = 4 for critical damping ⇒ ( s + 2) 2 =0 So the natural response is i ( t ) = A t e −2t + B e −2t i ( 0 ) = 0 ⇒ B = 0 and ∴ i ( t ) = 4 t e −2t A di ( 0 ) = 4 − R ( i( 0 ) ) = 4 − R ( 0 ) = 4 = A dt 9-22
• Section 9-7: Natural Response of an Underdamped Unforced Parallel RLC Circuit P9.7-1 After t = 0 KCL: vc ( t ) dv ( t ) + iL ( t ) + 5 ×10−6 c = 0 dt 250 KVL: vc ( t ) = 0.8 (1) diL ( t ) dt ( 2) d 2vc ( t ) dv ( t ) + 800 c + 2.5 ×105 v ( t ) = 0 ⇒ s 2 + 800 s + 250, 000 = 0, s1,2 = −400 ± j 300 c dt dt 2 The natural response is of the form v c (t ) Before t = 0 the circuit is at steady state: = e −400t  A1 cos 300t + A 2sin 300t    −6 A 500 vc ( 0+ ) = vc ( 0− ) = 250 −6 iL ( 0+ ) = iL ( 0− ) = ( 500 )+6 = 3 V From equation (1) : dvc ( 0+ ) dt = − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0 vc ( 0+ ) = 3 = A1 dvc ( 0+ ) dt = 0 = −400 A1 + 300 A2 ⇒ A2 = 4 ∴ vc ( t ) = e −400t [3cos 300t + 4sin 300t ] V 9-23
• P9.7-2 Before t = 0 v ( 0+ ) = v ( 0− ) = 0 V i ( 0+ ) = i ( 0− ) = 2 A After t = 0 KCL : t v (t ) 1 d v (t ) + + 2 ∫ v (τ ) dτ + i ( 0 ) = 0 0 1 4 dt Using the operator s = d we have dt 1  2 v (t ) +   s v (t ) +   v (t ) + i (0) = 0 ⇒  4 s (s 2 + 4s + 8) v ( t ) = 0 The characteristic equation and natural frequencies are: s 2 + 4s + 8 = 0 The natural response is of the form: v ( t ) = e v ( 0 ) = 0 = B1 and −2 t  B1 cos 2t + B 2 sin 2t    dv ( 0 ) = 4  −i ( 0 ) − v ( 0 )  = −4 [ 2] = −8 = 2 B2 or B2 = −4   dt so v ( t ) = −4e −2t sin 2t V P9.7-3 After t = 0 1 dvc ( t ) vc ( t ) + + iL ( t ) = 0 4 dt 2 4 diL ( t ) KVL : vc ( t ) = + 8 iL ( t ) dt KCL : Characteristic Equation: ⇒ s = −2 ± j 2 d 2i L ( t ) dt 2 +4 di L ( t ) dt (1) ( 2) + 5 i L ( t ) = 0 ⇒ s 2 + 4 s + 5 = 0 ⇒ s1,2 = −2 ± i Natural Response: i L ( t ) = e −2t  A1 cos t + A2 sin t    9-24
• Before t = 0 vc ( 0− )  48  + − = 7  4 8 + 2  ⇒ vc ( 0 ) = vc ( 0 ) = 8 V  2   8 iL ( 0+ ) = iL ( 0− ) = − = −4 A 2 diL ( 0+ ) dt = vc ( 0+ ) 4 − 2iL ( 0+ ) = 8 A − 2 ( −4 ) = 10 4 s iL ( 0+ ) = −4 = A1 diL ( 0+ ) dt ∴ iL ( t ) = e −2t [ −4 cos t + 2sin t ] A = 10 = − 2 A1 + A2 ⇒ A2 = 2 P9.7-4 The plot shows an underdamped response, i.e. v ( t ) = e − α t [ k1 cos ω t + k2 sin ω t ] + k3 . Examining the plot shows v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 . Therefore, v ( t ) = k2 e − α t sin ω t . Again examining the plot we see that the maximum voltage is approximately 260 mV the time is approximately 5 ms and that the minimum voltage is approximately −200 mV the time is approximately 7.5 ms. The time between adjacent maximums is approximately 5 ms so 2π ω≈ = 1257 rad/s . Then 5 ×10−3 −α ( 0.005 ) 0.26 = k2 e sin (1257 (.005 ) ) (1) −0.2 = k2 e −α ( 0.0075 ) sin (1257 (.0075 ) ) ( 2) To find α we divide (1) by (2) to get ( 6.29 rad )  ⇒    sin ( 9.43 rad )  α ( 0.0025 )  sin − 1.3 = e e0.0025α = 1.95 ⇒ α = 267 From (1) we get k2 = 544 . Then v ( t ) = 544e −267t sin1257t V 9-25
• P9.7-5 After t = 0 The characteristic equation is: s2 + 1 1 s+ = 0 or s 2 + 2 s + 5 = 0 RC LC The natural frequencies are: s1,2 = −1 ± j 2 The natural response is of the form: v(t ) = e − t  B1 cos 2t + B 2 sin 2t    v ( 0+ ) 2 1 ) = − 5 − i ( 0 ) = − 5 − 10 = − 1 V 2 s dv ( 0 ) 3  1 = 10 − = − B + 2 B ⇒ B = − v(0 ) = 2 = B1 . From KCL, ic ( 0 + + + so + dt    2 3 Finally, v ( t ) = 2e − t cos 2t − e− t sin 2t V 2 1 2 2 2 t≥0 9-26
• Section 9-8: Forced Response of an RLC Circuit P9.8-1 After t = 0 KCL : is ( t ) = v (t ) R KVL : v ( t ) = L is ( t ) = 2 d 2iL ( t ) dt 2 dv ( t ) diL ( t ) dt dt d 2iL ( t ) L diL ( t ) + iL ( t ) + LC R dt dt 2 d 2iL ( t ) dt + iL ( t ) + C + 1 diL ( t ) 1 1 + iL ( t ) = is ( t ) RC dt LC LC + ( 650 ) diL ( t ) dt + (105 ) iL ( t ) = (105 ) is ( t ) (a) Try a forced response of the form i f ( t ) = A . Substituting into the differential equations gives 0+0+ A 1 1 = ⇒ −3 (.01) (1×10 ) (.01) (1×10−3 ) A = 1 . Therefore i f ( t ) = 1 A . (b) Try a forced response of the form i f ( t ) = A t + B . Substituting into the differential equations gives 0 + A 65 1 + ( A t + B) = 0.5 t . Therefore A = 0.5 and ( 0.01)( 0.001) (100 ) ( 0.001) B = −3.25 × 10 −3 . Finally i f ( t ) = 5 t − 3.25 × 10−3 A . (c) Try a forced response of the form i f ( t ) = A e−250t . It doesn’t work so try a forced response of the form i f ( t ) = B t e−250t . Substituting into the differential equation gives ( −250 )2 B e−250t − 500 B e−250t  + 650 ( −250 ) B t e−250t + B e−250t  + 105 B t e−250t = 2 e−250t .     Equating coefficients gives ( 250 ) and 2 2 B + 650 ( −250 ) B + 105 B = 0 ⇒ ( 250 ) + 650 ( −250 ) + 105  B = 0 ⇒   −500 B + 650 B = 2 ⇒ [ 0] B = 0 B = 0.0133 Finally i f ( t ) = 0.0133 t e −250t A . 9-27
• P9.8-2 After t = 0 d 2v ( t ) dt d 2v ( t ) dt + v (t ) R dv ( t ) 1 + v (t ) = s L dt LC LC + 70 dv ( t ) dt + 12000v ( t ) = 12000 vs ( t ) (a) Try a forced response of the form v f ( t ) = A . Substituting into the differential equations gives 0 + 0 + 12000 A = 24000 ⇒ A = 2 . Therefore v f ( t ) = 2 V . (b) Try a forced response of the form v f ( t ) = A + B t . Substituting into the differential equations gives 70 A + 12000 A t + 12000 B = 2400 t . Therefore A = 0.2 and B = Finally v f ( t ) = ( −1.167 × 10−3 ) t + 0.2 V . −70 A = −1.167 × 10 −3 . 12000 (c) Try a forced response of the form v f ( t ) = A e −30t . Substituting into the differential equations gives 900 Ae −30t − 2100 Ae −30 t + 12000 Ae −30 t = 12000 e −30t . Therefore A = v f ( t ) = 1.11e−250 t V . 12000 = 1.11 . Finally 10800 9-28
• Section 9-9: Complete Response of an RLC Circuit P9.9-1 First, find the steady state response for t < 0, when the switch is open. Both inputs are constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. After a source transformation at the left of the circuit: i L ( 0) = and 11 − 4 = 2.33 mA 3000 v C ( 0) = 4 V After the switch closes Apply KCL at node a: vC R +C d vC + iL = 0 dt Apply KVL to the right mesh: L d d i L + Vs − vC = 0 ⇒ vC = L i L + Vs dt dt After some algebra: V 1 d 1 d2 i + iL + iL = − s 2 L dt R C dt LC R LC ⇒ d2 d i + ( 500 ) i L + (1.6 × 105 ) i L = −320 2 L dt dt The characteristic equation is s 2 + 500 s + 1.6 × 105 = 0 ⇒ s1,2 = −250 ± j 312 rad/s 9-29
• After the switch closes the steady-state inductor current is iL(∞) = -2 mA so i L ( t ) = −0.002 + e −250 t ( A1 cos 312 t + A2 sin 312 t ) d i L (t ) + 4 dt = 6.25 e −250 t  −250 ( A1 cos 312 t − A2 sin 312 t ) − 312 ( A1 sin 312 t − A2 cos 312 t )  + 4   v C ( t ) = 6.25 = 6.25 e −250 t ( 312 A2 − 250 A1 ) cos 312 t + ( 250 A2 + 312 A1 ) sin 312 t  + 4   Let t = 0 and use the initial conditions: iL ( 0+ ) = 0.00233 = −0.002 + A1 ⇒ 0.00433 = A1 vC ( 0+ ) = 4 = 6.25 ( 312 A2 − 250 A1 ) + 4 ⇒ A2 = 250 250 A2 = ( 0.00433) = 0.00347 312 312 Then i L ( t ) = −0.002 + e −250 t ( 0.00433cos 312 t + 0.00345sin 312 t ) = −0.002 + 0.00555 e −250 t cos ( 312 t − 36.68° ) A v C ( t ) = 4 + 13.9 e −250 t sin ( 312 t ) V i (t ) = vC (t ) 2000 = 2 + 6.95 e −250 t sin ( 312 t ) mA (checked using LNAP on 7/22/03) P9.9-2 First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = −1 = −0.2 A 1+ 4 and v (0) = 4 ( −1) = −0.8 V 1+ 4 9-30
• For t > 0 Apply KCL at node a: v − Vs d +C v+i = 0 R1 dt Apply KVL to the right mesh: R2 i + L d d i − v = 0 ⇒ v = R2 i + L iL dt dt After some algebra: L + R1 R 2C d R1 + R 2 d2 Vs i+ i+ i= 2 dt R1 L C dt R1 L C R1 L C The forced response will be a constant, if = B so 1 = ⇒ d2 d i +5 i +5i =1 2 dt dt d2 d B + 5 B + 5 B ⇒ B = 0.2 A . 2 dt dt To find the natural response, consider the characteristic equation: 0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 ) The natural response is in = A1 e −3.62 t + A2 e−1.38 t so i ( t ) = A1 e −3.62 t + A2 e−1.38 t + 0.2 Then d   v ( t ) =  4 i ( t ) + 4 i ( t )  = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8 dt   At t=0+ −0.2 = i ( 0 + ) = A1 + A2 + 0.2 −0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8 so A1 = 0.246 and A2 = −0.646. Finally i ( t ) = 0.2 + 0.246 e −3.62 t − 0.646 e−1.38 t A 9-31
• P9.9-3 First, find the steady state response for t < 0. The input is constant so the capacitors will act like an open circuits at steady state. v1 ( 0 ) = and 1000 (10 ) = 5 V 1000 + 1000 v2 ( 0 ) = 0 V For t > 0, Node equations: v1 − 10  1 v −v  d +  × 10−6  v1 + 1 2 = 0 1000  6 1000  dt 1  d ⇒ 2 v1 +  × 10−3  v1 − 10 = v2 6  dt v1 − v2  1  d =  × 10−6  v2 1000  16  dt 1  d ⇒ v1 − v2 =  × 10−3  v2  16  dt After some algebra: d2 d v + ( 2.8 × 104 ) v1 + ( 9.6 ×107 ) v1 = 9.6 × 108 2 1 dt dt The forced response will be a constant, vf = B so d2 d B + ( 2.8 × 104 ) B + ( 9.6 × 107 ) B = 9.6 × 108 2 dt dt ⇒ B = 10 V . To find the natural response, consider the characteristic equation: s 2 + ( 2.8 × 104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104 The natural response is 3 vn = A1 e −4×10 t + A2 e−2.4×10 4t so 3 v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10 4t + 10 At t = 0 9-32
• 5 = v1 ( 0 ) = A1 e −4×103 ( 0 ) + A2 e −2.4×104 ( 0 ) + 10 = A1 + A2 + 10 (1) Next 1  d 2 v1 +  ×10−3  v1 − 10 = v2 6  dt ⇒ d v1 = −12000v1 + 6000 v2 + 6 × 104 dt At t = 0 d v1 ( 0 ) = −12000v1 ( 0 ) + 6000 v2 ( 0 ) + 6 × 104 = −12000 ( 5 ) + 6000 ( 0 ) + 6 × 104 = 0 dt so 3 4 d v1 ( t ) = A1 −4 × 103 e −4×10 t + A2 −2.4 × 104 e−2.4×10 t dt ( ) ( ) At t = 0+ 0= d −4×103 ( 0 ) −2.4×104 ( 0 ) v1 ( 0 ) = A1 −4 × 103 e + A2 −2.4 × 104 e = A1 −4 × 103 + A2 −2.4 × 104 dt ( ) ( ) ( ) ( ) so A1 = −6 and A2 = 1. Finally v1 ( t ) = 10 + e −2.4 ×10 4t P9.9-4 For t > 0 3t − 6 e−4 ×10 V for t > 0 KCL at top node: diL ( t )   1 dv ( t ) − 5 cos t  + iL ( t ) + =0  0.5 dt 12 dt   (1) KVL for right mesh: 0.5 diL ( t ) 1 dv ( t ) = + v (t ) 12 dt dt ( 2) Taking the derivative of these equations gives: d 2iL ( t ) diL ( t ) 1 d 2 v ( t ) of (1) ⇒ 0.5 + + = −5 sin t dt 12 dt 2 dt 2 dt 2 2 d of ( 2 ) ⇒ 0.5 d iL ( t ) = 1 d v ( t ) + dv ( t ) dt 12 dt 2 dt 2 dt d (3) ( 4) 9-33
• d 2iL ( t ) di ( t ) Solving for in ( 4 ) and L in ( 2 ) & plugging into ( 3) gives 2 dt dt d 2v ( t ) dv ( t ) +7 + 12v ( t ) = −30sin t dt 2 dt The characteristic equation is: s 2 + 7s+12 = 0 . The natural frequencies are s1,2 = −3, −4 . The natural response is of the form vn (t ) = A1e −3t + A2 e −4t . Try a forced response of the form v f ( t ) = B1 cos t + B 2 sin t . Substituting the forced response into the differential equation and equating like terms gives B1 = 21 33 and B 2 = − . 17 17 v ( t ) = vn (t ) + v f ( t ) = A1e−3t + A2 e−4t + 21 33 cos t − sin t 17 17 We will use the initial conditions to evaluate A1 and A2. We are given iL ( 0 ) = 0 and v ( 0 ) = 1 V . Apply KVL to the outside loop to get 1 iC ( t ) + iL ( t )  + 1( iC ( t ) ) + v ( t ) − 5cos t = 0   At t = 0+ iC ( 0 ) = 5cos ( 0 ) + iL ( 0 ) − v ( 0 ) 5 + 0 − 1 = =2 A 2 2 dv ( 0 ) iC ( 0 ) 2 = = = 24 V/s dt 1 12 1 12     ⇒ + 33  dv(0 ) = 24 = −3 A1 − 4 A2 − dt 17   v(0+ ) = 1 = A1 + A2 + 21 17 A1 = 25 429 A2 = − 17 Finally, ∴ v(t ) = 25e −3t − 429e −4t − 21cos t + 33sin t V 17 9-34
• P9.9-5 Use superposition. Find the response to inputs 2u(t) and –2u(t-2) and then add the two responses. First, consider the input 2u(t): For 0< t < 2 s Using the operator s = d we have dt KVL: vc ( t ) + siL ( t ) + 4 iL ( t ) − 2  = 0   (1) KCL: 1 3 s vc ( t ) ⇒ vc ( t ) = iL ( t ) (2) s 3 Plugging (2) into (1) yields the characteristic equation: ( s 2 + 4 s + 3) = 0 . The natural frequencies are s1,2 = −1 , −3 . The inductor current can be expressed as iL ( t ) = iL (t ) = in (t ) + i f (t ) = ( A1 e −t + A2 e−3t ) + 0 = A1 e− t + A2 e −3t . Assume that the circuit is at steady state before t = 0. Then vc (0+ ) = 0 and iL (0+ ) = 0 . Using KVL we see that (1) diL (0+ ) = 4  2 − iL (0+ )  − vC (0+ ) = 8 A/s . Then   dt iL (0) = 0 = A1 + A2    A1 = 4 , A2 = −4 . diL (0) = 8 = − A1 − 3 A2  dt  Therefore iL (t ) = 4e − t − 4e−3t A . The response to 2u(t) is 0 t<0  v1 (t ) = 8 − 4 iL (t ) =  −t −3t 8 − 16 e + 16 e V t > 0 . = 8 − 16 e − t + 16 e −3t  u ( t ) V   The response to –2u(t-2) can be obtained from the response to 2u(t) by first replacing t by t-2 everywhere is appears and the multiplying by –1. Therefore, the response to –2u(t-2) is v2 (t ) =  −8 + 16e − (t − 2) − 16 e−3(t − 2)  u ( t - 2 ) V .   By superposition, v(t ) = v1 (t ) + v2 (t ) . Therefore v(t ) = 8 − 16e −t + 16e−3t  u (t ) +  −8 + 16e− (t − 2) − 16 e −3(t − 2)  u (t − 2) V     9-35
• P9.9-6 First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = − and After the switch opens 5 = −1.25 A 4 v (0) = 5 V Apply KCL at node a: v d + 0.125 v = i 2 dt Apply KVL to the right mesh: −10 cos t + v + 4 d i+4i =0 dt After some algebra: d2 d v + 5 v + 6 v = 20 cos t 2 dt dt The characteristic equation is s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3 Try vf = A cos t + B sin t d2 d A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t 2 ( dt dt 9-36
• ( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t ( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t Equating the coefficients of the sine and cosine terms yields A =2 and B =2. Then vf = 2 cos t + 2 sin t v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e −3 t Next v (t ) d + 0.125 v ( t ) = i ( t ) ⇒ 2 dt d v (t ) = 8 i (t ) − 4 v (t ) dt d V  5 v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8  −  − 4 ( 5 ) = −30 dt s  4 Let t = 0 and use the initial conditions: 5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e −0 = 2 + A1 + A2 d v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t dt −30 = d v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2 dt So A1 = −23 and A2 = 26 and v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t V for t > 0 9-37
• P9.9-7 First, find the steady-state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = 0 A and After t = 0 Apply KCL at node a: v (0) = 0 V C d v=i dt Apply KVL to the right mesh: d 8 i + v + 2 i + 4 (2 + i) = 0 dt d 12 i + v + 2 i = −8 dt 2  1  d d 4 After some algebra: v + ( 6) v +  v = − 2 dt dt C 2C The forced response will be a constant, vf = B so  1  d2 d 4 B + ( 6) B +  B = − 2 dt dt C 2C (a) ⇒ B = −8 V d2 d v + ( 6 ) v + ( 9 ) v = −72 . 2 dt dt 2 The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3 When C = 1/18 F the differential equation is Then v ( t ) = ( A1 + A2 t ) e −3t − 8 . Using the initial conditions: 9-38
• 0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒ 0 = i ( 0) = C So (b) A1 = 8 d v ( 0 ) = C  −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0  ⇒   dt A2 = 3 A1 = 24 v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0 d2 d v + ( 6 ) v + ( 5 ) v = −40 2 dt dt 2 The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5 When C = 1/10 F the differential equation is Then v ( t ) = A1 e − t + A2 e −5 t − 8 . Using the initial conditions: 0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8    ⇒ A1 = 10 and A2 = −2 d 0 0 0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0  dt  So v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0 (c) d2 d v + ( 6 ) v + (10 ) v = −80 2 dt dt 2 The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j When C = 1/20 F the differential equation is Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 . Using the initial conditions: 0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒ 0= So A1 = 8 d v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒ dt A2 = 3 A1 = 24 v ( t ) = e −3 t ( 8cos t + 24 sin t ) − 8 V for t > 0 9-39
• P9.9-8 The circuit will be at steady state for t<0: so iL(0+) = iL(0−) = 0.5 A and vC(0+) = vC(0−) = 2 V. For t>0: Apply KCL at node b to get: 1 1 d 1 1 d = i (t ) + v (t ) ⇒ i (t ) = − v (t ) L C L 4 4 dt 4 4 dt C Apply KVL at the right-most mesh to get: 4i L( t) + 2 d 1 d  i (t ) = 8  vc ( t )  + v ( t ) L dt  4 dt  c Use the substitution method to get d 1 1 d 1 1 d   1 d  4 − vC ( t )  + 2  − vC ( t )  = 8  vc ( t )  + v ( t ) dt  4 4 dt  4 4 dt    4 dt  c or d2 d v t + 6 v (t ) + 2 v (t ) 2 C ( ) C dt dt C d2 d The forced response will be a constant, vC = B so 2 = B + 6 B + 2B ⇒ B = 1 V . dt dt 2 2= To find the natural response, consider the characteristic equation: 0 = s 2 + 6 s + 2 = ( s + 5.65 )( s + 0.35 ) The natural response is vn = A1 e −5.65 t + A2 e −0.35 t vC ( t ) = A1 e −5.65 t + A2 e −0.35 t so +1 Then iL ( t ) = 1 1 d 1 −5.65 t −0.35 t + + 0.0875 A2 e vC ( t ) = + 1.41A1 e 4 4 dt 4 9-40
• At t=0+ 2 = vC ( 0 + ) = A1 + A + 1 2 1 1 = iL ( 0+ ) = + 1.41A1 + 0.0875 A 2 2 4 so A1 = 0.123 and A2 = 0.877. Finally vC ( t ) = 0.123 e −5.65 t + 0.877 e −0.35 t +1 V P9.9-9 The inductor current and voltage are related be After t = 0 v (t ) = L di ( t ) dt (1) Apply KCL at the top node to get C dv( t ) v( t ) + i (t ) + =5 dt 2 (2) d , and substituting (1) into (2) yields ( s 2 + 4s + 29 ) i ( t ) = 5 . dt The characteristic equation is s 2 + 4 s + 29 = 0 . The characteristic roots are s1,2 = − 2 ± j 5 . Using the operator s = The natural response is of the form in ( t ) = e −2t [ A cos 5t + B sin 5t ] . Try a forced response of the form i f ( t ) = A . Substituting into the differential equation gives A = 5 . Therefore i f ( t ) = 5 A . The complete response is i(t ) = 5 + e−2t [ A cos 5t + B sin 5t ] where the constants A and B are yet to be evaluated using the initial condition: i (0) = 0 = A + 5 ⇒ A = −5 di (0) di (0) 2A 0 = v ( 0) = L ⇒ = 0 = −2 A + 5 B ⇒ B = = −2 dt dt 5 Finally, i (t ) = 5 + e −2t [ −5cos 5t − 2sin 5t ] A . P9.9-10 9-41
• Assume that the circuit is at steady before t = 0. 2 ×9 = 6 A 2+1 1 v(0+ ) = v(0− ) = × 9× 1.5 = 4.5 V 2+1 i (0+ ) = i (0− ) = Apply KCL at the top node of the current source to get After t = 0: i ( t ) + 0.5 dv( t ) v( t ) + = is ( t ) dt 1.5 (1) Apply KVL and KCL to get dv( t ) v( t )  5di( t )  v ( t ) + 0.5 + + i (t )  0.5 = dt dt 1.5   (2) Solving for i(t) in (1) and plugging into (2) yields d 2 v( t ) 49 dv( t ) 4 di ( t ) 2 + + v ( t ) = is ( t ) + 2 s 2 dt dt 30 dt 5 5 where is ( t ) = 9 + 3e −2t A d 49 4 , the characteristic equation is s 2 + s + = 0 and the characteristic 30 5 dt roots are s1,2 =−.817 ± j.365 . The natural response has the form Using the operator s = vn (t ) = e−0.817t  A1 cos (0.365 t ) + A2 sin (0.365 t )    Try a forced response of the form v f (t ) = B0 + B1e −2 t . Substituting into the differential equations gives B0 = 4.5 and B1 = −7.04 . The complete response has the form v(t ) = e−.817 t  A1 cos(0.365 t ) + A2 sin (0.365 t )  + 4.5 − 7.04 e −2t   Next, consider the initial conditions: v (0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04 9-42
• d v(0) 4 4 = 2 is (0) - 2 i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6 dt 3 3 6= d v( 0 ) = −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −6.38 dt So the voltage is given by v(t ) = e−0.817 t 7.04 cos(0.365 t ) + A2 sin (0.365 t )  + 4.5 − 7.04 e−2t   Next the current given by i (t ) = is (t ) − Finally v(t ) d v(t ) − 0.5 1.5 dt i (t ) = e −0.817 t [ 2.37 cos(0.365t ) + 7.14sin(0.365t ) ] + 6 + 0.65e −2t A P9.9-11 First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. va ( 0− ) = −4 i ( 0− ) ( ) i ( 0− ) = 2va ( 0− ) = 2 −4 i ( 0− ) ⇒ i ( 0− ) = 0 A and v ( 0− ) = 10 V 9-43
• For t > 0 Apply KCL at node 2: va d + K va + C v=0 R dt KCL at node 1 and Ohm’s Law: va = − R i so d 1+ K R v= i dt C Apply KVL to the outside loop: L d i + R i + v − Vs = 0 dt After some algebra: d2 R d 1+ K R 1+ K R v+ v+ v= Vs 2 dt L dt LC LC ⇒ d2 d v + 40 v + 144 v = 2304 2 dt dt The forced response will be a constant, vf = B so d2 d B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V 2 dt dt The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 . v ( t ) = A1 e− 4 t + A2 e−36 t + 16 . Then Using the initial conditions: 10 = v ( 0+ ) = A1 e0 + A2 e0 + 16 ⇒ 0= d v ( 0+ ) = −4 A1 e0 − 36 A2 e0 dt    ⇒ ⇒ − 4 A1 − 36 A2 = 0   A1 + A2 = −6 A1 = −6.75 and A2 = 0.75 So v ( t ) = 0.75 e −36 t − 6.75 e−4 t + 16 V for t > 0 (checked using LNAP on 7/22/03) 9-44
• Section 9-10: State Variable Approach to Circuit Analysis P9.10-1 At t = 0− the circuit is source free ∴ iL (0) = 0 and v(0) = 0. Apply KCL at the top node to get After t = 0 iL ( t ) + 1 dv( t ) =4 5 dt (1) Apply KVL to the right mesh to get v( t ) −(1) Solving for i1 ( t ) in (1) and plugging into (2) ⇒ diL ( t ) − 6 iL ( t ) = 0 dt (2) d 2 v( t ) dv( t ) + 6 + 5v ( t ) = 120 . 2 dt dt The characteristic equation is s 2 + 6 s + 5= 0 . The natural frequencies are s1,2 =−1, −5 . The natural response has the form vn (t ) = A1 e−t + A2 e−5t . Try v f ( t ) = B as the forced response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete response has the form v (t ) = A1 e−t + A2 e−5t + 24 . Now consider the initial conditions. From (1) dv(0) = 20 − 5 iL (0) = 20 V . Then s dt v(0) = 0 = A1 + A2 + 24    ⇒ dv(0) = 20 = − A1 −5 A2  dt  A1= −25, A2 = 1 Finally v (t ) = − 25e −t + e −5t + 24 V . 9-45
• P9.10-2 Before t = 0 there are no sources in the circuit so iL (0) = 0 and v(0) = 0 . After t = 0 we have: Apply KCL at the top node to get iL ( t ) = 4 − 1 dv ( t ) 10 dt (1) Apply KVL to the left mesh to get v( t ) − diL ( t ) − 6iL ( t ) = 0 dt (2) Substituting iL ( t ) from (1) into (2) gives d 2 v( t ) dv( t ) +6 + 10v( t ) = 240 2 dt dt The characteristic equation is s 2 + 6 s + 10 = 0 . The natural frequencies are s1,2 = −3 ± j . The natural response has the form vn (t ) = e −3t  A1 cos t + A2 sin t  . Try v f ( t ) = B as the forced   response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete response has the form v(t ) = e −3t  A1 cos t + A2 sin t  + 24 .   Now consider the initial conditions. From (1) dv(0) = 40 − 10 iL (0) = 40 V . Then s dt v (0) = 0 = A1 + 24 ⇒ A1 = −24 dv(0) = 40 = −3 A1 + A2 = 72 + A2 ⇒ A2 = −32 dt Finally, v (t ) = e −3t [ −24 cos t −32sin t ] + 24 V 9-46
• P9.10-3 Assume that the circuit is at steady state before t = 0 so iL (0) = −3 A and v(0) = 0 V . After t = 0 we have KCL: i ( t ) + C KVL: v( t ) = L i (t ) + C d 2i ( t ) −6 1 di ( t ) 1 + + i (t ) = 2 dt R C dt LC LC ⇒ dv( t ) v( t ) + + 6=0 dt R di ( t ) dt d  di( t )  1  di ( t )  L  + L + 6=0 dt  dt  R  dt  d 2i ( t ) di ( t ) + 100 + 250i ( t ) = −1500 2 dt dt The characteristic equation is s 2 + 100 s + 250 = 0 . The natural frequencies are s1,2 = −2.57, − 97.4 . The natural response has the form in (t ) = A1 e−2.57 t + A2 e−97.4t . Try ( t ) = B as the forced response. Substituting into the differential equation gives B = −6 so i f ( t ) = −6 A . The complete response has the form i(t ) = A1 e−2.57 t + A2 e−97.4t − 6 . i f Now consider the initial conditions: i (0) = A1 + A2 − 6 = −3   A1 = 3 .081  di (0) = 0 = − 2.57 A1 −97.4 A2  A2 =−0.081 dt  Finally: i (t ) = 3. 081 e v(t ) = .2 −2.57 t −.081e di ( t ) = −1.58e dt −97.4 t −2.57 t −6 A +1.58e−97.4t V 9-47
• P9.10-4 Apply KCL to the supernode corresponding to the dependent voltage source to get ix ( t ) − 2ix ( t ) − 0.01 dv ( t ) vx ( t ) + =0 2 dt Apply KCL at node 1 to get i ( t ) − 2ix ( t ) + (Encircled numbers are node numbers.) vx ( t ) =0 2 Apply KVL to the top-right mesh to get vx ( t ) + v ( t ) − 0.1 di ( t ) =0 dt Apply KVL to the outside loop to get ix ( t ) = −2 vx ( t ) − v ( t ) . Eliminate ix ( t ) to get Then eliminate vx ( t ) to get dv ( t ) 5 =0 vx ( t ) + v ( t ) − 0.01 2 dt 9 i ( t ) + vx ( t ) + 2 v ( t ) = 0 2 d i (t ) vx ( t ) = −v ( t ) + 0.01 dt dv ( t ) di ( t ) + 0.25 =0 dt dt di ( t ) −2.5 v ( t ) + i ( t ) + 0.45 =0 dt −1.5 v ( t ) − 0.01 Using the operator s = d we have dt (−1.5 − .01s )v ( t ) + (.25s ) i ( t ) = 0 (−2.5)v( t ) + (1+.45s ) i ( y ) = 0 The characteristic equation is s 2 +13.33 s + 333.33 = 0 . The natural frequencies are s1 , s2 = −6.67 ± j 17 . The natural response has the form vn (t ) = [ A cos17 t + B sin17 t ] e−6.67 t . ( t ) = 0 . The complete response has the form + B sin 17 t ] e−6.67 t . The forced response is v v(t ) = [ A cos17 t f 9-48
• The given initial conditions are i (0) = 0 and v (0) = 10 V. Then v(0) =10 = A and dv(0) =−111= − 6.67 A +17 B ⇒ B =−2.6 dt Finally i (t ) = [3.27 sin 17 t ] e−6.67 t A . (Checked using LNAP on 7/22/03) P9.10-5 Assume that the circuit is at steady state before t = 0 so v(0) = 10 V and iL (0) = The switch is open when 0 < t < 0.5 s v(0) 10 = A. 3 3 For this series RLC circuit we have: α= R 1 = 3 and ω 02 = = 12 2L LC −α ± α 2 −ω02 = s1,2 = −3 ± j 3 The natural response has the form vn (t ) = e −3t ( A cos1.73 t + B sin1.72 t ) . There is no source so v f ( t ) =0 . The complete response has the form v (t ) = e −3t ( A cos1.73 t + B sin1.72 t ) . Next v(0) = 10 = A   A= 10  dv(0) i ( 0 ) 10 3  ⇒  =− =− = − 20 = −3 A +1.73 B   B =5.77 dt C 16  so v(t ) = e−3t (10 cos1.73 t + 5.77 sin1.73 t ) V i(t ) = e−3t ( 3.33 cos1.73 t − 5.77 sin1.73 t ) A In particular, 1.73 1.73   v (0.5) = e −1.5  10 cos + 5.77 sin  = 0.2231× ( 6.4864 + 4.3915 ) = 2.43 V 2 2   and 1.73 1.73   i (0.5) = e −1.5  3.33cos −5.77 sin  = 0.2231× ( 2.1600 − 4.3915 ) = −0.50 A 2 2   9-49
• The switch is closed when t > 0.5 s Apply KCL at the top node: v( t ) −30 1 dv( t ) + iL ( t ) + =0 6 6 dt dv( t )  1 ⇒ iL ( t ) = 5 −  v( t ) +  6 dt  ⇒ 2 d iL ( t ) 1  dv( t ) d v( t )  =−  +  dt 6  dt dt 2  Apply KVL to the right mesh: 1 diL ( t ) v( t ) = 3 iL ( t ) + 2 dt The circuit is represented by the differential equation d 2v ( t ) dv ( t ) +7 + 18 v ( t ) = 180 2 dt dt The characteristic equation is 0 = s 2 + 7 s + 18 . The natural frequencies are s1,2 = −3.7 ± j 2.4 . The natural response has the form vn (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) . The forced response is v f ( t ) = 10 V . The complete response has the form v (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) + 10 . Next v (0.5) = e −3.5×0.5 ( A cos1.2 + B sin1.2 ) = 0.063 A + 0.162 B dv ( 0.5 ) = e−3.5×0.5 ( −3.5 A + 2.4 B ) cos1.2 − ( 3.5B + 2.4 A ) sin1.2    dt = e−3.5×0.5 ( −3.5cos1.2 − 2.4sin1.2 ) A + e−3.5×0.5 ( 2.4 cos1.2 − 3.5sin1.2 ) B = −0.6091A − 0.4158B Using the initial conditions yields 2.43=v(0.5) = 0.063 A+ 0.162 B  A=−20.65  dv(0.5) i ( 0.5 ) −1 2  ⇒ B = 23.03 =− =− =3=−0.6091A− 0.4158B  dt C 16  Finally In summary v(t ) = e −3.5 t ( −20.65 cos 2.4 t + 23.03 sin 2.4 t ) + 10  e −3t (10 cos1.73 t + 5.77 sin1.73 t ) V 0<t <0.5  v(t ) =  −3.5 t e ( −20.65 cos 2.4 t + 23.03 sin 2.4 t )+10 V 0.5<t  9-50
• Section 9-11: Roots in the Complex Plane P9.11-1 After t = 0 i1 + i 2 + 2 ×10−3 d i1 dt 2 ×10−3 d i1 dt 2000 −6 =0 = 3000 i 2 + 2 ×10−3 d i2 dt d yields dt  2000 + 2 × 10−3 s   i1   6  2000 =   −3 −3   2 × 10 s 3000 + 2 × 10 s  i 2   0   Using the operator s = s 2 + 3.5 × 106 s + 1.5 ×1012 = 0 ⇒ s1,2 = −5 × 105 , −3 ×106 P9.11-2 From P9.7-1 s2 + 1 1 1 1 = 0 ⇒ s2 + =0 s+ s+ −6 RC LC ( 250 ) ( 5 ×10 ) ( 0.8) ( 5 ×10−6 ) ⇒ s 2 + 800 s + 250000 = 0 s1,2 = 400 ± j 300 P9.11-3 dv( t ) v( t ) 1 KCL: i L ( t ) = × 10−6 + 4 dt 4000 di ( t ) KVL: vs ( t ) = 4 L + v( t ) dt d 2v ( t ) dv ( t ) dv( t ) v( t )  d 1 v ( t ) = 4  ×10−6 + + v ( t ) = 10−6 + 10−3 + v (t )  2 s dt  4 dt 4000  dt dt 9-51
• Characteristic equation: s 2 + 103 s + 106 = 0 Characteristic roots: s1,2 = −500 ± j 866 P9.11-4 Before t = 0 the voltage source voltage is 0 V so vb (0+) = vb (0−) = 0 V and i (0+) = i (0−) = 0 A . Apply KCL at node a to get va (0+ ) −36 v (0+ ) − vb (0+ ) − i (0+) + a = 0 ⇒ va (0+ ) + 2 va (0+ ) = 36 ⇒ va (0) = 12 V 12 6 After t = 0 the node equations are: − va ( t ) − vs ( t ) 1 t v ( t ) − va ( t ) + ∫ ( vb (τ ) − va (τ ) ) dτ + b =0 0 12 6 L C Using the operator s = d vb ( t ) vb ( t ) − va ( t ) 1 t + + ∫ ( vb (τ ) − va (τ ) ) dτ = 0 6 dt L 0 d we have dt v (t ) 1 1 1  1 1 + +  va ( t ) +  − −  vb ( t ) = s  6 12 s  12  6 s 1 1  1 1 1  − −  va ( t ) +  s + +  vb ( t ) = 0  6 s  18 6 s  Using Cramer’s rule 9-52
• ( s 2 +5s + 6) vb ( t ) = ( s + 6) vs ( t ) =( s + 6) ( 36 ) The characteristic equation is s 2 + 5s + 6 = 0 . The natural frequencies are s1,2 = −2, −3 . The natural response has the form vn (t ) = A1 e −2 t + A2 e −3 t . Try v f ( t ) = B as the forced response. Substituting into the differential equation gives B = 36 so v f ( t ) = 36 V. The complete response has the form vb (t ) = A1 e −2 t + A2 e −3t + 36 . Next vb (0+ ) = 36 + A1 + A2 dvb + (0 ) = −2 A1 − 3 A2 dt Apply KCL at node a to get At t = 0+ 1 dvb ( t ) vb ( t ) − va ( t ) + + i (t ) = 0 18 dt 6 1 1 d vb ( 0 (−2 A1 − 3 A2 ) = 18 18 dt + ) = v ( 0 )−v ( 0 ) − i + + a b 6 (0+ ) = 126−0 − 0 = 2 So 0 = vb (0+ ) =36+ A1 + A 2    ⇒ 1 ( −2 A 1 − 3 A 2 ) = 2  18  Finally vb ( t ) = 36 − 72e −2 t A1 = −72, A2 = 36 + 36e −3t V for t ≥ 0 9-53
• PSpice Problems SP 9-1 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) 9-54
• V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom. 9-55
• SP 9-2 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) 9-56
• V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom. 9-57
• SP 9-3 9-58
• SP 9-4 9-59
• Verification Problems VP 9-1 This problem is similar to the verification example in this chapter. First, check the steady-state inductor current v ( t ) 25 i (t ) = s = = 250 mA 100 100 This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an underdamped response. That requires 12 ×10−3 = L < 4 R 2C = 4(100) 2 (2 × 10−6 ) = 8 × 10−2 This inequality is satisfied, which also agrees with the plot. The damped resonant frequency is given by ω = d 1 LC  1  −   2 RC  2 2   1 1 = − = 5.95 × 103 −6  −6 −3 2(100)(2 × 10 )  ( 2 ×10 )(12 ×10 )  The plot indicates a maxima at 550.6µs and a minima at 1078.7µs. The period of the damped oscillation is T d = 2 (1078.7 µ s − 550.6 µ s) = 1056.2 µ s Finally, check that 5.95 ×103 = ω d = 2π 2π = = 5.949 × 103 −6 1056.2 ×10 Td The value of ω d determined from the plot agrees with the value obtained from the circuit. The plot is correct. VP 9-2 This problem is similar to the verification example in this chapter. First, check the steady-state inductor current. v ( t ) 15 i (t ) = s = = 150 mA 100 100 9-60
• This agrees with the value of 149.952 mA shown on the plot. Next, the plot shows an underdamped response. This requires 8 ×10−3 = L < 4 R 2C = 4 (100)2 (0.2 × 10−6 ) = 8 × 10−3 This inequality is not satisfied. The values in the circuit would produce a critically damped, not underdamped, response. This plot is not correct. Design Problems DP 9-1 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = The specifications require that vC ( ∞ ) = 1 so 2 R2 1 = 2 R1 + R 2 R2 R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get 9-61
• vs ( t ) = L   v (t )  d  vC ( t ) d d + C vC ( t )  + R1  C + C vC ( t )  + vC ( t )    R2  dt  R 2 dt dt     = LC  L d  R1  d2 + R1 C  vC ( t ) +  1 + vC ( t ) +  v t  R2  dt  R2  C ( )  dt 2     The characteristic equation is R1   1+  1 R1  R2 + s+ s2 +   R 2 C L   LC         = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) = 0     Equating coefficients of like powers of s: 1 R2 C + R1 L 1+ = 6 and R1 R2 =8 LC Using R1 = R 2 = R gives 1 R + =6 ⇒ RC L 1 =4 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 4 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4 Pick R = 1.309 Ω. Then vc ( t ) = iL ( t ) = At t = 0+ vC ( t ) 1.309 + 1 + A1 e −2 t + A2 e−4 t V 2 d vC ( t ) = −1.236 A1 e−2 t − 3.236 A2 e−4 t + 0.3819 dt ( ) 0 = vc 0+ = A1 + A2 + 0.5 ( ) = −1.236 A − 3.236 A 0 = iL 0 + 1 2 + 0.3819 Solving these equations gives A1 = -1 and A2 = 0.5, so vc ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 9-62
• DP 9-2 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = The specifications require that vC ( ∞ ) = 1 so 4 R2 1 = 4 R1 + R 2 R2 R1 + R 2 1 ⇒ 3 R 2 = R1 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L   v (t )  d  vC ( t ) d d + C vC ( t )  + R1  C + C vC ( t )  + vC ( t )    R2  dt  R 2 dt dt      L d  R1  d2 v t + + R1 C  vC ( t ) +  1 + v t 2 C ( )   R2  dt  R2  C ( )  dt     The characteristic equation is R1    1+   1 R1   R2  2 2 2 + s+ = s + 4s + 4 = ( s + 2 ) = 0 s +  R 2 C L   LC          Equating coefficients of like powers of s: R 1+ 1 R1 R2 1 + = 4 and =4 R2 C L LC = LC 9-63
• Using R 2 = R and R1 = 3R gives 1 3R + =4 ⇒ RC L 1 =1 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3 Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω . vc ( t ) = iL ( t ) = vC ( t ) + At t = 0+ 1 + ( A1 + A2 t ) e −2 t V 4 d 1 vC ( t ) = + dt 4 ( ) (( A 0 = vc 0+ = A1 + ( ) 0 = iL 0+ = 2 ) − A1 ) − A2 t e −2 t 1 4 1 + A2 − A1 4 Solving these equations gives A1 = -0.25 and A2 = -0.5, so vc ( t ) = 1  1 1  −2 t −  + t e V 4 4 2  DP 9-3 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 4 The specifications require that vC ( ∞ ) = so 5 9-64
• R2 4 = 5 R1 + R 2 ⇒ 4 R1 = R 2 Next, represent the circuit by a 2nd order differential equation: vC ( t ) KCL at the top node of R2 gives: R2 vs ( t ) = L KVL around the outside loop gives: +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L   v (t )  d  vC ( t ) d d + C vC ( t )  + R1  C + C vC ( t )  + vC ( t )    R2  dt  R 2 dt dt     = LC  L d  R1  d2 v t + + R1 C  vC ( t ) +  1 + v t 2 C ( )   R2  dt  R2  C ( )  dt     The characteristic equation is R1   1+  1 R1  R2 + s+ s2 +   R 2 C L   LC         = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0     Equating coefficients of like powers of s: R1 1 + = 4 and R2 C L 1+ R1 R2 LC = 20 Using R1 = R and R 2 = 4 R gives 1 R + = 4 and 4R C L 1 = 16 LC 1 1 These equations do not have a unique solution. Try C = F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 1 = 0 ⇒ R = 1 Ω R Then R1 = 1 Ω and R 2 = 4 Ω . Next vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 9-65
• iL ( t ) = + vC ( t ) 4 + A 2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 At t = 0 ( ) 0 = vc 0+ = 0.8 + A1 ( ) 0 = iL 0+ = 0.2 + A2 2 Solving these equations gives A1 = −0.8 and A2 = −0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V DP 9-4 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = The specifications require that vC ( ∞ ) = 1 so 2 R2 1 = 2 R1 + R 2 R2 R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get 9-66
•   v (t )  d  vC ( t ) d d + C vC ( t )  + R1  C + C vC ( t )  + vC ( t )    R2  dt  R 2 dt dt     vs ( t ) = L = LC  L d  R1  d2 vC ( t ) +  + R1 C  vC ( t ) +  1 + v t  R2  dt  R2  C ( )  dt 2     The characteristic equation is R1   1+  1 R1  R2 + s+ s2 +   R 2 C L   LC         = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0     Equating coefficients of like powers of s: R1 1 + = 4 and R2 C L 1+ R1 R2 LC = 20 Using R1 = R 2 = R gives 1 R + = 4 and RC L Substituting L = 1 = 10 LC 1 into the first equation gives 10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1) 4 1 − ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied. DP 9-5 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions 9-67
• vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1 so 2 R2 1 R1 + R 2 1 = 2 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 ⇒ R1 = R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 vs ( t ) = d  d   d   L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC  L d  R1 + R 2  d2 + R1 C  vo ( t ) +  v t + v t 2 o( )   R2  dt  R2  o ( )  dt     The characteristic equation is R2   1+  1 R2   R1 + s2 +  s+  R1 C L   LC      Equating coefficients of like powers of s:    = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) = 0     R2 1 + = 6 and R1 C L 1+ R2 R1 LC =8 Using R1 = R 2 = R gives 1 R + =6 ⇒ RC L 1 =4 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 4 9-68
• 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4 Pick R = 1.309 Ω. Then 1 + A1 e −2 t + A2 e −4 t V 2 A1 −2 t A2 −4 t v (t ) 1 iL ( t ) = o e + e V = + 1.309 2.618 1.309 1.309 vo ( t ) = vC ( t ) = 1.309 iL ( t ) + At t = 0+ 1 d 1 iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2 ( ) 0 = iL 0+ = ( ) 0 = vC 0+ = A1 A2 1 + + 2.618 1.309 1.309 1 + 0.6167 A1 + 0.2361 A2 2 Solving these equations gives A1 = -1 and A2 = 0.5, so vo ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 DP 9-6 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 3 so 4 R2 3 = 4 R1 + R 2 ⇒ 3R1 = R 2 9-69
• Next, represent the circuit by a 2nd order differential equation: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 vs ( t ) = d  d   d   L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC  L d  R1 + R 2  d2 + R1 C  vo ( t ) +  vo ( t ) +  v t  R2  dt  R2  o ( )  dt 2     The characteristic equation is R2   1+  1 R2   R1 + s2 +  s+  R1 C L   LC      Equating coefficients of like powers of s:    = s 2 + 4 s + 4 = ( s + 2 )2 = 0     R2 1 + = 4 and R1 C L 1+ R2 R1 LC =4 Using R1 = R and R 2 = 3R gives 1 3R + = 4 and RC L 1 =1 LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3 Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω . vo ( t ) = 3 + ( A1 + A2 t ) e −2 t V 4 9-70
• iL ( t ) = vo ( t ) vC ( t ) = 3 iL ( t ) + = 3 1  A1 A2  −2 t t e V + + 4  3 3  3   A1 A2  A2  −2 t d iL ( t ) = +   + t e + 4  3 3  3  dt   At t = 0+ 0 = iL ( 0 + ) = A1 + 1 4 3 3 A1 A2 0 = vC ( 0 + ) = + + 4 3 3 Solving these equations gives A1 = −0.75 and A2 = −1.5, so vo ( t ) = 3  3 3  −2 t −  + t e V 4 4 2  DP 9-7 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = 1 The specifications require that vo ( ∞ ) = so 5 R2 1 = 5 R1 + R 2 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 ⇒ R1 = 4 R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L 9-71
• Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 vs ( t ) = d  d   d   L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC  L d  R1 + R 2  d2 + R1 C  vo ( t ) +  v t + v t 2 o( )   R2  dt  R2  o ( )  dt     The characteristic equation is R2   1+  1 R2   R1 + s2 +  s+  R1 C L   LC         = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0     Equating coefficients of like powers of s: R2 1 + = 4 and R1 C L 1+ R2 R1 LC = 20 Using R 2 = R and R1 = 4 R gives 1 R + = 4 and 4R C L These equations do not have a unique solution. Try C = 1 = 16 LC 1 1 F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω R Then R1 = 4 Ω and R 2 = 1 Ω . Next vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) = vo ( t ) vC ( t ) = iL ( t ) + At t = 0+ 1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 1 d iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t 2 dt 9-72
• ( ) ( 0 ) = 0.2 + 2 A 0 = iL 0+ = 0.2 + A1 0 = vC + 2 Solving these equations gives A1 = −0.8 and A2 = −0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V DP 9-8 When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vC ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 1 so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 d  d   d   L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt 9-73
• vs ( t ) = R1 R2 LC  L d  R1 + R 2  d2 + R1 C  vo ( t ) +  v t + v t 2 o( )   R2  dt  R2  o ( )  dt     The characteristic equation is R2   1+  1 R2   R1 + s2 +  s+  R1 C L   LC         = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0     Equating coefficients of like powers of s: R2 1 + = 4 and R1 C L 1+ R2 R1 LC = 20 Using R1 = R 2 = R gives 1 R + =4 ⇒ RC L Substituting L = 1 = 10 LC 1 into the first equation gives 10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1) 4 1 − ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied. 9-74
• DP 9-9 Let’s simulate the three copies of the circuit simultaneously. Each copy uses a different value of the inductance. The PSpice transient response shows that when L = 1 H the inductor current has its maximum at approximately t=0.5 s. Consequently, we choose L = 1 H. 9-75
• Chapter 10 – Sinusoidal Steady-State Analysis Exercises Ex. 10.3-1 (a) T = 2π / ω = 2π / 4 (b) v leads i by 30 − (−70) = 100° Ex. 10.3-2 v ( t ) = 3cos 4 t + 4sin 4 t = ( 3 )) = 5 cos(4 t −53° ) ( (3) 2 + (4) 2 cos 4 t − tan −1 4 Ex. 10.3-3    12    i ( t ) = −5 cos 5t + 12 sin 5t = (−5) 2 + (12) 2 cos  5 t − 180 + tan −1     = 13 cos (5 t −112.6°)  −5      Ex. 10.4-1 KCL: is ( t ) = v( t ) d I v (t ) + = m cos ω t dt RC C v( t ) d + C v (t ) ⇒ R dt Try v f ( t ) = A cos ω t + B sin ω t & plug into above differential equation to get −ω A sin ω t + ω B cos ω t + 1 I ( A cos ω t + B sin ω t ) = m cos ω t RC C Equating sin ω t & cos ω t terms yields R Im ω R2 C Im and B = A = 1+ω 2 R 2 C 2 1+ω 2 R 2 C 2 Therefore v f (t ) = R Im ω R2 C Im cos ω t + sin ω t = 1+ω 2 R 2 C 2 1+ω 2 R 2 C 2 R Im 1+ ω R C 2 2 2 cos ω t − tan −1 (ω RC )    10-1
• Ex. 10.4-2 KVL : − 10 + j 3 I + 2 I = 0 ⇒ I = 10 = 2+ j 3 10∠0 13 ∠56.3 = 10 ∠ − 56.3 A 13 Therefore i (t ) = 10 cos (3 t − 56.3 ) A 13 Ex. 10.5-1 10 = 4.24 e− j 45 = 3− j 3 2.36 e j 45 Ex. 10.5-2 j 32 32e j 90 32 j (90-111) = = = 3.75 e − j 21 e −3+ j 8 8.54 e j111 8.54 Ex. 10.6-1 (a) i = 4 cos(ω t − 80 ) = Re{4 e j ω t e − j 80° } ⇒ I = 4 e − j 80° = 4∠ − 80° A (b) i = 10 cos(ω t + 20° ) = Re{10 e j ω t e j 20° } ⇒ I = 10e j 20° = 10∠20° (c) i = 8sin (ω t − 20 ) = 8cos (ω t − 110 ) = 8 Re{e j ω t e − j110° } ⇒ I = 8e − j110° = 8∠ − 110° A Ex. 10.6-2 (a) V = 10∠ − 140° = 10 e − j140° V ⇒ v(t ) = Re{10 e − j140° e j ω t } = 10 cos (ω t − 140°) V (b) V = 80 + j 75 = 109.7∠43.2° = 109.7 e j 43.2° ⇒ v (t ) = Re{109.7 e j 43.2° e jω t } = 109.7 cos(ω t + 43.2°) V 10-2
• Ex. 10.6-3 d v + v = 10 cos 100 t dt ( 0.01)( j 100 )V + V =10 0.01 10 =7.071 ∠− 45° 1+ j v = 7.071 cos 100 t V V= Ex. 10.6-4 { vs = 40 cos100t = Re 4 e j100 t KVL: i (t ) + 10 × 10 −3 di (t ) 1 + dt 5×10 −3 ∫ t −∞ } i (t ) dt = vS Assume i (t ) = Ae j100 t where A is complex number to be determined. Plugging into the differential equation yields Ae j100 t + j Ae j100 t + (− j 2 A)e j100 t = 4 e j100 t ⇒ A= 4 = 2 2 e j 45° 1− j In the time domain: { } { } i (t ) = Re 2 2 e j100 t e j 45° = Re 2 2 e j (100 t -45°) = 2 2 cos (100 t + 45° ) A Ex. 10.7-1 (a) v = R i = 10 (5 cos100 t ) = 50 cos 100 t (b) di v = L = 0.01[5(−100) sin100 t ] = −5sin100 t = 5cos (100 t + 90° ) V dt (c) 1 v = ∫ i dt = 103 ∫ 5cos100 t dt = 50sin100 t = 50 cos (100 t − 90°) V C Ex. 10.7-2 i=C dv = 10 × 10−6 [100(−500) sin (500 t + 30°)] dt = − 0.5sin (500 t +30°) = 0.5sin (500 t + 210°) = 0.5cos(500t +120°) A 10-3
• Ex. 10.7-3 From Figure E10.7-3 we get i (t ) = I m sin ω t = I m cos(ω t − 90°) A ⇔ I = I m ∠ − 90°A v(t ) = Vm cos ω t ⇔ V = Vm ∠0° V The voltage leads the current by 90° so the element is an inductor: Z eq = V ∠0° V V = m = m ∠90° Ω I m ∠−90° Im I Also Z eq = j ω L = ω L ∠90° ⇒ ω L = Vm Im ⇒ L = Vm ω Im Ex. 10.8-1 1 ZR = 8 Ω, ZC = = 2.4 j 2.4 = = − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω, j j× j 1 12 ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V. j5 Ex. 10.8-2 ZR = 8 Ω, ZC = 1 = j4 4 = = − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω, j j× j 1 12 ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° V. j3 10-4
• Ex 10.9-1 V1 (ω ) = j10 5 e − j 90 = 3.9 e − j 51 8 + j10 V 2 (ω ) = j 20 5 e − j 90 = 5.68 e − j 90 j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 Ex 10.9-2 V1 (ω ) = V 2 (ω ) = 8 ( j6) 4 e j15 = 19.2 e j 68 8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75 j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 10-5
• Ex. 10.10-1 KCL at Va: Va V −V + a b =1 4− j 2 − j10 (4 − j12) Va + (−4 + j 2) Vb = −20 − j 40 KCL at Vb: Vb − Va V + b + 0.5∠ − 90° = 0 ⇒ (−2 − j 4) Va + (2 − j 6) Vb = 10 + j 20 − j10 2+ j 4 Cramer’s rule yields: (−20− j 40) (−4+ j 2) (10+ j 20) (2 − j 6) −200+ j100 = = 5∠296.5° V V = a (4 − j12) (−4+ j 2) −80− j 60 (−2− j 4) (2- j 6) Therefore v (t ) = 5 cos (100 t + 296.5° ) = 5 cos (100 t − 63.5° ) V a Ex. 10.10-2 The mesh equations are: j15 I1 + 10 (I1 − I 2 ) = 20 ⇒ (10 + j15) I1 − 10 I 2 = 20 − j 5 I 2 +10(I 2 − I1 ) = −30∠−90° ⇒ −10I1 + (10− j 5) I 2 = j 30 Cramer’s rule yields: 20 −10 j 30 10− j 5 200+ j 200 = = 2.263∠ − 8.1° A I1 = 10 + j15 −10 75+ j100 10 − j 5 −10 Next VL = ( j15) I1 = (15∠90°) (2.263∠−8.1°) = 24 2∠82° V Therefore vL (t ) = 24 2 cos (ω t + 82°) V 10-6
• Ex. 10.10-3 The mesh equations are: (10+ j 50) I1 −10 I 2 = j 30 −10 I1 + (10− j 20) I 2 + j 20 I 3 = j 50 j 20 I 2 + (30− j10) I 3 = 0 Solving the mesh equations gives: I1 = − 0.87 − j 0.09 A, I 2 = −1.32+ j 1.27 A, I 3 = 0.5+ j 1.05 A Then Va = 10 (I1 − I 2 ) = 14.3∠ − 72° V and Vb = Va + j 50 = 36.6 ∠83° V Ex 10.11-1 V1 = V2 = j10 5 e − j 90 = 3.9 e− j 51 8 + j10 j 20 5 e − j 90 = 5.68 e− j 90 j 20 − j 2.4 Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e− j 90 = 3.58 e j 47 10-7
• Zt = 8 ( j10 ) − j 2.4 ( j 20 ) + = 4.9 + j 1.2 8 + j10 − j 2.4 + j 20 Ex 10.11-2 V1 (ω ) = V 2 (ω ) = j10 5 e − j 90 = 3.9 e− j 51 8 + j10 j 20 5 e − j 90 = 5.68 e− j 90 j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 V1 (ω ) = V 2 (ω ) = 8 ( j6) 4 e j15 = 19.2 e j 68 8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75 j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t - 22° ) 10-8
• Ex. 10.11-3 Use superposition. First, find the response to the voltage source acting alone: Z eq = − j 10⋅10 = 5(1 − j ) Ω 10− j10 Replacing the parallel elements by the equivalent impedance. The write a mesh equation : −10 + 5 I1 + j15 I1 + 5(1 − j) I1 = 0 ⇒ I1 = 10 = 0.707∠ − 45° A 10+ j10 Therefore: i1 (t ) = 0.707 cos(10 t − 45° ) A Next, find the response to the dc current source acting alone: Current division: Using superposition: I2 = − 10 × 3 = −2 A 15 i (t ) = 0.707 cos(10 t − 45°) − 2 A Ex. 10.12-1 ω2 = 1 1 = = 106 −3 −3 LC (1×10 )(1×10 ) ⇒ ω = 1000 rad sec Ex. 10.12-2 Diagram drawn with relative magnitudes arbitrarily chosen: 10-9
• Ex. 10.12-3 Two possible phasor diagrams for currents: In both cases: I CL = I LC = ( 25 )−(15 ) 2 2 = 20 A In the first case: I LC = I L − I C ⇒ I C = 6 − 20 = −14 A That isn’t possible. Turning to the second case: I CL = I C − I L ⇒ I C = 20 + 6 = 26 A Ex. 10.14-1 Z1 = 1 1 R1 X 1 ( X 1 − jR1 ) and R1 = 1 kΩ , X 1 = = = 1 kΩ 2 2 ω C1 (1000 )(10−6 ) R1 + X 1 Z1 = (1)(1)(1− j1) 1 1 = − j kΩ and Z 2 = R 2 = 1 kΩ 1+1 2 2 Vo −1 Z =− 2 = = −1− j 1 1 Vs Z1 −j 2 2 10-10
• Problems Section 10-3: Sinusoidal Sources P10.3-1 (a) i (t ) = 2 cos(6t + 120° ) + 4 sin(6t − 60° ) = 2 (cos 6t cos120° −sin 6t sin120° ) + 4 (sin 6t cos 60° − cos 6t sin 60° ) = 2.46 cos 6t + 0.27 sin 6t = 2.47 cos(6t − 6.26° ) (b) v(t ) = 5 2 cos8t + 10 sin(8t + 45° ) = 5 2 cos8t +10[sin 8t cos 45° + cos8t sin 45° ] = 10 2 cos8t + 5 2 sin 8t v(t ) = P10.3-2 250 cos(8t − 26.56° ) = 5 10 sin(8t + 63.4° ) V 2π 2π = = 6283 rad sec T 1×10−3 v(t ) = Vm sin(ω t + φ ) = 100 sin(6283 t + φ ) ω = 2π f = v(0) = 10 = 100 sin φ ⇒ φ = sin −1 (0.1) = 6° v(t ) = 100 sin(6283 t + 6°) V P10.3-3 ω 1200π = = 600 Hz 2π 2π i (2 × 10−3 ) = 300 cos(1200 π (2 × 10−3 ) + 55°) = 3cos(2.4π + 55°) f =  180°  −3 2.4π ×   = 432° ⇒ i (2 × 10 ) = 300 cos(432°+55°) = 300 cos(127°) = −180.5 mA  π  P10.3-4 10-11
• P10.3-5 A = 18 V T = 18 − 2 = 16 ms ω= 2π 2π = = 393 rad/s T 0.016 16 = 18 cos (θ ) ⇒ θ = 27° v ( t ) = 18 cos ( 393 t + 27° ) V P10.3-6 A = 15 V T = 43 − 21 = 32 ms ω= 2π 2π = = 196 rad/s T 0.032 8 = 15 cos (θ ) ⇒ θ = 58° v ( t ) = 15 cos (196 t + 58° ) V 10-12
• Section 10-4: Steady-State Response of an RL Circuit for a Sinusoidal Forcing Function P10.4-1 L di + R i = − vs dt Try i f = A cos 300 t + B sin 300 t then equating coefficients gives Then ⇒ di f dt di + 120 i = −400 cos 300 t dt = −300 A sin 300 t + 300 B cos 300 t . Substituting and −300 A+120 B = 0  A = − 0.46  B = −1.15 300 B +120 A =−400  i (t ) = −0.46 cos 300 t − 1.15sin 300 t = 1.24 cos (300 t − 68°) A P10.4-2 v dv dv +C =0 ⇒ + 500 v = 500 cos1000 t dt dt 2 dv f Try v f = A cos1000 t + B sin1000 t then = −1000 A cos1000 t + 1000 B cos1000 t . dt Substituting and equating coefficients gives −is + −1000 A+ 500 B =0   ⇒ 1000 B +500 A=500  Then A = 0.2 B = 0.4 v (t ) = 0.2 cos1000 t + 0.4 sin1000 t = 0.447 cos (1000t − 63°) V P10.4-3 ( j 4) (.05) = j (0.2) I (ω ) = 12 e j 45° ~ 12 e j 45° j 45° = (2 ⋅10−3 ) e ⇒ i (t ) = 2 cos (4 t + 45°) mA 6000 + j (0.2) 6000 10-13
• Section 10.5: Complex Exponential Forcing Function P10.5-1 (5∠36.9° ) (10∠−53.1° ) 50∠−16.2° 10∠−16.2° = = = 2 5∠10.36° ° (4 + j3)(6− j8) 10− j5 5∠− 26.56 P10.5-2  3 2∠− 45°  3 5∠ + 81.87°  4− j 3+  = 5∠ + 81.87°[4 − j 3 + ∠ − 36.87°] 5 5 2∠−8.13°   = 5∠+81.87° (4.48− j 3.36) = 5∠+81.87° (5.6∠−36.87°) = 28∠+ 45°= 14 2 + j14 2 P10.5-3 A*C* (3− j 7) 5e − j 2.3 = = 0.65 − j 6.31 B 6 e j15 P10.5-4 (6∠120° ) (−4 + j 3 + 2e j15 ) = −12.1 − j 21.3 ⇒ a =−12.1 and b =−21.3 P10.5-5  3− b  (a) j tan −1   2 2 j120  −4  Ae = −4 + j (3 − b) = 4 + (3−b) e  3−b  ° 120 = tan −1   ⇒ b = 3 + 4 + tan (120 ) = −3.93  −4  A = 42 + (3−b) 2 = 42 + (3− (−3.93)) 2 = 8.00 (b) −4 + 8 cos θ + j (b + 8 sin θ ) = 3e − j120 = − 1.5 − j 2.6 2.5 −4+8 cos θ = −1.5 ⇒ θ = cos −1 = 72° 8 b + 8 sin (72° ) = − 26 ⇒ b = −10.2 (c) −10 + j 2a = Ae j 60 = A cos 60° − j A sin 60 A= −10 −20sin 60° = −20 and a = = −8.66 cos 60° 2 10-14
• P 10.5-6 d   5  0.1 v  + v = cos 2 t ⇒  dt  d v + 2 v = 2 cos 2 t dt Replace the real excitation by a complex exponential excitation to get d v + 2 v = 2 e j 2t dt Let ve = A e j 2t so d ve = j 2 A e j 2t and dt d ve + 2 ve = 2 e j 2t dt j 2 A e j 2 t + 2 A e j 2t = 2 e j 2 t ⇒ ( j 2 + 2 ) A e j 2t = 2 e j 2 t ⇒ A= 2 1 = ∠ − 45° 2 + j2 2 1 j ( 2t −45° )  1 − j 45°  j 2t ve =  e e e = 2  2  so Finally v ( t ) = Re ( ve ) = 1 cos ( 2t − 45° ) V 2 P 10.5-7 d d2 0.45 v + v + 0.15 2 v = 4 cos 5 t ⇒ dt dt d2 d 20 80 v+3 v+ v= cos 5 t 2 dt dt 3 3 Replace the real excitation by a complex exponential excitation to get d2 dt 2 v+3 d 20 80 j 5t v+ v= e dt 3 3 d d2 ve = j 5 A e j 5 t , and 2 ve = −25 A e j 5 t dt dt 2 d d 20 80 j 5t 20 80 j 5t v+3 v+ v= e ⇒ − 25 A e j 5t + 3 j 5 A e j 5t + A e j 5t = e 2 dt 3 3 3 3 dt 80 20  80 j 5t 80  j 5t 3 e ⇒ A= = = 1.126∠ − 141  −25 + j15 +  A e = 20 −55 + j 45 3  3  −25 + j15 + 3 Let ve = A e j 5 t so ( so ( ) ( ) ) j 5t −141° ) ve = 1.126 e− j141° e j 5t = 1.126 e ( Finally v ( t ) = Re ( ve ) = 1.126 cos ( 2t − 141° ) V 10-15
• Section 10-6: The Phasor Concept P10.6-1 Apply KVL 6i+2 d i − 15 cos 4 t = 0 dt or 2 d i + 6 i = 15 cos 4t dt j 4 t +θ ) Now use i = I m Re{e ( } and 15 cos 4 t = 15 Re{e 4 t } to write 2 ( ) ( ) d j 4 t +θ j 4 t +θ I m Re{e ( ) } + 6 I m Re{e ( ) } = 15 Re{e 4 t } dt  d  Re 2 ( I m e j 4 t e jθ ) + 6 ( I m e j 4 t e jθ )  = Re{15 e 4 t }  dt  { } Re 2 ( j 4 I m e j 4 t e jθ ) + 6 ( I m e j 4 t e jθ ) = Re{15 e 4 t } j8 ( I m e jθ ) + 6 ( I m e jθ ) = 15 I m e jθ = 15 15 = = 1.5∠ − 53° 6 + j8 10∠53° i ( t ) = 1.5 cos ( 4 t − 53° ) A Finally v (t ) = 2 d d i ( t ) = 2 (1.5 cos ( 4 t − 53° ) ) = 3 ( −4sin ( 4 t − 53° ) ) dt dt = −12 ( cos ( 4 t − 143° ) ) = 12 cos ( 4 t + 37° ) V 10-16
• P10.6-2 Apply KCL at node a: v − 4 cos 2 t d + 0.25 v + i = 0 1 dt Apply KVL to the right mesh: 4i + 4 d d i − v = 0 ⇒ v = 4 i + 4 iL dt dt After some algebra: d2 d i + 5 i + 5 i = 4 cos 2t 2 dt dt j 2 t +θ ) } and 4 cos 2 t = 4 Re{e 2 t } to write Now use i = I m Re{e ( d2  d j 2 t +θ j 2 t +θ j 2 t +θ I Re{e ( ) } + 5  I m Re{e ( ) } + 5  I m Re{e ( ) } = 4 Re{e 2 t } 2  m      dt dt  d2  d j 2 t +θ j 2 t +θ j 2 t +θ Re  2  I m e ( )  + 5  I m e ( )  + 5  I m e ( )   = Re{4 e 2 t }       dt  dt  { } Re −4 e jθ I m e j 2 t + 5 ( j 2 e jθ I m e j 2 t ) + 5 e jθ I m e j 2 t = Re{4 e2t } −4 e j θ I m + 5 ( j 2 e j θ I m ) + 5 e j θ I m = 4 I m e jθ = 4 4 4 = = = 0.398∠ − 84° −4 + 5 ( j 2 ) + 5 1 + j 10 10.05∠84 i ( t ) = 0.398 cos ( 2 t − 85° ) A 10-17
• P10.6-3 VS = 2∠ − 90° V Z R = R; Z C = −j −j = = − j 16000 Ω ω C (500)(0.125×10−6 )  − j 16000  (16000∠−90° )( 2∠−90° ) = 1.25∠ − 141° V V (ω ) =   ( 2∠−90° ) = 25612∠−39°  20000 − j 16000  therefore v(t) = 1.25 cos (500t −141° ) V Section 10-7: Phasor Relationships for R, L, and C Elements P10.7-1 P10.7-2 10-18
• P10.7.3 P10.7-4 10-19
• P10.7-5 (a) v = 15cos (400 t + 30°) V i = 3 sin(400 t+30°) = 3 cos (400 t − 60°) V v leads i by 90° ⇒ element is an inductor v 15 Z L = peak = = 5 = ω L = 400 L ⇒ L = 0.0125 H = 12.5 mH 3 ipeak (b) i leads v by 90° ⇒ the element is a capacitor (c) 8 1 1 =4= = ⇒ C = 277.77 µ F ω C 900 C ipeak 2 v = 20 cos (250 t + 60°) V Zc = vpeak = i = 5sin (250 t +150°) =5cos (250 t + 60°) A Since v & i are in phase ⇒ element is a resistor v 20 ∴ R = peak = =4Ω 5 ipeak P10.7-6 V1 = 150 cos(−30°) + j150sin(−30°) = 130 − j 75 V V2 = 200 cos 60°+ j 200sin 60° = 100+ j173 V V = V1 + V2 = 230+ j 98 = 250∠23.1° V Thus v(t ) = v1 (t ) + v2 (t ) = 250 cos (377 t + 23.1°) V 10-20
• Section 10-8: Impedance and Admittance P10.8-1 ω = 2π f = 2π (10 ×103 ) = 62830 rad sec Z R = R = 36 Ω ⇔ YR = 1 1 = 0.0278 S = Z R 36 Z L = jω L = j (62830)(160×10−6 ) = j10 Ω ⇔ YL = ZC = 1 = − 0.1 j S ZL −j −j 1 = = − j 16 Ω ⇔ YC = = 0.0625 j S −6 ZC ω C (62830)(1×10 ) Yeq = YR + YL + YC = 0.0278 − j0.00375 = 0.027 ∠9° S Z eq = P10.8-2 Z= 1 = 36.5∠ 9° = 36 − j5.86 Ω Yeq V −10 ∠40° = = − 5000∠ − 155°Ω = 4532 + 2113 j = R + j ω L −I 2×10−3 ∠195° so R =4532 Ω and L = 2113 ω = 2113 = 1.06 m H 2×106 P10.8-3 j L R ( R + jω L) −j C ωC = Z(ω ) = ω C j 1  − + ( R + jω L) R + j  ω L −  ωC ωC    − 1  R  L   −j  R − j  ω L −  ωC    C ω C    = 2 1   R 2 + ω L − ωC    = Z(ω ) will be purely resistive when 1   R2 L  1  RL R  − + ω L− ω L− − j  ωC   ωC C  ωC   C ωC   1   R 2 + ω L − ωC    2 10-21
• R2 L  1  1 R 2 + ω L− −  = 0 ⇒ ω = CL  L ωC C ωC 2 when R =6 Ω , C = 22 µ F, and L = 27 mH, then ω = 1278 rad/s. P10.8-4 R Zc R R + j (ω L −ω R 2 C +ω 3 R 2 L C 2 ) jω C = jω L + = 1 R +Z c 1+ (ω R C ) 2 R+ jω C Set real part equal to 100 Ω to get C Z = ZL + R = 100 ⇒ C = 0.158 µF 1+ (ω R C ) 2 Set imaginary part of numerator equal to 0 to get L ( ω = 2π f = 6283 rad sec ) L − R 2C + ω 2 R 2 LC 2 = 0 ⇒ L = 0.1587 H P10.8-5 Z L = j ω L = j (6.28×106 ) (47×10−6 ) = j 300 Ω  1   ( 300 + j 300 )  jω C  Z eq = Z c ||(Z R +Z L ) = = 590.7 Ω 1 + 300+ j 300 jω C 300+300 j 590.7 = ⇒ 590.7 −(590.7)(300 ω C ) + j (590.7)(300ω C ) = 300 + j 300 1+300 j ω C −300 ω C ( Equating imaginary terms ω =2π f = 6.28×106 rad sec ) (590.7) (300ω C ) = 300 ⇒ C = 0.27 nF 10-22
• Section 10-9: Kirchhoff’s Laws Using Phasors P10.9-1 (a) (b) (c) P10.9-2 Z1 =3+ j 4 = 5∠53.1° Ω and Z 2 =8− j8 = 8 2 ∠− 45° Ω Total impedance = Z1 + Z 2 = 3 + j 4 + 8 − j8 = 11 − j 4 = 11.7∠− 20.0° Ω I= 100∠0° 100 100 = = ∠20.0° ⇒ Z1 +Z 2 11.7 ∠− 20° 11.7 i (t ) = 8.55 cos (1250 t + 20.0°) A V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠47° − 1.59∠125° = ( 5.23 + j 5.62 ) − ( −0.91 + 1.30 ) = ( 5.23 + 0.91) + j ( 5.62 − 1.30 ) = 6.14 + j 4.32 = 7.51∠35° v1 ( t ) = 7.51 cos ( 2 t + 35° ) V P10.9-3 I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 ) = ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = −0.443 − j 0.125 = 0.460∠196° i ( t ) = 460 cos (2 t + 196°) mA P10.9-4 Vs = 2 ∠30° V and I = 2 ∠30° = 0.185 ∠ − 26.3° A 6+ j12+ 3 / j i (t ) = 0.185 cos (4 t − 26.3°) A 10-23
• P10.9-5 j 15 = j (2π ⋅ 796) (3 ⋅10−3 ) 12 = 0.48 ∠ − 37° A 20 + j15 i (t ) = 0.48 cos (2π ⋅ 796 t − 37°) A I= P10.9-6 Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A 8 I B ∠−51.87° Z1 = = = = 2 ∠−15° Is Z1 + Z 2 8+ j 3L 8 ∠0°  3L  82 + (3L) 2 ∠ tan −1    8  Equate the magnitudes and the angle.  3L  angles: + 36.87 = + tan −1   ⇒ L = 2 H  8  8 B = ⇒ B =1.6 magnitudes: 64+ 9 L2 2 P10.9-7 The voltage V can be calculated using Ohm's Law. V = (1.72 ∠ - 69°) (4.24∠45°) = 7.29∠ - 24° V The current I can be calculated using KCL. 10-24
• I = (3.05 ∠ - 77°) - (1.72∠ - 69°) = 1.34∠ - 87° A Using KVL to calculate the voltage across the inductor and then Ohm's Law gives: j 2L = 24 - 4(1.34∠-87°) ⇒ L=4 H 3.05∠-77° P10.9-8  10  10   V10 = Vs   = 20∠0°    10 2∠− 45°   10 − j10  = 0 2∠45° v10 (t ) = 10 2 cos (100 t + 45°) V P10.9-9 (a) (b) 160 ∠0° 160 ∠0° = (−1326) (300 + j 37.7) 303 ∠−5.9° − j1326 + 300 + j 37.7 = 0.53 ∠5.9° A i(t ) = 0.53cos (120π t +5.9°) A I= I= 160∠0° 160∠0° = (− j199)(300 + j 251) 256∠−59.9° − j199+ 300+ j 251 = 0.625∠59.9° A i(t ) =0.625 cos (800π t +59.9°) A 10-25
• Section 10-10: Node Voltage and Mesh Current Analysis Using Phasors P10.10-1 Draw frequency domain circuit and write node equations: VA VA − VC + = 0 ⇒ (2 + j )VA − 2VC = j 20 10 j5 VC − VA VC KCL at C: + − (1+ j ) = 0 ⇒ 4VA + VC = 20− j 20 j5 − j4 KCL at A − 2 + Solve using Cramers rule: (2 + j ) j 20 4 20 − j 20 60 − j100 116.6 ∠−59° Vc = = = = 11.6 ∠ − 64.7° V (2 + j ) −2 10 + j 101 ∠5.7° 4 1 P10.10-2 KCL: V V V (V −100) + + = 0 ⇒ V = 57.6 ∠22.9° V + − j125 j80 250 150 IS = 100− V = 0.667 − 0.384 ∠22.9° = 0.347 ∠− 25.5° A 150 IC = V = 0.461 ∠112.9° A 125 ∠−90° 10-26
• IL = V = 0.720 ∠− 67.1° A 80∠90° IR = V = 0.230∠22.9° A 250 P10.10-3 KCL at node A: Va Va − Vb + =0 200 j 100 (1) KCL at node B: Vb − Va V V −1.2 + b + b = 0 j 100 − j 50 j 80 1 3 ⇒ Va = Vb − 4 2 (2) Substitute Eqn (2) into Eqn (1) to get Vb = 2.21 ∠ − 144° V Then Eqn (2) gives Va = ( 0.55∠−144° ) − 1.5 = 1.97∠ − 171° V Finally va (t ) = 1.97 cos (4000 t − 171°) V and vb (t ) = 2.21cos (4000 t − 144°) V 10-27
• P10.10-4 ω = 104 rad s I s = 20∠53° A The node equations are:  1 1 j   1  KCL at a:  + +  Va +  −  Vb = 20∠53.13°  20 40 60   40   1   1 j j  KCL at b:  −  Va + − +  Vb − j Vc = 0 80  40   40 40 80  −j  1 j  KCL at c: Vb + +  Vc = 0 80  40 80  Solving thes equation yields Va = 2 ⋅ 240∠45° V ⇒ P10.10-5 va (t ) = 339.4 cos (ω t + 45°) V vs = sin (2π ⋅ 400 t ) V R = 100 Ω LR = 40 mH  40 mH LS =   60 mH door opened door closed With the door open VA − VB = 0 since the bridge circuit is balanced. With the door closed Z LR = j (800π )(0.04) = j100.5 Ω and Z LS = j (800π )(0.06) = j150.8 Ω. The node equations are: KCL at node B: VB − VC VB j100.5 + = 0 ⇒ VB = VC R j100.5+100 Z LR KCL at node A : VA − VC VA + =0 Z LS R 10-28
• Since VC = Vs =1 V VB =0.709∠44.86° V and VA = 0.833∠33.55 V Therefore VA − VB = 0.833∠33.55° − 0.709∠44.86° = (0.694 + j.460) − (0.503 + j 0.500) = 0.191 − j 0.040 = 0.195∠ − 11.83° V P10.10-6 The node equations are: V1 −(−1+ j ) V1 V1 − V2 + + =0 j2 2 −j2 V2 − V1 V2 + −I C = 0 − j2 − j2 Also, expressing the controlling signal of the dependent source in terms of the node voltages yields  −1+ j  −1 + j Ix = ⇒ IC = 2 I x = 2   = −1 − j A -2 j  -2 j  Solving these equations yields V2 = −3− j = 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 135°) V 1+ j 2 10-29
• P10.10-7 V2 = 0.7571∠66.7° V V3 = 0.6064∠ − 69.8° V  I1 = I 2 + I 3    I 3 = 0.3032 ∠20.2° A V3 − V2   I2 =  yields  I 2 = 0.1267∠−184° A j 10   I =0.195∠36° A  1 V3  I3 =  −j2  therefore i1 (t ) =0.195cos (2 t + 36°) A P10.10-8 The mesh equations are (4 + j 6) I1 − j 6 I 2 = 12 + j12 3 - j 6 I1 + (8 + j 2) I 2 = 0 Using Cramer’s rule yields I1 = (12+ j 12 3) (8+ j 2) = 2.5∠29° = 2.2 + j 1.2 A (4+ j 6) (8+ j 2) − (− j 6) (− j 6) I2 = j6 6∠90° (2.5∠29°) = (2.5∠29°) = 1.82∠105° A 8+ j 2 68∠14° Then and VL = j 6(I1 − I 2 ) = (6∠90°) (2.5∠29° − 1.82∠105°) = (6∠90°) (2.71∠ − 11.3°) = 16.3∠78.7° V Finally V = − j 4I 2 = (4∠ − 90°)(1.82∠105°) = 7.28∠15° V c 10-30
• P10.10-9 The mesh equations are: (10 − j ) I1 + ( j ) I 2 + 0 I 3 = 10 j I1 − j I 2 + j I 3 = 0 0 I1 + j I 2 + (1− j ) I 3 = j10 Solving these mesh equations using Cramer’s rule yields: (10− j ) j 0 I2 = (10− j ) j 0 10 0 j 0 j 10 (1− j ) 90 − j 20 = = 8.38∠77.5° A ⇒ i (t ) = 8.38cos (103 t + 77.5° ) A j 0 −11 j −j j j (1− j ) (checked using LNAPAC on 7/3/03) P10.10-10 The mesh equations are: −1 − j 4   I1  10∠30°  (2 + j 4)  −1 (2 +1/ j 4) −1   I 2  =  0        − j4 (3+ j 4)   I 3   0  −1      Using Cramer’s rule yields I3 = Then 2 + j8 (10∠30° ) = 3.225∠44° A 12+ j 22.5 V = 2 I 3 = 2 ( 3.225∠44° ) = 6.45∠44° V ⇒ v(t ) = 6.45cos (105 t + 44° ) V 10-31
• P10.10-11 Mesh Equations: j 75 I1 − j 100 I 2 = 375 − j 100 I1 + (100+ j 100) I 2 = 0 Solving for I 2 yields I 2 = 4.5 + j 1.5 A ⇒ i 2 (t ) = 4.74 ∠18.4° A 10-32
• Section 10-11: Superposition, Thèvenin and Norton Equivalents and Source Transformations P10.11-1 Use superposition I1 = 12∠45° = 3.3∠11.3° mA 3000 + j 2000 I2 = −5∠0° = 1.5∠153° mA 3000+ j1500 i (t ) = 3.3cos (4000 t + 11.3°) + 1.5cos (3000 t + 153°) mA P10.11-2 Use superposition I1 = 3 = 0.5 mA 6000 I 2 (ω ) = −1∠45° = −0.166 ×10−3 ∠45° A 6000+ j 0.2 i (t ) = i 2 (t ) + i1 (t ) = − 0.166 cos (4 t + 45°) + 0.5 mA = 0.166 cos (4 t − 135°) + 0.5 mA P10.11-3 Use superposition 12∠ 45° = 2∠45° mA I1 (ω ) = 6000 + j 0.2 5∠−90° = 0.833∠ − 90° mA I 2 (ω ) = 6000 + j 0.15 i (t ) = i1 (t ) − i 2 (t ) = 2 cos (4 t + 45°) − 0.833cos (3 t − 90°) mA 10-33
• P10.11-4 Find Voc :  80 + j80  Voc = ( 5 ∠−30° )    80 + j80 − j 20   80 2∠− 21.9°  = ( 5 ∠−30° )    100∠36.90°  = 4 2∠ − 21.9° V Find Z t : Zt = ( − j 20 )( 80 + j80 ) = 23 ∠ − 81.9° − j 20 + 80 + j80 Ω The Thevenin equivalent is 10-34
• P10.11-5 First, determine Voc : The mesh equations are 600 I1 − j 300 (I1 − I 2 ) = 9 ⇒ (600 − j 300) I1 + j 300 I 2 = 9∠0° −2 V + 300 I 2 − j 300 (I1 − I 2 ) = 0 and V = j 300 (I1 − I 2 ) ⇒ j 3 I1 + (1 − j 3) I 2 = 0 Using Cramer’s rule: I 2 = 0.0124∠ − 16° A Then Voc = 300 I 2 = 3.71∠ − 16° V Next, determine I sc : −2 V − V = 0 ⇒ V = 0 ⇒ I sc = The Thevenin impedance is ZT = 9∠0° = 0.015∠0° A 600 Voc 3.71∠−16° = = 247∠ − 16° Ω 0.015∠0° I sc The Thevenin equivalent is 10-35
• P10.11-6 First, determine Voc : The node equation is: Voc Voc − (6 + j8) 3  Voc − (6+ j8)  + −  =0 − j4 j2 j2 2  Voc =3+ j 4=5∠53.1° V Vs = 10∠53° = 6 + j 8 V Next, determine I sc : The node equation is: V V V − (6 + j8) 3  V − (6 + j8)  + + −  =0 2 − j4 j2 2 j2  V= I sc = 3 + j4 1− j V 3+ j 4 = 2 2− j 2 Vs = 10∠53° = 6 + j 8 V The Thevenin impedance is ZT =  2− j 2  Voc = 3 + j4   = 2 − j2 Ω I sc  3+ j 4  The Thevenin equivalent is 10-36
• P10.11-7 Y = G + YL + YC 1 Y = G when YL + YC = 0 or + jω C = 0 jω L 1 1 1 , fO = ωO = = 2π LC 2π 39.6×10−15 LC = 0.07998×107 Hz =800 KHz (80 on the dial of the radio) P10.11-8 In general: I= Voc ZL and V = Voc Z t +Z L Z t +Z L In the three given cases, we have Z1 = 50 Ω ⇒ Z2 = I1 = V1 25 = = 0.5 A Z1 50 1 1 = = − j 200 Ω ⇒ jω C j (2000)(2.5×10 −6 ) Z 3 = jω L = j (2000)(50 × 10−3 ) = j100 Ω ⇒ I2 = I3 = V2 100 = = 0.5 A Z 2 200 V3 50 = = 0.5 A Z 3 100 Since |I| is the same in all three cases, Z t +Z1 = Z t +Z 2 = Z t +Z3 . Let Z t = R + j X . Then ( R + 50) 2 + X 2 = R 2 + ( X − 200) 2 = R 2 + ( X + 100) 2 This requires ( X − 200) 2 = ( X + 100) 2 ⇒ X = 50 Ω Then ( R + 50) 2 + (50) 2 = R 2 + (−150) 2 ⇒ R = 175 Ω so Z t =175+ j 50 Ω and Voc = I1 Z t + R1 =(0.5) (175+ 50) 2 + (50) 2 =115.25 V 10-37
• P10.11-9 Z1 = (− j 3)(4) = 2.4∠ − 53.1° Ω − j 3+ 4 =1.44 − j1.92 Ω Z 2 = Z1 + j 4 = 1.44 + j 2.08 = 2.53∠55.3° Ω Z3 = 3.51∠ − 37.9° Ω = 2.77 − j 2.16 Ω  3.51∠−37.9°  ( 3.51∠−37.9° ) = 1.9∠ − 92° A I = ( 2.85∠− 78.4° )   = ( 2.85∠− 78.4° ) ( 5.24∠− 24.4° )  2.77 − j 2.16 + 2  10-38
• P10.11-10 Z2 = I= (200)(− j 4) = 4∠ − 88.8° Ω 200− j 4 0.4∠− 44° = 4∠ − 44° mA −4 j +100+ j 4 i (t ) = 4 cos (25000 t − 44°) mA 10-39
• Section 10-12: Phasor Diagrams P10-12-1 V = V1 − V2 + V3 = ( 3+ j 3) − ( 4 + j 2 ) + ( 3+ j 2 ) = −4 + j 3 * * P10.12-2 I= 10∠0° = 0.74∠42° A 10+ j1− j10 VR = R I = 7.4∠42° V VL = Z L I = (1∠90°)(0.74∠42°) = 0.74∠132° V VC = Z C I = (10∠−90°)(0.74∠42°) = 7.4∠− 48° V VS = 10∠0° V P10.12-3 I = 72 3 + 36 3∠(140° − 90°) + 144∠210° + 25∠φ = 40.08 − j 24.23 + 25∠φ = 46.83∠ − 31.15° + 25∠φ To maximize I , require that the 2 terms on the right side have the same angle ⇒ φ = −31.15°. 10-40
• Section 10-14: Phasor Circuits and the Operational Amplifier P10.14-1 Vo (ω )  104 || − j104  10 − j 225 −j e = − =  = −10 Vs (ω ) 1− j 2  1000   10 − j 225  − j 225 Vs (ω ) = 2 ⇒ Vo (ω ) =  e  2 = 10e  2  vo ( t ) = 10 cos (1000t − 225°) V H (ω ) = P10.14-2 Node equations: V1 − VS VS + j ω C 1 V1 = 0 ⇒ V1 = 1 + j ω C1 R1 R1  R3  V1 V1 − V0 V + = 0 ⇒ V0 =  1 +  R2  1  R2 R3   Solving: 1+ R3 R2 V0 = VS 1 + j ω C 1 R1 P10.14-3 Node equations: j ω C 1 VS V1 + j ω C 1 ( V1 − VS ) = 0 ⇒ V1 = R1 1 + j ω C 1 R1  R3  V1 V1 − V0 + = 0 ⇒ V0 =  1 + V  R2  1  R2 R3   Solving:  R  j ω C 1 1 + 3   R2  V0   = 1 + j ω C 1 R1 VS 10-41
• P10.14-4 Node equations: V1 − VS V VS + 1 = 0 ⇒ V1 = 175 − j1.6 1 + j 109 V − V0 V1 + 1 = 0 ⇒ V0 = 11 V1 1000 10000 Solving: V0 = 11 11 VS = ( 0.005∠0° ) 1 + j 109 110∠89.5° = 0.5∠89.5° mV Therefore v0 (t ) = 0.5cos (ω t − 89.5°) mV 10-42
• PSpice Problems SP10-1 10-43
• SP10-2 10-44
• SP 10-3 10-45
• SP 10-4 The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of Vm and Im are Vm = 12.5 V and Im = -1.667 A. 10-46
• Verification Problems VP 10-1 Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the division in rectangular form. Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40 KCL at node 1: 2− V1 10 − V1 − V 2 j 10 = 2− − j 20 − j 20 − ( 20 − j 40 ) − = 2+ j2−2− j2 = 0 10 j 10 KCL at node 2: V1 − V 2 j 10 −  V1  − j 20 − ( 20 − j 40 ) 20 − j 40  − j 20  + 3  = − + 3  = ( 2 + j 2) − ( 2 − j4) − j 6 = 0 10 j 10 10  10   10  V2 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. VP 10-2 I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180° Generally, it is more convenient to multiply complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the multiplication in rectangular form. Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284 KVL for mesh 1: 8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − (− j 5) = j 10 Since KVL is not satisfied for mesh 1, the mesh currents are not correct. Here is a MATLAB file for this problem: 10-47
• % Impedance and phasors for Figure VP 10-2 Vs = -j*5; Z1 = 8; Z2 = j*10; Z3 = -j*2.4; Z4 = j*20; % Mesh equations in matrix form Z = [ Z1+Z2 0; 0 Z3+Z4 ]; V = [ Vs; -Vs ]; I = ZV abs(I) angle(I)*180/3.14159 % Verify solution by obtaining the algebraic sum of voltages for % each mesh. KVL requires that both M1 and M2 be zero. M1 = -Vs + Z1*I(1) +Z2*I(1) M2 = Vs + Z3*I(2) + Z4*I(2) VP 10-3 V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V KCL at node 1 : 19.2 ∠ 68° 19.2 ∠ 68° + − 4∠15 = 0 2 j6 KCL at node 2: 24 ∠105° 24 ∠105° + + 4∠15 = 0 − j4 j12 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. Here is a MATLAB file for this problem: % Impedance and phasors for Figure VP 10-3 Is = 4*exp(j*15*3.14159/180); Z1 = 8; Z2 = j*6; Z3 = -j*4; 10-48
• Z4 = j*12; % Mesh equations in matrix form Y = [ 1/Z1 + 1/Z2 0; 0 1/Z3 + 1/Z4 ]; I = [ Is; -Is ]; V = YI abs(V) angle(V)*180/3.14159 % Verify solution by obtaining the algebraic sum of currents for % each node. KCL requires that both M1 and M2 be zero. M1 = -Is + V(1)/Z1 + V(1)/Z2 M2 = Is + V(2)/Z3 + V(2)/Z4 VP 10-4 First, replace the parallel resistor and capacitor by an equivalent impedance ZP = (3000)(− j 1000) = 949 ∠ − 72° = 300 − j 900 Ω 3000− j 1000 The current is given by I= VS 100 ∠0° = = 0.2∠53° A j 500+ Z P j 500+ 300− j 900 Current division yields  − j 1000  I1 =   ( 0.2 ∠53° ) = 63.3 ∠ − 18.5° mA  3000 − j 1000    3000 I2 =   ( 0.2 ∠53° ) = 190∠71.4° mA  3000 − j 1000  The reported value of I1 is off by an order of magnitude. 10-49
• Design Problems DP 10-1 R2 1 jω C = R2 1 + jω CR 2 R2 R2 1 + jω CR 2 R1 Vo (ω ) =− =− Vi (ω ) R1 1 + jω CR 2 R2 R1 Vo (ω ) j (180 − tan −1 ω CR 2 ) = e 2 Vi (ω ) 1 + (ω CR 2 ) In this case the angle of Vo (ω ) tan (180 − 76 ) is specified to be 104° so CR 2 = = 0.004 and the 1000 Vi (ω ) Vo (ω ) 8 so is specified to be 2.5 Vi (ω ) R2 R1 = R2 8 ⇒ 2.5 = 132 . One set of values R1 1 + 16 that satisfies these two equations is C = 0.2 µ F, R1 = 1515 Ω, R 2 = 20 kΩ . magnitude of DP 10-2 R2 Vo (ω ) = Vi (ω ) where K = 1 jω C = R2 1 + jω CR 2 R2 1 + jω CR 2 K = R2 1 + jω CR p R1 + 1 + jω CR 2 R1 R1 + R 2 and R p = R1 R 2 R1 + R 2 Vo (ω ) K − j tan −1 ω CR p = e 2 Vi (ω ) 1 + (ω CR p ) 10-50
• In this case the angle of C Rp = C R1 R 2 R1 + R 2 =− Vo (ω ) is specified to be -76° so Vi (ω ) tan ( −76 ) V (ω ) 2.5 = 0.004 and the magnitude of o so is specified to be 12 1000 Vi (ω ) R2 2.5 K = ⇒ 0.859 = K = . One set of values that satisfies these two equations is R1 + R 2 1 + 16 12 C = 0.2 µ F, R1 = 23.3 kΩ, R 2 = 142 kΩ . DP 10-3 jω L R 2 L R 2 + jω L R1 Vo (ω ) = =− jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = jω R1 R 2 R1 + R 2 Vo (ω ) Vi (ω ) ω = L R1  L  1+ ω  Rp     2 e  L  j  90 − tan −1 ω   Rp    Vo (ω ) L L ( R1 + R 2 ) tan ( 90 − 14 ) is specified to be 14° so = = = 0.1 Rp R1 R 2 Vi (ω ) 40 L 40 R1 V (ω ) 2.5 L 2.5 = ⇒ = 0.0322 . One so is specified to be and the magnitude of o 8 R1 Vi (ω ) 8 1 + 16 In this case the angle of set of values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω . 10-51
• DP 10-4 jω L R 2 L R 2 + jω L R1 Vo (ω ) = =− jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = R1 R 2 R1 + R 2 Vo (ω ) Vi (ω ) In this case the angle of jω ω = L R1  L  1+ ω  Rp     2 e  L  j  90 − tan −1 ω   Rp    Vo (ω ) is specified to be -14°. This requires Vi (ω ) L L ( R1 + R 2 ) tan ( 90 + 14 ) = = = −0.1 Rp R1 R 2 40 This condition cannot be satisfied with positive DP 10-5 Z1 =10 Ω 1 Z2 = jω C Z3 = R + jω L 1 S 10 Y2 = jω C 1 Y3 = R + jω L Y1 = v(t ) = 80 cos (1000 t − θ ) V ⇒ V = 8∠ − θ V iS (t ) = 10 cos 100 t A ⇒ I s =10∠0° A so 1  1 + + jω C  = 10∠0° ⇒ R + 10 − 10 ω 2 LC + j (ω L + 10 ω RC ) = 1.25 R + j1.25 ω L 10 R + jω L  (80∠−θ )  Equate real part: 40 − 40ω 2 LC = R where ω = 1000 rad sec Equate imaginary part: 40 RC = L Solving yields R = 40(1− 4×107 RC 2 ) 10-52
• Now try R = 20 Ω ⇒ 1 − 2(1 − 4 × 107 (20)C 2 ) which yields C = 2.5×10−5 F= 25 µ F so L = 40 RC = 0.02 H=20 mH Now check the angle of the voltage. First Y1 = 1/10 = 0.1 S Y2 = j 0.25 S Y3 = 1/(20+ j 20) = .025− j.025 S then Y = Y1 + Y2 + Y3 = 0.125 , so V =YI s = (0.125∠0°)(10∠0°) = 1.25∠0° V So the angle of the voltage is θ =0° , which satisfies the specifications. 10-53
• Chapter 11: AC Steady State Power Exercises: Ex. 11.3-1 Z = j3 + 4(− j 2) = 0.8 + j1.4 4− j 2 = 1.6 ∠60.3° Ω ∴I = V 7∠0° = = 4.38 ∠ − 60.3° A ° Z 1.6∠60.3 i (t ) = 4.38cos (10 t − 60.3° ) A The instantaneous power delivered by the source is given by (7)(4.38) [cos (60.3°) + cos (20 t −60.3°)] 2 = 7.6 + 15.3cos (20 t − 60.3°) W p(t ) = v(t ) ⋅ i (t ) = (7 cos 10 t )(4.38cos (10 t − 60.3°)) = The inductor voltage is calculated as VL = I ⋅ Z L = (4.38 ∠ − 60.3°)(j3) = 13.12 ∠29.69° V vL (t ) = 13.12 cos (10 t + 29.69°) V The instantaneous power delivered to the inductor is given by pL (t ) = vL (t ) ⋅ i (t ) =  (13.12 cos (10t + 29.69°)(4.38cos (l0t − 60.3° )    57.47 [ cos (29.69°+ 60.3°)+ cos (20 t + 29.69°−60.3°)] 2 = 28.7 cos (20t − 30.6°) W = 11-1
• Ex. 11.3-2 (a) When the element is a resistor, the current has the same phase angle as the voltage: i (t ) = v(t ) Vm cos (ω t + θ ) A = R R The instantaneous power delivered to the resistor is given by 2 2 2 V V Vm V pR (t ) = v(t ) ⋅ i (t ) = Vm cos (ω t + θ ) ⋅ cos (ω t + θ ) = m cos 2 (ω t + θ ) = m + m cos (2ω t + θ ) R R 2R 2R (b) When the element is an inductor, the current will lag the voltage by 90°. Z L = jω L = ω L∠90° Ω ⇒ I= V V ∠θ V = m = m ∠ (θ −90° ) Z ω L∠90° ω L The instantaneous power delivered to the inductor is given by 2 V V pL (t ) = i (t ) ⋅ v(t ) = m cos (ω t + θ − 90° ) ⋅Vm cos (ω t + θ ) = m cos ( 2ω t + 2θ −90° ) W ωL 2ω L 11-2
• Ex. 11.3-3 The equivalent impedance of the parallel resistor (1)( j ) = 1 1+ j Ω . Then and inductor is Z = ( ) 1+ j 2 10∠0° 20 I= = ∠ − 18.4° A 1 10 1+ (1+ j ) 2 (a) Psource I V = cos θ = 2  20    10  cos −18.4° = 30.0 W ( ) 2 (10 )  2  20  (1) 2 I max R1  10    (b) PR 1 = = = 20 W 2 2 Ex. 11.4-1 I eff = 3 1 t 2 1 2 i (t ) dt = ∫0 (10)2 dt + ∫2 (5)2 dt  = 8.66  T ∫0 3   Ex. 11.4-2 (a) I 2 i (t ) = 2 cos 3 t A ⇒ I eff max = = 2A 2 2 (b) i (t ) = cos (3 t − 90° ) + cos (3 t + 60° ) A 1 3 +j = 0.518∠ − 15° A 2 2 0.518 i (t ) = 0.518 cos (3t − 15°) A ⇒ I eff = = 0.366 A 2 I = (1∠−90° ) + (1∠60° ) = − j + (c) 2  2   3  2 I eff =   +   2  2 2 ⇒ I eff = 2.55 A 11-3
• Ex. 11.4-3 Use superposition: V1 = 5∠0° V V2 = 2.5 V (dc) V3 = 3∠−90° V V1 and V2 are phasors having the same frequency, so we can add them: V1 + V3 = ( 5∠0° ) + ( 3∠ − 90° ) = 5 − j3 = 5.83∠ − 31.0° V Then vR (t ) = v1 (t ) + ( v2 (t ) + v3 (t ) ) = 2.5 + 5.83cos (100 t − 31.0°) V Finally 2 2 R eff V  5.83  = (2.5) +   = 23.24 V ⇒ VR eff = 4.82 V  2  2 Ex. 11.5-1 Analysis using Mathcad (ex11_5_1.mcd): Enter the parameters of the voltage source: Enter the values of R and L R := 10 A := 12 L := 4 The impedance seen by the voltage source is: The mesh current is: I := ω := 2 Z := R + j ⋅ω ⋅L A Z 11-4
•  I ⋅( I ⋅Z) The complex power delivered by the source is: Sv := Sv = 4.39 + 3.512i 2  I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := Sr = 4.39 2  I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := Sl = 3.512i 2 Verify Sv = Sr + Sl : Sr + Sl = 4.39 + 3.512i Sv = 4.39 + 3.512i Ex. 11.5-2 Analysis using Mathcad (ex11_5_2.mcd): Enter the parameters of the voltage source: A := 12 Enter the values of R, L an C d R := 10 ω := 2 L := 4 C := 0.1 The impedance seen by the voltage source is: Z := R + j ⋅ω ⋅L + The mesh current is: I := A Z is: The complex power delivered by the source The complex power delivered to the resistor is: The complex power delivered to the inductor is: Sv := Sr := Sl := 1 j ⋅ω ⋅C  I ⋅( I ⋅Z) Sv = 6.606 + 1.982i 2  I ⋅( I ⋅R) Sr = 6.606 2  I ⋅( I ⋅j ⋅ω ⋅L) 2  I ⋅ I ⋅  j ⋅ω ⋅C  The complex power delivered to the capacitor is: Sc :=  1 2 Verify Sv = Sr + Sl + Sc : Sr + Sl + Sc = 6.606 + 1.982i Sl = 5.284i Sc = −3.303i Sv = 6.606 + 1.982i 11-5
• Ex. 11.5-3 Analysis using Mathcad (ex11_5_3.mcd): Enter the parameters of the voltage source: A := 12 ω := 2 Enter the Average and Reactive Power delivered to the RL circuit: P := 8 The complex power delivered to the RL circuit is: Q := 6 S := P + j ⋅Q 2 The impedance seen by the voltage source is: Z := Calculate the required values of R and L R := Re( Z) The mesh current is: I := A  2 ⋅S L := Im( Z) A ω L = 2.16 Z  I ⋅( I ⋅Z) The complex power delivered by the source is: Sv := 2  I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := 2  I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := 2 Verify Sv = Sr + Sl : R = 5.76 Sr + Sl = 8 + 6i Sv = 8 + 6i Sr = 8 Sl = 6i Sv = 8 + 6i Ex. 11.6-1 ( ) (377) (5)   pf = cos (∠ Z ) = cos  tan −1 ω L  = cos  tan −1 = 0.053  R  100      Ex. 11.6-2 ( ) ( ) pf = cos (∠ Z) = cos  tan −1 X  = cos  tan −1 80  = 0.53 lagging   50  R      XC = (50)2 + (80)2 = −111.25 Ω ⇒ ZC = − j 111.25 Ω 50 tan (cos −1 1) −80 11-6
• Ex. 11.6-3 PT = 30 + 86 = 116 W and QT = 51 VAR S T = PT + j QT = 116+ j 51 = 126.7 ∠23.7° VA pf plant = cos 23.7° = 0.915 Ex. 11.6-4 P = V I cos θ ⇒ I = P 4000 = = 44.3 A V cos θ (110)(.82) V ∠ cos −1 (0.82) = 2.48 ∠34.9° = 2.03+ j 1.42 = R + j X I To correct power factor to 0.95 requires Z= R2 + X 2 (2.03) 2 + (1.42) 2 = = − 8.16 Ω R tan (cos −1 pfc ) − X (2.03) tan (18.19° ) −1.42 −1 C= =325 µ F ω X1 X1 = Ex. 11.7-1 (a) I = I1 + I 2 = ( 0.4714∠135° ) + (1.414∠ − 45° ) = 0.9428∠ − 45° A 0.94282 ( 6 ) = 2.66 W 2 (b) 1.22 I1 = 1.2∠53° A ⇒ p1 = ( 6 ) = 4.32 W 2 0.47142 I 2 = 0.4714∠135° A ⇒ p1 = ( 6 ) = 0.666 W 2 ∴ p = p1 + p2 = 4.99 W ⇒ p= Ex. 11.8-1 For maximum power, transfer Z L = Z* = 10 − j14 Ω t 100 I= =5 A (10+ j14) (10− j14) 2  5  PL =   Re(10 − j14) = 125 W  2 11-7
• Ex. 11.8-2 If the station transmits a signal at 52 MHz then ω = 2π f = 104π ×106 rad/sec so the received signal is vs (t ) = 4 cos (104π × 106 t ) mV (a) If the receiver has an input impedance of Zin =300 Ω then 2 300 1 2  1   2.4×10−3  Zin Vin = Vs = × 4 ×10−3 = 2.4 mV ⇒ P = Vin  =  = 9.6 nW R + Zin 200+300 2  R L   2(300)    (b) If two receivers are connected in parallel then Zin = 300||300 = 150 Ω and Vin = total P = (c) Zin 150 VS = (4 ×10−3 ) =1.71× 10−3 V R + Zin 200+150 2 Vin  1  (1.71×10−3 ) 2 = 9.7 nW or 4.85 nW to each set  = 2  Zin  2(150) In this case, we need Zin = R || R = 200 Ω ⇒ R = 400 Ω , where R is the input impedance of each television receiver. Then Ptotal = Vm 2 (2×10−3 ) 2 = = 10 nW ⇒ 5 nW to each set 2 Zin 2(200) 11-8
• Ex 11.9-1 Coil voltages: V1 = j 24 I 1 + j 16 I 2 = j 40 I V 2 = j 16 I 1 + j 40 I 2 = j 56 I Mesh equation: 24 = V1 + V 2 = j 40 I + j 56 I = j 96 I 24 1 =−j 4 j 96  1 Vo = V 2 = ( j 56 )  − j  = 14  4 vo = 14 cos 4t V I= Ex 11.9-2 Coil voltages: V1 = j 24 I 1 − j 16 I 2 = j 8 I V 2 = − j 16 I 1 + j 40 I 2 = j 24 I Mesh equation: 24 = V1 + V 2 = j 8 I + j 24 I = j 32 I 24 3 =−j 4 j 32  3 Vo = V 2 = ( j 24 )  − j  = 18  4 vo = 18 cos 4t V I= Ex 11.9-3 0 = V 2 = j 16 I 1 + j 40 I 2 40 I 2 = −2.5 I 2 16 V s = V1 = j 24 I 1 + j 16 I 2 ⇒ I1 = − = j (24(−2.5) + 16) I 2 = − j 44 I 2 24 6 = j − j 44 11 I o = I 1 − I 2 = (−2.5 − 1) I 2 I2 = = −3.5 I 2  6 = −3.5  j  = − j 1.909  11  io = 1.909 cos ( 4t - 90° ) A 11-9
• Ex 11.9-4 0 = V 2 = − j 16 I 1 + j 40 I 2 40 I 2 = 2.5 I 2 16 V s = V1 = j 24 I 1 − j 16 I 2 ⇒ I1 = = j (24(2.5) − 16) I 2 = j 44 I 2 24 6 =−j 11 j 44 I o = I 1 − I 2 = (2.5 − 1) I 2 I2 = = 1.5 I 2 6  = 1.5  − j  = − j 0.818  11  io = 0.818 cos ( 4t - 90° ) A Ex. 11.10-1 I1 = 5∠0° 5∠0° = = 2∠0° A 100 − j 75 (1 + j 3) + ( 4 − j 3) (1 + j 3) + 52 V1 = ( 4 − j 3) 2∠0° = 10∠ − 36.9° V 11-10
• Ex. 11.10-2 I1 = 5∠0° 5∠0° 5∠0° = = = 0.68∠42° A 2+ j 0.2 5.5 + j 4.95 7.4∠− 42° 5− j 5 ) + ( 22 1 I 2 = I1 = 0.34∠42° A 2 V2 = ( 2 + j 0.2 ) I 2 = ( 2.01∠5.7° )( 0.34∠42° ) = 0.68∠47.7° V so v2 (t ) = 0.68cos (10 t + 47.7°) V and i 2 (t ) = 0.34 cos (10 t + 42°) A Ex. 11.10-3 Z1 = 1  Z Z 1 Z Z  = , Z 2 = 2  Z + 2  = 9  Z +  and Z 3 = 2 (Z + Z 2 ) 2   n3 4 n1 n2  n3  4  then 1 Z   Z ab = Zin = Z + Z3 = Z +  Z +9 Z+   = 4.0625 Z 4 4   11-11
• Problems Section 11-3: Instantaneous Power and Average Power P11.3-1 1∠0° = V V V + + ⇒ V = 14.6∠ − 43° V 20 j 63 − j16 I= V = 0.23∠ − 133° A j 63 p(t ) = i (t )v(t ) = 0.23(cos (2π ⋅103 t − 133° )) × 14.6 cos (2π ⋅103 t − 43° ) = 3.36 cos (2π ⋅103 t −133° ) cos (2π ⋅103 t − 43° ) =1.68 (cos (90° ) + cos (4π ⋅103 t −176° )) =1.68 cos (4π ⋅103 t −176° ) P11.3-2 Current division:  1800 − j 2400  I=4 5  1800− j 2400+ 600  =5 5 ∠ − 8.1° mA 2 2 P600Ω I 600 5 = = 300(25)   = 1.875 ×104 µW = 18.75 mW 2  2 Psource =  5 V I cos θ 1 = (600)  5  4 5 cos(−8.1° ) = 2.1× 104 µW = 21 mW 2 2  2 ( ) P11.3-3 11-12
• Node equations: 20I X −100 20I X − V + IX + = 0 ⇒ I X (20 − j15) − V = − j 50 − j5 10 V − 20I X V − 3I X + = 0 ⇒ I X (−40 + j 30) + V (−2 − j ) = 0 − j5 10 Solving the node equations using Cramer’s rule yields IX = j 50(2 − j ) 50 5∠63.4° = = 2 5∠10.3° A (40 − j 30) − (20 − j15)(2 − j ) 25∠53.1° Then 2 PAVE ( I = X (20) = 10 2 5 2 ) 2 = 200 W P11.3-4 A node equation: (V − 16) V + − (2 2∠45°) = 0 j4 8  2 ⇒ V =  16  ∠18.4° V  3   Then I= 16 − V = 3.2 ∠ − 116.6° A j4 2 PAVE 8Ω PAVE current source = − 1 V = × 2 8 2  2 16  5 1  = × = 6.4 W absorbed 2 8 1 1 2 V 2 2 cos θ = − 16  2 2 cos ( 26.6° ) = −12.8 W absorbed 2 2 5 ( ) ( ) PAVE inductor = 0 1 1 PAVE voltage source = − (16) I cosθ = − (16)( 3.2)cos( − 116.6°) = 6.4 W absorbed 2 2 11-13
• P11.3-6 A node equation: −20 + V1 V1 + (3 / 2)V1 + = 0 ⇒ V1 = 50 5 ∠ − 26.6° V 10 15 − j 20 Then I= V1 + ( 3/2 ) V1 ( 5 / 2 ) V1 = 5 5 ∠26.6° A = 15 − j 20 25∠ − 53.1° Now the various powers can be calculated: 2 2 PAVE 10Ω PAVE current source = − 1 1 V (20) cos θ = − (50 5)(20) cos (−26.6°) = −1000 W absorbed 2 2 PAVE 15Ω = PAVE voltage source = − 1 V1 1 (50 5) = = = 625 W absorbed 2 10 2 10 I 2 2 (5 5 ) (15 ) = − 2 2 ( )( (15 ) = 937.5 W absorbed ) 1 3 1 I V1 cos θ = − 5 5 75 5 cos ( −53.1° ) = −562.5 W absorbed 2 2 2 PAVE capacitor = 0 W P11.3-7 Z= I = 200 ( j 200 ) 200 ∠90° 200 = = ∠45° Ω 200 (1 + j ) 2 ∠45° 2   120∠0 200 = 0.85 ∠ − 45° A, I R =   I = 0.6∠0° A 200  200 + j 200  ∠45 2 P = I R = ( 0.6 ) ( 200 ) = 72 W and w = ( 72 )(1) = 72 J 2 2 11-14
• Section 11-4: Effective Value of a Periodic Waveform P11.4-1 (a) ( Treat i as two sources of different frequencies.) i = 2 − 4 cos 2t = i1 + i 2 2A source: I eff = lim T →∞ 1 T ∫ T o 2 (2) dt = 2 A and 4 cos 2t source: 4 A 2 Ieff = The total is calculated as 2 I eff (b) 2  4  = ( 2) +   = 12 A ⇒ I rms = I eff = 12 = 2 3 A  2 2 i ( t ) = 3cos (π t − 90° ) + 2 cos π t ⇒ I = ( 3∠ − 90° ) + ( 2 ∠0° ) = 2 − j 3 = 3.32∠ − 64.8° A I rms = 3.32 = 2.35 A 2 i ( t ) = 2 cos 2t + 4 2 cos ( 2t + 45° ) + 12 cos ( 2 t − 90° ) (c) ( ) I = ( 2∠0° ) + 4 2 ∠45° + (12∠ − 90° ) = ( 2 + 4 ) + ( j 4 − j12 ) =10∠ − 53.1° A I rms = P11.4-2 (a) 1 Vrms = 5 (b) (c) 10 =5 2 A 2 ( ∫ 6 dt +∫ 2 dt ) = 1 ( ∫ 36 dt +∫ 4 dt ) = 5 1 84 = 4.10 V ( 72 + 12 ) = 5 5 2 2 0 5 2 5 0 2 2 2 Vrms = 1 5 ( ∫ 2 dt +∫ 6 dt ) = 1 ( ∫ 4 dt +∫ 36 dt ) = 5 1 116 = 4.81 V (8 + 108) = 5 5 Vrms = 1 5 ( ∫ 2 dt +∫ 6 dt ) = 1 ( ∫ 4 dt +∫ 36 dt ) = 5 1 84 = 4.10 V (12 + 72 ) = 5 5 2 2 0 3 0 5 2 2 5 3 2 2 5 0 2 2 3 5 0 3 11-15
• P11.4-3 2 (a) 1 4 4 2  4 4 4 4 2 2 Vrms = ∫1  3 t + 3  dt = 27 ∫1 ( 2 t + 1) dt = 27 ∫1 ( 4 t + 4 t + 1) dt 3   4  4 t3  = 27  3  4 1 4t 2 + 2 4 1  4 +t 1    = = (b) 4 ( (85.33 − 1.33) + ( 2 )(16 − 1) + 3) 27 4 (117 ) = 4.16 V 27 2 Vrms 1 4  4 22  4 4 4 4 2 2 = ( −2 t + 11) dt = ( 4 t − 44 t + 121) dt  − t +  dt = 3 ∫1  3 3  27 ∫1 27 ∫1 4  4 t3  = 27  3  4 1 44t 2 − 2 4 1  4 + 121 t 1    = = (c) 4 (84 + ( −22 )15 + (121) 3) 27 4 (117 ) = 4.16 V 27 2 Vrms 1 3 4 4 3 4 3 2 2  = ∫ 0  3 t + 2  dt = 27 ∫0 ( 2 t + 3) dt = 27 ∫ 0 ( 4 t + 12 t + 9 ) dt 3   4  4 t3  = 27  3  3 12t 2 + 2 0  3 + 9t 0   0  3 = 4 ( 36 + 54 + 27 ) 27 = 4 (117 ) = 4.16 V 27 11-16
• P11.4-4 (a)  2π v ( t ) = 1 + cos   T vdc eff 2 t 1 T  =  ∫ 0 1dt  = T  T T 0  t  = vdc + vac  T  2 =  − 0  = 1 V and vac eff = T  1 V 2 2 2 2 veff = vdc eff + vac eff (b) ω= I rms 2 1 T /2 A2 2 = ∫ 0 ( A sin ω t ) dt = T T I rms = ∫ 2 = 2π , I rms = T T /2 0  1  1 +  = 1.225 V  2 2 1 T 2 i ( t ) dt T ∫0 2 T /2 1 A2  T / 2 = A A (1−cos 2ω t ) dt =  ∫0 dt − ∫0 cos 2ω t dt   4 2 2T  A2 A = , where A = 10 mA ⇒ 4 2 I rms = 5 mA P11.4-5 90 t  v ( t ) = 90 ( 0.2−t ) 0  2 Vrms = 0 ≤ t ≤ 0.1 0.1≤ t ≤ 0.2 0.2 ≤ t ≤ 0.3 0.2 2 1  0.1 902 2 ( 90t ) dt + ∫ 0.1 90 ( 0.2−t )  dt  =    .3   .3  ∫ 0 902 = .3 V rms  0.1 t 2 dt + 0.2 0.2−t 2 dt  )  ∫0.1 ( ∫0    .001 .001   3 + 3  = 18 V   = 18 = 4.24 V 11-17
• Section 11-5: Complex Power P11.5-1 I* = R+ j4 L = 2 ( 3.6 + j 7.2 ) 2S = = 0.6 + j 1.2 = 1.342∠63.43° A 12∠0° 12∠0° 12∠0° = 8.94∠63.43 = 4 + j 8 ⇒ R = 4 Ω and L = 2 H 1.342∠ − 63.43 P 11.5-2 I* = 2 (18 + j 9 ) 2S = = 3 + j 1.5 = 3.35∠26.56° A 12∠0° 12∠0° 1 1 1 1 3.35∠ − 26.56° + = −j = = 0.2791∠ − 26.56 = 0.250 + j 0.125 R j4 L R 4L 12∠0° ⇒ R = 4 Ω and L = 2 H P11.5-3 Let Zp = 8 ( j 8) j 8 1− j 8 + j 8 = × = = 4+ j4 V 8 + j 8 1+ j 1− j 2 Next 12∠0° 12∠0° I= = = 1.342∠ − 26.6° A 4 + Z p 4 + (8 + j 8) Finally (12∠0° )(1.342∠ − 26.6° ) S= * 2 = 7.2 + j 3.6 VA 11-18
• P11.5-4 Before writing node equations, we can simplify the circuit using a source transformation: The node equations are: V1 V1 − V2 + = 0 ⇒ V1 (1 + j ) − V2 = 10 2 j2 V2 − V1 1 V2 + V1 + = 0 ⇒ V1 ( −4 + j ) + V2 ( j 8 ) = 0 j2 8 0.8 + j 0.4 −5 + Using Cramers’s rule V1 = 80 = (16 / 3) ∠126.9° V (4 − j ) − j8(1+ j ) then 1 1 I = − V1 − V2 = − V1 − V1 (1+ j )+ j 10= 2.66∠126.9° A 8 8 Now the complex power can be calculated as ( )( 2.66∠−126.9° −(2 / 3)∠36.9° I* ( −(1/ 8)V1 ) S= = 2 2 ) =−j 8 VA 9 Finally S = P + jQ = j 8 8 ⇒ P = 0, Q = VAR 9 9 11-19
• P11.5-5 I= S = VI* = 50∠120° 50∠120° = = 2.5∠83.13° A 16+ j12 20∠36.87° ( 50∠120° )( 2.5∠−83.13° ) = 125 ∠36.87° = 100 + j 75 VA P11.5-6 KVL: (10+ j 20 ) I1 = 5∠0° − j 2 I 2 ⇒ (10 + j 20 ) I1 + j 2 I 2 = 5∠0° KCL: I1 + I 2 = 6∠0° Solving these equations using Cramer’s rule: ∆= I1 = 10 + j 20 1 j2 = 10 + j18 1 1 5 j2 5 − j12 = = 0.63∠232° A = −0.39 − j 0.5 A ∆ 6 1 10 + j18 I 2 = 6 − I1 = 6 + 3.9 + j.5 = 6.39 + j.5 = 6.41∠4.47° A Now we are ready to calculate the powers. First, the powers delivered: ( ) 1 ( 5∠0° ) −I 2* = 2.5 ( 6.41∠(180− 4.47) ) = −16.0 + j1.1 VA 2 1 S 6∠ 0° = [5− j 2 I 2 ]( 6∠0° ) = 5− j 2( 6.39+ j.5 )  3 = 18.0− j 38.3 VA   2 S Total = S5∠ 0° + S 6∠ 0° = 2.0− j 37.2 VA S5∠ 0° = delivered 11-20
• Next, the powers absorbed: 1 10 2 2 10 I1 = (.63) = 2.0 VA 2 2 j 20 2 S j20Ω = I1 = j 4.0 VA 2 1 2 2 S − j2Ω = ( − j 2 ) I 2 = − j ( 6.41) = − j 41.1 VA 2 S Total = 2.0 − j 37.1 VA S10Ω = absorbed To our numerical accuracy, the total complex power delivered is equal to the total complex power absorbed. P11.5-7 (a) (b) Z= V 100∠20° = = 4∠30° Ω I 25∠ − 10° P= I V cos θ (100 )( 25 ) cos 30° = = 1082.5 W 2 2 1 = 0.25∠ − 30° = 0.2165 − j 0.125 S . To cancel the phase angle we add a capacitor Z having an admittance of YC = j 0.125 S . That requires ω C =0.125 ⇒ C = 1.25 mF . (c) Y = P11.5-8 Apply KCL at the top node to get 10 − V1 V1 3 − V1 + = 0 ⇒ V1 = 4∠36.9° V 3 3 4 1− j 2 Then V I1 = 1 = 1∠36.9° A 4 The complex power delivered by the source is calculated as − S= Finally (1∠36.9° )* (10∠0° ) = 2 5∠ − 36.9° VA pf = cos ( −36.9° ) = .8 leading 11-21
• Section 11.6: Power Factor P11.6-1 Heating: P = 30 kW Motor: θ = cos −1 ( 0.6 ) = 53.1°   S = 150 kVA Total (plant):  P = 150 cos 53.1° = 90 kW  ⇒   Q = 150sin 53.1° = 120 kVAR  P = 30 + 90 = 120 kW   ⇒ S = 120 + j120 = 170∠45° VA Q = 0 + 120 = 120 kVAR  The power factor is pf = cos 45° = 0.707 lagging. The current required by the plant is I = P11.6-2 Load 1: S 170 kVA = = 42.5 A . V 4 kV P = S cosθ = (12 kVA )( 0.7 ) = 8.4 kW 1 Q1 = S sin ( cos −1 (.7 ) ) = (12 kVA ) sin ( 45.6° ) = 8.57 kVAR Load 2: P2 = (10 kVA )( 0.8 ) = 8 kW Q2 = 10sin ( cos −1 ( 0.8 ) ) = 10sin ( 36.9° ) = 6.0 kVAR Total: S = P + jQ = 8.4 + 8 + j ( 8.57 + 6.0 ) = 16.4 + j14.57 = 21.9∠41.6° kVA The power factor is pf = cos ( 41.6° ) = 0.75 . The average power is P = 16.4 kW. The apparent power is |S| = 21.9 kVA. 11-22
• P11.6-3 The source current can be calculated from the apparent power: -1 Vs I s∗ ∗ = 2 S = 2( 50∠ cos 0.8 ) = 5∠36.9° A S= ⇒I s Vs 20∠0° 2 I s = 5∠ − 36.9° = 4 − j3 A Next I1 = Vs 20∠0° = = 2∠ − 53.1° = 1.2 − j1.6 A 6 + j8 10∠53.1° I 2 = I s − I1 = 4 − j 3 − 1.2 + j1.6 = 2.8 − j1.4 = 3.13∠ − 26.6° A Finally, Z= Vs 20∠0° = = 6.39∠26.6° Ω I z 3.13∠ − 26.6° P11.6-4 (Using all rms values.) 2 (a) V 2 2 P = I R= ⇒ V = P ⋅ R = ( 500 )( 20 ) ⇒ R (b) Is = I + I L = V = 100 Vrms V V 100∠0° 100∠0° + = + = 5 − j 5 = 5 2∠ − 45° A 20 j 20 20 j 20 (c) Zs = − j 20 + ( 20 )( j 20 ) = 10 20 + j 20 pf = cos ( −45° ) = 2∠ − 45° Ω 1 leading 2 11-23
• (d) No average power is dissipated in the capacitor or inductor. Therefore, PAVE = PAVE = 500 W ⇒ 20Ω source P11.6-5 Load 1: Vs I s cos θ = 500 ⇒ Vs = 500 = I s cos θ ( 500 = 100 V  1  5 2    2 ) V = 100∠160° V I = 2∠190° A = −1.97 − j 0.348 A P = 23.2 W, Q1 = 50 VAR 1 S1 = P + jQ1 = 23.2 + j 50 = 55.12∠65.1° VA 1 pf1 = cos 65.1° = 0.422 lagging ∗ I1 = Load 2: S1 55.12∠65.1° = = 0.551∠ − 94.9°, so I1 = 0.551∠94.9° A 100∠160° Vs I 2 = I - I1 = −1.97 − j 0.348 + 0.047 − j.549 = 2.12∠ − 155° A S 2 = VI 2 = (100∠160° )( 2.12∠155° ) = 212∠ − 45° = 150 − j150 VA ∗ pf 2 = cos ( −45° ) = 0.707 leading S = S1 + S 2 = ( 23.2 + j50 ) + (150 − j150 ) = 173.2 − j100 = 200∠ − 30° VA Total: pf = cos ( −30° ) = 0.866 leading P11.6-6 Z refrig = 120 = 14.12 Ω 8.5 Z refrig = 14.12∠45° = 10 + j10 Ω V 2 (120 ) = = = 144 Ω P 100 2 R lamp R range (a) Ι refrig = ( 240 ) = 2 12, 000 = 4.8 Ω 120∠0° 120∠0° = 8.5∠ − 45° Arms , I lamp = = 0.83∠0° Arms 10 + j10 144 and 11-24
• I range = From KCL: 240∠0° = 50∠0° A 4.8 I1 = I refrig + I range = 56 − j 6 = 56.3∠ − 6.1° A I 2 = −I lamp − I range = 50.83∠180° A I N = −I1 − I 2 = 7.92∠ − 49° A (b) Prefrig = I refrig 2 2 Rrefrig = 722.5 W and Qrefrig = I refrig X refrig = 722.5 VAR Plamp = 100 W and Q lamp = 0 Ptotal = 722 + 100 + 12, 000 = 12.82 kW   ⇒ S = 12,822 + j 722 = 12.84∠3.2° kVA Qtotal = 722 + 0 + 0 = 722 VAR  The overall power factor is pf = cos ( 3.2° ) = 0.998 , (c) Mesh equations: − 20  30 + j10  − 20 164   −10 − j10 − 144  Solve to get: − 10 − j10  I A  120∠0°  − 144   I B  = 120∠0°      158.8 + j10   I C   0     I A = 54.3 − j1.57 = 54.3∠ − 1.7° Arms I B = 51.3 − j 0.19 = 51.3∠ − 0.5°Arms I C = 50 + j 0 = 50∠0° Arms The voltage across the lamp is Vlamp = Rlamp I B − IC = 144 1.27∠ − 8.6° = 183.2 V 11-25
• P11.6-7 (a) VI=220 ( 7.6 ) = 1672 VA P 1317 = = .788 VI 1672 θ =cos −1pf = 38.0° ⇒ Q=VIsinθ =1030VAR To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing –Q, then V2 (220)2 ⇒ X c = 47Ω 1030 = = Xc Xc pf = (b) ∴ C= 1 1 ω X = (377)(47) = 56.5µ F (c) P = VI cosθ where θ = 0° then 1317 = 220I ∴ I = 6.0A for corrected pf * Note I = 7.6A for uncorrected pf P11.6-8 First load: S1 = P + jQ = P (1 + j tan (cos −1 (.6))) = 500(1 + j tan 53.1°) = 500 + j 677 kVA Second load: S 2 = 400 + j 600 kVA Total: S = S1 + S 2 = 900 + j1277 kVA S desired = P + jP tan (cos −1 (.90)) = 900 + j 436 VA From the vector diagram: S desired = S + Q . Therefore 900 + j 436 = 900 + j1277 + Q ⇒ Q = − j841 VAR 2 2 V V (1000) 2 j = − j841 ⇒ Z* = = = j1189 ⇒ Z = − j1189 = − * − j841 − j841 377 C Z Finally, C= 1 = 2.20 µ F (1189)(377) 11-26
• P11.6-9 (a) S = P + jQ = P + jP tan (cos −1 pf ) = 1000 + j1000 tan (cos −1 0.8) = 1000 + j 750 VA S 1000 + j750 = = 10 + j7.5 ⇒ I = 10 − j 7.5 A Let VL = 100∠0° Vrms. Then I* = VL 100∠0° V 100∠0° ZL = L = = 8∠36.9° = 6.4 + j 4.8 V I 12.5∠−36.9° (b) VL =[6.4 + j (200)(.024) + Z L )(I ) = (12.8 + j 9.6)(10 − j 7.5) = 200∠0° V 1 For maximum power transfer, we require ( 6.4 + j 4.8 )* = Z L || Z new = . YL +Ynew 1 1 1 = YL + Ynew ⇒ Ynew = − = j 0.15 S (6.4 − j4.8) 6.4 − j 4.8 6.4 + j 4.8 Then Z new = − j 6.67 Ω so we need a capacitor given by 1 1 = 6.67 ⇒ C = = 0.075 µ F ωC (6.67)(200) Section 11-7: The Power Superposition Principle P11.7-1 Use superposition since we have two different frequency sources. First consider the dc source (ω = 0):  12  I1 = 14   = 12 A  12 + 2  2 P1 = I1 R = (12) 2 (2) = 288 W Next, consider the ac source (ω = 20 rad/s): 11-27
• After a source transformation, current division gives − j 60     25 (12− j 5) ∠116.6° A I 2 = −9.166   = − j 60 5   + 2+ j 4  (12 − j 5)    Then 2 I (125)(2) P2 = 2 (2) = =125 W 2 2 Now using power superposition P = P + P2 = 288 + 125 = 413 W 1 P11.7-2 Use superposition since we have two different frequency sources. First consider ω = 2000 rad/s source: Current division yields  8   −j2  5 I1 = 5  ∠63.4° A = 5  8 +8  −j2    Then 2 I 8 P= 1 = 20 W 1 2 Next consider ω = 8000 rad/s source. Current division yields 11-28
•  8   j7  5 I 2 = − j5  ∠ − 171.9° A = 8 50  +8   j7    Then 2 I 8 P2 = 2 =2W 2 Now using power superposition P = P + P2 = 22 W 1 P11.7-3 Use superposition since we have two different frequency. First consider the dc source (ω = 0): 1 i 2 (t ) = 0 and i1 (t ) = 10   = 1 A  10  PR1 = i12 R1 = 12 (10) = 10 W PR 2 = 0 W Next consider ω = 5 rad/s sources. Apply KCL at the top node to get −6 I 2 + I1 + I 2 − ( 4 ∠−30° ) + (10 I1 −10∠40°) =0 j10 Apply KVL to get −10 I1 + (5− j 2) I 2 = 0 Solving these equations gives I1 = −0.56 ∠ − 64.3° A and I 2 = −1.04∠ − 42.5° A Then 11-29
• PR1 = 2 I1 R1 2 ( 0.56 ) = 2 (10) 2 = 1.57 W and PR 2 = 2 I2 R2 2 = (1.04 ) 2 (5) 2 = 2.7 W Now using power superposition PR = 10 + 1.57 = 11.57 W and PR 2 = 0 + 2.7 = 2.7 W 1 P11.7-4 Use superposition since we have two different frequency. First consider the ω = 10 rad/s source: V1 4∠0° = = 0.28 + j 0.7 A Z 2− j5 V = 2 I1 = 2(0.28 + j.7) = 0.56 + j1.4 = 1.51 ∠68.2° V I1 = R1 V = − j 5 I1 = 3.77 ∠ − 21.8° V C1 Next consider ω = 5 rad/s source. V2 6∠−90° = = 0.577 − j 0.12 A Z 2 − j10 VR = 2 I 2 = 2(.577 − j 0.12) =1.15− j 0.24 I2 = 2 V C2 =1.17∠−11.8° V = − j10I 2 =5.9∠258.3° V Now using superposition vR (t) =1.51cos (10 t + 68.2°) + 1.17 cos (5 t − 11.8°) V vC (t) = 3.77 cos (10 t − 21.8°) +5.9 cos (5 t − 258.3°) V Then 2 2 2 Reff  1.51   1.17  =  +  = 1.82 ⇒ VReff =1.35 V  2  2  2 Ceff  3.77   5.9  =  +  = 24.52 ⇒ VCeff = 4.95 V  2   2 V 2 V 2 11-30
• 11-31
• Section 11-8: Maximum Power Transfer Theorem P11.8-1 Z t = 4000 || − j 2000 = 800 − j1600 Ω Z L = Z* =800 + j1600 Ω t  R =800 Ω R + j1000 L = 800 + j1600 ⇒   L =1.6 H P11.8-2 Z t = 25, 000 || − j 50, 000 = 20, 000 − j10, 000 Ω Z L =Z* = 20,000+ j10,000 Ω t  R = 20 kΩ  R + jω L = 20, 000 + j10, 000 ⇒ 100 L =10,000  L =100 H  After selecting these values of R and L, 2 I = 1.4 mA and Pmax  0.14×10−2  3 =   ( 20×10 ) = 19.5 mW 2   Since Pmax > 12 mW , yes, we can deliver 12 mW to the load. P11.8-3  −j  R  2  ω C  = R − jω R C Z t = 800 + j1600 Ω and Z L = j 1 + (ω RC ) 2 R− ωC  −j  R  2  ω C  = R − jω R C = 800 − j1600 Ω Z L = Z* ⇒ t j 1+ (ω RC ) 2 R− ωC Equating the real parts gives 800 = R 4000 = 2 1+ (ω RC ) 1+[(5000)(4000)C ]2 ⇒ C = 0.1 µ F 11-32
• P11.8-4 Z t = 400 + j 800 Ω and Z L = 2000 || − j1000 = 400 − j 800 Ω Since Z L = Z* the average power delivered to the load is maximum and cannot be increased by t adjusting the value of the capacitance. The voltage across the 2000 Ω resistor is VR = 5 ZL = 2.5 − j 5 = 5.59e− j 63.4 V Zt + ZL So 2  5.59  1 P= = 7.8 mW   2  2000 is the average power delivered to the 2000 Ω resistor. P11.8-5 Notice that Zt,not ZL, is being adjusted .When Zt is fixed, then the average power delivered to the load is maximized by choosing ZL = Zt*. In contrast, when ZL is fixed, then the average power delivered to the load is maximized by minimizing the real part of Zt. In this case, choose R = 0. Since no average power is dissipated by capacitors or inductors, all of the average power provided by source is delivered to the load. Section 11-9: Mutual Inductance P11-9-1 Vs + I jω L1 + I jω M + I jω L2 − I jω M = 0 ⇒ jω ( L1 + L2 − 2 M ) = Vs I Therefore Lab = L1 + L2 − 2M 11-33
• P11.9-2 KCL: I1 + I 2 = I s The coil voltages are given by: V = I1 jω L1 + I 2 jω M V = I 2 jω L2 + I1 jω M Then I2 = and V − jω L1I s jω ( M − L1 ) V = I 2 jω L2 + ( I s − I 2 ) jω M Then V= (V − jω L1I s )  jω ( L2 − M )    jω ( M − L1 ) + jω MI s ⇒  L L −M 2  V = jω  1 2  Is  L1 + L2 − 2M  Finally Lab = L1 L2 − M 2 L1 + L2 − 2M P11.9-3 Mesh equations: −141.4∠0° + 2 I1 + j 40 I1 − j 60 I 2 = 0 l200 I 2 + j 60 I 2 − j 60 I1 = 0 ⇒ I 2 = (0.23∠51°) I1 Solving yields I1 = 4.17∠ − 68° A and I 2 = 0.96∠ − 17°A Finally i1 ( t ) = 4.2 cos(100t −68°) A and i 2 ( t ) = 1.0 cos(100t −17°) A 11-34
• P11.9-4 Mesh equations: (10+ j5) I 1 − j50 I 2 = 10 − j 50 I1 +( 400+ j 500 ) I 2 = 0 Solving the mesh equations using Cramer’s rule: I2 = (10+ j5)( 0 )−( − j50 )(10 ) 2 (10+ j5) ( 400+ j500 ) − ( − j50 ) = 0.062 ∠29.7° A Then V2 400 I 2 = = 40 I 2 = 40 ( 0.062 ∠29.7° ) = 2.5 ∠29.7° V1 10∠0° P11.9-5 Mesh equations: −10∠0° − j 5 I1 + j 9 I1 + j 3 I 2 = 0 28 I 2 + j 6 I 2 + j 3 I1 + j 9 I 2 − j 3 I 2 = 0 Solving the mesh equations yields I1 = 0.25∠161° A and I 2 = 2.55∠ − 86° A then Finally V = j 9(I1 − I 2 ) = j 9 ( 2.6 ∠−81° ) = 23 ∠9° V v (t ) = 23cos (30t + 9°) V 11-35
• P11.9-6 (a) I 2 = 0 ⇒ I1 = 10 ∠0° A ⇒ i1 (0) = 10 A 2 w= L1 i1 (0) 2 = (0.3) (10) 2 = 15 J 2 Mesh equations: (b) j 6 I 2 - j 3 I1 = 0 ⇒ I1 = 2 I 2 I1 = 10∠0° A ⇒ I 2 = 5∠0° A Then w= 1 1 1 1 2 2 L1i1 (0) + L 2i1 (0) − M i1 (0) i 2 (0) = (0.3)(10) 2 + (1.2)(5) 2 − (0.6) (10)(5) = 0 2 2 2 2 (7 + j 6) I 2 − j 3 I1 = 0 (c) I 2 = 3.25 ∠49.4° A i 2 (t ) = 3.25cos(5t + 49.4°) A i 2 (0) = 2.12 A Finally 1 1 w = (0.3) (10) 2 + (1.2) (2.12) 2 − (0.6) (10) (2.12) = 5.0 J 2 2 P11.9-7 Mesh equations: −VT + j8 I1 + j 5(I1 − I 2 ) − j 6 I1 + j 6 (I1 − I 2 ) + j 5 I1 = 0 3 I 2 + j 6 (I 2 − I1 ) − j 5 I1 = 0 11-36
• Solving yields I 2 = (1.64 ∠27° ) I1 I1 ( j 18) + I 2 (− j 11) = VT Then Z = VT = 8.2 + j 2 = 8.4 ∠14° Ω I1 P11.9-8 The coil voltages are given by V1 = j 6 I1 − j 2 (I1 − I 2 ) − j 4 I 2 = j 4 I1 − j 2 I 2 V2 = j 4 ( I1 − I 2 ) − j 2 I1 + j 2 I 2 = j 2 I1 − j 2 I 2 V3 = j8 I 2 − j 4 I1 + j 2 ( I1 − I 2 ) = − j 2 I1 + j 6 I 2 The mesh equations are 5 I1 + V1 + 6 (I1 − I 2 ) + V2 = 10∠0° −V2 + 6 (I 2 − I1 ) + 2 I 2 + V3 − j 5 I 2 = 0 Combining and solving yields 11 + j 6 10 −6 − j 4 0 60 + j 40 = = 1.2 ∠0.28° A I2 = 11 + j 6 −6 − j 4 50 + j 33 −6 − j 4 8 + j 3 Finally V = − j 5 I 2 = 6.0 ∠ − 89.72° A ⇒ v(t ) = 6sin(2 t − 89.7°) V 11-37
• Section 11-10: The Ideal Transformer P11.10-1 Z = (2 + j 3) + I1 = (100− j 75) =6Ω 52 12∠0° 12∠0° = =2A Z 6  100− j 75   100− j 75  V1 = I1   = (2)   = 10∠ − 36.9° V 2 n    25  V2 = nV1 = 5 (10∠−36.9°) = 50 ∠−36.9° V I2 = I1 2 = A n 5 P11.10-2 (a) V0 = (5 ×10−3 )(10, 000) = 50 V n= (b) (c) V N2 50 = 0 = = 5 N1 V1 10 Rab = Is = 1 1 R2 = (10 × 103 ) = 400 Ω 2 n 25 10 10 = = 0.025 A = 25 mA Rab 400 11-38
• P11.10-3 Z1 = 1 Z 2 = 9 Z 2 = 9(5 − j8) = 45 − j 72 Ω n2 Using voltage division, the voltage across Z 1 is   45− j 72 V1 = ( 80∠ − 50° )   = 74.4 ∠ − 73.3° V  45− j 72+ 30 + j 20  then V2 = nV1 = 74.4 ∠ − 73.3° = 24.8 ∠ − 73.3° V 3 Using voltage division again yields  − j8   8∠−90°  Vc = V2   = ( 24.8∠−73.3° )   = 21.0∠ − 105.3° V  89∠−58°   5− j 8  P11.10-4 n = 5, Z1 = 200 ( 5) 2 = 8 Ω ⇒ V1 = 8 ( 50∠0° ) = 40∠0° V ⇒ V2 = n V1 = 200∠0° V 8+ 2 11-39
• P11.10-5 Z= 320 jω L + 2 n2 n Maximum power transfer requires jω L = j160 kΩ and n2 320 = 80 n2 so n = 2. Then ω L = 640 kΩ so 640×103 L = = 6.4 H 105 P11.10-6 Z= 1 ( 2 + 6) = 2 Ω 22   6   2  Voc =   ( 2)   16∠0°  = 12∠0° V  6 + 2   2 + 2   Z= 1 1 2 = Ω 2 ( ) 2 2    16∠0°  1 1 I sc = −I 2 = I1 =   = 3.25∠0° A 2 2 2+ 1   2  Then Zt = 12∠0° = 3.75∠0° Ω 3.2∠0° 11-40
• P11.10-7 1 V1 2 V − V2 V1 I3 = 1 = 2 4 V V I 2 = I3 − 2 = 1 6 6 V 1 I1 = − I 2 = − 1 2 12 V I T = I 3 − I1 = 1 6 V Z= 1 =6 IT V2 = P11.10-8 * Maximum power transfer requires Z L = Z t . First  1  1 2 1 X C1  2  = X L1 , 2 = = n  10 5 n2  2  ⇒ n2 = 5 then   1   1   RL  2  + 1  2  = 100 Ω ⇒     n2    n1    1 100 3 = ⇒ n1 = 2 3 10 n1 11-41
• P11.10-9 ZL = 1  20 (1+ j 7.54)  8.1∠23° = = 0.3 + j 0.13 Ω 52  20+10+ j 7.54  25   V V ( 230 ) = 88 kW/home PL = L = 2 = 2 R 2 2 R L 2( 0.3) 2 2 2 Therefore, 529 kW are required for six homes. P11.10-10 (a) Coil voltages: V1 = j16 I1 V2 = j12 I 2 Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute the coil voltages into the mesh equations and do some algebra: 11-42
• 8 I1 + j16 I1 = 5∠45° ⇒ I1 = 0.28∠ − 18.4° 12 I 2 + j12 I 2 = 0 ⇒ I 2 = 0 V2 = −12 I 2 = 0 (b) Coil voltages: V1 = j16 I1 + j8 I 2 V2 = j12 I 2 + j8 I1 Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute the coil voltages into the mesh equations and do some algebra: 8 I1 + ( j16 I1 + j8 I 2 ) = 5∠45° 12 I 2 + ( j12 I 2 + j8 I1 ) = 0 I1 = − 12 + j12 3 I 2 = ( j − 1) I 2 j8 2   3 ( 8 + j16 )  2  ( j − 1) + j8 I 2 = 5∠45° ⇒ I 2 = 0.138∠ − 141°     V2 = −12 I 2 = 1.656∠39° 11-43
• (c ) Coil voltages and currents: 10 V2 8.66 8.66 I1 = − I2 10 V1 = Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute into the second mesh equation and do some algebra: 2  10  8.66  10  −12  − I1  = V1 ⇒ V1 = 12   I1  8.66  10  8.66  2  10  8 I1 + 12   I1 = 5∠45° ⇒ I1 = 0.208∠45°  8.66   10  12 (10 ) 0.208∠45° = 2.88∠45° V2 = −12 I 2 = −12  − I1  = 8.66  8.66  11-44
• PSpice Problems SP 11-1 The coupling coefficient is k = 3 = 0.94868 . 2×5 11-45
• SP 11-2 Here is the circuit with printers inserted to measure the coil voltages and currents: Here is the output from the printers, giving the voltage of coil 2 as 2.498∠107.2°, the current of coil 1 as 0.4484∠-94.57°, the current of coil 2 as 0.6245∠-72.77° and the voltage of coil 1 as 4.292∠-58.74°: FREQ 7.958E-01 VM(N00984) 2.498E+00 VP(N00984) 1.072E+02 FREQ IM(V_PRINT1)IP(V_PRINT1) 7.958E-01 4.484E-01 -9.457E+01 FREQ IM(V_PRINT2)IP(V_PRINT2) 7.958E-01 6.245E-01 -7.277E+01 FREQ 7.958E-01 VM(N00959) VP(N00959) 4.292E+00 -5.874E+01 The power received by the coupled inductors is p= ( 4.292 )( 0.4484 ) cos 2 = 0.78016 − .78000 ≈ 0 ( −58.74 − ( −94.57 ) ) + ( 2.498)( 0.6245) cos 2 (107.2 − ( −72.77 ) ) 11-46
• SP 11-3 The inductance are selected so that L2 L1 = N2 N1 = 3 and the impedance of these inductors 2 are much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit while providing a connected circuit. Here is the output from the printers, giving the voltage of coil 2 as 24.00∠114.1°, the current of coil 1 as 4.000∠114.0°, the current of coil 2 as 2.667∠-65.90° and the voltage of coil 1 as 16.00∠114.1°: FREQ 6.366E-01 VM(N00984) 2.400E+01 VP(N00984) 1.141E+02 FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 4.000E+00 1.140E+02 FREQ IM(V_PRINT2)IP(V_PRINT2) 6.366E-01 2.667E+00 -6.590E+01 FREQ 6.366E-01 VM(N00959) 1.600E+01 VP(N00959) 1.141E+02 The power received by the transformer is p= (16 )( 4 ) cos (114 − 114 ) + 2 = 32 − 32.004 ≈ 0 ( 24 )( 2.667 ) cos 2 (114 − ( −66 ) ) 11-47
• SP 11-4 The inductance are selected so that L2 L1 = N2 N1 = 2 and the impedance of these inductors 5 are much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit while providing a connected circuit. FREQ 6.366E-01 VM(N00921) 1.011E+02 VP(N00921) 7.844E+01 VR(N00921) 2.025E+01 VI(N00921) 9.903E+01 The printer output gives the voltage across the current source as 20.25 + j 99.03 = 101.1∠78.44° V The input impedance is Zt = 20.25 + j 99.03 = 20.25 + j 99.03 Ω = 101.1∠78.44° Ω 1  52  (We expected Z t = 8 +  2  ( 2 + j ( 4 )( 4 ) ) = 20.5 + j 100 Ω . That’s about 1% error.) 2  11-48
• Verification Problems VP 11-1 The average power supplied by the source is Ps = (12 )( 2.327 ) cos 2 ( 30° − ( −25.22° ) ) = 7.96 W Capacitors and inductors receive zero average power, so the sum of the average powers received by the other circuit elements is equal to the sum of the average powers received by the resistors: 2.327 2 1.1292 PR = ( 4) + ( 2 ) = 10.83 + 1.27 = 12.10 W 2 2 The average power supplied by the voltage source is equal to the sum of the average powers received by the other circuit elements. The mesh currents cannot be correct. (What went wrong? It appears that the resistances of the two resistors were interchanged when the data was entered for the computer analysis. Notice that PR = 2.327 2 1.1292 ( 2) + ( 4 ) = 5.41 + 2.55 = 7.96 W 2 2 The mesh currents would be correct if the resistances of the two resistors were interchanged. The computer was used to analyze the wrong circuit.) VP 11-2 The average complex supplied by the source is Ss = (12∠30° )(1.647∠ − 17.92° ) * = (12∠30° )(1.647∠17.92° ) = 9.88∠47.92° = 6.62 + j 7.33 2 2 The complex power received by the 4 Ω resistor is S4Ω = ( 4 ×1.647∠ − 17.92° )(1.647∠ − 17.92° ) * = 5.43 + j 0 2 VA The complex power received by the 2 Ω resistor is S2Ω = ( 2 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 1.20 + j 0 2 VA 11-49 W
• The current in the 2 H inductor is (1.647∠ − 17.92° ) − (1.094∠ − 13.15° ) = 0.5640∠ − 27.19° The complex power received by the 2 H inductor is S 2H = ( j 8 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 1.27 VA 2 The complex power received by the 4 H inductor is S 4H = ( j 16 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 9.57 2 VA S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 1.27 ) + ( 0 + j 9.57 ) = 6.63 + j 10.84 ≠ Ss The complex power supplied by the voltage source is equal to the sum of the complex powers received by the other circuit elements. The mesh currents cannot be correct. (Suppose the inductances of the inductors were interchanged. Then the complex power received by the 4 H inductor would be S 4H = ( j 16 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 2.54 2 VA The complex power received by the 2 H inductor would be S 2H = ( j 8 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 4.79 2 VA S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 2.54 ) + ( 0 + j 4.79 ) = 6.63 + j 7.33 ≈ S s The mesh currents would be correct if the inductances of the two inductors were interchanged. The computer was used to analyze the wrong circuit.) VP 11-3 The voltage across the right coil must be equal to the voltage source voltage. Notice that the mesh currents both enter the undotted ends of the coils. In the frequency domain, the voltage across the right coil is 11-50
• ( j 16 )(1.001∠ − 47.01°) + ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° + 5.092∠75° = (11.715 + j 10.923) + (1.318 + j 4.918 ) = 13.033 + j 15.841 = 20.513∠50.55° This isn’t equal to the voltage source voltage so the computer analysis isn’t correct. What happened? A data entry error was made while doing the computer analysis. Both coils were described as having the dotted end at the top. If both coils had the dot at the top, the equation for the voltage across the right coil would be ( j 16 )(1.001∠ − 47.01°) − ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° − 5.092∠75° = (11.715 + j 10.923) − (1.318 + j 4.918 ) = 10.397 + j 6.005 = 12.007∠30.01° This is equal to the voltage source voltage. The computer was used to analyze the wrong circuit. VP 11-4 First check the ratio of the voltages across the coils. n1 2 12∠30° = 2.5 ≠ = n2 5 ( 75 )( 0.064∠30° ) The transformer voltages don’t satisfy the equations describing the ideal transformer. The given mesh currents are not correct. That’ enough but let’s also check the ratio of coil currents. (Notice that the reference direction of the i2(t) is different from the reference direction that we used when discussing transformers.) n1 2 0.064∠30° = 2.5 ≠ = n2 5 0.0256∠30° The transformer currents don’t satisfy the equations describing the ideal transformer. n1 1 . This suggests that a data 2.5 n2 entry error was made while doing the computer analysis. The numbers of turns for the two coils was interchanged. In both case, we calculated to be 2.5 instead of 0.4 = 11-51
• 11-52
• 11-53
• Design Problems DP 11. 1 P 100  S= = =125 kVA P =100 W   pf 0.8  ⇒  pf = 0.8  Q = S sin (cos −1 0.8) =125sin (36.9°) =75 kVAR  (a) Now pf = 0.95 so P 100 = =105.3 kVA pf 0.95 Q = S sin (cos −1 0.95) =105.3sin (18.2°) =32.9 kVAR S= so an additional 125 − 105.3 = 19.7 kVA is available. (b) Now pf = 1 so P 100 = =100 kVA pf 1 Q = S sin (cos −1 1) = 0 S= and an additional 125-100= 25 kVA is available. (c) In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In part (b), the capacitors are required to reduce Q by 75 – 0 = 0 kVAR. (d) Corrected power factor Additional available apparent power Reduction in reactive power 0.95 19.7 kVA 1.0 25 kVA 42.1 kVAR 75 kVAR 11-54
• DP 11-2 This example demonstrates that loads can be specified either by kW or kVA. The procedure is as follows: First load: Second load: Total load:  P = S1 pf =( 50 )( 0.9 ) = 45 W S1 =50 VA   1  ⇒  −1 pf =0.9  Q1 = S1 sin (cos 0.9) =50sin (25.8°) = 21.8 kVAR  P 45  = 49.45 kVA S2 = 2 = P2 = 45 W   pf 0.91  ⇒  pf = 0.91  Q = S sin (cos −1 0.91) = 49.45sin (24.5°) = 20.5 kVAR  2 2 S L = S1 + S 2 = (45 + 45) + j(21.8+20.5) = 90 + j 42.3 kVA Specified load: P 90  Ss = s = =92.8 kVA Ps =90 W   pf 0.97  ⇒  pf = 0.97  Q = S sin (cos −1 0.97) =92.8sin (14.1°) = 22.6 kVAR  s s The compensating capacitive load is Qc = 42.3 − 22.6 = 19.7 kVAR . The required capacitor is calculated as 2 Vc (7.2 × 103 ) 2 1 Xc = = = 2626 Ω ⇒ C = = 1.01 µ F 3 Qc 19.7 × 10 377 (2626) 11-55
• DP 11-3 Find the open circuit voltage: −10 + 5I + j10I − 0.5 Voc = 0 and I= 10 − Voc 5 so Voc = 8∠36.9° = 6.4 + j 4.8 V Find the short circuit current: I sc = 10∠0° = 2∠0° A 5 The the Thevenin impedance is: Zt = The short circuit forces the controlling voltage to be zero. Then the controlled voltage is also zero. Consequently the dependent source has been replaced by a short circuit. Voc = 3.2 + j 2.4 Ω I sc (a) Maximum power transfer requires Z L = Z t * = 3.2 − j 2.4 Ω . (c) ZL can be implemented as the series combination of a resistor and a capacitor with R = 3.2 Ω and C = (b) Pmax = 1 = 4.17 mF . (100 ) (2.4) | Voc |2 64 = = 2.5 W 8R 8 ( 3.2 ) 11-56
• DP 11-4 Using an equation from section 11.8, the power is given as 2  R  Vs  2 n  2 P= 2 2 R  3   3 + 2  +  2 + 4 n  n   When R = 4 Ω, 2 n 2 R Vs P= 25n 4 + 48n 2 + 25 4 2 2 3 dP 2  2 n(25n + 48n + 25) − n (100n + 96n )  = R Vs   dn (25n 4 + 48n 2 + 25) 2   5 4 ⇒ − 50n + 50n = 0 ⇒ n = 1 ⇒ n = 1 0= When R = 8 Ω, a similar calculation gives n = 1.31. 11-57
• DP 11-5 Maximum power transfer requires 10 + j 6.28 = Z1 = (1 + j 0.628 ) * n2 Equating real parts gives 10 = 1 ⇒ n = 3.16 n2 Equating imaginary parts requires jX = − j 0.628 ⇒ 3.162 X = −6.28 This reactance can be realized by adding a capacitance C in series with the resistor and inductor that comprise Z2. Then −6.28 = X = − 1 1 + 6.28 ⇒ C = = 0.1267 µ F 5 5 ( 2π ×10 ) C ( 2π ×10 ) (12.56 ) 11-58
• DP 11-6 Maximum power transfer requires 1 || R = (100 + j107 ×10−6 ) * j107 C R = 100 − j10 1 + j107 R C R = (100 − j10) (1 + j107 R C ) = 100 + 108 R C + j (109 R C − 10 ) Equating real and imaginary parts yields R = 100 + 108 R C and 109 R C − 10 = 0 then RC = 10−8  10−8  10−8 ⇒ R = 100 + 108 R  = 99 Ω ⇒ C = = 0.101 nF  99  R  11-59
• Chapter 12: Three-Phase Circuits Exercises Ex. 12.3-1 VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120° Vbc = 3 (120 ) ∠ − 90° Ex. 12.4-1 Four-wire Y-to-Y Circuit Mathcad analysis (12v4_1.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: Calculate the line currents: IaA = 1.079 − 0.674i IaA = 1.272 180 π ⋅ arg( IaA ) = −32.005 IaA := π 180 ⋅ − 120 Vc := Va⋅ e ZA := 80 + j⋅ 50 Va IbB := ZA j⋅ Vb ZB IbB = 1.061 π 180 ⋅ 120 ZB := 80 + j⋅ 80 IbB = −1.025 − 0.275i 180 π ⋅ arg( IbB) = −165 IcC := ZC := 100 − j⋅ 25 Vc ZC IcC = −0.809 + 0.837i IcC = 1.164 180 π ⋅ arg( IcC) = 134.036 12-1
• Calculate the current in the neutral wire: INn := IaA + IbB + IcC INn = −0.755 − 0.112i Calculate the power delivered to the load:   SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 129.438 + 80.899i  SC := IcC⋅ IcC⋅ ZC SB = 90 + 90i SC = 135.529 − 33.882i Total power delivered to the load: SA + SB + SC = 354.968 + 137.017i Calculate the power supplied by the source:  Sa := IaA ⋅ Va  Sb := IbB⋅ Vb  Sc := IcC⋅ Vc Sa = 129.438 + 80.899i Sb = 90 + 90i Sc = 135.529 − 33.882i Total power delivered by the source: Sa + Sb + Sc = 354.968 + 137.017i Ex. 12.4-2 Four-wire Y-to-Y Circuit Mathcad analysis (12x4_2.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: Calculate the line currents: IaA = 1.92 − 1.44i IaA = 2.4 180 π ⋅ arg( IaA ) = −36.87 IaA := π 180 ⋅ − 120 Vc := Va⋅ e ZA := 40 + j⋅ 30 Va IbB := ZA j⋅ Vb ZB IbB = 2.4 π ⋅ 120 ZB := ZA IbB = −2.207 − 0.943i 180 π 180 ⋅ arg( IbB) = −156.87 IcC := ZC := ZA Vc ZC IcC = 0.287 + 2.383i IcC = 2.4 180 π ⋅ arg( IcC) = 83.13 12-2
• Calculate the current in the neutral wire: INn := IaA + IbB + IcC INn = 0 Calculate the power delivered to the load:   SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 230.4 + 172.8i  SC := IcC⋅ IcC⋅ ZC SB = 230.4 + 172.8i SC = 230.4 + 172.8i Total power delivered to the load: SA + SB + SC = 691.2 + 518.4i Calculate the power supplied by the source:  Sa := IaA ⋅ Va  Sb := IbB⋅ Vb  Sc := IcC⋅ Vc Sa = 230.4 + 172.8i Sb = 230.4 + 172.8i Sc = 230.4 + 172.8i Total power delivered by the source: Sa + Sb + Sc = 691.2 + 518.4i Ex. 12.4-3 Three-wire unbalanced Y-to-Y Circuit with line impedances Mathcad analysis (12x4_3.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: π 180 ⋅ − 120 ZA := 80 + j⋅ 50 j⋅ Vc := Va⋅ e π 180 ⋅ 120 ZB := 80 + j⋅ 80 ZC := 100 − j⋅ 25 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e 2 j⋅ ⋅ π 3 + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC VnN = −25.137 − 14.236i VnN = 28.888 ⋅ Vp 180 π ⋅ arg( VnN) = −150.475 12-3
• Calculate the line currents: IaA := Va − VnN ZA IaA = 1.385 − 0.687i π Check: Vb − VnN IcC := ZB IbB = −0.778 − 0.343i IaA = 1.546 180 IbB := 180 π ZC IcC = −0.606 + 1.03i IbB = 0.851 ⋅ arg( IaA ) = −26.403 Vc − VnN IcC = 1.195 180 ⋅ arg( IbB) = −156.242 π ⋅ arg( IcC) = 120.475 IaA + IbB + IcC = 0 Calculate the power delivered to the load:   SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 191.168 + 119.48i  SC := IcC⋅ IcC⋅ ZC SB = 57.87 + 57.87i Total power delivered to the load: SC = 142.843 − 35.711i SA + SB + SC = 391.88 + 141.639i Ex. 12.4-4 Three-wire balanced Y-to-Y Circuit with line impedances Mathcad analysis (12x4_4.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 Describe the three-phase load: j⋅ Vb := Va⋅ e π 180 ⋅ − 120 ZA := 40 + j⋅ 30 j⋅ Vc := Va⋅ e ZB := ZA π 180 ⋅ 120 ZC := ZA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e 2 j⋅ ⋅ π 3 + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC ⋅ Vp 12-4
• − 14 VnN = −1.31 × 10 − 14 Calculate the line currents: IaA := VnN = 2.301 × 10 Va − VnN ZA IaA = 1.92 − 1.44i π ⋅ arg( VnN) = 124.695 Vb − VnN IcC := ZB 180 π − 15 IaA + IbB + IcC = 1.055 × 10 180 ⋅ arg( IbB) = −156.87 π − 15 ⋅ arg( IcC) = 83.13 − 2.22i × 10 SB = 230.4 + 172.8i Total power delivered to the load: ZC IcC = 2.4 Calculate the power delivered to the load:   SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 230.4 + 172.8i Vc − VnN IcC = 0.287 + 2.383i IbB = 2.4 ⋅ arg( IaA ) = −36.87 Check: IbB := π IbB = −2.207 − 0.943i IaA = 2.4 180 180 − 14 + 1.892i× 10  SC := IcC⋅ IcC⋅ ZC SC = 230.4 + 172.8i SA + SB + SC = 691.2 + 518.4i Ex. 12.6-1 Balanced delta load: (See Table 12.5-1) Z ∆ = 180∠ − 45° VAB 360∠0° = = 2∠45° A Z ∆ 180∠− 45 I BC = phase currents: I AB = VBC 360∠−120° = = 2∠− 75° A ° Z ∆ 18∠− 45 I CA line currents: VCA 360∠120° = = = 2∠165° A ° Z ∆ 180∠− 45 I A = I AB − I CA = 2∠45° − 2∠165° = 2 3∠15° A I B = 2 3∠−105° A I C = 2 3∠135° A 12-5
• Ex. 12.7-1 Three-wire Y-to-Delta Circuit with line impedances Mathcad analysis (12x4_4.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 110 ⋅0 j⋅ Vb := Va⋅ e Describe the delta connected load: π 180 ⋅ − 120 Z1 := 150 + j⋅ 270 j⋅ Vc := Va⋅ e π 180 Z2 := Z1 ⋅ 120 Z3 := Z1 Convert the delta connected load to the equivalent Y connected load: ZA := Z1⋅ Z3 Z1 + Z2 + Z3 ZA = 50 + 90i Describe the three-phase line: ZB := Z2⋅ Z3 Z1 + Z2 + Z3 ZB = 50 + 90i ZaA := 10 + j⋅ 25 ZC := Z1⋅ Z2 Z1 + Z2 + Z3 ZC = 50 + 90i ZbB := ZaA ZcC := ZaA 12-6
• Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e − 14 − 14 IaA := IaA = 0.392 − 0.752i IaA = 0.848 Check: − 14 + 1.784i× 10 Calculate the line currents: π + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) VnN = −1.172 × 10 180 4 j⋅ ⋅ π 3 ⋅ arg( IaA ) = −62.447 VnN = 2.135 × 10 Va − VnN IbB := ZA + ZaA ⋅ arg( VnN) = 123.304 IcC := ZB + ZbB IbB = 0.848 π π Vb − VnN IbB = −0.847 + 0.036i 180 180 ⋅ arg( IbB) = 177.553 ⋅ Vp Vc − VnN ZC + ZcC IcC = 0.455 + 0.716i IcC = 0.848 180 π ⋅ arg( IcC) = 57.553 IaA + IbB + IcC = 0 Calculate the phase voltages of the Y-connected load: VAN := IaA ⋅ ZA VAN = 87.311 180 π ⋅ arg( VAN) = −1.502 VBN := IbB⋅ ZB VBN = 87.311 180 π ⋅ arg( VBN) = −121.502 VCN := IcC⋅ ZC VCN = 87.311 180 π ⋅ arg( VCN) = 118.498 Calculate the line-to-line voltages at the load: VAB := VAN − VBN VAB = 151.227 180 π ⋅ arg( VAB) = 28.498 VBC:= VBN − VCN VBC = 151.227 180 π ⋅ arg( VBC = −91.502 ) VCA := VCN − VAN VCA = 151.227 180 π ⋅ arg( VCA) = 148.498 Calculate the phase currents of the ∆ -connected load: IAB := VAB Z3 IAB = 0.49 180 π ⋅ arg( IAB) = −32.447 IBC := VBC Z1 IBC = 0.49 180 π ⋅ arg( IBC) = −152.447 ICA := VCA Z2 ICA = 0.49 180 π ⋅ arg( ICA) = 87.553 12-7
• Ex. 12.8-1 Continuing Ex. 12.8-1: Calculate the power delivered to the load:   SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 35.958 + 64.725i  SC := IcC⋅ IcC⋅ ZC SB = 35.958 + 64.725i Total power delivered to the load: SC = 35.958 + 64.725i SA + SB + SC = 107.875 + 194.175i Ex. 12.9-1 P1 = VAB I A cos(θ +30° ) + VCB I C cos(θ − 30° )= P + P2 1 pf = .4 lagging ⇒ θ = 61.97 ° So P T = 450(24)  cos 91.97° + cos 31.97°  = 8791 W   ∴ P 1 = − 371 W P2 = 9162 W Ex. 12.9-2 Consider Fig. 12.9-1 with P = 60 kW P2 = 40 kW . 1 (a.) P = P + P2 = 100 kW 1 (b.) use equation 12.9-7 to get tan θ = 3 P2 − P 40− 60 1 = 3 = − .346 ⇒ θ = − 19.11° PL + P2 100 then pf = cos ( − 19.110°) = 0.945 leading 12-8
• Problems Section 12-3: Three Phase Voltages P12.3-1 Given VC = 277 ∠45° and an abc phase sequence: VA = 277 ∠ ( 45−120 ) ° = 277 ∠ − 75° VB = 277 ∠( 45° +120 )° = 277 ∠165° VAB = VA − VB =( 277 ∠− 75° )−( 277 ∠165° ) =( 71.69 − j 267.56 ) −( −267.56+ j 71.69 ) =339.25− j 339.25 = 479.77 ∠− 45° 480 ∠− 45° Similarly: VBC = 480 ∠ − 165° and VCA = 480 ∠75° P12.3-2 VAB 3∠30° = 12470 ∠145° V VAB = VA × 3∠30° ⇒ VA = VAB = −VBA = − (12470 ∠−35° ) In our case: VA = So 12470 ∠145° = 7200∠115° 3∠30° Then, for an abc phase sequence: VC = 7200 ∠ (115 + 120 ) ° = 7200 ∠235° = 7200 ∠ − 125° VB = 7200 ∠ (115 − 120 ) ° = 7200 ∠ − 5° V P12.3-3 Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the line-to-line voltage is So the phase voltage is Vab = 1500 ∠30° V 1500 ∠30° Va = = 866∠0° V 3∠30° 12-9
• Section 12-4: The Y-to-Y Circuit P12.4-1 Balanced, three-wire, Y-Y circuit: where Z A = Z B = Z C = 12∠30 = 10.4 + j 6 MathCAD analysis (12p4_1.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 208 3 ⋅0 j⋅ Vb := Va⋅ e Describe the balanced three-phase load: π 180 ⋅ − 120 j⋅ Vc := Va⋅ e ZA := 10.4 + j⋅ 6 π 180 ⋅ 120 ZB := ZA ZC := ZB Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e IaA := Va − VnN ZA IaA = 8.663 − 4.998i Check: − 14 ⋅ Vp IbB := VnN = 2.762 × 10 Vb − VnN IbB = −8.66 − 5.004i IaA = 10.002 π + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC Calculate the line currents: 180 2 j⋅ ⋅ π 3 IbB = 10.002 180 ⋅ arg( IaA ) = −29.982 π − 15 IaA + IbB + IcC = 4.696 × 10 IcC := ZB Vc − VnN ZC −3 IcC = −3.205 × 10 + 10.002i IcC = 10.002 ⋅ arg( IbB) = −149.982 180 π ⋅ arg( IcC) = 90.018 − 14 − 1.066i× 10 12-10
• Calculate the power delivered to the load:   SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB 3 SA = 1.04 × 10 + 600.222i Total power delivered to the load:  SC := IcC⋅ IcC⋅ ZC 3 SB = 1.04 × 10 + 600.222i 3 3 SC = 1.04 × 10 + 600.222i 3 SA + SB + SC = 3.121 × 10 + 1.801i× 10 Consequently: (a) The phase voltages are Va = 208 ∠0° = 120∠0° V rms, Vb = 120∠ − 120° V rms and Vc = 120∠120° V rms 3 (b) The currents are equal the line currents (c) I a = I aA = 10∠ − 30° A rms, I b = I bB = 10∠ − 150° A rms and I c = I cC = 10∠90° A rms (d) The power delivered to the load is S = 3.121 + j1.801 kVA . P12.4-2 Balanced, three-wire, Y-Y circuit: where Va = 120∠0° Vrms, Vb = 120∠ − 120° Vrms and Vc = 120∠120° Vrms Z A = Z B = Z C = 10 + j ( 2 × π × 60 ) (100 × 10−3 ) = 10 + j 37.7 Ω and Z aA = Z bB = Z cC = 2 Ω Mathcad Analysis (12p4_2.mcd): 12-11
• Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ − 120 π j⋅ 180 Vc := Va⋅ e ⋅ 120 Describe the three-phase load: ZA := 10 + j⋅ 37.7 ZB := ZA ZC := ZB Describe the three-phase line: ZaA := 2 ZbB := ZaA ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e VnN := 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 15 VnN = −8.693 × 10 − 14 Calculate the line currents: IaA = 0.92 − 2.89i IaA := VnN = 2.396 × 10 Va − VnN ZA + ZaA π 180 π − 15 Calculate the phase voltages at the load: VA = 118.301 180 π π Vc − VnN ZC + ZcC IcC = 2.043 + 2.242i 180 π ⋅ arg( IcC) = 47.656 − 15 − 3.109i× 10 VA := ZA ⋅ IaA 180 IcC := IcC = 3.033 VB = 118.301 ⋅ arg( VA) = 2.801 ⋅ arg( VnN) = 111.277 ZB + ZbB ⋅ arg( IbB) = 167.656 IaA + IbB + IcC = −1.332 × 10 π Vb − VnN IbB = 3.033 ⋅ arg( IaA ) = −72.344 Check: IbB := IbB = −2.963 + 0.648i IaA = 3.033 180 180 − 14 + 2.232i× 10 ⋅ Vp ⋅ arg( VB) = −117.199 VB := ZB⋅ IbB VC := ZC⋅ IcC VC = 118.301 180 π ⋅ arg( VC) = 122.801 Consequently, the line-to-line voltages at the source are: Vab = Va × 3∠30° = 120∠0°× 3∠30° = 208∠30° Vrms, Vbc = 208∠ − 120° Vrms and Vca = 208∠120° Vrms The line-to-line voltages at the load are: VAB = VA × 3∠30° = 118.3∠3°× 3∠30° = 205∠33° Vrms, Vbc = 205∠ − 117° Vrms and Vca = 205∠123° Vrms and the phase currents are I a = I aA = 10∠ − 72° A rms, I b = I bB = 3∠168° A rms and I c = I cC = 3∠48° A rms 12-12
• P12.4-3 Balanced, three-wire, Y-Y circuit: where Va = 10∠0° V = 7.07∠0° V rms, Vb = 7.07∠ − 120° V rms and Vc = 7.07∠120° V rms and Z A = Z B = Z C = 12 + j (16 )(1) = 12 + j16 Ω MathCAD analysis (12p4_3.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 10 2 ⋅0 j⋅ Vb := Va⋅ e Describe the balanced three-phase load: π 180 ⋅ − 120 j⋅ Vc := Va⋅ e ZA := 12 + j⋅ 16 π 180 ⋅ 120 ZB := ZA ZC := ZB Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e IaA := IaA = 0.212 − 0.283i IaA = 0.354 π + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC Calculate the line currents: 180 2 j⋅ ⋅ π 3 ⋅ arg( IaA ) = −53.13 Va − VnN ZA − 15 ⋅ Vp IbB := VnN = 1.675 × 10 Vb − VnN IbB = −0.351 − 0.042i IbB = 0.354 180 π ⋅ arg( IbB) = −173.13 Calculate the power delivered to the load:   SB := IbB⋅ IbB⋅ ZB SA := IaA ⋅ IaA ⋅ ZA SA = 1.5 + 2i Total power delivered to the load: SB = 1.5 + 2i IcC := ZB Vc − VnN ZC IcC = 0.139 + 0.325i IcC = 0.354 180 π ⋅ arg( IcC) = 66.87  SC := IcC⋅ IcC⋅ ZC SC = 1.5 + 2i SA + SB + SC = 4.5 + 6i 12-13
• Consequently (a) The rms value of ia(t) is 0.354 A rms. (b) The average power delivered to the load is P = Re {S} = Re {4.5 + j 6} = 4.5 W P12.4-4 Unbalanced, three-wire, Y-Y circuit: where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω, Z B = 40 + j ( 377 ) ( 40 × 10−3 ) = 40 + j 15.1 Ω Z C = 60 + j ( 377 ) ( 20 ×10−3 ) = 60 + j 7.54 Ω and Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω Mathcad Analysis (12p4_4.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 100 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ 120 j⋅ Vc := Va⋅ e π 180 ⋅ − 120 Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load: ZA := 20 + j⋅ ω⋅ 0.06 Describe the three-phase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZB := 40 + j⋅ ω⋅ 0.04 ZC := 60 + j⋅ ω⋅ 0.02 ZcC := ZaA 12-14
• Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) VnN = 12.209 − 24.552i Calculate the line currents: IaA := Va − VnN ZA + ZaA IaA = 2.156 − 0.943i IaA = 2.353 180 π 180 VnN = 27.42 IbB := π IcC := ZB + ZbB 180 π ⋅ arg( IbB) = 100.492 ( SA = 55.382 + 62.637i Total power delivered to the load: ZC + ZcC IcC = 1.244 Calculate the power delivered to the load:   IbB⋅ IbB IaA ⋅ IaA ⋅ ZA SB := ⋅ ZB SA := 2 2 ) Vc − VnN IcC = −0.99 − 0.753i IbB = 2.412 ⋅ arg( IaA ) = −23.619 ( ⋅ arg( VnN) = −63.561 Vb − VnN IbB = −0.439 + 2.372i ⋅ Vp ) SB = 116.402 + 43.884i 180 π ⋅ arg( IcC) = −142.741 SC :=  (IcC⋅ IcC) 2 ⋅ ZC SC = 46.425 + 5.834i SA + SB + SC = 218.209 + 112.355i The average power delivered to the load is P = Re {S} = Re {218.2 + j112.4} = 218.2 W P12.4-5 Balanced, three-wire, Y-Y circuit: where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = Z B = Z C = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω and Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω 12-15
• Mathcad Analysis (12p4_5.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 100 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ 120 π j⋅ 180 Vc := Va⋅ e ⋅ − 120 Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load: ZA := 20 + j⋅ ω⋅ 0.06 ZB := ZA ZC := ZA Describe the three-phase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 15 VnN = −8.982 × 10 − 14 Calculate the line currents: IaA := IaA = 1.999 − 1.633i IaA = 2.582 180 π − 14 + 1.879i× 10 ⋅ arg( IaA ) = −39.243 VnN = 2.083 × 10 Va − VnN ZA + ZaA IbB := 180 π ⋅ arg( VnN) = 115.55 Vb − VnN IcC := ZB + ZbB IbB = 0.415 + 2.548i π ⋅ arg( IbB) = 80.757 Vc − VnN ZC + ZcC IcC = −2.414 − 0.915i IbB = 2.582 180 ⋅ Vp IcC = 2.582 180 π ⋅ arg( IcC) = −159.243 Calculate the power delivered to the load: SA :=  (IaA⋅ IaA) ⋅ ZA 2 SA = 66.645 + 75.375i Total power delivered to the load: SB :=  (IbB⋅ IbB) ⋅ ZB 2 SB = 66.645 + 75.375i SC :=  (IcC⋅ IcC) ⋅ ZC 2 SC = 66.645 + 75.375i SA + SB + SC = 199.934 + 226.125i The average power delivered to the load is P = Re {S} = Re {200 + j 226} = 200 W 12-16
• P12.4-6 Unbalanced, three-wire, Y-Y circuit: where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = 4 + j ( 4 )(1) = 4 + j 4 Ω, Z B = 2 + j ( 4 )( 2 ) = 2 + j 8 Ω and Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω Mathcad Analysis (12p4_6.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 10 ⋅ − 90 j⋅ Vb := Vp⋅ e π 180 ⋅ 150 j⋅ Vc := Vp⋅ e π 180 ⋅ 30 Enter the frequency of the 3-phase source: ω := 4 Describe the three-phase load: ZA := 4 + j⋅ ω⋅ 1 ZB := 2 + j⋅ ω⋅ 2 ZC := 4 + j⋅ ω⋅ 2 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC VnN = 1.528 − 0.863i Calculate the line currents: IaA := IaA = −1.333 − 0.951i IaA = 1.638 180 π 180 VnN = 1.755 ⋅ arg( IaA ) = −144.495 Va − VnN ZA IbB := π Vb − VnN IbB = 0.39 + 1.371i IbB = 1.426 180 π ⋅ arg( VnN) = −29.466 ⋅ arg( IbB) = 74.116 IcC := ZB Vc − VnN ZC IcC = 0.943 − 0.42i IcC = 1.032 180 π ⋅ arg( IcC) = −24.011 12-17
• Calculate the power delivered to the load:   IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2 ( ) ( SA = 5.363 + 5.363i ) SC := SB = 2.032 + 8.128i Total power delivered to the load:  (IcC⋅ IcC) 2 ⋅ ZC SC = 2.131 + 4.262i SA + SB + SC = 9.527 + 17.754i The average power delivered to the load is P = Re {S} = Re {9.527 + j17.754} = 9.527 W P12.4-7 Unbalanced, three-wire, Y-Y circuit: where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = Z B = Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω Mathcad Analysis (12p4_7.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 10 ⋅ − 90 j⋅ Vb := Vp⋅ e π 180 ⋅ 150 j⋅ Vc := Vp⋅ e π 180 ⋅ 30 Enter the frequency of the 3-phase source: ω := 4 Describe the three-phase load: ZA := 4 + j⋅ ω⋅ 2 ZB := ZA ZC := ZA The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC − 15 VnN = 1.517 × 10 12-18
• Calculate the line currents: IaA := IaA = −1 − 0.5i π IbB := ZA Vb − VnN 180 π ⋅ arg( IbB) = 86.565 SA = 2.5 + 5i Total power delivered to the load: ( ZC IcC = 1.118 Calculate the power delivered to the load:   IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2 ) Vc − VnN IcC = 0.933 − 0.616i IbB = 1.118 ⋅ arg( IaA ) = −153.435 ( IcC := ZB IbB = 0.067 + 1.116i IaA = 1.118 180 Va − VnN ) SB = 2.5 + 5i 180 π ⋅ arg( IcC) = −33.435 SC :=  (IcC⋅ IcC) 2 ⋅ ZC SC = 2.5 + 5i SA + SB + SC = 7.5 + 15i The average power delivered to the load is P = Re {S} = Re {7.5 + j15} = 7.5 W Section 12-6: The ∆- Connected Source and Load P12.5-1 Given I B = 50∠ − 40° A rms and assuming the abc phase sequence we have I A = 50∠80° A rms and I C = 50∠200° A rms From Eqn 12.6-4 I A = I AB × 3∠ − 30° ⇒ I AB = so IA 3∠ − 30° 50∠80° = 28.9∠110° A rms 3∠−30° = 28.9∠ − 10° A rms and ICA = 28.9∠ − 130° A rms I AB = I BC 12-19
• P12.5-2 The two delta loads connected in parallel are equivalent to a single delta load with Z ∆ = 5 || 20 = 4 Ω The magnitude of phase current is 480 Ip = = 120 A rms 4 The magnitude of line current is I L = 3 I p = 208 A rms Section 12-6: The Y- to ∆- Circuit P12.6-1 We have a delta load with Z = 12∠30° . One phase current is I AB  208   208  ∠−30°  − ∠−150°   V V −V 3   3  = 208∠0° = 17.31∠ − 30° A rms = AB = A A =  Z Z 12∠30° 12∠30° The other phase currents are I BC = 17.31∠ − 150° A rms and I CA = 17.31∠90° A rms One line currents is I A = I AB × 3∠ − 30° = (17.31∠ − 30° ) × ( ) 3∠ − 30° = 30∠0° A rms The other line currents are I B = 30∠ − 120° A rms and I C = 30∠120° A rms The power delivered to the load is P = 3( 208 ) (30) cos ( 0 − 30° ) = 9360 W 3 12-20
• P12.6-2 The balanced delta load with Z ∆ = 39∠− 40° Ω is equivalent to a balanced Y load with ZY = Z∆ = 13∠ − 40° = 9.96 − j 8.36 Ω 3 Z T = Z Y + 4 = 13.96 − j 8.36 = 16.3∠ − 30.9 Ω 480 ∠−30° 3 then I A = = 17∠0.9° A rms ° 16.3 ∠−30.9 P12.6-3 Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the given line-to-line voltage is Vab = 380 ∠30° V rms 380 ∠30° So one phase voltage is Va = = 200∠0° V rms 3∠30° So VAB = 380∠30° V rms VA = 220∠0° V rms VBC = 380∠-90° V rms VB = 220∠−120° V rms VCA = 380∠150° V rms VC = 220∠120° V rms One phase current is IA = VA 220∠0° = 44∠ − 53.1° A rms Z 3+ j4 The other phase currents are I B = 44∠−173.1° A rms amd I C = 44∠66.9° A rms 12-21
• P12.6-4 Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the given line-to-line voltage is Vab = 380 ∠0° V rms Va = So one phase voltage is So 380 ∠0° = 200∠ − 30° V rms 3∠30° Vab = 380∠0° V rms Va = 220∠ − 30° V rms Vbc = 380∠-120° V rms Vb = 220∠−150° V rms Vca = 380∠120° V rms Vc = 220∠90° V rms One phase current is IA = Va 220∠−30° = = 14.67∠ − 83.1° A rms Z 9 + j12 The other phase currents are I B = 14.67∠ − 203.1° A rms and I C = 14.67∠36.9° A rms 12-22
• 12-23
• Section 12-7: Balanced Three-Phase Circuits P12.7-1 Va = IA 25 ×103 ∠0° Vrms 3 25 ×103 ∠0° Va = = 3 = 96∠ − 25° A rms Z 150 ∠25°  25  ×103  96 cos(0 − 25°) = 3.77 mW P = 3 Va I A cos (θ v -θ I ) = 3   3  P12.7-2 Convert the delta load to an equivalent Y connected load: ˆ Z ∆ ⇒ Z Y = 50 Ω 3 To get the per-phase equivalent circuit shown to the right: The phase voltage of the source is Z ∆ = 50 Ω Va = 45×103 ∠0° = 26∠0° kV rms 3 The equivalent impedance of the load together with the line is 50 3 + 2 = 12 + j 5 = 13∠22.6° Ω Z eq = 50 10 + j 20 + 3 (10 + j 20 ) The line current is Ι aA = Va 26 × 103 ∠0° = = 2000∠ − 22.6° A rms Z eq 13∠22.6° The power delivered to the parallel loads (per phase) is 50    (10 + j 20 ) 3  2 PLoads = I aA × Re  = 4 ×106 × 10 = 40 MW 50  10 + j 20 +  3   The power lost in the line (per phase) is 12-24
• PLine = I aA × Re {Z Line } = 4 × 106 × 2 = 8 MW 2 The percentage of the total power lost in the line is PLine 8 × 100% = × 100% = 16.7% PLoad + PLine 40 +8 P12.7-3 Ia = Va 5∠30° = = 0.5∠ − 23° A ∴ I a = 0.5 A Z T 6 + j8 2 PLoad I = 3 a Re {Z Load } = 3 × 0.125 × 4 = 1.5 W 2 also (but not required) : PSource = 3 (5) (0.5) cos(−30 − 23) = 2.25 W 2 2 Pline I = 3 a Re{Z Line } = 3×0.125× 2 = 0.75 W 2 12-25
• Section 12-8: Power in a Balanced Load P12.8-1 Assuming the abc phase sequence: VCB = 208∠15° V rms ⇒ VBC = 208∠195° V rms ⇒ VAB = 208∠315° V rms Then VA = VAB 208∠315° 208 = = ∠285° V rms 3∠30° 3∠30° 3 also I B = 3∠110° A rms ⇒ I A = 3∠230° A rms Finally P = 3 VAB I A cos (θ V − θ I ) = 3( 208 ) (3) cos(285° − 230°) = 620 W 3 P12.8-2 Assuming a lagging power factor: cos θ = pf = 0.8 ⇒ θ = 36.9° The power supplied by the three-phase source is given by Pin = Pout η = Pin = 3 I A VA pf 20 ( 745.7 ) = 17.55 kW where 1 hp = 745.7 W 0.85 ⇒ Pin 17.55 ×103 IA = = = 26.4 A rms 3 VA pf  480  3  ( 0.8 )  3 480 ° I A = 26.4∠ − 36.9° A rms when VA = ∠0 V rms 3 12-26
• P12.8-3 (a) For a ∆-connected load, Eqn 12.8-5 gives PT 1500 = = 4.92 A rms 3 VP I L pf 3( 220 )(.8) 3 The phase current in the ∆-connected load is given by PT = 3 VP I L pf ⇒ I L = I IL 4.92 ⇒ IP = L = = 2.84 A rms 3 3 3 The phase impedance is determined as: IP = Z= V 220 VL VL = ∠ (θ V − θ I ) = L ∠ cos −1 pf = ∠ cos −1 0.8 = 77.44∠36.9° Ω IP IP IP 2.84 (b) For a ∆-connected load, Eqn 12.8-4 gives PT = 3 VP I L pf ⇒ I L = PT 1500 = = 4.92 A rms 3 VP I L pf 3( 220 )(.8) 3 The phase impedance is determined as: 220 VP VP V Z= P = ∠ (θ V − θ I ) = ∠ cos −1 pf = 3 ∠ cos −1 0.8 = 25.8∠36.9° Ω IP IP IP 4.92 P12.8-4 Parallel ∆ loads Z1Z 2 (40∠30° ) (50∠−60° ) Z∆ = = = 31.2 ∠−8.7° Ω ° ° Ζ1 + Ζ 2 40∠30 + 50∠− 60 VL = VP , Ι P = VP 600 = = 19.2 A rms, Z∆ 31.2 IL = 3 Ι P = 33.3 A rms So P = 3 VL I L pf = 3 (600) (33.3) cos ( − 8.7° ) = 34.2 kW 12-27
• P12.8-5 We will use In our case: S = S ∠θ = S cosθ + S sin θ = S pf + S sin ( cos −1 pf ) S1 = 39 (0.7) + j 39 sin ( cos −1 ( 0.7 ) ) = 27.3 + j 27.85 kVA 15 sin ( cos −1 ( 0.21) ) = 15 − j 69.84 kVA 0.21 S S3φ = S1 + S 2 = 42.3 − j 42.0 kVA ⇒ Sφ = 3φ = 14.1− j 14.0 kVA 3 The line current is S 2 = 15 + *  S  (14100+ j 14000) S = Vp I L ⇒ I L =   = = 117.5 + j 116.7 A rms = 167 ∠45° A rms V  208  p 3 208 ∠0° = 120∠0° V rms. The source must The phase voltage at the load is required to be 3 provide this voltage plus the voltage dropped across the line, therefore * VSφ = 120∠0° + (0.038 + j 0.072)(117.5 + j 116.7) = 115.9 + j 12.9 = 116.6 ∠6.4° V rms Finally VSφ = 116.6 V rms P12.8-6 The required phase voltage at the load is VP = 4.16 ∠0° = 2.402∠0° kVrms . 3 Let I1 be the line current required by the ∆-connected load. The apparent power per phase 500 kVA required by the ∆-connected load is S1 = = 167 kVA . Then 3 S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 167 ∠ cos−1 ( 0.85) = 167∠31.8° kVA and * 3  S1   (167 ×10 ) ∠31.8°   = 69.6∠ − 31.8° = 59 − j36.56 A rms ⇒ I1 =   =  3  VP   ( 2.402 ×10 ) ∠0°    * S1 = VP I1 * 12-28
• Let I2 be the line current required by the first Y-connected load. The apparent power per phase 75 kVA required by this load is S 2 = = 25 kVA . Then, noticing the leading power factor, 3 S 2 = S 2 ∠θ = S 2 ∠ cos −1 ( pf ) = 25 ∠ cos −1 ( 0 ) = 25∠ − 90° kVA and * 3  S 2   ( 25 ×10 ) ∠ − 90°    = 10.4∠90° = j10.4 A rms ⇒ I2 =   = VP   ( 2.402 ×103 ) ∠0°     * S 2 = VP I 2 * Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3 to be 2402∠0° 2402∠0° I3 = + = 16 − j 10.7 A rms 150 j 225 The line current is I L = I1 + I 2 + I 3 = 75− j 36.8 A rms 4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 2402∠0° + (8.45 + j 3.9) (75 − j 36.8) = 3179 ∠−0.3° Vrms Finally VSL = 3 (3179) = 5506 Vrms P12.8-7 The required phase voltage at the load is VP = 4.16 ∠0° = 2.402∠0° kVrms . 3 Let I1 be the line current required by the ∆-connected load. The apparent power per phase 1.5 MVA required by the ∆-connected load is S1 = = 0.5 MVA . Then 3 S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 0.5 ∠ cos −1 ( 0.75) = 0.5∠41.4° MVA and * 6  S1   ( 0.5 ×10 ) ∠41.4°    = 2081.6∠ − 41.4° = 1561.4 − j1376.6 A rms ⇒ I1 =   = VP   ( 2.402 ×103 ) ∠0°     * S1 = VP I1 * 12-29
• Let I2 be the line current required by the first Y-connected load. The complex power, per phase, is 0.67 S 2 = 0.67 + sin ( cos −1 ( 0.8 ) ) = 0.67 + j 0.5 MVA 0.8 * 6  S 2   ( 0.67 + j 0.5 ) ×106   ( 0.833 ×10 ) ∠ − 36.9°   =  I2 =   =  3 3  VP   ( 2.402 ×10 ) ∠0°   ( 2.402 ×10 ) ∠0°      = 346.9∠ − 36.9° = 277.4 − j 208.3 A rms The line current is I L = I1 + I 2 = 433.7 − j 345.9 = 554.7∠ − 38.6 A rms * * 4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 2402∠0° + (0.4 + j 0.8) (433.7 − j 345.9) = 2859.6 ∠ − 38.6° Vrms Finally VSL = 3 (2859.6) = 4953 Vrms The power supplied by the source is PS = 3 (4953) (554.7) cos (4.2° + 38.6° ) = 3.49 MW The power lost in the line is PLine = 3 × ( 554.7 2 ) × Re {0.4+ j 0.8} = 0.369 MW The percentage of the power consumed by the loads is 3.49 − 0.369 ×100% = 89.4% 3.49 12-30
• P12.8–8 The required phase voltage at the load is VP = 600 ∠0° = 346.4∠0° Vrms . 3 θ = cos −1 (0.8) = 37 ° Let I be the line current required by the load. The complex power, per phase, is S = 160 + 160 sin ( cos −1 ( 0.8 ) ) = 160 + j 120 kVA 0.8 The line current is  S   (160 + j 120 ) × 103  I=  =  = 461.9 − j 346.4 A rms 346.4∠0°  VP    * * 600 ∠0° = 346.4∠0° Vrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 346.4∠0° + (0.005 + j 0.025) (461.9 − j 346.4) = 357.5 ∠1.6° Vrms Finally VSL = 3 (357.5) = 619.2 Vrms The power factor of the source is pf = cos (θ V − θ I ) = cos (1.6 − ( − 37)) = 0.78 12-31
• Section 12-9: Two-Wattmeter Power Measurement P12.9-1 W = 14920 W hp P 14920 Pin = out = = 20 kW 0.746 η Pout = 20 hp × 746 Pin = 3 VL I L cos θ Pin 20 × 103 = = 0.50 3 VL I L 3 (440) (52.5) ⇒ cos θ = ⇒ θ cos -1 ( 0.5 ) = 60° The powers read by the two wattmeters are P = VL I L cos (θ + 30° ) = (440) (52.5)cos ( 60° + 30° ) = 0 1 and P2 = VL I L cos (θ − 30° ) = (440) (52.5)cos ( 60° − 30° ) = 20 kW P12.9-2 VP = VL = 4000 V rms IP = VP 4000 = = 80 A rms Z∆ 50 Z ∆ = 40 + j 30 = 50 ∠36.9° Ι L = 3 I P = 138.6 A rms pf = cos θ = cos (36.9° ) = 0.80 P1 = VL I L cos (θ + 30° ) = 4000 (138.6) cos 66.9° = 217.5 kW P2 = VL I L cos (θ −30° ) = 4000 (138.6) cos 6.9° = 550.4 kW PT = P1 + P2 = 767.9 kW Check : PT = 3 Ι L VL cos θ = 3 (4000) (138.6) cos 36.9° = 768 kW which checks 12-32
• P12.9–3 Vp = Vp = 200 = 115.47 Vrms 3 VA =115.47∠0° V rms, VB = 115.47∠−120° V rms and VC = 115.47∠120° V rms IA = VA 115.47∠0° = = 1.633∠ − 45° A rms Z 70.7∠45° I B = 1.633 ∠ − 165° A rms and I C = 1.633 ∠75° A rms PT = 3 VL I L cos θ = 3 (200) (1.633) cos 45° = 400 W PB = VAC I A cos θ1 = 200 (1.633) cos (45° − 30° ) = 315.47 W PC = VBC I B cos θ 2 = 200 (1.633) cos (45° + 30° ) = 84.53 W P12.9-4 ZY = 10∠ − 30° Ω and Z ∆ = 15∠30° Ω Convert Z ∆ to Z Y → Z Y = ˆ ˆ then Zeq = Z∆ = 5∠30° Ω 3 (10∠−30° ) ( 5∠30° ) = 10∠−30°+5∠30° 208 Vp = Vp = = 120 V rms 3 VA = 120∠0° V rms ⇒ I A = 50∠0° = 3.78∠10.9° Ω 13.228 ∠−10.9° 120∠0° = 31.75 ∠−10.9° 3.78 ∠10.9° I B = 31.75∠−130.9° I C = 31.75∠109.1° PT = 3VL I L cos θ = 3 ( 208 ) ( 31.75 ) cos (10.9 ) =11.23 kW W1 = VL I L cos (θ −30°) = 6.24 kW W2 = VL I L cos (θ + 30°) = 4.99 kW 12-33
• P12.9-5 PT = PA + PC = 920 + 460 = 1380 W tan θ = 3 ( −460 ) = −0.577 ⇒ θ = −30° PA − PC = 3 1380 PA + PC PT = 3 VL I L cos θ so I L = IP = P12.9-6 IL = 4.43 A rms 3 ∴ Z∆ = Z = 0.868 + j 4.924 = 5∠80° VL = 380 V rms, VP = I L = I P and I P = 1380 PT = =7.67 A rms 3 VL cos θ 2 ×120×cos( −30 ) ⇒ 120 = 27.1 Ω ο r Z ∆ = 27.1 ∠−30° 4.43 θ = 80° 380 = 219.4 V rms 3 VP = 43.9 A rms Z P = ( 380 ) ( 43.9 ) cos (θ −30° ) = 10,723 W 1 P2 = ( 380 ) ( 43.9 ) cos (θ + 30° ) = −5706 W PT = P + P2 = 5017 W 1 12-34
• PSpice Problems SP 12-1 FREQ 6.000E+01 IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01 FREQ 6.000E+01 IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 3.142E+00 -4.443E+01 2.244E+00 -2.200E+00 FREQ 6.000E+01 VM(N01496) 2.045E-14 FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 3.142E+00 7.557E+01 7.829E-01 3.043E+00 VP(N01496) 2.211E+01 VR(N01496) 1.895E-14 VI(N01496) 7.698E-15 3.1422 20 = 98.7 W 2 3.1422 I B = 3.142∠75.57° A and RB = 20 Ω ⇒ PB = 20 = 98.7 W 2 3.1422 I C = 3.142∠ − 164.4° A and RC = 20 Ω ⇒ PC = 20 = 98.7 W 2 I A = 3.142∠ − 43.43° A and RA = 20 Ω ⇒ PA = P = 3 ( 98.7 ) = 696.1 W 12-35
• SP 12-2 FREQ 6.000E+01 IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00 FREQ 6.000E+01 IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 2.537E+00 -3.748E+01 2.013E+00 -1.544E+00 FREQ 6.000E+01 VM(N01496) VP(N01496) 1.215E+01 -1.439E+01 FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 2.858E+00 1.084E+02 -9.023E-01 2.712E+00 VR(N01496) VI(N01496) 1.177E+01 -3.018E+00 2.537 2 20 = 64.4 W 2 2.8582 I B = 2.858∠108.4° A and RB = 30 Ω ⇒ PB = 30 = 122.5 W 2 1.6122 I C = 1.612∠ − 133.6° A and RC = 600 Ω ⇒ PC = 60 = 78 W 2 I A = 2.537∠ − 37.48° A and RA = 20 Ω ⇒ PA = P = 64.4 + 122.5 + 78 = 264.7 V 12-36
• Verification Problems VP 12-1 416 = 240 V = VA 3 Z = 10 + j4 = 10.77 ∠21.8° Ω VA = VA 240 = = 22.28 A rms ≠ 38.63 A rms Z 10.77 38.63 = 22.3 . It appears that the line-to-line voltage was The report is not correct. (Notice that 3 mistakenly used in place of the phase voltage.) IA = VP 12-2 VL = VP = 240∠0° Vrms Z = 40 + j 30 = 50 ∠36.9° Ω IP = VP 240∠0° = = 4.8 ∠−36.9° A rms Z 50∠36.9° The result is correct. Design Problems DP 12-1 P = 400 W per phase, 0.94 = pf = cos θ ⇒ θ = cos-1 ( 0.94 ) =20° 208 I L 0.94 ⇒ I L = 3.5 A rms 3 I I ∆ = L = 2.04 A rms 3 V 208 = 101.8 Ω Z = L = 2.04 I∆ 400 = Z = 101.8 ∠20° Ω 12-37
• DP 12-2 VL = 240 V rms PA = VL I L cos (30° + θ ) = 1440 W PC = VL I L cos (30° − θ ) = 0 W ⇒ 30−θ = 90° or θ = −60° then 1440 = 240 I L cos (−30° ) I L = 6.93 A rms IL = IP = VP Z ⇒ ⇒ 240 V Z = P = 3 = 20 Ω IP 6.93 Finally, Z = 20 ∠ − 60° Ω DP 12-3 Pin = Pout η 100 hp × (746 = W ) hp 0.8 = 93.2 kW, P = φ Pin = 31.07 kW 3 VL = 480 V rms, pfc = 0.9 and pf = 0.75. We need the impedance of the load so that we can use Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor. 0.75 = pf = cos θ ⇒ θ = cos-1 ( 0.75) = 41.4° 480 I P 0.75 ⇒ I P = 149.5 A rms 3 480 VP 3 = 1.85 Ω Z = = IP 149.5 31070 = Z = 1.85 ∠41.4° Ω = 1.388 + j1.223 Ω The capacitance required to correct the power factor is given by  tan (cos −1 0.75) − tan (cos −1 0.9)  1.365  = 434 µ F C= × 2 2 1.365 +1.204 377 (Checked using LNAPAC 6/12/03) 12-38
• DP 12-4 VL = 4∠0° kV rms n2 25 VL = 4000∠0° = 100∠0° kVrms n1 1 Try n2 = 25 then V2 = VL 4×103∠0° = 3∠0° kA rms = IL = 4 ZL 3 3000∠0° The line current in 2.5 Ω is I = = 120∠0° A rms 25 Thus V1 = ( R + j X ) I + V2 = (2.5 + j 40) (120∠0°) + 100×103 = 100.4 ∠2.7° kV Step need : n1 = Ploss = I 2 100.4 kV = 5.02 ≅ 5 20 kV R = 120 2 (2.5) = 36 kW, P = (4×103 ) (3× 103 ) = 12 MW 12 − .036 × 100% = 99.7 % of the power supplied by the source 12 is delivered to the load. ∴η = 12-39
• Chapter 13: Frequency Response Exercises Ex. 13.3-1 Vo (ω ) 1 = Vs (ω ) 1 + jω C R 1 H (ω ) = gain = 1 + (ω C R) 2 phase shift = − tan −1 ω C R 1 When R = 104 , ω = 100, and C = 10−6 , then gain = = 0.707 and phase shift = − 45o 2 Ex. 13.3-2 H (ω ) = gain = Vo (ω ) R = Vs (ω ) R + jω L R R 2 + (ω L) 2 2 30 0.6 = 30 + (2ω ) 2 2  30  2   − 30  .6  ⇒ ω= = 20 rad s 2 Ex. 13.3-3 H (ω ) = gain = I (ω ) 1 = R + jω L Vs (ω ) 1 R 2 + (ω L) 2 phase shift = − tan −1 ωL R When R = 30 Ω, L = 2 H, and ω = 20 rad/s, then gain = 1 30 + 40 2 2 = 0.02 A  40  and phase shift = − tan −1   = − 53.1° V  30  13-1
• Ex. 13.3-4 H (ω ) = Vo (ω ) 1 = Vs (ω ) 1 + jω C R gain = 1 1 + (ω C R) 2 phase shift = − tan −1 ω C R −45° = − tan −1 (20 ⋅10−6 ⋅ R ) ⇒ R= tan (45° ) = 50 ⋅103 Ω −6 20⋅10 Ex. 13.3-5 H (ω ) = gain = Vo (ω ) 1 = 1 + jω C R Vs (ω ) 1 1 + (ω C R) 2 ω , C , and R are all positive, or at least nonnegative, so gain ≤ 1. These specifications cannot be met. Ex. 13.4-1 (a) dB = 20 log (.5) = −6.02 dB (b) dB = 20 log 2 = 6.02 dB Ex. 13.4-2  1  20 log H = 20 log  2  = 20 log (ω )-2 = −40 log ω ω  ω  slope = 20 log H (ω 2 ) − 20 log H (ω1 ) = 40 log ω 2 + 40 log ω1 = -40 log 2   ω1  let ω 2 = 10 ω1 to consider 1 decade, then slope = − 40 log10 = − 40 dB decade 13-2
• Ex. 13.4-3 When ω C >> B, H (ω ) − (d ) (b) jω A A = jω C C  A H (ω ) in dB = 20 log10 H (ω ) = 20 log10   C H (ω ) does not depend on ω so slope = 0 When ω C << B, H (ω ) − jω A  A = jω   B  B  A H (ω ) in dB = 20 log10 H (ω ) = 20 log10ω +20 log10    B (c) The slope is the coefficient of 20 log10 ω , that is, slope = 20 dB decade B (a) The break frequency is the frequency at which ω C = B, that is, ω = C Ex. 13.4-4  R  Vo (ω ) =  1+ 1  Vc (ω )  R2   R   1 =  1+ 1    Vs (ω )  R2   1 + jω C R  H (ω ) =  Vo (ω )  R1  1 =1+   Vs (ω )  R2  1+ jω C R  When R C = 0.1 and then H (ω ) = 4 1+j R1 = 3, R2 ω 10 13-3
• Ex. 13.4-5 a) Zo = R2 + 1 jω C ω Vo Zo ω1 jω C = = = 1 ω R1 + Z o Vs R1 + R2 + 1+j jω C ω2 R2 + where ω1 = and ω 2 = 1 R2 C 1 1+ j = 16.7 rad/s 1 = 5.56 rad/s ( R1 + R2 )C vs ( t ) = 10 cos 20 t or Vs = 10∠0° ( ( ) ) 1+ j 20 Vo 16.7 ∴ = 20 Vs 1+ j 5.56 1+ j 1.20 = = 0.417 ∠− 24.3° 1+ j 3.60 b) So Vo = 4.17 ∠ − 24.3° vo (t ) = 4.17 cos(20t − 24.3°) V 13-4
• Ex. 13.4-6 Ex. 13.5-1 a) Q ωo RC = R C b) BW = ωo Q = 2.5 × 10−7 = 8000 = 20 L 40 × 10−3 1 1 = = 500 rad s −3 −7 Q LC 20 (40×10 ) (2.5×10 ) Ex. 13.5-2 Q= ω0 BW Now ωo = 7 = 10 1 LC 2 × 105 = 50 ⇒ L = 1 2 ωo C = 1 = 1 mH (10 ) (10 × 10−12 ) 7 2 Ex. 13.5-3 ωo = 1 Q = ωo LC BW 1 =  (10−3 )(10−5 )    4 = 10 2π (15.9) 1 = 104 rad s 2 = 100 (104 )(10−3 ) R= = = 0.1 Ω Q 100 ωo L 13-5
• Ex. 13.5-4 a) ωo = 1 ω Q= o b) Q = H= H= ωo ⇒ C = LC BW 1 (106 ) 2 (0.01) BW 103 = = 10 Ω (106 ) 2 (10−10 ) C BW = 1 ωo RC ⇒ R = ω 2 o 6 = 10 = 100 pF 103 1 ω ω  1+ j Q − o   ωo ω  = = 1000 1 1.05×106 106  − 1+ j 1000   6 1.05×104   10   1 1+ j 97.6 13-6
• Problems Section 13-3: Gain, Phase Shift, and the Network Function P13.3-1 R 2 || H (ω ) = 1 jω C = Vo (ω ) = Vi (ω ) R2 1 + jω C R 2 R2 1 + jω C R 2 R2 R1 + 1 + jω C R 2 R2 = 1 + jω C R p where Rp = R1 || R2. When R1 = 40 W, R2 = 10 W and C = 0.5 F H (ω ) = R1 + R 2 0.2 1 + j 4ω (checked using ELab on 8/6/02) P13.3-2 H (ω ) = 1 R2 + Vo (ω ) jω C = Vi (ω ) R + R + 1 1 2 jω C 1 + jω C R 2 = 1 + jω C R1 + R 2 ( ) When R1 = 40 kW, R2 = 160 kW and C = 0.025 µF H (ω ) = 1 + j ( 0.004 ) ω 1 + j ( 0.005 ) ω (checked using ELab on 8/6/02) 13-7
• P13.3-3 H (ω ) = R2 Vo (ω ) = Vi (ω ) R1 + R 2 + jω L R2 = R1 + R 2 L 1 + jω R1 + R 2 When R1 = 4 W, R2 = 6 W and L = 8 H H (ω ) = 0.6 1 + j ( 0.8 ) ω (checked using ELab on 8/6/02) P13.3-4 H (ω ) = R 2 + jω L Vo (ω ) = Vi (ω ) R + R 2 + jω L L   1 + jω R  R2  2 =   R + R2  L    1 + jω R + R2        Comparing the given and derived network functions, we require L   1 + jω R  R2  2    R + R2  L    1 + jω R + R2  Since R2 = 60 W, we have L =   1+  = ( 0.6 )  1+   j j ω 12 ω 20  R2 = 0.6   R + R2  R  2 ⇒  = 12  L  R + R2 = 20  L   60 = 5 H , then R = ( 20 )( 5 ) − 60 = 40 Ω . 12 (checked using ELab on 8/6/02) 13-8
• P13.3-5 R 2 || H (ω ) = 1 jω C = R2 1 + jω C R 2 Vo (ω ) = Vi (ω ) R2 1 + jω C R 2 R2 R+ 1 + jω C R 2 R2 = R + R2 1 + jω C R p where Rp = R || R2. Comparing the given and derived network functions, we require R2 R + R2 0.2 = 1 + jω C R p 1 + j 4 ω Since R2 = 2 W, we have Finally, C =  R2 = 0.2  ⇒  R + R2  CR =4 p  ( 2 )(8 ) = 1.6 Ω . 2 = 0.2 ⇒ R = 8 Ω . Then R p = R+2 2+8 4 = 2.5 F . 1.6 (checked using ELab on 8/6/02) P13.3-6 Vi (ω )   R + jω L   ⇒ 1 Vo (ω ) = ( A I a (ω ) )   jω C  I a (ω ) = A Vo (ω ) CR = L Vi (ω ) ( j ω ) 1 + j ω    R  13-9
• When R = 20 W, L = 4 H, A = 3 A/A and C = 0.25 F H (ω ) = 0.6 ( j ω ) (1 + j ( 0.2 ) ω ) (checked using LNAP on 12/29/02) P13.3-7 In the frequency domain, use voltage division on the left side of the circuit to get: 1 1 jω C VC (ω ) = Vi (ω ) = Vi (ω ) 1 1 + jω C R1 R1 + jω C Next, use voltage division on the right side of the circuit to get: 2 A R3 2 3 Vo (ω ) = A VC (ω ) = A VC (ω ) = Vi (ω ) R 2 + R3 3 1 + jω C R1 Compare the specified network function to the calculated network function: 2 2 A A 2 1 3 3 = = ⇒ 4 = A and = 2000 C ω 1 + jω C R1 1 + jω C 2000 3 100 1+ j 100 4 Thus, C = 5 µF and A = 6 V/V. (checked using ELab on 8/6/02) 13-10
• P13.3-8 H (ω ) = Vo (ω ) =− Vi (ω ) R2 1 jω C R1  R2  −  R1     = 1+ jω C R2 When R 1 = 10 kΩ, R 2 = 50 kΩ , and C = 2 µ F, then R2 R1 = 5 and R2 C = 1 −5 so H (ω ) = ω 10 1+ j 10 P13.3-9 H (ω ) = Vo (ω ) =− Vi (ω ) R2 1 j ω C2 R1 1 j ω C1 R2 1 + jω C2 R2 = − R1 1 + jω C1R1  R   1 + jω C1R1  H (ω ) = −  2     R1   1 + jω C2 R2  When R1 = 10 kΩ, R2 = 50 kΩ , C1 = 4 µ F and C2 = 2µ F, then so R2 R1 = 5 , C1R1 = 1 1 and C2 R 2 = 25 10 ω    1 + j 25  H (ω ) = − 5  ω   1+ j    10   13-11
• gain = H(ω ) = ( 5 ) 1+ 1+ ω2 625 ω2 100 ω  ω  phase shift = ∠Η (ω ) = 180 + tan −1   − tan −1    25   10  P13.3-10 1 R1 jω C = 1 1 + jω C R3 R3 + jω C R3 1 = jω C R3 H (ω ) = − R2 + R3 1+ jω C R3 R + R + jω R2 R3C =− 2 3 R1 R1 + jω R1 R3C 5 = lim H (ω ) = R2 + R3 R1 2 = lim H (ω ) = R2 ⇒ R2 = 2 R1 = 20 kΩ R1 ω →0 ω →∞ then R3 = 5 R1 − R2 = 30 kΩ P13.3-11 H (ω ) = − R2 + 1 jω C R1 =− 1 + jω C R 2 jω C R1 ∠H (ω ) = 180° + tan −1 ω C R 2 − 90° ∠H (ω ) = 135° ⇒ tan −1 ω CR2 = 45° ⇒ ω C R 2 = 1 ⇒ R2 = 10 = lim H (ω ) = ω →∞ 1 = 10 kΩ 10 10−7 3 R2 R ⇒ R1 = 2 = 1 kΩ R1 10 13-12
• P13.3-12 −R2 jω C R 2 = − 1 1+ jω C R1 jω C R 10 = lim H(ω ) = 2 ⇒ R2 = 10 R1 R1 ω →∞ H (ω ) = ∠ H(ω ) = 180°+90°− tan −1 ω C R1 ⇒ R1 = tan (270°−∠ H(ω )) = 104 ⋅tan(270°−∠H(ω )) = 104 = 10 kΩ ωC ⇒ R2 = 100 kΩ P13.3-13 1 jω C2 V (ω ) =− H (ω ) = o 1 Vs (ω ) R1 + jω C1 R2 ( −C1R2 ) jω = (1 + jω R1C1 ) (1 + jω R2C2 ) When R1 = 5 kΩ, C1 = 1 µ F, R2 = 10 kΩ and C2 = 0.1 µ F, then ( −0.01) jω H (ω ) = ω   ω   1 + j  1 + j  200   1000   so ω 0 500 2500 H (ω ) 0 1.66 0.74 ∠H (ω ) −90° 175° 116° Then v ( t ) = (0) 50 + (1.66) ( 30 ) cos(500t + 115° + 175°) − (0.74) ( 20 ) cos(2500t + 30° + 116°) o =49.8cos(500t − 70°) −14.8cos (2500t +146°) mV When R1 =5 kΩ, C1 =1 µ F, R 2 =10 kΩ and C2 = 0.01 µ F, then 13-13
• H (ω )= − 0.01 jω ω  ω    1+ j   1+ j 200   10,000    So ω 0 500 2500 H (ω ) 0 1.855 1.934 ∠H (ω ) −90° −161° 170° Then v (t ) = (0) ( 50 ) + (1.855) ( 30 ) cos(500t + 115° − 161°) − (1.934) ( 20 ) cos(2500t + 30° + 170°) o = 55.65 cos(500t − 46°) − 38.68cos(2500t + 190°) mV P13.3-14 a) 2 V (8 div)    div  = 8 V Vs = 2 2 V (6.2 div)    div  = 6.2 V Vo = 2 V 6.2 gain = o = = 0.775 8 Vs b) 1 H (ω ) = Vo (ω ) 1 jω C = = 1 Vs (ω ) 1 + jω C R R+ jω C 1 1 Let g = H (ω ) = then C = ωR 1 + ω 2C 2 R 2 2 1   −1 g In this case ω = 2π ⋅ 500 = 3142 rad s , H (ω ) = 0.775 and R = 1000 Ω so C = 0.26 µ F. c) tan( −∠ H (ω )) RC Recalling that R = 1000 Ω and C=0.26µF, we calculate ∠ H (ω )= − tan −1 ω R C so ω = 13-14
• ω H (ω ) 0.95 0.26 2π (200) 2π (2000) ∠H (ω ) −18° −73° ( ) tan  − −45°      = 3846 rad s ∠ H (ω )= − 45° requires ω = 1000 .26×10−6 ( ∠ Η (ω ) = −135° requires ω = ) ( ) tan ( −(−135°)) = − 3846 rad s (1000)(0.26×10−6 ) A negative frequency is not acceptable. We conclude that this circuit cannot produce a phase shift equal to −135 . ° d)  tan (−(60° )) = 0.55µ F C=  (2π ⋅ 500) (1000) tan (−∠H (ω ))  ⇒  C= ° ωR C = tan (−(−300 )) = −0.55µ F  (2π ⋅ 500 ) (1000)  A negative value of capacitance is not acceptable and indicates that this circuit cannot be designed to produce a phase shift at −300 at a frequency of 500 Hz. ° e) tan( − (−120° )) C = = −0.55 µ F (2π ⋅ 500)(100) This circuit cannot be designed to produce a phase shift of −120 at 500 Hz. ° 13-15
• Section 13-4: Bode Plots P13.4-1 ω  20  1 + j  5 H (ω )=  ω  1 + j  50   P13.4-2 H1 (ω ) = 1+ j 1+ j    20     20  j ω     ≈  5     ω  20  j 5   = 200    jω   50     ω <5 5 < ω < 50 50 < ω ω 5 ω H 2 (ω ) = 10 50 1+ j 1+ j ω 5 ω 50 Both H1(ω) and H2(ω) have a pole at ω = 50 rad/s and a zero at ω = 5 rad/s. The slopes of both magnitude Bode plots increase by 20 dB/decade at ω = 5 rad/s and decrease by 20 dB/decade at ω = 50rad/s. The difference is that for ω < 5rad/s H1 (ω ) 1 = 0 dB and H 2 (ω ) 10 = 20 dB 13-16
• P13.4-3 R2 jω 1+ jω C2 R2 H (ω ) = − = −C1 R2 1 (1+ jω R1C1 )(1+ jω R2C2 ) R1 + jω C1 This network function has poles at p1 = 1 1 = 2000 rad s and p2 = = 1000 rad s R1C1 R2C2 so H (ω )  (C R ) jω  1 2  jω R = 2 =2 (C1 R2 ) jω C1 R1 R1   1 jω = (C1 R2 ) ( jω C1 R1 )( jω C2 R2 ) jω C2 R1  ω < p1 p1 < ω < p2 ω > p2 13-17
• P13.4-4 R2 R (1+ jω C1 R1 ) R 1 1 1+ jω C2 R2 H (ω ) = − =− 2 so K = − 2 , z = and p = R1 R1 (1+ jω C2 R2 ) R1 C1 R1 C2 R2 1+ jω C1 R1 When z < p When z > p P13.4-5 Using voltage division twice gives: V2 (ω ) = Vi (ω ) jω L R 2 R 2 + jω L jω L R 2 L jω = = jω L R 2 R1 R 2 + jω L ( R1 + R 2 ) R1 L ( R1 + R 2 ) R1 + 1 + jω R 2 + jω L R1 R 2 and Vo (ω ) = V2 (ω ) R4 A R4 R 3 + jω C R 4 A R4 R3 + R 4 A= = R4 C R3 R 4 R3 + R 4 + jω C R3 R 4 R3 + 1 + jω R 3 + jω C R 4 R3 + R 4 Combining these equations gives 13-18
• H (ω ) = ALR 4 Vo (ω ) jω = Vi (ω ) R1 ( R 3 + R 4 )  L ( R1 + R 2 )   CR 3 R 4  1 + jω   1 + jω    R1 R 2 R3 + R 4      The Bode plot corresponds to the network function: H (ω ) = k jω k jω = ω  ω   ω  ω   1 + j  1 + j  1 + j 200   1 + j 20000    p1   p2        k jω = k jω  1 ⋅1   k jω H (ω ) ≈  = k p1 jω  ⋅1 p1   k jω k p1 p2  = jω  jω ⋅ jω  p p  1 2 ω ≤ p1 p1 ≤ ω ≤ p2 ω ≥ p2 This equation indicates that |H(ω)|=k p1 when p1 ≤ ω ≤ p2. The Bode plot indicates that |H(ω)|=20 dB = 10 when p1 ≤ ω ≤ p2. Consequently k= Finally, H (ω ) = 10 10 = = 0.05 p1 200 0.05 jω ω  ω   1 + j  1 + j  200   20000   Comparing the equation for H(ω) obtained from the circuit to the equation for H(ω)obtained from the Bode plot gives: 0.05 = ALR 4 R1 ( R 3 + R 4 ) , 200 = R1 R 2 L ( R1 + R 2 ) and 20000 = R3 + R 4 C R3 R 4 Pick L = 1 H, and R1 = R2 , then R1 = R2 = 400 Ω. Let C = 0.1 µF and R3 = R4 , then R3 = R4 = 1000 Ω. Finally, A=40. (Checked using ELab 3/5/01) 13-19
• P13.4-6 From Table 13.4-2: R2 R1 = k = 32 dB = 40 R 2 = 40 (10 × 103 ) = 400 kΩ 1 1 = p = 400 rad/s ⇒ C 2 = = 6.25 nF C 2 R2 ( 400 ) ( 400 ×103 ) 1 1 = z = 4000 rad/s ⇒ C 1 = = 25 nF C 1 R1 ( 4000 ) (10 ×103 ) P13.4-7 H (ω ) = R 2 + jω L Vo (ω ) = Vi (ω ) R + R 2 + jω L L   1 + jω R  R2  2 =   R + R2  L    1 + jω R + R2  H (ω ) = ( 0.2 ) (1 + j ( 0.25 ) ω ) 1 + j ( 0.05 ) ω         k = 0.2  1  ⇒  z= =4 0.25  1   p = 0.05 = 20  P13.4-8 • The slope is 40dB/decade for low frequencies, so the numerator will include the factor (jω)2 . • The slope decreases by 40 dB/decade at ω = 0.7rad/sec. So there is a second order pole at ω 0 = 0.7 rad/s. The damping factor of this pole cannot be determined from the asymptotic Bode plot; call it δ 1. The denominator of the network function will contain the factor 1 + 2 δ1 j • ω  ω  −  0.7  0.7  2 The slope increases by 20 dB/decade at ω = 10 rad/s, indicating a zero at 10 rad/s. 13-20
• • • The slope decreases by 20 dB/decade at ω = 100 rad/s, indicating a pole at 100 rad/s. The slope decreases by 40 dB/decade at ω = 600 rad/s, indicating a second order pole at ω 0 = 600rad/s. The damping factor of this pole cannot be determined from an asymptotic Bode plot; call it δ 2. The denominator of the network function will contain the factor ω  ω  − 1 + 2δ 2 j  600  600  H (ω ) = K (1+ j ω 10 )( jω ) 2 2 2  ω  ω   ω  ω   ω  − −  1+ 2δ1 j  1+ 2δ 2 j   1+ j   0.7  0.7   600  600   100     To determine K , notice that H (ω ) = 0 dB=1 1= P13.4-9 (a) 2 K (1)( jω ) 2 2  ω  −  (1)(1)  0.7  when 0.7 < ω < 10. That is = K (0.7) 2 ⇒ K = 2 ω  K 1+ j  z  H (ω ) = jω H (ω ) = K ω ω  1+   z H (ω ) dB = 20 log10 K ω 2 ω  1+  z 2 = 20 log10 K − 20 log10 ω + 20 log10 Let ω  1+  z 2 H L (ω ) dB = 20 log10 K − 20 log10 ω K z  H (ω ) dB _  L Then H (ω ) dB ~   H H (ω ) dB  and H H (ω ) dB = 20 log10 < ω< z ω>z > 13-21
• So H L (ω ) dB and H H (ω ) dB are the required low and high-frequency asymptotes. The Bode plot will be within 1% of |H(ω)| dB both for ω << z and for ω >> z. The range when ω << z is characterized by H L (ω ) = 0.99 H (ω ) (gains not in dB) or equivalently 20 log10 ( 0.99 ) = H L (ω ) dB − H (ω ) dB = 20 log10 K − 20 log10 ω − 20 log10 2 = − 20 log10 ω  1+  = 20 log10 z (gains in dB) ω  1+  ω z K 2 1 ω  1+   z 2 Therefore 0.99 = 2 1 ω  1+  z 2 z  1  ⇒ ω= z   −1 = 0.14 z − 7  .99  The range when ω >> z is characterized by H H (ω ) = .99 H (ω ) (gains not in dB) or equivalently 13-22
• 20 log10 0.99 = H H (ω ) dB − H (ω ) dB (gains in dB) ω  = 20 log10 K − 20 log10 z − 20 log10 1+   ω z K 2 2 1 ω  = − 20 log10 1+   = 20 log10 2 ω z z   +1 ω  z Therefore 2  1  =   −1 ⇒ ω  .99  z The error is less than 1% when ω < ω= z 2  1    −1  .99  = z − 7z 0.14 z and when ω > 7z. 7 P13.4-10 H (ω ) = = V0 (ω ) = Vs (ω ) Rt R t + R1 1 jω C = Rt +  Rt = R1 + R t + jω C R1 R t  R1 + R t  R t (1+ jω C R1 ) Rt R1 1+ jω C R1  1+ jω C R1    1+ jω  C R1 R t   R1 + R t      When R1 = 1 kΩ, C = 1 µ F and R t = 5 kΩ ω    1+ j 1000  5 H (ω ) =   ⇒ 6  1+ j ω  1200   5 ω <1000 6   5  ω 1000<ω <1200 H (ω ) ≅   j  6  1000 1 ω <1200   13-23
• P13.4-11 Mesh equations: Vin (ω ) = I (ω ) [ R1 + ( jω L1 − jω M ) + (− jω M + jω L2 ) + R2 ] Vo (ω ) = I (ω ) [(− jω M + jω L2 ) + R2 ] Solving yields: H (ω ) = V0 (ω ) R2 + jω ( L2 − M ) = Vin (ω ) R1 + R2 + jω ( L1 + L2 − 2M ) Comparing to the given Bode plot yields: lim K1 = ω →∞ |H (ω )| = z = L2 − M R2 = 0.75 and K 2 = lim | H (ω ) | = = 0.2 ω →0 L1 + L2 − 2 M R1 + R2 R2 R1 + R2 = 333 rad s and p = =1250 rad s L2 − M L1 + L2 − 2 M 13-24
• P13.4-12 1 1+ jω R1 C1 1 (1+ jω R1 C1 ) jω C2 H (ω ) = − =− =− jω R1 C2 R1 C2 jω 1 R1 jω C1  1  1  −    R C jω  H (ω ) −  1 2   − 1 ( R C ) = − C1  R1 C2 1 1 C2  ω< ω> 1 R1 C1 1 R1 C1 With the given values: C1 1 = = −6 dB, C2 2 1 = 4000 rad / s R1 C1 13-25
• P13.4-13 Pick the appropriate circuit from Table 13.4-2. We require 200 = z = 1 1 p C , 500 = p = and 14 dB = 5 = k = 1 C 1 R1 C2 R 2 z C2 Pick C1 = 1 µ F, then C2 = 0.2 µ F, R1 = 5 kΩ and R 2 = 10 kΩ. P13.4-14 Pick the appropriate circuit from Table 13.4-2. We require 500 = p = R2 1 and 34 dB = 50 = C R2 R1 Pick C = 0.1 µ F, then R 2 = 20 kΩ and R1 = 400 Ω. 13-26
• P13.4-15 Pick the appropriate circuit from Table 13.4-2. We require 500 = z = 1 1 p C , 200 = p = and 14 dB = 5 = k = 1 C 1 R1 C2 R 2 z C2 Pick C1 = 0.1 µ F, then C2 = 0.05 µ F, R1 = 20 kΩ and R 2 = 100 kΩ. 13-27
• P13.4-16 Pick the appropriate circuit from Table 13.4-2. We require 200 = p 1 = 1 1 , 200 = p 2 = and 34 dB = 50 = k = C 1 R 2 C 1 R1 C2 R 2 Pick C1 = 1 µ F, then C2 = 0.04 µ F, R1 = 5 kΩ and R 2 = 50 kΩ. P13.4-17 H (ω ) = 10(1+ jω 50) (1+ jω 2)(1+ jω 20)(1+ jω 80) 13-28
• ϕ = ∠H (ω ) = tan −1 (ω 50 ) − ( tan −1 (ω 2 ) + tan −1 (ω 20 ) + tan −1 (ω 80 ) ) P13.4-18 (a) H (ω ) = Vo (ω ) R2 R1 =− Vs (ω ) 1+ jω R2C 10 =− 1+ j (b) 10,000 10 = 20 dB (c) ω 10,000 rad/s 13-29
• P13.4-19      ⇒ Vo (1 + jω C 1 R1 )(1 + jω C 2 R 2 ) = jω C 1 R1Vo + Vs   Va (ω ) − Vs (ω ) 0= + jω C 1 (Va (ω ) − Vo (ω ))  R1  1 jω C 2 Vo (ω ) = Va (ω ) 1 R+ jω C 2 T(ω ) = Vo (ω ) 1 1 = = 2 2 Vs (ω ) 1+ C 2 R 2 jω −ω C 1C 2 R1 R 2 −ω + 0.8 jω +1 This is a second order transfer function with ωo = 0 and δ = 0.4 . 13-30
• Section 13-5: Resonant Circuits P13.5-1 1 1 = = 60 k rad sec LC  1  1  x 10−6    120   30   ω0 = 1 × 10−6 C = 10, 000 30 = 20 Q= R 1 L 120 ω0 2 2 ω  ω  ω 2 2 +  0  +ω 0 = 58.52 k rad s and ω 2 = 0 +  0  +ω 0 = 61.52 k rad s ω1 = − 2Q 2Q  2Q   2Q  1 1 = = 3 krad s BW = RC  1 −6  (10000 ) ×10   30  Notice that BW = ω 2 − ω1 = ω0 Q . P13.5-2 H (ω ) = k ω ω  1+ Q  − 0   ω0 ω  2 2 so R = k = H (ω 0 ) = At ω = 897.6 rad s , H (ω ) = 8 = 400 Ω and ω 0 = 1000 rad s 20⋅10−3 4 = 200, so 20.10−3 200 = 400  897.6 1000  − 1+ Q 2    1000 897.6  2 ⇒ Q =8 Then 1  = ω 0 = 1000  LC  C = 20 µ F  ⇒ L = 50 mH C =Q=8  400  L  13-31
• P13.5-3 ω0 = 1 1 L R = 105 rad s , Q = = 10, BW = = 104 rad s R C L LC ω0 = 1 1 L R = 104 rad s , Q = = 10, BW = = 103 rad s R C L LC P13.5-4 P13.5-5 R = Z (ω 0 ) = 100 Ω 1 = BW = 500 ⇒ C = 20 µ F 100 C 1 = ω 0 = 2500 ⇒ L = 8 mH 20⋅10−6 ) L ( P13.5-6 R= 1 Y (ω 0 ) = 100 Ω 100 = BW = 500 ⇒ L = 0.2 H L 1 = ω 0 = 2500 ⇒ C = 0.8 µ F ( 0.2 )C 13-32
• P13.5-7 Y (ω ) = jω C + = = (R +R 1 2 1 1 + R1 + jω L R 2 −ω 2C L R 2 ) + jω ( L +C R1 R 2 ) R1 − jω L × R1 − jω L R 2 ( R1 + jω L ) R1 ( R1 + R 2 −ω 2C L R 2 ) −ω 2 L( L +C R1 R 2 )+ jω R1 ( L C R1 R 2 ) − jω L( R1 + R 2 −ω 2C L R 2 ) R 2 ( R1 −ω 2 L2 ) ω = ω 0 is the frequency at which the imaginary part of Y (ω ) is zero : R1 ( L C R1 R 2 ) − L ( R1 + R 2 −ω C L R 2 ) = 0 2 0 ⇒ ω0 = L R 2 −C R12 R 2 C L2 R 2 = 12.9 M rad sec 13-33
• P13.5-8 (a) Using voltage division yields (100 )( − j100 ) ( Vo = 1000∠0° 100 j100 ) (100)( − j−100) + j100 100 − j100 105 100 2 ∠−135° = 1000∠0° = 2∠−135° = 1000∠90° V 100 2 ∠−135°+ j100 50 2∠−135° ( ) ∴|Vo | = 1000 V (b) Do a source transformation to obtain This is a resonant circuit with ω 0 = 1 LC = 400 rad/s. That’s also the frequency of the input, so this circuit is being operated at resonance. At resonance the impedances of the capacitor and inductor cancel each other, leaving the impedance of the resistor. Increasing the resistance by a factor of 10 will increase the voltage Vo by a factor of 10. This increased voltage will cause increased currents in both the inductance and the capacitance, causing the sparks and smoke. 13-34
• P13.5-9 Let G 2 = 1 . Then R2 Z = R1 + jω L + (R G = 1 2 1 G 2 + jω C + 1 − ω 2 L C ) + j (ω LG 2 + ω C R1 ) G 2 + jω C At resonance, ∠Z = 0° so tan −1 ω L G2 +ω C R1 ωC = tan −1 2 G2 ( R1G 2 +1−ω L C ) so C − L G 22 ω L G 2 +ω C R1 ωC 2 = ⇒ ω = LC2 ( R1 G 2 +1−ω 2 L C ) G 2 and C > G 22 L C−L . Then choose C and calculate L: LC2 C = 10 mF ⇒ L = 5 mH 2 With R1 = R 2 = 1 Ω and ω 0 = 100 rad s , ω0 = 104 = Since C > G 22 L , we are done. 13-35
• P13.5-10 (a) R ( R −ω 2 R L C )+ jω L jω C Z in = jω L + = 1 1+ jω R C R+ jω C Consequently, | Zin | = (b) ( R −ω 2 R L C ) + (ω L ) 2 1+(ω R C ) 2 2 (c) ω= 1 LC ⇒ | Zin | = 1 C  R2 C  1 +  L L  P13.5-11 Let V (ω ) = A∠0 and V2 (ω ) = B∠θ . Then I (ω ) = Y (ω ) = V (ω ) V (ω ) − V2 (ω ) A − B∠θ A − B cos θ − j B sin θ R = = AR AR V (ω ) ( A − B cosθ ) + ( B sin θ ) 2 | Y (ω ) | = 2 AR 13-36
• PSpice Problems SP13.1 Here are the magnitude and phase frequency response plots: From the magnitude plot, the low frequency gain is k = 200m = 0.2. From the phase plot, the angle is -45° at p = 2π ( 39.891) = 251 rad/s . 13-37
• SP13-2 Here is the magnitude frequency response plot: The low frequency gain is 0.6 = lim H (ω ) = k ⇒ k = 0.6 . ω →0 The high frequency gain is 1 = lim H (ω ) = k ω →∞ At ω = 2π ( 2.8157 ) = 17.69 rad/s , p z ⇒ z = ( 0.6 ) p ⇒ 16 p 2 + 869 = 2 9 p + 313 ⇒ 16 2 p + 313 = p 2 + 869 9 ⇒ ( 0.77778 ) p 2 = 312.56 ⇒ p = 20 rad/s ⇒ z = 12 rad/s 2  17.69  1+    0.6 p  0.8 = 0.6 2  17.69  1+    p  ( ) 13-38
• SP13-3 From the magnitude plot, the low frequency gain is k = 4.0. From the phase plot, the angle is -45° at p = 2π (15.998 ) = 100.5 rad/s . 13-39
• SP13-4 From the magnitude plot, the low frequency gain is k = 5.0. From the phase plot, the angle is 180°-45°=135° at p = 2π (1.5849 ) = 9.958 rad/s . 13-40
• SP13-5 104 104 R R H (ω ) = − = ∠ − tan −1 (ω C 104 ) 4 2 1 + jω C 10 1 + (ω C 104 ) When ω = 200 rad/sec = 31.83 Hertz 1.8565∠158° = 104 R 1 + (ω C 10 ) 4 2 ∠ − tan −1 (ω C 104 ) Equating phase shifts gives ω C 104 = 103 C R 104 = tan(22°) = 0.404 ⇒ C = 0.2 µ F R + 104 Equating gains gives 1.8565 = 104 R 1 + (ω C 10 ) 4 2 = 104 R 1 + ( 0.404 ) 2 ⇒ R = 5 kΩ SP13-6 104 1 + jω C R 2 104 104  C R 104  R + 104 R + 104 = = ∠ − tan −1  ω H (ω ) = 4  2 104 C R 104  R + 10   C R 104  + R 1 + jω 1 + ω 1 + jω C 104 R + 104 4   R + 10  When ω = 1000 rad/sec = 159.1 Hertz 104  C R 104  R + 104 ∠ − tan −1  ω 0.171408∠ − 59° = 4  2  R + 10   C R 104  1+ ω 4   R + 10  Equating phase shifts gives C R 104 C R 104 ω = 103 = tan(59°) = 1.665 R + 104 R + 104 Equating gains gives 13-41
• VP13-4 The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode plot indicates a pole at 200 rad/s and a zero at 800 rad/s. Consequently, the Bode plot and network function don’t correspond to each other. Design Problems DP13-1 Pick the appropriate circuit from Table 13.4-2. We require 2π × 1000 < z = R2 1 1 p C , 2π ×10000 > p = , 2=k = and 5 = k = 1 C 1 R1 C2 R 2 z C2 R1 Try z = 2π × 2000. Pick C1 = 0.05 µ F. Then R1 = Check: p = 1 C C = 1.592 kΩ, R 2 = 2 R1 = 3.183 kΩ and C2 = 1 = 1 = 0.01 µ F p 2 C1 z k z 1 = 31.42 k rad s < 2π ⋅10, 000 rad s. C 2 R2 13-43
• DP13-2 1 1 V (ω ) jω C 1+ jω C R LC H (ω ) = o = = = R 1 1 Vs (ω )  1  −ω 2 + jω + || R  jω L + jω L +  1+ jω C R RC LC  jω C  Pick || R R 1 = ω 0 = 2π (100 ⋅103 ) rad s . When ω = ω 0 LC 1 LC H 0 (ω ) = 1 1 1 1 − +j + LC LC RC LC So H (ω 0 ) = R C . We require L −3 dB = 0.707 = H (ω 0 ) = R C C = 1000 L L Finally 1  = 2π (100⋅103 )  LC C =1.13 nF  ⇒ C  L = 2.26 mH 0.707 =1000  L  13-44
• DP13-3 R1 = 10 kΩ R 2 = 866 kΩ R 3 = 8.06 kΩ R 4 = 1 MΩ R 5 = 2.37 MΩ R 6 = 499 kΩ C 1 = 0.47 µ F C 2 = 0.1 µ F Va = − Circuit A R3 R2 Vc − R3 R1 Vs = −H 1 Vc − H 2 Vs R5 Circuit B Vo = − Circuit C Vc = − R4 1 + jω C 1 R 5 Va = − H 3 Va 1 Vo = −H 4 Vo jω C 2 R 6 Then Vc = H 3 H 4 Va Va = −H 2 Vs − H1 H 3 H 4 Va Vo = −H 3 Va = ⇒ Va = −H 2 Vs 1 + H1 H 3 H 4 H 2 H3 Vs 1 + H1 H 3 H 4 After some algebra jω Vo = R3 R1 R 4 C 1 R3 R 2 R 4 R 6 C1 C 2 −ω + j 2 ω Vs R5 C1 This MATLAB program plots the Bode plot: R1=10; R2=866; R3=8.060; R4=1000; % units: kOhms and mF so RC has units of sec 13-45
• R5=2370; R6=449; C1=0.00047; C2=0.0001; pi=3.14159; fmin=5*10^5; fmax=2*10^6; f=logspace(log10(fmin),log10(fmax),200); w=2*pi*f; b1=R3/R1/R4/C1; a0=R3/R2/R4/R6/C1/C2; a1=R5/C1; for k=1:length(w) H(k)=(j*w(k)*b1)/(a0-w(k)*w(k)+j+w(k)*a1); gain(k)=abs(H(k)); phase(k)=angle(H(k)); end subplot(2,1,1), semilogx(f, 20*log10(gain)) xlabel('Frequency, Hz'), ylabel('Gain, dB') title('Bode Plot') subplot(2,1,2), semilogx(f, phase*180/pi) xlabel('Frequency, Hz'), ylabel('Phase, deg') 13-46
• DP13-4 Pick the appropriate circuits from Table 13.4-2. We require 10 = − k1k2 = R 2C1 Pick C 1 = 1 µ F. Then R1 = R4 R3 , 200 = p1 = 1 1 and 500 = p2 = R1 C 1 C 2 R4 1 1 = 5 kΩ. Pick C 2 = 0.1 µF. Then R 4 = = 20 kΩ. p1 C1 p2C2 Next 10 = R2 R3 (10−6 )(20 ⋅103 ) ⇒ R2 R3 = 500 Let R 2 = 500 kΩ and R 3 = 1 kΩ. 13-47
• DP13-5 Pick the appropriate circuits from Table 13.4-2. We require 20 dB = 10 = − k1k2 = R 2C1 Pick C 1 = 20 µ F. Then R1 = R4 R3 , 0.1 = p1 = 1 1 and 100 = p2 = R1 C 1 C 2 R4 1 1 = 500 kΩ. Pick C 2 = 1 µF. Then R 4 = = 10 kΩ. p1 C 1 p2C2 Next 10 = R2 R3 (20 ⋅10−6 )(10 ⋅103 ) ⇒ R2 R3 = 50 Let R 2 = 200 kΩ and R 3 = 4 kΩ. 13-48
• DP13-6 1+ The network function of this circuit is H (ω ) = R2 R3 1+ jω R1C The phase shift of this network function is θ = − tan −1 ω R1C 1+ The gain of this network function is G = H (ω ) = R3 R2 1+ (ω R1C ) 2 1+ = R3 R2 1+ ( tan θ ) 2 Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is tan(−θ ) . Next pick R2 = 10kΩ (a convenient value) and calculated R3 using calculated from R1 = ωC R 3 = (G ⋅ 1+ (tan θ ) 2 − 1) ⋅ R 2 . Finally θ = −45 deg, G = 2, ω = 1000 rad s ⇒ R1 = 10 kΩ, R 2 = 10 kΩ, R 3 = 18.284 kΩ, C = 0.1 µF DP13-7 From Table 13.4-2 and the Bode plot: 800 = z = 1 ⇒ R1 = 2.5 kΩ R1 (0.5×10−6 ) 32 dB = 40 = 200 = p = (Check: 20 dB = 10 = k R2 R1 ⇒ R 2 = 100 kΩ 1 1 ⇒ C = = 0.05µ F R2C (200)(100×103 ) p 0.5×10−6 0.5×10−6 = = ) z C 0.05×10−6 DP13-8 −R2 jω C R 2 = − 1 1+ jω C R1 1+ jω C tan(270°−195°) 195° = 180 + 90 − tan −1 ω C R1 ⇒ R1 = = 37.3 kΩ (1000)(0.1×10−6 ) R 10 = lim H (ω ) = 2 ⇒ R 2 = 10 R1 = 373 kΩ ω →∞ R1 H (ω ) = 13-49
• Chapter 14: The Laplace Transform Exercises Ex. 14.3-1 f ( t ) = cos ω t = e + jω t + e − jωt 1 and L  e at  =   s −a 2 F ( s ) = L[cos ω t ] = 1 1 1  s  s − jω + s + jω  = s 2 +ω 2 2  Ex. 14.3-2 F ( s ) = L [e −2 t s2 + s + 3 1 1 e −2t  + L [sin t ] = + sin t ] = L   + = s + 2 s 2 +1 ( s + 2)( s 2 + 1) Ex. 14.4-1 F ( s ) = L [2u (t ) + 3e −4t u (t )] = 2L [u (t )] + 3L [e−4t u (t )] = 2 3 + s s+4 Ex. 14.4-2  1  F ( s ) = L [sin(t − 2)u (t − 2)] = e −2 s L [sin t ] = e −2 s  2   s +1  Ex. 14.4-3 F ( s ) = L[t e − t ] = L[t ] s → s +1= 1 s2 = s → s +1 1 ( s + 1) 2 Ex. 14.4-4  5   5  f ( t ) =  − t + 5  u ( t ) −  − ( t − 4.2 )  u ( t − 4.2 )  3   3  −4.2 s − 1)  5 5  −4.2 s  5  15 s + 5 ( e − 2= F (s) =  − 2 +  − e  s 3 s2  3s  3s  14-1
• Ex. 14.4-5 ∞ F (s) = ∫0 3 e − st f (t ) e dt = ∫ 0 3 e dt = −s − st 2 2 − st = 0 3(1−e −2 s ) s Ex. 14.4-6 5 2 t 0<t < 2 f (t ) =  otherwise 0 5 5 5 5 t u ( t ) −u ( t − 2 )  = t u ( t ) − t u ( t − 2 ) = t u ( t ) −( t − 2 )u ( t − 2 )− 2u (t − 2)     2 2 2 2 5  1 e −2 s 2e −2 s  5 1 −2 s −2 s ∴F ( s ) = L  f ( t )  =  2 − 2 −  = 2 s 2 1− e − 2 se    2 s   s s   f (t ) = Ex. 14.5-1 F (s) = c + jd c − jd me jθ me jθ where m = c 2 + d 2 , θ = tan −1 d + = + c s + a − jω s + a + jω s + a − jω s + a + jω ∴ f ( t ) = e − at [ c cos ω t − d sin ω t ] u ( t ) = e− at  c 2 + d 2 cos(ω t +θ )  u ( t ) = m e− at cos(ω t +θ ) u ( t )   Ex. 14.5-2 (a) F (s) = 8s −3 1 2( 8s −3) = × s + 4s +13 2 ( s + 2 )2 + 9 2 ∴ a = 2, c =8, ω =3 & ca −ω d =−3 ⇒ d = −3( 8 )( 2 ) = 6.33 −3 2 2  6.33  ° ∴ θ = tan −1   =38.4 , m = ( 8 ) + ( 6.33) =10.2  8  ⇒ f ( t ) =10.2 e −2 t cos( 3 t +38.40 ) u ( t ) 14-2
• (b) Given F ( s ) = ( 2( 3 ) ) . 3e − s 3 1 = × , first consider F1 ( s ) = 2 2 s + 2s +17 s + 2s +17 2 ( s +1)2 +16 Identify a =1, c = 0, ω = 4 and −ω d =3 ⇒ d =−3 4. Then m =|d |=3 4, θ = tan −1 ( −3 ( 4 0 ) )=−90° So f1 (t ) = (3 4)e −t sin 4t u ( t ) . Next, F ( s ) = e − s F1 ( s ) ⇒ f ( t ) = f1 ( t −1) . Finally ∴ f ( t ) =(3 4)e − (t −1) sin  4( t −1)  u ( t −1)   Ex. 14.5-3 (a) F (s) = where s 2 −5 s ( s +1) = 2 A B C + + s s +1 ( s +1)2 −5 1−5 2 = −5 and C = ( s +1) F ( s ) |s = −1 = =4 1 −1 A = sF ( s ) |s = 0 = Multiply both sides by s ( s + 1) 2 s 2 − 5 = −5 ( s +1) + Bs ( s +1) + 4 s ⇒ 2 Then F (s) = Finally B=6 −5 6 4 + + s s +1 ( s +1)2 f ( t ) = ( −5 + 6 e − t + 4 t e − t ) u ( t ) (b) F (s) = 4s 2 ( s + 3) 3 = A B C + + 2 ( s +3) ( s +3) ( s+3)3 Where A= 1 d2  d  2 3 s + 3) F ( s )  s + 3) F ( s )  = 4, B = = −24 2 (  s =−3 (  s =−3 2 ds ds and C = ( s + 3) F ( s ) s =−3 = 36 3 Then F (s) = Finally 4 −24 36 + + 2 ( s + 3 ) ( s + 3 ) ( s + 3 )3 f ( t ) = ( 4 − 24 t + 18t 2 ) e −3t u ( t ) 14-3
• Ex. 14.6-1 (a) F ( s ) = 6s +5 s + 2s +1 2  s( 6s +5)  f ( 0 ) = lim sF ( s ) = lim  2 =6 s →∞ s →∞  s + 2 s +1   s( 6s +5)  f ( ∞ ) = lim sF ( s ) = lim  2 =0 s →0 s →0  s + 2 s +1  (b) F ( s ) = 6 s − 2s +1 2  6s  =0 f ( 0 ) = lim  2 s →∞  s − 2s +1    6s  = undefined ⇒ no final value f ( ∞ ) = lim  2 s →0  s − 2s +1   Ex. 14.7-1 KCL: v1 5 + i = 7 e −6 t di di + 3 i − v1 = 0 ⇒ v1 = 4 + 3 i dt dt di 4 + 3i 35 di Then dt + i = 7 e −6 t ⇒ + 2 i = e −6 t dt 5 4 KVL: 4 Taking the Laplace transform of the differential equation: s I ( s ) − i (0) + 2 I ( s ) = 35 1 35 1 ⇒ I (s) = 4 s +6 4 ( s + 2)( s + 6) Where we have used i (0) = 0 . Next, we perform partial fraction expansion. 1 A B 1 where A = = + ( s + 2) ( s + 6) s + 2 s + 6 s+6 = s =−2 1 1 1 and B = =− 4 s + 2 s = −6 4 Then I ( s) = 35 35 35 1 35 1 − ⇒ i (t ) = e −2t − e −6t 16 s + 2 16 s + 6 16 16 14-4
• Ex. 14.7-2 Apply KCL at node a to get 1 d v1 v 2 − v1 = 48 dt 24 ⇒ 2 v1 + d v1 dt = 2 v2 Apply KCL at node b to get v 2 − 50 cos 2 t 20 + v 2 − v1 24 + v2 30 + d v2 1 d v2 = 0 ⇒ − v1 + 3 v 2 + = 60 cos 2t 24 dt dt Take the Laplace transforms of these equations, using v1 (0) = 10 V and v2 (0) = 25 V , to get ( 2+ s ) V1 ( s) − 2V2 ( s) = 10 and − V1 ( s) + ( 3+ s ) V2 (s) = 25s 2 + 60s +100 s2 + 4 Solve these equations using Cramer’s rule to get  25s 2 + 60s +100  ( 2+ s )   +10 ( 2+ s ) ( 25s 2 + 60 s +100 ) +10 ( s 2 + 4 ) s2 +4   V2 ( s ) = = ( 2+ s ) (3+ s) − 2 ( s 2 + 4 ) ( s +1)( s + 4 ) = 25s 3 +120s 2 + 220 s + 240 ( s 2 + 4 ) ( s +1)( s + 4 ) Next, partial fraction expansion gives V2 ( s ) = A A* B C + + + s + j 2 s − j 2 s +1 s + 4 where A = 25s 3 +120 s 2 + 220 s + 240 ( s +1) ( s + 4 ) ( s − j 2 ) s =− j 2 = −240− j 240 = 6 + j6 −40 A* = 6 − j 6 B = 25s 3 +120 s 2 + 220 s + 240 ( s2 +4) ( s+4) C = 25s 3 +120 s 2 + 220 s + 240 ( s 2 + 4 ) ( s +1) s =−1 s =−4 = 115 23 = 15 3 = −320 16 = −60 3 Then 14-5
• V2 ( s ) = 6+ j 6 6− j 6 23 3 16 3 + + + s + j 2 s − j 2 s +1 s + 4 Finally v2 (t ) = 12 cos 2 t + 12 sin 2 t + 23 − t 16 −4t e + e V t≥0 3 3 Ex. 14.7-3 Taking Laplace Transform of the differential equation: s 2 F ( s ) = s f ( 0 ) − f ' ( 0 ) + 5  s F ( s )− f ( 0 )  + 6 F ( s ) =   10 s +3 Using the given initial conditions ( s 2 + 5s + 6) F ( s ) = 10 2 s 2 +16s + 40 + 2s + 10 = s +3 s +3 2 s 2 +16s + 40 A B C F (s) = = + + 2 ( s + 3) ( s + 2 ) ( s + 3) ( s + 3) ( s + 3) ( s + 2 ) where A = − 10, B = − 14, and C = 16 . Then F ( s) = −10 ( s + 3) 2 + −14 16 + ⇒ s +3 s + 2 f (t ) = − 10te−3t − 14e −3t + 16e−2t for t ≥ 0 Ex. 14.8-1 KCL at top node: VC ( s ) 3 s 2 + VC ( s ) = + 2 s 2 VC ( s ) = ( 6 2 − s s+ 2 3 v C (t ) = 6 − 2 e 2 VC ( s ) I C (s) = −2= 3 2 2 s+ 3 s ⇒ iC (t ) = − ( 2 / 3) t ) u(t ) V 2 − ( 2 / 3) t e u (t ) A 3 14-6
• Ex. 14.8-2 Mesh Equations: 4 1 4  1  − − I C ( s ) − 6 ( I ( s ) − I C ( s )) = 0 ⇒ − =  6 +  I C ( s ) + 6 I ( s ) 2s  s 2s s  10 6 ( I ( s ) − I C ( s )) + 3 I ( s ) + 4 I C ( s ) = 0 ⇒ I ( s ) = − I C ( s ) 9 Solving for Ic(s): 4  2 1  6 − =  − +  I C (s) ⇒ I C (s) = 3 s  3 2s  s− 4 So Vo(s) is 24 Vo ( s ) = 4 I C ( s ) = 3 s− 4 Back in the time domain: v o ( t ) = 24 e0.75t u (t ) V Ex. 14.8-3 KVL: 8  20  + 4 =  + 8 + 4s  I L ( s ) s  s  so I L ( s) = ( s + 1) + 1 2+ s = s + 2 s + 5 ( s + 1) 2 + 4 2 Taking the inverse Laplace transform: 1   i ( t ) =  e − t cos 2 t + e − t sin 2 t  u ( t ) A 2   14-7
• Ex. 14.9-1 10   5 − 5t −10 t (a) impulse response = L−1  − ) u(t )  = ( 5 e − 10 e  s + 5 s + 10  1   1 −10 t −5t (b) step response = L−1  −  = ( e − e ) u (t )  s + 10 s + 5  Ex. 14.9-2 H ( s ) = L 5 e − 2 t sin ( 4 t ) u ( t )  =   5 ( 4) ( s + 2) 2 +4 2 = 20 s + 4 s + 20 2  H (s)   s+4 1 -1  1 − 2t step response = L-1  =L  − 2  = (1 − e (cos 4t − 2 sin 4t )) u (t )  s s + 4 s + 20   s  Ex. 14.10-1 Voltage division yields ( 8 )  2  s 2 8+ V ( s) s = H (s) = c 2 V1 ( s ) ( 8 )    s 2+   2 8+ s 16 16 1 s = = = 4 16 16 s + 20 s +1.25 16 + + s s so h ( t ) = L-1  H ( s )  = 1.25 e −1.25 t u ( t )   14-8
• Ex. 14.10-2 h ( t ) = e −2 t f (t ) = u(t ) 1 s+2 1 F(s) = s H (s) = ⇒ ⇒  1  −1 1 2 −1 2   1 −2t  + =  −e  u ( t ) h( t ) ∗ f ( t ) = L−1  H ( s ) F ( s )  = L−1  = L    s( s + 2 )  s+2   2  s    Ex. 14.11-1 − (3 − k ) ± The poles of the transfer function are p1,2 = a.) When k = 2 V/V, the poles are p1,2 = (3 − k ) 2 −8 2 . −1 ± −7 so the circuit is stable. The transfer function is 2 H (s) = Vo ( s ) Vi ( s ) = 2s s +s+2 2 The circuit is stable when k =2 V/V so we can determine the network function from the transfer function by letting s = jω. V o (ω ) V i (ω ) = H (ω ) = H ( s ) s = j ω = 2s 2 jω = s + s + 2 s = j ω ( 2 − ω 2 ) + jω 2 The input is v i ( t ) = 5cos 2 t V . The phasor of the steady state response is determine by multiplying the phasor of the input by the network function evaluated at ω = 2 rad/s.  2 jω V o (ω ) = H (ω ) ω = 2 × V i (ω ) =   2 − ω 2 ) + jω (   ( 5∠0° ) =  j 4  ( 5∠0° ) = 7.07∠ − 45°     −2 + j 2  ω =2  The steady state response is vo ( t ) = 7.07 cos ( 2 t − 45° ) V . b. When k = 3 − 2 2 , the poles are p1,2 = −2 2 ± 0 = − 2, − 2 so the circuit is stable. The 2 transfer function is H (s) = 0.17 s = 0.17 − 0.17 2 (s + 2) (s + 2) (s + 2) 2 2 The impulse response is 14-9
• h ( t ) = L -1 H ( s )  = 0.17 e −   2t (1 − ) 2 t u (t ) We see that when k = 3 − 2 2 the circuit is stable and lim h(t ) = 0 . t →∞ c. When, k = 3 + 2 2 the poles are p1,2 = 2 2± 0 = 2, 2 so the circuit is not stable. The 2 transfer function is H (s) = 5.83s = 5.83 + 5.83 2 (s − 2) (s − 2) (s − 2) 2 The impulse response is h ( t ) = L -1 H ( s )  = 5.83 e   2t (1 + 2 ) 2 t u (t ) We see that when k = 3 + 2 2 the circuit is unstable and lim h(t ) = ∞ . t →∞ Ex. 14.12-1 For the poles to be in the left half of the s-plane, the s-term needs to be positive. 2   s +5 10 2s + 20 2  + V0 ( s ) = 0.1  − 2   = 0.1  − 2 2 2   ( s +5 ) +102 ( s +5 ) +102    s s +10 s + 125  s    = 0.1 2( s 2 +10 s +125 ) −( 2s + 20 ) s s 2 +10s +125 250 25 = 0.1 2 = 2 s +10s +125 s +10s +125 These specifications are consistent. 14-10
• Problems Section 14-3: Laplace Transform P14.3-1 L  A f1 ( t )  = A F1 ( s )   f1 ( t ) = cos (ω t )  As  s  ⇒ F ( s ) = s2 + ω 2 ⇒ F1 ( s ) = 2 2  s +ω  P14.3-2 L−1 t n  =   P14.3-3 n! s n +1 F ( s ) = L−1 t1  =   1 1! = 2 1+1 s s Linearity: L  a1 f1 ( t ) + a2 f 2 ( t )  = a1 F1 ( s ) + a2 F2 ( s )   Here a1 = a2 = 1 L  f1 ( t )  = L e −3t  =     L  f 2 ( t )  = L[t ] =   so F ( s ) = P14.3-4 1 = F1 ( s ) s+3 1 = F2 ( s ) s2 1 1 + 2 s +3 s f ( t ) = A (1−e −bt ) u ( t ) = L [ Af1 (t ) ] = AF1 ( s ) f1 ( t ) = A(1− e − bt ) u ( t ) = 1u ( t ) −e − bt u ( t ) = f 2 ( t ) + f3 ( t ) 1 −1 , F3 ( s ) = s s +b 1  Ab 1 ∴ F (s) = A  −  = s( s +b )  s s +b  F2 ( s ) = 14-11
• Section 14-4: Impulse Function and Time Shift Property P14.4-1 f ( t ) = A u ( t ) − u ( t −T )    (1−e A Ae − sT F ( s ) = AL u ( t )  − AL u ( t −T )  = − =A     s s s − sT P14.4-2 f ( t ) = 1 u ( t ) −u ( t −T )  eat   L u ( t ) −u ( t −T )  =   ⇒ F ( s ) = L e at u ( t )−u ( t −T )      1− e− sT s L eat g ( t )  =G ( s − a )     P14.4-3 (a) F (s) = )   1− e( s − a )T  ⇒ F (s) =  ( s −a )    2 ( s +3) 3 (b) f ( t ) = δ ( t −T ) u ( t −T ) ⇒ F ( s ) = e − sT L δ ( t )  = e − sT   (c) F (s) = P14.4-4 5 ( s + 4 ) +( 5) 2 2 = 5 5 = 2 ( s + 8 s + 16 ) + 25 s + 8 s + 41 2 g ( t ) = e − t u ( t −0.5 ) = e − (t + (0.5−0.5))u ( t − 0.5 ) = e−0.5 e− ( t − 0.5)u ( t − 0.5 ) L e −0.5 e − (t −0.5)u ( t − 0.5 )  = e−0.5 L e− (t − 0.5)u ( t − 0.5 )  = e−0.5 e−0.5 s L  e− t u ( t )  =       e0.5 e −0.5 s e0.5 − 0.5 s = s +1 s +1 P14.4-5 − sT e − sT  t −T  − sT  t  e L − L − t u( t ) = u( t − T ) = e L − u( t ) =   T s2 T  T   T  14-12
• Section 14-5: Inverse Laplace Transform P14.5-1 F (s) = s +3 s +3 A Bs + C = = + 2 2 2 s + 3s + 6s + 4 ( s +1) ( s +1) + 3 s +1 s + 2s + 4   3 where A= s +3 ( s +1) 2 +3 s =−1 = 2 3 Then 2 2 8 Bs + C 4  = 3 + 2 ⇒ ( s + 3) = ( + B ) s 2 +  + B + C  s + + C 2 3 3 3  ( s +1) ( s + 2s + 4 ) s +1 s + 2s + 4 ( s + 3) Equating coefficient yields 2 2 + B ⇒ B= − 3 3 4 2 1 s : 1= − + C ⇒ C = 3 3 3 s2 : 0 = Then 1 2 2 1 2 2 3 − s+ − ( s +1) 3 3 + 3 3 = 3 + 3 F (s) = + s +1 ( s +1)2 + 3 s +1 ( s +1)2 + 3 ( s +1)2 + 3 Taking the inverse Laplace transform yields f (t ) = 2 −t 2 −t 1 −t e − e cos 3t + e sin 3 3 3 3 14-13
• P14.5-2 F (s) = s 2 − 2s + 1 s 2 − 2s + 1 a a* b = = + + 3 2 s + 3s + 4s + 2 ( s +1) ( s +1− j )( s +1+ j ) s +1− j s +1+ j s +1 where s 2 − 2s +1 b= ( s +1) a= 2 +1 s =−1 =4 3− j 4 3 s 2 − 2s +1 = =− + j 2 ( s +1) ( s +1+ j ) s =−1+ j −2 2 3 a* = − − j 2 2 Then 3 3 − + j2 − − j2 4 F (s) = 2 + 2 + s +1− j s +1+ j s +1 Next m= From Equation 14.5-8 ( -3 2 )2 + ( 2 ) 2    2  5 = = 126.9° and θ = tan −1  3 2 −   2 f ( t ) = 5 e − t cos ( t + 127° ) + 4 e −t  u ( t )   P14.5-3 F (s) = 5 s −1 ( s +1) ( s − 2 ) 2 = A B C + + 2 s +1 ( s +1) s −2 where B= 5 s −1 s −2 = 2 and C = s =−1 5 s −1 ( s +1) 2 =1 s=2 Then A= Finally F (s) = d  2 s +1) F ( s )  (  ds −1 2 1 + + ⇒ 2 s +1 ( s +1) s−2 s =−1 = −9 ( s −2) = −1 2 s =−1 f ( t ) =  −e − t + 2 t e − t + e 2t  u ( t )   14-14
• P14.5-4 Y (s) = 1 1 A Bs +C = = + 2   ( s +1) ( s + 2s + 2 ) ( s +1) ( s +1) + 1 s +1 ( s +1)2 +1 2 where A= Next 1 s + 2s + 2 2 =1 s =−1 1 1 Bs +C = + 2 ⇒ 1 = s 2 + 2 s + 2 + ( Bs + C ) ( s + 1) ( s +1) ( s + 2s + 2 ) s +1 s + 2s + 2 2 ⇒ 1 = ( B +1) s 2 + ( B + C + 2 ) s + C + 2 Equating coefficients: s 2 : 0 = B + 1 ⇒ B = −1 s : 0 = B + C + 2 ⇒ C =−1 Finally Y (s) = P14.5-5 F (s) = 1 s +1 − ⇒ s +1 ( s +1)2 +1 2( s + 3) ( s +1) ( s 2 + 2 s +5) = y ( t ) =  e− t − e − t cos t  u ( t )   −( s +1) 1 2 + + 2 s +1 ( s +1) + 4 ( s +1)2 + 4 f ( t ) = e− t − e − t cos ( 2t ) + e −t sin ( 2t )  u ( t )   P14.5-6 F (s) = where A = sF ( s ) s =0 = 2( s+3) A B C = + + s( s+1) ( s + 2 ) s s +1 s + 2 2( s + 3 ) ( s +1) ( s + 2 ) s =0 = 3, B = ( s +1) F ( s ) s =−1 = and ( s + 2 ) F ( s ) s = −2 = Finally 3 −4 1 + ⇒ F (s) = + s s +1 s + 2 2( s + 3) s( s +1) s =−2 2( s + 3) s( s + 2 ) s =−1 =−4 = C =1 f ( t ) = ( 3 − 4e− t + e −2t ) u ( t ) 14-15
• Section 14-6: Initial and Final Value Theorems P14.6-1 (a) 2s 2 −3s + 4 2s 2 f ( 0 ) = lim sF ( s ) = lim = 2 =2 2 s s →∞ s →∞ s +3s + 2 (b) 4 f ( ∞ ) = lim sF ( s ) = = 2 2 s →0 P14.6-2 Initial value: s( s +16 ) s 2 + 16 s v ( 0 ) = lim sV ( s ) = lim 2 = lim 2 =1 s →∞ s →∞ s + 4 s + 12 s →∞ s + 4 s + 12 Final value:   s +16 s 2 + 16 s = lim 2 =0 v ( ∞ ) = lim s  2  s→ 0  s + 4 s + 12  s → 0 s + 4 s +12 (Check: V(s) is stable because Re { pi } < 0 since pi = − 2 ± 2.828 j . We expect the final value to exist.) P14.6-3 Initial value: s 2 +10 s v(0) = lim sV ( s ) = lim = 0 s →∞ s →∞ 3 s 3 + 2 s 2 +1 Final value: v ( ∞ ) = lim sV ( s ) = lim s →0 s →0 s ( s +10 ) s ( 3s 2 + 2s +1) = 10 (Check: V(s) is stable because pi = −0.333 ± 0.471 i . We expect the final value to exist.) P14.6-4 Initial value: f ( 0 ) = lim s F ( s ) = lim s →∞ Final value: s →∞ −2 s 2 −14 s = −2 s 2 − 2 s + 10 F(s) is not stable because Re { p1} > 0 since pi = 1 ± 3i . No final value exists. 14-16
• Section 14-7: Solution of Differential Equations Describing a Circuit P14.7-1 KVL: 4 di + v = 2 e−2×10 t dt The capacitor current and voltage are related by 50 i + 0.001 i = ( 2.5 × 10−6 ) dv dt 4t v1 = 2 e −2×10 V , i (0) = 1 A, v(0) = 8 V Taking the Laplace transforms of these equations yields 50 I ( s) + 0.001 [ s I ( s) − i (0) ] + V ( s) = I ( s) = ( 2.5 × 10−6 )  sV ( s ) − v( 0 )    2 s + 2×104 Solving for I(s) yields I (s) = s 2 + 1.4× 104 s − 1.6 × 108 ( s +104 ) ( s + 2 × 104 ) ( s + 4 ×104 ) = A B C + + 4 4 s +10 s + 2× 10 s + 4×104 where A = ( s + 10 ) I ( s ) 4 s = −10 B = ( s + 2×104 ) I ( s ) C = ( s + 4×104 ) I ( s ) 4 s 2 + 1.4× 104 s −1.6× 108 = ( s + 2 × 104 )( s + 4 × 104 ) s = − 2 ×10 s = − 4 ×10 4 = 4 = s = −10 s 2 + 1.4× 104s −1.6 × 108 ( s+ 104 ) ( s + 4×104 ) s 2 + 1.4× 104s −1.6 × 108 ( s+104 ) ( s+ 2× 104 ) 4 −2×108 −2 = = 3× 108 3 = s = − 2 ×10 4 = s = − 4 ×10 4 .4 × 108 1 = 8 2 × 10 5 8.8 × 108 22 = 8 6 × 10 15 Then I (s) = − 23 15 22 15 + + 4 s +10 s + 2×10 s + 4 × 104 ⇒ i (t ) = 4 4 4 1 −10 e−10 t + 3e−2x10 t + 22 e−4x10 t  u ( t ) A   15 14-17
• P14.7-2 We are given v ( t ) = 160 cos 400t . The capacitor is initially uncharged, so v C ( 0 ) = 0 V . Then i ( 0) = 160 cos ( 400 × 0 ) − 0 = 160 A 1 KCL yields 10−3 dvC dt + vC =i 100 Apply Ohm’s law to the 1 Ω resistor to get v −v C i= ⇒ vC = v− i 1 Solving yields di + 1010 i = 1600 cos 400t − ( 6.4 × 104 ) sin 400t dt Taking the Laplace transform yields s I ( s ) − i (0) + (1010 ) I ( s ) = ( 6.4×10 ) ( 400 ) 2 1600s s 2 + ( 400 ) 2 − s 2 + ( 400 ) 2 so I (s) = 160 1600s − 2.5×107 + s + 1010 ( s + 1010 )  s 2 + (400) 2    Next A B B* 1600s − 2.5×107 = + + ( s + 1010 )  s 2 + (400)2  s + 1010 s + j 400 s − j 400   where A = B = 1600 s − 2.5 x 107 ( s +1010 ) ( s − j 400 ) s 2 + ( 400 ) = s = − j 400 Then I (s) = Finally 1600s − 2.5×107 = − 23.1 , 2 s = −1010 2.56 x 107 ∠1.4° 8.69 x 10 ∠68.4 5 ° = 11.5 − j 27.2 and B* = 11.5 + j 27.2 136.9 11.5− j 27.2 11.5 + j 27.2 + + s + 1010 s + j 400 s − j 400 i ( t ) = 136.9e−1010t + 2 (11.5 ) cos 400t − 2 ( 27.2 ) sin 400t for t > 0 = 136.9e−1010t + 23.0 cos 400t − 54.4sin 400t for t > 0 14-18
• P14.7-3 vC (0) = 0 vc +15 i = 10 cos 2t    ⇒ 1 d vc i=  30 dt  d vc + 2 vc = 20 cos 2t dt Taking the Laplace Transform yields: sVc ( s ) − vc ( 0 ) + 2Vc ( s) = 20 s A B B* 20s ⇒ Vc ( s ) = = + + s2+ 4 ( s + 2 )( s 2 + 4 ) s + 2 s + j 2 s − j 2 where A= 20 s s2 +4 = s = −2 −40 20s = −5, B = 8 ( s + 2 )( s − j 2 ) = s = − j2 j5 5 5 5 5 = + j and B* = − j 1+ j 2 2 2 2 Then 5 5 5 5 +j −j −5 2 2 2 2 Vc ( s ) = + + s+2 s+ j2 s− j2 ⇒ vc ( t ) = −5e −2t + 5 ( cos 2t + sin 2t ) V P14.7-4 vc + 12i L + 2 diL dt = −8 and i L = C d vc dt Taking the Laplace transform yields Vc ( s ) + 12 I L ( s ) + 2  sI L ( s ) −iL ( 0 )  = −   I L ( s ) = C  sVc ( s ) −vc ( 0 )    8 s vc (0) = 0, iL (0) = 0 Solving yields 14-19
• −4 C Vc ( s ) = (a) C = 1 F 18 C  s  s 2 + 6s +  2  Vc ( s ) = −72 s ( s + 3) 2 = c a b + + s s + 3 ( s + 3) 2 a = − 8, b = 8, and c = 24 ⇒ Vc ( s ) = 24 −8 8 + + s s + 3 ( s + 3) 2 vc ( t ) = −8 + 8 e −3t + 24 t e −3t (b) C = 1 F 10 Vc ( s ) = c a b −40 = + + s ( s +1) ( s +5 ) s s +1 s + 5 a = − 8, b = 10, and c = −2 ⇒ Vc ( s ) = −2 −8 10 + + s s +1 s + 5 vc ( t ) = − 8 + 10 e − t − 2 e −5 t P14.7-5 vc (0− ) = 10 V, i L( 0− ) = 0 A i = ( 5 × 10−6 ) d vc di and 400 i + 1 + vc = 0 dt dt Taking Laplace transforms yields  1   −  400  I ( s ) = ( 5 ×10 ) ( sVc ( s ) − 10 )  −10 40  =   ⇒ I (s) = 2 2 5 s + 400 s + 2 ×10 400 I ( s ) + ( s I ( s ) − 0 ) + Vc ( s ) = 0  ( s + 200 ) + 4002  −6 so i (t ) = − 1 −200t e sin ( 400t ) u ( t ) A 40 14-20
• Section 14-8: Circuit Analysis Using Impedance and Initial Conditions P14.8-1 6 6 − 0.010 − 0.002 s .003 .005 s s I L ( s) = = = = 5s + 2000 s ( s + 400) s s + 400 2 mA iL (t ) =  −400 t mA  3−5e t <0 t >0 P14.8-2 10 8 VL ( s ) − ( −.015 ) s + VL ( s ) + 3 ⇒ VL ( s ) = 0= 4000 2000 4000 5s s+ 15 8 V ( s ) + 0.15 0.005 0.002 15 I L ( s) = L = + 0.015 = − 4000 5s 5  4000  s+ s  s+  15 15   VL ( s ) − iL (t ) = 5 − 3e − 4000 t 15 mA, t > 0 14-21
• P14.8-3 Vc ( s ) − 8 s =0 − 0.006 Vc ( s ) + + s 2000 − 6000 8  + 500 Vc ( s ) + 0.5s  Vc ( s ) −  = 0 s s  Vc ( s ) = 106 .5s 8 s + 12000 12 4 = − s ( s +1000) s s +1000 Vc (t ) = 12 − 4e −1000t V, t > 0 P14.8-4 6 s + Vc ( s ) +  0.5s   V ( s ) − 8  = 0   c  s 2000 4000  106   Vc ( s ) − 6 8   500  Vc ( s ) −  + 250Vc ( s ) + 0.5s  Vc ( s ) −  = 0 s s   Vc ( s ) = 6000 + 8s 4 4 = + s ( s + 1500 ) s s + 1500 vc (t ) = 4 + 4e−1500t V, t > 0 14-22
• P14.8-5 Node equations: Va ( s ) − VC ( s ) Va ( s ) 1 6 6 + = ⇒ Va ( s ) = VC ( s ) + s 6 s s+6 s+6 6   6 6 VC ( s ) −  VC ( s ) + VC ( s ) −  3 s+6 1 s  s+6 s+3 + + + VC ( s ) − = 0 4 2 s s 4 After quite a bite of algebra: VC ( s ) = 6 s 2 + 56s + 132 ( s + 2 )( s + 3)( s + 5) Partial fraction expansion: 44 1 6 s + 56 s + 132 9 V (s) = = 3 − + 3 c ( s +3)( s + 2 )( s +5) s + 2 s +3 s +5 Inverse Laplace transform: v (t ) = 44 3 e −2t − 9e −3t + (1 3)e −5t V c 2 14-23
• P14.8–6 Write a node equation in the frequency domain: 10 s = Vo ( s ) − 5 C + Vo ( s ) ⇒ V ( s ) = o 1 R1 R2 Cs 10 +5s R1C  1  ss +   R 2C    10 = R2 R1 s 5 − 10 + R2 R1 1 s+ R 2C Inverse Laplace transform: vo ( t ) = 10  R 2  −t +  5 − 10 e R1  R1    R2 R 2C = 10 − 5 e−1000t V for t > 0 14-24
• P14.8-7 Here are the equations describing the coupled coils: di1 di +M 2 dt dt di di v2 (t ) = L2 2 + M 1 dt dt v1 (t ) = L1 ⇒ V1 ( s) = 3 ( s I1 ( s ) − 2 ) + ( sI 2 ( s) − 3) = 3s I1 ( s ) + sI 2 ( s ) − 9 ⇒ V2 ( s) = s ( I1 ( s ) − 2 ) + 2( sI 2 ( s ) −3) = sI1 ( s) + 2sI 2 ( s ) − 8 Writing mesh equations: 5 5 = 2 ( I1 ( s ) + I 2 ( s ) ) + V1 = 2 ( I1 ( s ) + I 2 ( s ) ) + 3s I1 ( s ) + sI 2 ( s ) − 9 ⇒ ( 3s + 2 ) I1 + ( s + 2 ) I 2 = 9 + s s V1 ( s ) = V2 ( s ) + 1I 2 ( s ) ⇒ 3s I1 ( s ) + sI 2 ( s ) − 9 = sI1 ( s ) + 2 sI 2 ( s ) − 8+ I 2 ( s ) ⇒ 2s I1 − ( s +1) I 2 =1 Solving the mesh equations for I2(s): I2 ( s ) = 15s + 8 3s + 1.6 0.64 2.36 = + = s + 0.26 s + 1.54 5s 2 + 9s + 2 ( s + 0.26 )( s +1.54 ) Taking the inverse Laplace transform: i2 (t ) = 0.64e −0.26t + 2.36e −1.54t A for t > 0 P14.8-8 t<0 time domain frequency domain Mesh equations in the frequency domain: 6 I 1 ( s ) + 6 ( I 1 ( s ) − I 2 ( s )) + 6 I 1 ( s ) + 12 2 2 = 0 ⇒ I1 (s) = I 2 (s) − s 3 3s 14-25
• 2 6 2 6  I 2 ( s ) − − 6 ( I 1 ( s ) − I 2 ( s )) = 0 ⇒  6 +  I 2 ( s ) − 6 I 1 ( s ) = s s s s  Solving for I2(s): 1 2 2 2  6  ⇒ I 2 (s) = 2  6 +  I 2 (s) − 6 I 2 (s) −  = 1 s 3 3s  s   s+ 2 Calculate for Vo(s): 1 6 Vo ( s ) = I 2 ( s ) − = 2 s  1  1 2  6 4 −2 −  − = 2 s+ 1  s s+ 1 s 2 2  Take the Inverse Laplace transform: vo ( t ) = − ( 4 + 2 e− t / 2 ) V for t > 0 (Checked using LNAP, 12/29/02) P14.8-9 t<0 frequency domain time domain Writing a mesh equation: 2    −6  s +  3 12 3  5 ( 4 + 5 s ) I ( s ) + 30 + = 0 ⇒ I ( s ) =  4 = −  + 4  s   s s+  ss +    5  5  Take the Inverse Laplace transform: i ( t ) = −3 (1 + e−0.8 t ) A for t > 0 (Checked using LNAP, 12/29/02) 14-26
• P14.8-10 Steady-state for t<0: From the equation for vo(t): vo ( ∞ ) = 6 + 12 e − 2 (∞) Steady-state for t>0: =6 V From the circuit: vo ( ∞ ) = 3 (18) R+3 Therefore: 6= 3 (18) ⇒ R = 6 Ω R+3  −6 1  18 6 I (s) 2 +  + − = 0 ⇒ I (s) = 1 Cs s s  s+ 2C    −6  18 −12 1 18 1 12 18 12 6 + = + + = + Vo ( s ) = I (s) + =  1  s 1 1 Cs s Cs  s + s s s+ s  s+   2C  2C 2C  Taking the inverse Laplace transform: vo ( t ) = 6 + 12 e − t / 2C V for t > 0 Comparing this to the given equation for vo(t), we see that 2 = 1 2C ⇒ C = 0.25 F . (Checked using LNAP, 12/29/02) 14-27
• Section 14-9: Transfer Function and Impedance P14.9-1 R1 R1 R2 C1s and Z 2 = = R 1 R1C1s + 1 R2C2 s +1 R1 + C1s R 2(τ 1 s + 1) Let τ 1 = R1C1 and τ 2 = R2C2 then H ( s ) = R 1 (τ 2s + 1) + (τ 1 s + 1) R 2 H (s) = Z2 Z1 + Z 2 When τ 2 = τ 2 = τ ⇒ H ( s ) = where Z1 = R2(τ 1s + 1) R2 = = constant, as required. ( R1 + R2 ) (τ s + 1) R1 + R2 ∴ we require R1C 1 = R2C2 P14.9-2 1 and Z 2 = R + Ls then the input impedance is Cs   1  L   2  R + ( R + Ls )  LCs +  RC + R  s +1  ZZ Cs     Z (s) = 1 2 =  = R 2 1 Z1 + Z 2  LCs + 2 RCs +1  R+ + R + Ls   Cs   L Now require : RC + = 2 RC ⇒ L = R 2C then Z = R R Let Z1 = R + P14.9-3 The transfer function is R2 H (s) = 1 R1 C R1 + R 2 R 2 C s +1 = R2 R1 + s+ R 2 C s +1 R1 R 2 C Using R1 = 2 Ω, R 2 = 8 Ω and C = 5 F gives H (s) = 0.1 s + 0.125 The impulse response is h (t ) = L -1  H ( s ) = 0.1 e−0.125 t u (t ) V . The step response is 14-28
•    H ( s) 0.1 0.8  -1  0.8 −0.125 t   = L -1  L -1  u (t ) V  s ( s + 0.125) = L  s − s + 0.125  = 0.8 1− e  s        ( ) (Checked using LNAP, 12/29/02) P14.9-4 The transfer function is: H ( s ) = L 12 t e −4 t u ( t )  =   12 ( s + 4) 2 = 12 s + 8s + 16 2 The Laplace transform of the step response is: 3 H (s) 12 k −3 = = 4+ + 2 2 s s ( s + 4) s+4 s ( s + 4) 2 The constant k is evaluated by multiplying both sides of the last equation by s ( s + 4) . 3  3 3 2 12 = ( s + 4) − 3s + ks ( s + 4) =  + k  s 2 + (3 + 4k ) s + 12 ⇒ k = −    4   4 4 The step response is  H ( s )   3 −4 t     =  − e 3 t + 3  u (t ) V  L −1      s  4  4     P14.9-5 The transfer function can also be calculated form the circuit itself. The circuit can be represented in the frequency domain as We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R and the op amp comprise a non-inverting amplifier. Thus  R  V s Va ( s ) = 1 +     10000  c ( )  Now, writing node equations, 14-29
• Vc ( s ) −Vi ( s ) Vo ( s ) −Va ( s ) Vo ( s ) + CsVc ( s ) = 0 and + =0 1000 5000 Ls Solving these node equations gives R  5000 1     1 +   10000  L 1000 C  H (s) =  1  5000      s +  s +    1000C   L  Comparing these two equations for the transfer function gives   1  1       = ( s + 2000) or  s +  = ( s + 5000) s +    1000C     1000C    5000  5000       = ( s + 2000) or  s +  = ( s + 5000) s +       L  L  1  R  5000 = 15 × 10 6 1 +  1000C  10000  L The solution isn’t unique, but there are only two possibilities. One of these possibilities is    s + 1  = ( s + 2000) ⇒ C = 0.5 µ F     1000C   5000   s +  = ( s + 5000) ⇒  L  L =1H   1 + R  5000 = 15×106 ⇒ R = 5 kΩ     1000 0.5×106  10000  1 ( 1 ) (Checked using LNAP, 12/29/02) 14-30
• P14.9-6 The transfer function of the circuit is R2 1 1+ R2 C s R1 C H (s) = − =− 1 R1 s+ R2 C The give step response is vo ( t ) = −4 (1 − e−250 t ) u ( t ) V . The correspond transfer function is calculated as H (s) 4  −1000 −1000 4 = L −4 (1 − e − 250 t ) u ( t ) = −  − ⇒ H (s) = = s s + 250  s s + 250  s ( s + 250 ) { } Comparing these results gives 1 1 1 = 250 ⇒ R 2 = = = 40 kΩ 250 C 250 ( 0.1×10 − 6 ) R2 C 1 1 1 = 1000 ⇒ R1 = = = 10 kΩ 1000 C 1000 ( 0.1×10 − 6 ) R1 C (Checked using LNAP, 12/29/02) P14.9-7  4   2  Va ( s ) =   Vi ( s ) =   Vi ( s )  s+2  4+ 2s   12  Vo ( s ) =  s 12 6+  s     2   2   2   2   Vb ( s ) =   Vb ( s ) =   5 Va ( s ) =  5   Vi ( s )  s+2  s+2  s+2  s+2    The transfer function is: H (s) = Vo ( s ) 20 = Vi ( s ) ( s + 2 )2 14-31
• The Laplace transform of the step response is: 20 5 −5 −10 Vo ( s ) = = + + 2 s s + 2 ( s + 2 )2 s ( s + 2) Taking the inverse Laplace transform: vo ( t ) = 5 − 5 e −2 t (1 + 2t )  u ( t ) V   (checked using LNAP 8/15/02) P 14.9-8 From the circuit:  1   1   4   6C   Cs   L   4   H (s) =  = (k ) (k )  4 + Ls  s+ 1   6+ 1   s+ 4     Cs   L 6C     From the given step response: H (s) 2 4 6 12 = L  ( 2 + 4 e −3 t − 6 e − 2 t ) u ( t )  = +   s s + 3 − s + 2 = s ( s + 3)( s + 2 ) s so 12 H (s) = s ( s + 3)( s + 2 ) Comparing the two representations of the transfer functions let 1 1 =3 ⇒ C = F, 6C 18 4 = 2 ⇒ L = 2 H and 2 × 3 × k = 12 ⇒ k = 2 V/V . L (Checked using LNAP, 12/29/02) P 14.9.9 From the circuit: 14-32
• R V (s) R+ Ls L = = H (s) = o Vi ( s ) 12 + R + L s s + 12 + R L From the given step response: s+ H (s) s+2 0.5 0.5 = L  0.5 (1 + e −4 t ) u ( t )  = + = ⇒   s s s + 4 s ( s + 4) H (s) = s+2 s+4 Comparing these two forms of the transfer function gives: R  =2  12 + 2 L  L = 4 ⇒ L = 6 H, R = 12 Ω  ⇒ 12 + R L  =4   L (Checked using LNAP, 12/29/02) P14.9-10 Mesh equations: 1 1  1  V ( s ) =  R1 + +  I1 ( s ) − I2 ( s ) Cs Cs  Cs  1  1  0 =  R+ R+  I2 ( s ) − I1 ( s ) Cs  Cs  Solving for I2(s): Then Vo ( s ) = R I 2 ( s ) gives H (s) =  1  V (s)    Cs  I2 ( s ) = 2  1  1   R1 +   2 R +  − (Cs) 2 Cs   Cs   V0 ( s ) RCs = = V (s) [ R1Cs + 2][ 2 RCs +1] − 1 s   1  s 2 + 4 RC + R1C s +  2 R1C 2 2  2 RR1C 2 ( 2RR1C )   14-33
• P14.9-11 Let  1  R  R Cs Z2 =   = 1 RCs + 1 R+ Cs Z1 = Rx + Lx s Then R RCs + 1 V2 Z2 = = V1 Z1 + Z 2 Rx + Lx s + V2 V1 R RCs + 1 = R Lx RCs + ( Lx + Rx RC ) s + Rx + R 2 1 LC = ( L + R RC ) s + Rx + R s2 + x x Lx RC Lx RC P14.9-12 Node equations: (V1 − Vin ) sC1 + V1 − Vout =0 ⇒ R1 ( R C s + 1)V 1 1 1 = R1C1sVin + Vout V0 + Vout + sC2 = 0 ⇒ V1 = − R 2C2 sVout R2 Solving gives: H (s) = − R1C1s Vout = = Vin R1 R 2C1C2 s 2 + R 2C2 s +1 s2 + − 1 s R 2 C2 1 1 s+ R1C1 R1 R 2C1C2 14-34
• P14.9-13 Node equations in the frequency domain: V1 − Vi V1 − V2 V1 − V0 + + =0 R1 R2 R3  1 V 1 1  V ⇒ V1 + + − 0 = i  R1 R2 R3  R3 R1 V2 − V1 − sC2V0 = 0 ⇒ V1 = − sC2 R2 V0 R2 After a little algebra: H ( s )= V0 − R3 = Vi sC2 R2 R3 + sC2 R1 R3 + sC2 R1 R2 + R1 P14.9-14 1 1 Vo ( s ) Cs LC H (s) = = = R 1 1 Vi ( s ) Ls + R + s2 + s + Cs L LC L, H C, F R, Ω H(s) 2 0.025 18 20 20 = s + 9 s + 20 ( s + 4 )( s + 5 ) 2 0.025 8 20 20 = s + 4s + 20 ( s + 2 )2 + 44 1 0.391 4 2.56 2.56 = s + 4 s + 2.56 ( s + 0.8 )( s + 3.2 ) 2 0.125 8 20 20 = s + 4s + 4 ( s + 2 )2 2 2 2 2 14-35
• a) H ( s ) = 20 ( s + 4 )( s + 5) 20 20 − ⇒ h ( t ) = ( 20e −4t − 20e −5t ) u (t ) s + 4 s +5 20 1 −5 4 H ( s) L {step response} = = = + + ⇒ s s ( s + 4) ( s + 5) s s + 4 s +5 L {h(t )} = H ( s ) = step response = (1+ 4e −5t −5e −4t ) u (t ) b) H ( s ) = 20 ( s + 2) 2 +4 4 5(4) ⇒ h ( t ) = 5e −2t sin 4t u (t ) ( s + 2) 2 + 42 H ( s) 20 1 K s + K2 = = + 21 L {step response} = 2 s s ( s + 4s + 20) s s + 4s + 20 L {h( t )} = H (s) = 20 = s 2 + 4s + 20 + s ( K1s + K 2 ) = s 2 (1+ K1 ) + s ( 4+ K 2 ) + 20 ⇒ K1 = −1, K 2 = − 4 1 − ( 4) −( s + 2 ) 1 2 L {step response} = + + s ( s + 2 )2 + 42 ( s + 2 ) + 42 1    step response = 1− e −2t  cos 4t + sin 4t   u (t ) 2    c) 2.56 H (s) = ( s + 0.8)( s + 3.2 ) L {h( t )} = H ( s ) = 1.07 1.07 − ⇒ h ( t ) = 1.07 ( e −.8t − e −3.2t ) u(t) s + .8 s + 3.2 1 −4 2.56 1 H (s) L {step response} = = = + 3 + 3 s s ( s + .8) ( s + 3.2) s s + .8 s + 3.2 4  1  step response = 1+ e −3.2t − e −.8t  u (t ) 3  3  d) H ( s ) = 20 ( s + 2) 2 h( t ) = 4te −2t u (t ) step response = (1−(1+ 2t )e −2t ) u (t ) 14-36
• P14.9-15 For an impulse response, take V1 ( s ) = 1 . Then 3( s + 2 ) A B B* V0 ( s ) = = + + s ( s + 3− j 2 ) ( s + 3+ j 2 ) s s + 3− j 2 s + 3+ j 2 Where A = sV0 ( s ) s =0 =.462, B = (s + 3 − j 2) V0 ( s ) s =−3+ j 2 = 0.47∠ − 119.7 o and B* = 0.47 ∠119.7 o Then V0 ( s ) = 0.462 0.47 ∠−119.7 o 0.47 ∠119.7 o + + s s +3− j 2 s +3+ j 2 The impulse response is v0 (t ) = 0.462 + 2(0.47)e −3t cos ( 2 t − 119.7 o )  u ( t ) V   Section 14-10: Convolution Theorem P14.10-1 1 e− s 1 − e− s f ( t ) = u ( t ) − u ( t − 1) ⇒ F ( s ) = L u ( t ) − u ( t − 1)  = −   s s = s −s 2    1 − 2 e − s + e−2 s  2 −1 −1  1 − e f ( t ) * f ( t ) = L  F ( s )  = L   = L −1     s2  s       = t u ( t ) − 2 ( t − 1) u ( t − 1) + ( t − 2 ) u ( t − 2 ) P14.10-2 f ( t ) = 2 u ( t ) − u ( t − 2 )  ⇒ F ( s ) =   2 2e −2 s − s s  4 8e −2 s 4e −4 s  ∗ f = L−1  F ( s ) F ( s )  = L−1  2 − 2 + 2  = 4t u ( t ) − 8 ( t − 2 ) u ( t − 2 ) + 4 ( t − 4 ) u ( t − 4 ) f   s s  s 14-37
• P14.10-3 v1 ( t ) = t u ( t ) ⇒ V1 ( s ) = 1 s2 1 1 H (s) = = Cs = RC 1 V1 ( s ) R + 1 s+ Cs RC V2 ( s ) v2 ( t ) = h ( t ) ∗ v1 ( t ) = L−1 V1 ( s ) H ( s )     1    1 V2 ( s ) = V1 ( s ) H ( s ) =  2   RC   s   s+ 1  RC   v2 ( t ) = t − RC (1 − e − t / RC ), t ≥ 0 P14.10-4 1 1 h ( t ) ∗ f ( t ) = L−1  H ( s ) F ( s )  where H ( s ) = 2 and F ( s ) =   s s+a  1   1  A B C So H ( s ) F ( s ) =    + 2 +  = s s s+a  s 2   s +a Solving the partial fractions yields: A = −1 a 2 , B = 1 a, C = 1 a 2 So h( t ) ∗ f ( t ) = −1 t e − ( at ) + + 2 , a2 a a t≥0 14-38
• Section 14-11: Stability P14.11-1 a. From the given step response: H (s) =L s 3 −100 t ) u ( t ) = s ( s 75 )  4 (1 − e  + 100   From the circuit: R H (s) = R + 5 + Ls ⇒ R H (s) L = R+5 s  ss +  L   Comparing gives R  = 75  R = 15 Ω  L  ⇒ R+5 L = 0.2 H = 100   L  b. The impulse response is  75  h ( t ) = L -1  = 75 e −100 t u ( t ) s + 100    c. H (ω ) ω =100 = 75 3 = ∠45° j 100 + 100 4 2 15  3  ∠45°  ( 5∠0° ) = ∠45° V Vo (ω ) =  4 2 4 2  vo ( t ) = 2.652 cos (100 t − 45° ) V (Checked using LNAP, 12/29/02) P14.11-2 The transfer function of this circuit is given by H (s) 5 −5 −10 20 = L ( 5 − 5 e −2 t (1 + 2t ) ) u ( t )  = +   s s+2+ s+2 2 = s+2 2 s ( ) ( ) ⇒ H (s) = 20 s ( s + 2) 2 This transfer function is stable so we can determine the network function as H (ω ) = H ( s ) s = j ω = 20 ( s + 2) = 2 s= jω 20 (2 + jω ) 2 14-39
• The phasor of the output is Vo (ω ) = 20 ( 2 + j 2) 2 The steady-state response is ( 5∠45° ) = (2 20 2∠45° ) 2 ( 5∠45° ) = 12.5∠ − 45° V vo ( t ) = 12.5cos ( 2 t − 45° ) V (Checked using LNAP, 12/29/02) P 14.11-3 The transfer function of the circuit is H ( s ) = L −130 t e −5t u (t )  =   30 ( s + 5) 2 . The circuit is stable so we can determine the network function as H (ω ) = H ( s ) s = j ω = 30 ( s + 5) = 2 s= jω 30 (5 + j ω ) 2 The phasor of the output is Vo (ω ) = 30 ( 5 + j 3) The steady-state response is 2 (10∠0° ) = 30 ( 5.83∠31° ) 2 (10∠0° ) = 8.82∠ − 62° V vo ( t ) = 8.82 cos ( 3 t − 62° ) V 14-40
• PSpice Problems SP 14-1 14-41
• SP 14-2 v(t ) = A + B e −t / τ for t > 0 ⇒ 7.2 = A + B    ⇒ B = −0.8 V −∞ ⇒ A = 8.0 V  8.0 = v(∞) = A + B e  0.05  8 − 7.7728  7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⇒ − = ln   = −1.25878 0.8 τ   0.05 ⇒ τ= = 39.72 ms 1.25878 7.2 = v(0) = A + B e 0 Therefore v(t ) = 8 − 0.8 e −t / 0.03972 for t > 0 14-42
• SP 14-3 i (t ) = A + B e −t / τ 0 = i (0) = A + B e 0 for t > 0 ⇒ 0 = A+ B 4 × 10−3 = i (∞) = A + B e −∞ ⇒ A = 4 × 10−3   −3  ⇒ B = −4 × 10 A A  2.4514 ×10−3 = v(5 ×10−6 ) = ( 4 × 10−3 ) − ( 4 × 10−3 ) e ⇒ − 5 × 10−6 ⇒ τ= Therefore τ ( ) − 5×10−6 / τ  ( 4 − 2.4514 ) × 10−3  = ln   = −0.94894 4 × 10−3   5 × 10−6 = 5.269 µ s 0.94894 i (t ) = 4 − 4 e −t / 5.269×10 −6 for t > 0 14-43
• SP 14-4 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) 14-44
• V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom. 14-45
• SP 14-5 Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) 14-46
• V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom. 14-47
• Verification Problems VP 14-1 v L (t ) = 3 iC (t ) = d i L ( t ) = −6 e − 2.1t − 2 e −15.9 t dt 1 d v C ( t ) = −0.092 e − 2.1t − 0.575 e −15.9 t 75 dt v R1 ( t ) = 12 − v L ( t ) = 12 + 6 e − 2.1t + 2 e −15.9 t i R2 (t ) = 12 − ( v L ( t ) + v C ( t ) ) i R3 (t ) = Thus, 6 vC (t ) 6 = 1 + 0.456 e − 2.1t − 0.123 e −15.9 t = 1 + 0.548 e − 2.1t + 0.452 e −15.9 t −12 + v L ( t ) + v R1 ( t ) = 0 and i R 2 ( t ) = i C ( t ) +i R 3 ( t ) as required. The analysis is correct. 14-48
• VP 14-2 18 20 and I 2 ( s ) = 3 3 s− s− 4 4     12 1  18   18 20  +  −  + 6  = 0 (ok) s 2s  s − 3   s − 3 s − 3  4  4 4         18   20   18  20 −6  − +3 −4 = 0 (ok) 3 3  3  3 s− s−  s−  s−  4 4  4  4  I1 (s) = KVL for left mesh: KVL for right mesh: The analysis is correct. VP 14-3 Initial value of IL (s): lim s+2 s 2 = 1 (ok) s→∞ s +s+5 Final value of IL (s): lim s+2 s 2 = 0 (ok) s→0 s +s+5 Initial value of VC (s): lim −20 ( s + 2 ) s = 0 (not ok) s→∞ s ( s 2 + s + 5) Final value of VC (s): lim −20 ( s + 2 ) s = −8 (not ok) s→0 s ( s 2 + s + 5) 14-49
• Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s) was calculated as − VC ( s ) = − 20 20 8 I L ( s ) instead of − I L ( s ) + . After correcting this error s s s 20  s + 2  8  + . s  s2 + s+5 s Initial value of VC (s):  −20 ( s + 2 ) 8  s +  = 8 (ok) s → ∞  s ( s 2 + s + 5) s    Final value of VC (s):  −20 ( s + 2 ) 8  +  = 0 (ok) s s → 0  s ( s 2 + s + 5) s    lim lim 14-50
• Design Problems DP 14-1 Equating the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response: kR 5 L = Vo ( s ) = 2 R 1 ( s + 4) s2 + s+ L LC Equating the poles: 2 R 4 R − ±   − L  L  LC s 1,2 = = −4 ± 0 2 Summarizing the results of these comparisons: R = 4, R = 2L 2 kR and =5 L LC Pick L = 1 H, then k = 0.625 V/V, R = 8 Ω and C = 0.0625 F. DP 14-2 Equating the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response: 14-51
• kR 10 10 L Vo ( s ) = = = 2 2 R 1 ( s + 4 ) + 4 s + 8 s + 20 s2 + s+ L LC Equating coefficients: R 1 kR = 8, = 20, and = 10 L LC L Pick L = 1 H, then k = 1.25 V/V, R = 8 Ω and C = 0.05 F. DP 14-3 Equating the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response: kR 5 5 10 L Vo ( s ) = = − = 2 R 1 ( s + 2) ( s + 4) s + 6 s + 8 s2 + s+ L LC Equating coefficients: R 1 kR = 6, = 8, and = 10 L LC L Pick L = 1 H, then k = 1.667 V/V, R = 6 Ω and C = 0.125 F. DP 14-4 14-52
• Comparing the Laplace transform of the step response of the give circuit to the Laplace transform of the given step response: kR 5 5 10 s + 30 L Vo ( s ) = ≠ + = 2 R 1 ( s + 2) ( s + 4) s + 6 s + 8 s2 + s+ L LC These two functions can not be made equal by any choice of k, R, C and L because the numerators have different forms. DP 14-5 a) Use voltage division to get R2 Vo ( s ) = V1 ( s ) sC2 R 2 +1 R1 R2 + sC1 R1 +1 sC2 R 2 +1 1   s+   C1 R1 C1   = R1 + R2 C1 +C2  s +   R1 R 2 ( C1 + C2 )    b) To make the natural response be zero, we eliminate the pole by causing it to cancel the zero. − R1 + R 2 1 =− C1 R1 R1 R 2 (C1 +C2 ) ⇒ C2 R1 = C1 R 2 1 c) Let v1 (t ) = u (t ) ⇒ V1 ( s ) = . Then s   1   s+ R1C1 C1  K2  K1 Vo ( s ) =   = s + R1 + R 2 C1 + C2   R1 + R 2   s s + s+  R1 R 2 (C1 +C2 )      R1 R 2 (C1 +C2 )   where K1 = R2 R1 + R 2 and K 2 = R2 C1 − C1 + C2 R1 + R 2 Then  R2  C R2  −t  1 vo (t ) =  + −  e τ  u (t )  R1 + R 2  C1 + C2 R 1 + R 2       14-53
• where τ = R1 R 2 (C1 + C2 ) R1 + R 2 . To make the step response be proportional to the step in put, we require R2 C1 = C1 + C2 R1 + R 2 Then vo (t ) = R2 R1 + R 2 u (t ) DP 14-6 The initial conditions are vc (0) = −0.4 V and i (0) = 0 A . Consider the circuit after t = 0. A source transformation yields The mesh equations are Solving for I 2 ( s ) yields 8 0.4  8  2 +  I1 ( s ) − I 2 ( s ) = s s  s 8 8  − I1 ( s ) +  Ls +  I 2 ( s ) = 0 s s  I2 ( s ) = 1.6 s ( Ls + 4 Ls +8) 2 Therefore, the characteristic equation is s 2 + 4s + 8 =0 L We require complex roots with significant damping. Try L = 1 H. Then I ( s) = 1.6 0.2 −0.2( s + 4) 0.2 −0.2( s + 2) 0.8 = + 2 = + − 2 ( s + 2) + 4 ( s + 2) 2 + 4 s ( s + 4 s +8) s s + 4 s +8 s 2 Finally i (t ) = 0.2 − 0.2e −2t cos 2t − 0.4e−2t sin 2t  u ( t ) A   14-54
• Chapter 15: – Fourier Series Exercises Ex. 15.3-1 Notice that f (t − T ) = f1 (t − T ) + f 2 (t − T ) = f1 (t ) + f 2 (t ) = f ( t ) Therefore, f(t) is a periodic function having the same period, T. Next f ( t ) = k1 f1 ( t ) + k2 f 2 ( t ) ∞   = k1  a10 + ∑ ( a1n cos ( n ω 0 t ) + b1n sin ( n ω 0 t ) )  n =1   ∞   + k2  a20 + ∑ ( a2 n cos ( n ω 0 t ) + b2 n sin ( n ω 0 t ) )  n =1   ∞ = ( k1 a10 + k2 a20 ) + ∑ ( ( k1 a1n + k2 a2 n ) cos ( n ω 0 t ) + ( k1 b1n + k2 b2 n ) sin ( n ω 0 t ) ) n =1 Ex. 15.3-1 f(t) = K is a Fourier Series. The coefficients are a0 = K; an = bn = 0 for n ≥ 1. Ex. 15.3-2 f(t) = Acosw0t is a Fourier Series. a1 = A and all other coefficients are zero. 15-1
• Ex. 15.4-1 2π π  π T = 4   = , ω0 = = 4 rad s T 8 2 Set origin at t = 0, so have an odd function; then an = 0 for n = 0,1, . . . Also, f(t) has half wave symmetry, so bn = 0 for n = even. For odd n, we have bn = 2 T2 2 0 2 T2 ∫−T 2 f (t ) sin( n ω0 t ) dt = − T ∫−T 2 sin( n ω0 t ) dt + T ∫0 sin ( n ω0 t ) dt T 4 T = ∫0 2 sin ( n ω 0 t ) dt T 4 4 (1−cos( n ω 0 T )) = = n = 1, 3, 5, . . . 2π nf 0T nπ Finally, f (t ) = 4 N 1 ∑ sin nω 0t; π n n n odd and ω 0 = 4 rad s Ex. 15.4-2 T = π , ω0 = 2π =2 T  a0 = 0 , an = 0 for all n odd function with quarter wave symmety ⇒  n = even bn = 0  −2t 0<t <π π 6 8 π4  bn = ∫0 f (t ) sin nω 0t dt where f (t ) =  6 π  π ≤t <π  −2 6 4  −24 1  nπ  Thus bn = 2 2 sin   π n  3  −24 N 1  nπ  so f (t ) = 2 ∑ 2 sin   sin (2nt ) π n =1 n  3  odd n 15-2
• Ex. 15.4-3 a) is neither even nor odd. f(t) will contain both sine and cosine terms 1 b) wave symmetry ⇒ no even harmonics 4 c) average value of f(t) = 0 ⇒ a0 = 0 Ex. 15.5-1 T = 2 s, ω 0 = Cn = 2π = π rad/s T − jnπ t 1 2 1 1 − jnπ t 1 2 f (t )e dt = ∫0 e dt − ∫1 e − jnπ t dt 2 ∫0 2 2 1 1  −e − jnπ +1+ e − j 2 nπ − e− jnπ  = (1− e− jnπ ) =   2 jnπ jnπ  2  Cn =  jnπ  0  n odd n even Finally, f (t ) = 2 jπ 1 1  jπ t 1 j 3π t 1 j 5π t  e + e + e + ... − e − jπ t − e− j 3π t − e− j 5π t −...   3 5 3 5   Ex. 15.5-3 Cn = 1 T 4 − jω0nt 1  T  − j 2π nt T ∫−T 4 e dt = T  − j 2π n  e T   (n −1)   (−1) 2   πn  Cn =  0  12     T 4 −T = 4 1 e − jπ n / 2 − e jπ n / 2   − j 2π n  n odd n even , n ≠ 0 n =0 15-3
• Ex. 15.6-1 Use the “stem plot” in Matlab to plot the required Fourier spectra: % Fourier Spectrum of a Pulse Train A = 8; T = 4; d = T/8; pi = 3.14159; w0 = 2*pi/T; % pulse amplitude % period % pulse width %fundamental frequency N = 49; n = linspace(-N,N,2*N+1); x = n*w0*d/2; % Eqn.15.6-3. Division by zero when n=0 causes Cn(N+1) to be NaN. Cn = (A*d/T)*sin(x)./x; Cn(N+1)=A*d/T; % Fix Cn(N+1); sin(0)/0 = 1 % Plot the spectrum using a stem plot stem(n,Cn,'filled'); xlabel('n'); ylabel('|Cn|'); title('Fourier Spectrum of Pules Train with d = T/8'); 15-4
• Ex. 15.8-1 ω 0 = 4 rad/s From Example 15.4-1: N vs (t ) = 3.24 ∑ n =1 odd n 1  nπ  sin n2  2 1 1 1    sin nω 0t = 3.24  sin 4 t − sin12 t + sin 20 t − sin 28 t 9 25 49      The network function of the circuit is Vo (ω ) 1 1 1 jω C = = Vs (ω ) R + 1 1 + jω C R 1 + jω 4 jω C Evaluating the network function at the frequencies of the input series 1 H ( n4 ) = n = 1,3,5... 1 + j 16 n n H(n4) 1 0.062∠-86° 3 0.021∠-89° 5 0.012∠-89° 7 0.0009∠-89° Using superposition 0.021 0.012 0.0009   vo (t ) = 3.24  ( 0.062 ) sin ( 4 t −86° )− sin (12 t −89° ) + sin ( 20 t −89° )− sin ( 28 t −89° )  9 25 49   H (ω ) = = vo (t ) = ( 0.2009 ) sin ( 4 t − 86° ) − (.00756 ) sin (12 t − 89° ) + ( 0.00156 ) sin ( 20 t − 89° ) − ( 5.95 × 10−5 ) sin ( 28 t − 89° ) Discarding the terms that are smaller than 25 of the fundamental term leaves vo (t ) = ( 0.2009 ) sin ( 4 t − 86° ) − (.00756 ) sin (12 t − 89° ) 15-5
• Ex. 15.9-1 f (t ) = e − at u (t ) F (ω ) = ∫ +∞ −∞ f (t ) e jω t ∞ dt = ∫ e e − at jω t 0 ∞ e( ) 1 = dt = − ( a + jω ) 0 a + jω − a + jω t Ex. 15.10-1 F { f ( at )} = ∫ ∞ −∞ f ( at ) e− jω t dt Let τ = at ⇒ t = F { f ( at )} = ∫ ∞ −∞ τ a f (τ ) e − jωτ a d τ a = 1 ∞ 1 ω  − j (ω a )τ dτ = F   ∫−∞ f (τ ) e a a a Ex. 15.10-2 f (t ) = 1 2π 1 ω ∫ ( 2πδ (ω ) A) e dt = 2π ∫ ( 2πδ (ω ) A) dt = A ∞ 0+ j t 0− −∞ Ex. 15.11-1 F -1 {δ (ω − ω 0 )} = 1 2π ∫ ∞ −∞ δ (ω − ω 0 ) e jω t dt = Take the Fourier Transform of both sides to get:   e jω0t + e − jω0t  F { A cos ω 0t} = F  A  2    ( 1 jω0t e 2π ) F e jω0t = 2πδ (ω − ω 0 )  A A  F e jω0t + F e − jω0t = ( 2πδ (ω − ω 0 ) + 2πδ (ω + ω 0 ) )  = 2  2  = Aπδ (ω − ω 0 ) + Aπδ (ω + ω 0 ) ( ( ) ( )) 15-6
• Ex. 15.12-1 a) b) Vin (ω ) = Win = 1 Wout = ∴η = ∫ π ∞ 0 1 π ∫ 120 24 + jω ⇒ Vin (ω ) = 1202 14400 = 2 2 24 + ω 576 + ω 2 ∞ 14400 14000  1 −1  ω   dω =  24 tan  24   = 300 J 2 576 + ω π    0 48 24 48 14400 14000  1 −1  ω   dω =  24 tan  24   = 61.3 J 2 576 + ω π    24 Wout 61.3 ×100% = ×100% = 20.5% Win 300 Ex. 15.13-1 f + ( t ) = te − at f − ( t ) = te at ∴ F + (s) = Then F (ω ) = F + ( s ) s = jω + F − (s) f − ( −t ) = −te− at ⇒ 1 (s + a) s =− jω = = 2 and F − ( s ) = 1 (s + a) + 2 s = jω 1 ( a + jω ) 2 − −1 (s + a) 2 −1 (s + a) 2 1 ( a − jω ) 2 s =− jω = − j 4aω (a 2 +ω2 ) 2 15-7
• Problems Section 15.3: The Fourier Series P15.3-1 T = 2 s ⇒ ω0 = 2π = π rad/s and f (t ) = t 2 for 0 ≤ t ≤ 2 . The coefficients of the Fourier 2 series are given by: 1 2 2 t dt = 4 3 2 ∫0 2 2 4 an = ∫0 t 2 cos nπ t dt = 2 2 ( nπ ) a0 = bn = ∴ f (t ) = −4 2 2 2 ∫0 t sin nπ t dt = nπ 2 4 4 N 1 4 ∞ 1 + 2 ∑ 2 cos nπ t − ∑ sin nπ t π n =1 n 3 π n =1 n 15-8
• P15.3-2 an = T  2π   2π   2 T 4 2  ∫0 cos  n t  dt + ∫T 2 cos  n t  dt  T  T   T   4 1 = nπ =   2π  sin  n T t     T 4 0 T   2π  2  + 2sin  n t   T T  4 1    nπ   nπ (sin  2  − 0) + 2 (sin nπ ) −sin  2 nπ             (−1) n +1 1 2 odd n   nπ  sin  = −  =  nπ nπ  2   even n  0 ( )  2π   2 sin  n t  dt   T   T T  1  2π  4   2π  2  cos  n t  + 2 cos  n t  = − nπ   T 0  T T   4 nπ  1  =− (2 cos (nπ ) −1) − cos 2  nπ    3 n is odd  nπ   2 n = 2,6,10,… = −  nπ n = 4,8,12,… 0   bn = 2 T 4 2π  ∫ sin n T t dt + T 0 T 2 T 4 ∫ 15-9
• P15.3-3 a 0 = average value of f ( t ) = t  f ( t ) = A 1 −   T an = 2 T ∫ T 0 t   2π A 1 −  cos  n  T  T A 2 for 0 ≤ t ≤ T 2A T   2π t  dt =  ∫ 0 cos  n T T    1 T   2π t  dt − ∫ t cos  n T 0   T   t  dt    T    2π   2π   2π    cos  n t+n t  sin  n t 2A 1  T   T   T   = 2 0 − T  T  2π    n    T  0   −A = 2 2 cos ( 2nπ ) − cos ( 0 ) + 2nπ sin ( 2nπ ) − 0   2n π  =0 bn = 2 T ∫ T 0 t   2π A  1 −  sin  n  T  T 2 A  T  2π  t  dt = sin  n T ∫0   T  1 T   2π t  dt − ∫ t sin  n 0 T   T   t  dt    T    2π   2π   2π    t−n t  cos  n t sin  n 2A 1  T   T   T   = 0− 2  T  T  2π    n     T  0   −A = 2 2 ( sin ( 2nπ ) − sin ( 0 ) ) − ( 2nπ cos ( 2nπ ) − 0 )   2n π  A = nπ f (t ) = A ∞ A  2π sin  n +∑ 2 n =1 n π  T  t  15-10
• P15.3-4 T = 2 s, ω 0 = 2π = π rad/s , a 0 = average value of f ( t ) = 1 , 2 f (t ) = t 2 an = 2 ∫ 2 0 t cos ( n π t ) dt = = for 0 ≤ t ≤ 2 cos ( n π t ) + ( n π t ) sin ( n π t ) ( nπ ) 2 2 0 1  cos ( 2nπ ) − cos ( 0 ) + 2nπ sin ( 2nπ ) − 0   n π2  2 =0 2 bn = 2 ∫ 2 0 t sin ( n π t ) dt = sin ( n π t ) − ( n π t ) cos ( n π t ) (nπ ) 2 2 0 1 ( sin ( 2nπ ) − sin ( 0 ) ) − ( 2nπ cos ( 2nπ ) − 0 )   n π2  −2 = nπ = 2 ∞ f (t ) = 1 − ∑ n =1 2  2π sin  n nπ  T  t  Use Matlab to check this answer: % P15.3-4 pi=3.14159; A=2; T=2; % input waveform parameters % period w0=2*pi/T; tf=2*T; dt=tf/200; t=0:dt:tf; % % % % a0=A/2; v1=0*t+a0; % avarage value of input % initialize input as vector fundamental frequency, rad/s final time time increment time, s for n=1:1:51 % for each term in the an=0; % specify series bn=-A/pi/n; cn=sqrt(an*an + bn*bn); % convert thetan=-atan2(bn,an); v1=v1+cn*cos(n*w0*t+thetan); % add the Fourier series end Fourier series ... coefficients of the input to magnitude and angle form next term of the input 15-11
• plot(t, v1,'black') % plot the Fourier series grid xlabel('t, s') ylabel('f(t)') title('P15.3-4') 15-12
• Section 15-4: Symmetry of the Function f(t) 2π π = rad/s . 4 2 The coefficients of the Fourier series are: 15.4-1 T = 4 s ⇒ ωo = a0 = average value of vd ( t ) = 0 an = 0 because vd(t) is an odd function of t. bn = 1 4  π  ∫0 ( 6 − 3 t ) sin  n 2 t  dt 2   4 3 4  π   π  = 3∫ sin  n t  dt − ∫ t sin  n t  dt 0 2 0  2   2  4   π     − cos  n 2 t   3  1  π   π   π    − = 3  2 2 sin  n t  −  n t  cos  n t   π  2   2   2    2n π n    4 0 2    0 6 = ( −1 + cos ( 2nπ ) ) − n26 2 ( sin ( 2nπ ) − 0 ) − ( 2 n π cos ( 2nπ ) ) nπ π 12 = nπ 4 ( ) The Fourier series is: ∞ vd ( t ) = ∑ n =1 12  π  sin  n t  nπ  2  P15.4-2 ∞ vc ( t ) = vd ( t − 1) − 6 = −6 + ∑ n =1 ∞ 12 12 π  π   π sin  n ( t − 1)  = −6 + ∑ sin  n t − n  nπ 2  2   2 n =1 n π 15-13
• P15.4-3 2π 1000 π π = rad/s = krad/s .006 3 3 The coefficients of the Fourier series are: 3× 2 1 V a0 = average value of va ( t ) = 2 = 6 2 bn = 0 because va(t) is an even function of t. T = 6 ms = 0.006 s ⇒ ω o =  2  0.001  1000π  an = 2  t  dt  ∫0 ( 3 − 3000 t ) cos  n 3  0.006    0.001 1  1000π   1000π t  dt − ( 2 ×106 ) ∫ t cos  n = 2000 ∫ cos  n 0 0 3 3      1000π  sin  n 3 = 2000    n 1000π  3   t  − 1000 n 2106 π 2 9  t  dt  0.001    1000π   1000π   1000π    t+n t  sin  n t   cos  n  3 3 3          0   3 9   π   π     π   π    = 2000   sin  n 3  − 0  − n 2103 π 2   cos  n 3  − 1 +   n 3  sin  n 3  − 0                 n1000 π   6  π   18    π  6  π sin  n  −  2 2   cos  n  − 1 − sin  n  = nπ  3   n π   3   nπ  3  18    π  = −  2 2   cos  n  − 1  n π   3  The Fourier series is va ( t ) = 1 ∞ 18   nπ    1000 π + ∑ 2 2 1 − cos    cos  n 2 n =1 n π  3  3    t  P15.4-4 1 ∞ 18   nπ    π  vb ( t ) = va ( t − 2 ) − 1 = −1 + + ∑ 2 2  1 − cos    cos  n ( t − 2 )  2 n =1 n π   3   3  1 ∞ 18   nπ = − + ∑ 2 2  1 − cos  2 n =1 n π   3 2π    1000 π t−n   cos  n  3 3    15-14
• P15.4-5 Choose t 0 = − π 2π =1 2π average value: a0 = 0 T = 2π , ω 0 = 2 T f ( t ) sin nω 0 t dt T ∫0 an = 0 since have odd function bn = −π < t < π f (t ) = t bn = 2 2π π ∫ π t sin nt dt − π 1  sin nt t cos nt  =  2 −  n π n  −π 1 π π  b1 =  +  = 2 π 1 1 b2 = −1 b3 = 2 3 P15.4-6 T = 8 s, ω 0 = π 4 rad/s bn = 0 because f ( t ) is an even functon a0 = average = ( 2×2 ) − 2×1 = 1 4 8 4 T2 f ( t ) cos n ω 0 t dt T ∫0 π π 2  4  1 =  ∫0 2 cos n t dt − ∫1 cos n t dt  4 4 8   2  nπ nπ  = 3 sin 4 −sin 2  nπ   a1 = .714, a2 = .955, a3 = .662 an = 15-15
• P15.4-7 ω 0 = 2ω , T = π ω π ω 2ω 2A a0 = ∫−π A cos ω t dt = π π 2ω an = 2ω π π ∫ 2ω −π 2ω A cos ω t cos 2nω t dt π 2ω A  sin ( 2n −1)ω t sin ( 2n +1)ω t  2ω = +   2( 2n +1)ω  − π π  2( 2n −1)ω 2ω π π  sin 2n −1) sin ( 2n +1)  2A  ( 2+ 2 =   2n +1  π  2n −1   2A  π π = 2 ( 2n +1) sin ( 2n −1) 2 −( 2n −1) sin ( 2n −1) 2  π ( 4n −1)   =− 4A cos( nπ ) π ( 4n 2 −1) =− 4 A( −1) n π ( 4n 2 −1) bn = 0 due to symmetry P15.4-8 2π = 5 π rad s T 0 ≤ t ≤ .1  A cos ω 0t  .1 ≤ t < .3 f (t ) = 0  A cos ω t .3 ≤ t ≤ .4 0  T = 0.4 s, ⇒ ω 0 = Choose period − .1 ≤ t ≤ .3 for integral 1 .1 A cos ω 0t = A π T ∫−.1 2 .1 an = ∫−.1 A cos ω 0t cos nω 0t dt T a0 = 15-16
• .1 a1 = 5 A∫−.1 cos 2 ω 0t dt = A 2 .1 an = 5 A∫−.1 cos ω 0t cos nω 0t dt = 5 A∫−.1 1 [ cos 5π (1+ n)t + cos 5π (1− n)t ] dt 2 2 A cos (nπ / 2) n ≠1 = 1− n 2 π bn =0 because the function is even. .1 P15.4-9 a0 = 0 because the average value is zero an = 0 because the function is odd bn = 0 for even due to Next:  nπ 8 sin  T 4  2 bn = ∫−T 4 t sin ( nω 0 t ) dt = 1 wave symmetry 4   nπ   8  − 4nπ cos    2 2   2  = n π  n 2π 2 − 8  n2 π 2  for n = 1,5,9, ... for n = 3, 7,11, ... Section 15.5: Exponential Form of the Fourier Series P15.5-1 2π = 2π , the coefficients of the complex Fourier series are given by: 1 1  e jπ t − e − jπ t  − j 2π nt 1 1 Cn = ∫ A sin (π t ) e − j 2π nt dt = ∫ A  dt e  0  1 0 2j   A 1 − jπ ( 2 n−1) t − jπ ( 2 n+1) t e dt = −e 2 j ∫0 T = 1 ⇒ ωo = ( ) 1 − jπ 2 n −1) t − jπ 2 n +1) t  Ae ( e ( −2 A = −   = 2 2 j  − jπ ( 2n − 1) − jπ ( 2n + 1)    0 π (4n − 1) where we have used e± j 2π n = 1 and e j π = e− j π . 15-17
• P15.5-2 2π 1 1 A  − j Cn = ∫  t  e T T 0 T  Recall the formula for integrating by parts: t2 ∫t 1 dv 2π − j nt = e T dt . nt dt = A 1 T2 ∫0 t e −j 2π nt T dt t2 t u dv = u v t2 − ∫ v du . Take u = t and t1 1 When n ≠ 0 , we get  2π  t e − j T nt A Cn = 2  T  − j 2π n  T   T + 0 T −j 1 e 2 π ∫0 j n T  2π  − j nt − j 2π n AT e e T =  + T  − j 2 π n  2 π 2 n j   T   2π nt  T dt           0 T    − j 2π n  AT e e − j 2π n − 1  = + T  − j 2 π n  2 π 2   n  j    T    A = j 2π n Now for n = 0 we have C0 = 1 TA A ∫0 T t dt = 2 T Finally, f (t ) = A A +j 2 2π n =∞ 1 jn ∑ ne n =−∞ 2π t T n≠0 15-18
• P15.5-3 d /2 2π − jn t d /2 T dt e −d / 2 A Cn = ∫ T π  − j n 2π t   j nπ d − jn d T  A e A e T e T =  =  −  2π T  − j n 2π  T  j n 2π jn T  −d / 2 T T        π  jnπ d − jn d T −e T A e =  nπ  2j  A  nπ d  = sin   nπ  T   nπ d  sin    Ad   T  =   T  nπ d T      P15.5-4 Cn = Let τ = t − td , then t = τ + td . 1 T 1 = T Cn = t0 +T −td ∫t −t 0 d t0 +T −td ∫t −t 0 d e − j n ω o td = T ( ( a f (τ ) + b ) e− j nω (τ +t )dτ o But ∫t −t 0 d be − j n ω oτ d ( a f (τ ) + b ) e− j nω τ e− j nω t dτ o t0 +T −td ∫t −t = a e − j n ω o td t0 +T −td 1 t0 +T ( a f ( t − td ) + b ) e− j nωot dt T ∫t0 0 d 1 )T ∫ o d ( a f (τ ) + b ) e− j nω τ dτ o t0 +T −td t 0 −t d ( f (τ ) e − j n ωoτ dτ + e− j n ωotd t +T −td  e − j nωoτ  0 dτ = b    − j n ω o  t −t  0 d 1 )T ∫ t0 +T −td t0 − t d b e− j n ωoτ dτ 0 n ≠ 0 so = b = 0 C0 = a C0 + b and C n = a e − j n ω o td C n n≠0 15-19
• P15.5-5 T = 8 s, ω 0 = 2 × 2 − 2 (1× 1) 1 2π π = rad/s, C 0 = average value = = T 4 8 4 The coefficients of the exponential Fourier series are calculated as nπ nπ nπ t t t −j −j −j 1 2  1  −1 C n =  ∫ − 1× e 4 dt + ∫ 2 × e 4 dt + ∫ − 1× e 4 dt  −1 1 8  −2  −1 1 2   nπ nπ nπ −j −j −j t t t   4 4 4 1 e e e  =  −1 × + 2× + ( −1) × nπ nπ nπ  8 −j −j −j  4 −2 4 −1 4 1   −j = 2 nπ nπ π nπ π nπ  j n4π −j j j   − j n4   − j n2 2 4 −e +e −e 4  e − e  − 2  e           and C −n − nπ − nπ − nπ −j t −j t −j t 1 2  1  −1 4 4 =  ∫ − 1× e dt + ∫ 2 × e dt + ∫ − 1× e 4 dt  −2 −1 1 8  −1 1 2   nπ nπ nπ j t j t j t  4 4 4 1 e e e  =  −1× + 2× + ( −1) × nπ nπ nπ  8 j j j  4 −2 4 −1 4 1   = j 2 nπ nπ  − j n4π −j −e 2 e    nπ −j   j nπ − 2 e 4 −e 4    π nπ j   j n2 +e −e 4        = −C n The function is represented as ∞ −∞ n =1 n =−1 f ( t ) = C0 + ∑ Cn e j n ω0 t + ∑ C− n e − j n ω0 t 15-20
• This result can be checked using MATLAB: pi = 3.14159; N=100; T = 8; t = linspace(0,2*T,200); c0 = 1/4; w0 = 2*pi/T; % % % % period time average value fundamental frequency for n = 1: N C(n) = -j*((exp(+j*n*pi/4)-exp(+j*n*pi/2))-2*(exp(-j*n*pi/4)exp(+j*n*pi/4))+(exp(-j*n*pi/2)-exp(-j*n*pi/4)))/(2*pi*n); end for i=1:length(t) f(i)=c0; for n=1:length(C) f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+C(n)*exp(-j*n*w0*t(i)); end end plot(t,f,'black'); xlabel('t, sec'); ylabel('f(t)'); 15-21
• Alternately, this result can be checked using Mathcad: N := 15 n := 1 , 2 .. N T := 8 ω := 2 d := T C := π T i := 1 , 2 .. 400 200 ⌠  ⌡ m := 1 , 2 .. N −1 t := d ⋅ i i ⌠ −1⋅ exp( −j⋅ n ⋅ ω⋅ t ) dt +  ⌡ 1 2 −1 −2 ⌠ 2⋅ exp( −j ⋅ n ⋅ ω⋅ t) dt +  −1 exp( −j ⋅ n ⋅ ω⋅ t ) dt ⌡ 1 n T ⌠  ⌡ C := −1 ⌠ −1⋅ exp( j ⋅ m⋅ ω⋅ t ) dt +  1 ⌡ −2 ⌠ 2⋅ exp( j ⋅ m⋅ ω⋅ t) dt +  ⌡ −1 −1 exp( j ⋅ m⋅ ω⋅ t ) dt 1 m T N f ( i) := 2 ∑ N ( ) ∑ C ⋅ exp j ⋅ n ⋅ ω⋅ t + n i n =1 ( C ⋅ exp −1⋅ j ⋅ m⋅ ω⋅ t m ) i m=1 C 0.357 0.357 1.685 0.477 0.477 1.745 0.331 0.331 1.807 0 0 1.856 0.199 0.199 1.88 0.159 0.159 1.872 0.051 0.051 1.831 0 0 1.767 0.04 0.04 1.693 0.095 0.095 1.628 0.09 0.09 1.589 0 0 1.589 0.076 0.076 1.633 0.068 0.068 1.717 0.024 0.024 1.825 n 2 f ( i) 0 2 100 200 i 300 400 m = f ( i) = C = 1.643 15-22
• P15.5-6 The function shown at right is related to the given function by v ( t ) = −v1 ( t + 1) − 6 (Multiply by –1 to flip v1 upside-down; subtract 6 to fix the average value; replace t by t+1 to shift to the left by 1 s.) From Table 15.5-1 v1 ( t ) = ∞ ∑ n =−∞ ∞ j A ( −1) j n ω 0 t j 6 ( −1) j n π t e = ∑ e 2 nπ nπ n =−∞ n n Therefore v ( t ) = −6 − ∞ ∑ n =−∞ n n ∞  j 6 ( −1) j n π (t +1) j 6 ( −1) j n π  j n π t 2 e e 2 e 2 = −6 − ∑   nπ  nπ n =−∞   The coefficients of this series are: j 6 ( −1) j n π C0 = −6 and Cn = − e 2 nπ This result can be checked using Matlab: n pi = 3.14159; N=100; A = 6; T = 4; t = linspace(0,2*T,200); c0 = -6; w0 = 2*pi/T; % amplitude % period % time % average value % fundamental frequency for n = 1: N C(n) = (-j*A*(-1)^n/n/pi)*exp(+j*n*pi/2); D(n) = (+j*A*(-1)^n/n/pi)*exp(-j*n*pi/2); end for i=1:length(t) f(i)=c0; for n=1:length(C) f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i)); end end plot(t,f,'black'); xlabel('t, sec'