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• Dr. Holbert Lecture 12 EEE 202
• ### Unit i

1. 1. THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS
2. 2. A Resistor in the s Domain+ R iv v=Ri (Ohm’s Law). + R I V V(s)=RI(s
3. 3. An Inductor in the s Domain diInitial current of I 0 v=L dt + V(s)=L[sI(s)-i(0-)]=sLI(s)-LI0 v L i V ( s) I0 I ( s) = + sL s I + sL I + V sL I0 V s + LI0
4. 4. A Capacitor in the s Domain dv Initially charged to V0 volts. i=C dt − I ( s ) = C [ sV ( s ) − v (0 )] = sCV ( s ) − CV0  1  V0V ( s) =  ÷I ( s ) +  sC  I s + I 1/sC + i + 1/sC CV0v C V V + V0/s
5. 5. The Natural Response of an RC Circuit t=0 + 1/sC I + C + R i v R V0 V0/s + V V0 1 = I ( s ) + RI ( s ) s sC V0 CV0 R I ( s) = = RCs + 1 s + ( 1 RC ) V0 − − i = e u( t ) ⇒ v = Ri = V0e u( t ) t t RC RC R
6. 6. The Step Response of a Parallel Circuit
7. 7. ?
8. 8. The Step Response of a Parallel Circuit V V I dcsCV + + = R sL s I dc CV= s +( 2 1 RC ) s + ( 1 LC ) I dc LCIL = s[ s + ( 2 1 RC ) s + ( 1 LC ) ]
9. 9. 384 × 105IL = s( s 2 + 64000 s + 16 × 108 ) 384 × 105IL = s( s + 32000 − j 24000)( s + 32000 + j 24000) * K1 K2 K2IL = + + s s + 32000 − j 24000 s + 32000 + j 24000 384 × 105K1 = = 24 × 10−3 16 × 108 384 × 105K2 = = 20 × 10−3 126.870 ( −32000 + j 24000)( j 48000)i L ( t ) = [24 + 40e −32000 t cos(24000t + 126.870 )]u( t )mA
10. 10. Transient Response of a Parallel RLC CircuitReplacing the dc current source with a sinusoidal current source sI m i g = I m cos ω t A ⇒ I g ( s ) = 2 2 s +ω V ( s) = ( 1C ) s I g ( s) 2 s + ( 1 RC ) s + ( 1 LC ) V ( s) = ( ) Im C s2 ( s 2 + ω 2 )[ s 2 + ( 1 RC ) s + ( 1 LC )] I L ( s) = V ( s) = 2 LC s( Im ) sL ( s + ω 2 )[ s 2 + ( 1 RC ) s + ( 1 LC )]
11. 11. I m = 24mA, ω = 40000rad / s 384 × 105 sI L (s) = 2 ( s + 16 × 108 )( s 2 + 64000 s + 16 × 108 ) * * K1 K1 K2 K2I L (s) = + + + s − j 40000 s + j 40000 s + 32000 − j 24000 s + 32000 + j 24000 384 × 105 ( j 40000)K1 = = 7.5 × 10−3 − 900 ( j 80000)(32000 + j16000)(32000 + j 64000) 384 × 105 ( −32000 + j 24000)K2 = = 12.5 × 10−3 900 ( −32000 − j16000)( −32000 + j 64000)( j 48000)i L ( t ) = (15sin 40000t − 25e −32000 t sin 24000t )u( t )mAi Lss = 15sin 40000t mA
12. 12. Mesh Analysis
13. 13. ?
14. 14. Mesh Analysis (cont.)336 = (42 + 8.4 s ) I1 − 42 I 2 s 0 = −42 I1 + (90 + 10 s ) I 2 40( s + 9) 15 14 1 I1 = = − − s( s + 2)( s + 12) s s + 2 s + 12 168 7 8.4 1.4 I2 = = − + s( s + 2)( s + 12) s s + 2 s + 12
15. 15. −2 t −12 ti1 = (15 − 14e −e )u( t ) A, −2 t −12 ti2 = (7 − 8.4e + 1.4e )u( t ) A. 336(90)i1 (∞ ) = = 15 A, 42(48) 15(42)i2 ( ∞ ) = =7A 90
16. 16. Thevenin’s TheoremUse the Thevenin’s theorem to find v c (t)
17. 17. ?
18. 18. Thevenin’s Theorem (cont.) (480 / s )(0.002 s ) 480VTh = = 20 + 0.002 s s + 10 4 0.002 s(20) 80( s + 7500)ZTh = 60 + = 20 + 0.002 s s + 10 4
19. 19. 4 480 /( s + 10 )IC = 4 5 [80( s + 7500) /( s + 10 )] + [(2 × 10 ) / s] 6s 6sIC = 2 6 = 2 s + 10000 s + 25 × 10 ( s + 5000) −30000 6IC = + ( s + 5000) 2 s + 5000 −5000 t −5000 tic ( t ) = ( −30000te + 6e )u( t ) A 5 5 1 2 × 10 6s 12 × 10Vc = IC = 2 = sC s ( s + 5000) ( s + 5000)2 5 −5000 tvc ( t ) = 12 × 10 te u( t )V
20. 20. MUTUAL INDUCTANCE EXAMPLE i2(t)=? − 60 − i1 (0 ) = = 5 A, i2 (0 ) = 0 12
21. 21. ?
22. 22. MUTUAL INDUCTANCE EXAMPLE Using the T-equivalent of the inductors, and s-domain equivalent gives the following circuit
23. 23. (3 + 2 s ) I1 + 2 sI 2 = 10 2 sI1 + (12 + 8 s ) I 2 = 10 2.5 1.25 1.25I2 = = − ( s + 1)( s + 3) s + 1 s + 3 −t −3 ti2 ( t ) = 1.25(e −e )u( t ) A
24. 24. THE TRANSFER FUNCTION The transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input when all the initial conditions are zero. Y ( s)H ( s) = Y(s) is the Laplace transform of the output, X ( s) X(s) is the Laplace transform of the input.
25. 25. THE TRANSFER FUNCTION (cont.) I ( s) 1 H1 ( s ) = = Vg ( s ) R + sL + 1/ sC sC = 2 s LC + RCs + 1 V ( s) 1 H 2 ( s) = = 2 Vg ( s ) s LC + RCs + 1
26. 26. EXAMPLEFind the transfer function V 0 /V g anddetermine the poles and zeros ofH(s).
27. 27. ?
28. 28. EXAMPLEV0 − Vg V0 V0 s + + 6 =0 1000 250 + 0.05 s 10 1000( s + 5000)V0 = 2 V 6 g s + 6000 s + 25 × 10 V0 1000( s + 5000)H ( s) = = 2 Vg s + 6000 s + 25 × 106p1 = −3000 + j 4000,p2 = −3000 − j 4000z1 = −5000
29. 29. Assume that v g (t)=50tu(t). Find v 0 (t). Identify the transient and steady-state components of v 0 (t). 1000( s + 5000) 50V0 ( s ) = H ( s )V g ( s ) = 2 6 2 ( s + 6000 s + 25 × 10 ) s * K1 K1 K2 K3= + + 2 + s + 3000 − j 4000 s + 3000 + j 4000 s s −4 0 −4K1 = 5 5 × 10 79.7 , K 2 = 10, K 3 = −4 × 10 −4 −3000 t 0v0 = [10 5 × 10 e cos(4000t + 79.7 ) −4 + 10t − 4 × 10 ]u( t )V
30. 30. The transient component is generated by the poles of the transfer function and it is: −4 −3000 t 010 5 × 10 e cos(4000t + 79.7 ) The steady-state components are generated by the poles of the driving function (input): −4 (10t − 4 × 10 )u( t )
31. 31. Time Invariant Systems If the input delayed by a seconds, thenL{ x ( t − a )u( t − a )} = e − as X (s) − asY ( s ) = H ( s ) X ( s )e y( t ) = L−1 { Y ( s )} = y( t − a )u( t − a )Therefore, delaying the input by a seconds simply delays theresponse function by a seconds. A circuit that exhibits thischaracteristic is said to be time invariant.
32. 32. Impulse ResponseIf a unit impulse source drives the circuit, the response of thecircuit equals the inverse transform of the transfer function. x( t ) = δ (t ) ⇒ X ( s ) = 1 Y ( s) = H ( s) y( t ) = L −1 { H ( s )} = h(t ) Note that this is also the natural response of the circuit because the application of an impulsive source is equivalent to instantaneously storing energy in the circuit.
33. 33. CONVOLUTION INTEGRAL x(t) y(t) x(t) y(t) t N a b t a bCircuit N is linear with no initial stored energy. If we know theform of x(t), then how is y(t) described? To answer thisquestion, we need to know something about N. Suppose we knowthe impulse response of the system. y( t ) = h( t ) x(t ) = δ (t ) (1) x(t) y(t) t N t
34. 34. Instead of applying the unit impulse at t=0, let us suppose that it isapplied at t=λ. The only change in the output is a time delay. δ (t − λ ) N h( t − λ ) Next, suppose that the unit impulse has some strength other than unity. Let the strength be equal to the value of x(t) when t= λ. Since the circuit is linear, the response should be multiplied by the same constant x(λ) x (λ )δ ( t − λ ) N x (λ )h( t − λ )
35. 35. Now let us sum this latest input over all possible values of λ and use the result as a forcing function for N. From the linearity, the response is the sum of the responses resulting from the use of all possible values of λ +∞ +∞ ∫−∞ x( λ )δ ( t − λ )d λ N ∫−∞ x(λ )h( t − λ )d λFrom the sifting property of the unit impulse, we see that the input issimply x(t) +∞ X(t) N ∫−∞ x(λ )h( t − λ )d λ
36. 36. Our question is now answered. When x(t) is known, and h(t), theunit impulse response of N is known, the response is expressed by +∞ y( t ) = ∫−∞ x ( λ )h( t − λ )d λ This important relation is known as the convolution integral . It is often abbreviated by means of y( t ) = x ( t ) * h( t ) Where the asterisk is read “convolved with”. If we let z=t-λ, then dλ=-dz, and the expression for y(t) becomes −∞ +∞ y( t ) = ∫∞ − x ( t − z )h( z )dz = ∫−∞ x ( t − z )h( z )dz ∞ ∞ y( t ) = x ( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫−∞ x ( t − z )h( z )dz
37. 37. ∞ ∞y( t ) = x( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫−∞ x ( t − z )h( z )dz
38. 38. Convolution and Realizable SystemsFor a physically realizable system, the response of the systemcannot begin before the forcing function is applied. Sinceh(t) is the response of the system when the unit impulse is applied att=0, h(t) cannot exist for t<0. It follows that, in the second integral, theintegrand is zero when z<0; in the first integral, the integrand is zerowhen (t-z) is negative, or when z>t. Therefore, for realizablesystems the convolution integral becomes t ∞y( t ) = x ( t ) * h( t ) = ∫−∞ x ( z )h( t − z )dz = ∫0 x ( t − z )h( z )dz
39. 39. EXAMPLE x(t) −t 1 t h(t) y(t) h( t ) = 2e u( t )x ( t ) = u( t ) − u( t − 1) ∞y( t ) = x ( t ) * h( t ) = ∫0 x ( t − z )h( z )dz ∞ −z= ∫0 [u( t − z ) − u( t − z − 1)][2e u( z )]dz
40. 40. Graphical Method of Convolution
41. 41. Since h(z) does not exist prior to t=0 and vi(t-z) does not existfor z>t, product of these functions has nonzero values only inthe interval of 0<z<t for the case shown where t<1. t y( t ) = ∫0 2e − z dz = 2(1 − e − t ) 0≤ t ≤1When t>1, the nonzero values for the product are obtained in theinterval (t-1)<z<t. t −z −t y( t ) = ∫t −1 2e dz = 2(e − 1)e t >1
42. 42. EXAMPLEApply a unit-step function, x(t)=u(t), as the input to a system whoseimpulse response is h(t) and determine the corresponding outputy(t)=x(t)*h(t).
43. 43. ?
44. 44. h(t)=u(t)-2u(t-1)+u(t-2),
45. 45. When t<0, there is no overlap and y(t)=0 for t<0For 0<t<1, the curves overlap from z=0 to z=t and product is 1.Thus, t y( t ) = ∫0 1dz = t 0< t <1 When 1<t<2, h(t-z) has slid far enough to the right to bring under the step function that part of the negative square extending from 0 to z=t-1. Thus, t −1 t t −1 ty( t ) = ∫0 −1dz + ∫t −11dz = − z 0 + z t −1 = 2 − t , 1< t < 2
46. 46. Finally, when t>2, h(t-z) has slid far enough to the right so that it liesentirely to the right of z=0 t −1 t y( t ) = ∫t − 2 −1dz + ∫t −11dz = 0 t>2
47. 47. Convolution and the Laplace TransformLet F 1 (s) and F 2 (s) be the Laplace transforms of f1(t) and f2(t),respectively. Now, consider the laplace transform of f1(t)*f2(t), L{ f1 ( t ) * f 2 ( t )} = L {∫ ∞ f (λ ) f 2 (t −∞ 1 − λ )d λ } Since we are dealing with the time functions that do not exist prior to t=0-, the lower limit can be changed to 0- ∞ ∞ − st L{ f1 ( t ) * f 2 ( t )} = ∫0 [ ∫ − 0− e f1 ( λ ) f 2 ( t − λ )dt ]d λ
48. 48. f1(λ) does not depend on t, and it can be moved outside the innerintegral ∞ ∞ − st L{ f1 ( t ) * f 2 ( t )} = ∫0 f1 ( λ )[ ∫ − 0− e f 2 (t − λ )dt ]d λ ∞ ∞ − s( x + λ ) = ∫0 f1 (λ )[ ∫ − 0− e f 2 ( x )dx ]d λ ∞ − sλ ∞ − sx = ∫0 f1 (λ )e − [∫ 0− e f 2 ( x )dx ]d λ ∞ = ∫0 f1 (λ )e − sλ [ F2 ( s )]d λ − ∞ = F2 ( s )∫0 f1 (λ )e − sλ d λ − = F1 ( s ) ×F2 ( s )
49. 49. STEADY-STATE SINUSOIDAL RESPONSEIf the input of a circuit is a sinusoidal function x ( t ) = A cos(ω t + θ ) x ( t ) = A cos ω t cosθ − A sin ω t sin θ ( A cosθ ) s ( A sin θ )ω X ( s) = 2 2 − 2 2 s +ω s +ω A(s cosθ − ω sinθ ) = 2 2 s +ω A(s cosθ − ω sinθ ) Y ( s) = H ( s) 2 2 s +ω
50. 50. The partial fraction expansion of Y(s) is * K1 K1Y ( s) = + + ∑ terms generated by the poles of H(s) s − jω s + jω If the poles of H(s) lie in the left half of the s plane, the corresponding time-domain terms approach zero as t increases and they do not contribute to the steady-state response. Thus only the first two terms determine the steady-state response.
51. 51. H ( s ) A( s cosθ − ω sinθ )K1 = s + jω s = jω H ( jω ) A( jω cosθ − ω sinθ ) = 2 jω H ( jω ) A(cosθ + j sinθ ) 1 = = H ( jω ) Ae jθ 2 2H ( jω ) = H ( jω ) e jφ (ω ) A j[θ +φ (ω )]K1 = H ( jω ) e 2yss ( t ) = A H ( jω ) cos[ω t + θ + φ (ω )]
52. 52. EXAMPLEIf the input is 120 cos(5000t+300)V, find the steady-state expression for v0 1000( s + 5000) H ( s) = 2 s + 6000 s + 25 × 106 1000(5000 + j 5000) H ( j 5000) = 6 6 −25 × 10 + j 5000(6000) + 25 × 10 1 + j1 2 = = − 450 j6 6 120 2 v0 ss = cos(5000t + 300 − 450 ) 6 =20 2 cos(5000t − 150 )V
53. 53. THE IMPULSE FUNCTION IN CIRCUIT ANALYSIS The capacitor is charged to an initial voltage V0 at the time the switch is closed. Find the expression for i(t) as R 0 V0 V0 s I= = R R + ( 1 sC ) + ( 1 sC ) 1 2 s+( 1 RC e ) C1C 2  V0 − t / RC  Ce = i (t ) =  e e ÷u( t ) C1 + C 2  R  As R decreases, the initial current (V0/R) increases and the time constant (RCe) decreases. Apparently i is approaching an impulse function as R approaches zero.
54. 54. The total area under the i versus t curve represents the total chargetransferred to C2 after the switch is closed. ∞ V0 Area=q = ∫0 − e − t / RC dt = V0C e e RThus, as R approaches zero, the current approaches an impulsestrength V0Ce. i → V0C eδ ( t )
55. 55. Series Inductor Circuit Find v0. Note that opening the switch forces an instantaneous change in the current L2. i1 (0− ) = 10 A, i2 (0− ) = 0 V0 V0 − [(100 s ) + 30] + =0 2 s + 15 3 s + 10 40( s + 7.5) 12( s + 7.5) V0 = + s( s + 5) s+5 60 10 V0 = + 12 + s s+5 −5 t v0 ( t ) = 12δ ( t ) + (60 + 10e )u( t )V
56. 56. Does this solution make sense? To answer this question, first let us determine the expression for the current. (100 s ) + 30 4 2 I= = + 5 s + 25 s s+5 −5 t i ( t ) = (4 + 2e )u( t ) ABefore the switch is opened, current through L1 is 10A and in L2 is 0 A,after the switch is opened both currents are 6A. Then the current in L 1changes instantaneously from 10 A to 6 A, while the current in L 2 changesinstantaneously from 0 to 6 A. How can we verify that these instantaneousjumps in the inductor current make sense in terms of the physical behaviorof the circuit?
57. 57. Switching operation places two inductors in series. Any impulsive voltageappearing across the 3H inductor must be balanced by an impulsivevoltage across the 2H inductor. Faraday’s law states that the inducedvoltage is proportional to the change in flux linkage ( v = d λ )before switching dt λ = L1i1 + L2i2 = 3(10) + 2(0) = 30 Wb-turnsAfter switching λ = ( L1 + L2 )i (0+ ) = 5i (0+ ) + 30 i (0 ) = = 6A 5 Thus the solution agrees with the principle of the conservation of flux linkage.
58. 58. Impulsive Sources When the voltage source is applied, the initial energy in the inductor is zero; therefore the initial current is zero. There is no voltage drop across R, so the impulsive source appears directly across L 1 t + V0 i = ∫0 V0δ ( x )dx ⇒ i (0 ) = − A L LThus, in an infinitesimal Current in the circuit decays tomoment, the impulsive voltage zero in accordance with thesource has stored natural response of the circuit 2 V0 − ( 1  V0  1 V02 i= e R )t u( t )w= L ÷ = L J 2  L 2 L L
59. 59. EXAMPLE Find i(t) and v0(t) for t>0 50 + ( 100 s ) + 30 I= 25 + 5 s 12 4 = + s+5 s i ( t ) = (12e −5 t + 4)u( t ) A 60 60 V0 = (15 + 2 s ) I = 32 + + s+5 s −5 t v0 ( t ) = 32δ ( t ) + (60e + 60)u( t )V
60. 60. Transfer Functions1. The Laplace transform2. Solution of linear differential equations3. Transient response example
61. 61. The Laplace Transform• Definition ∞ F ( s ) = L[ f (t )] = ∫ f (t )e − st dt 0 – Time (t) is replaced by a new independent variable (s) – We call s the Laplace transform variable• The Laplace domain – Often more convenient to work in Laplace domain than time domain – Time domain  ordinary differential equations in t – Laplace domain  algebraic equations in s• General solution approach – Formulate model in time domain – Convert model to Laplace domain – Solve problem in Laplace domain – Invert solution back to time domain
62. 62. Laplace Transform of Selected Functions • Constant function: f(t) = a ∞ ∞ a  a a F ( s ) = L(a ) = ∫ ae − st dt = − e − st = 0 −−  = 0 s 0  s s • Exponential function: f(t) = e-btF ( s ) = L (e −bt ∞ ) = ∫ e e dt = ∫ e 0 −bt − st 0 ∞ −( b + s ) t dt = − 1 b+s [ e −( b + s ) t ] ∞ 0 = 1 s +b • Derivatives and integrals  df  ∞ df − st ∞ ∞ L  = ∫ e dt = ∫ f (t )e − st sdt + f (t )e − st = sF ( s ) − f (0)  dt  0 dt 0 0 dn f  L n  = s n F ( s ) − s n −1 f (0) − s n −2 f (1) (0) −  − sf ( n −2 ) (0) − f ( n −1) (0)  dt     t f (t*)dt * = ∞  t f (t*)dt *e − st dt = 1 F ( s ) L ∫ 0   ∫0 ∫0     s
63. 63. Properties of Laplace Transforms• Superposition L[af (t ) + bg (t )] = aF ( s ) + bG ( s )• Final value theorem If lim y (t ) exists ⇒ lim y (t ) = lim[ sY ( s )] t →∞ t →∞ s →0• Initial value theorem lim y (t ) = lim[ sY ( s )] t →0 s →∞• Time delay f d (t ) = f (t − t0 ) S (t − t0 ) L → Fd ( s ) = e − st0 F ( s ) − st 0 L−1 Fd ( s ) = e F ( s) → f d (t ) = f (t − t0 ) S (t − t0 )
64. 64. Linear ODE Example• ODE dy 5 + 4y = 2 y ( 0) = 1 dt• Laplace transform 2 5[ sY ( s ) − y (0)] + 4Y ( s ) = s• Substitute y(0) & rearrange 5s + 2 s + 0.4 Y (s) = = s (5s + 4) s ( s + 0.8)• Inverse Laplace transform −1  s + 0.4  −1 y (t ) = L [Y ( s )] = L   s ( s + 0.8)  
65. 65. Linear ODE Example cont.• Table 3.1  −1 s + b3  b3 − b1 −b1t b3 − b2 −b2t L  = e + e  ( s + b1 )( s + b2 )  b2 − b1 b1 − b2• Our problem −1 s + 0.4  y (t ) = L   ⇒ b1 = 0 b2 = 0.8 b3 = 0.4  s ( s + 0.8) • Substitute and simplify  −1 s + 0.4  0.4 − 0 −0t 0.4 − 0.8 −0.8t L  = e + e = 0.5 + 0.5e −0.8t  ( s + 0)( s + 0.8)  0.8 − 0  0 − 0.8
66. 66. General ODE Solution Procedure • Procedure – Transform to Laplace domain – Solve resulting algebraic equations – Transform solution back to time domain • Partial fraction expansion – Necessary when inverse Laplace transform not tabularized – Break complex functions into simpler tabularized functions
67. 67. Partial Fraction Example• Partial fraction expansion s +5 s +5 α α Y (s) = = = 1 + 2 s 2 + 5s + 4 ( s + 1)( s + 4) s + 1 s + 4• Determination of coefficients s +5 4 s +5 1 α1 = = α2 = =− s + 4 s =−1 3 s + 1 s =−4 3• Inverse Laplace transform −1 −1 4 / 3 1 / 3  4 −t 1 − 4 t y (t ) = L [Y ( s )] = L  −  = 3e −3e  s +1 s + 4 
68. 68. Repeated Factor Example s +1 α1 α2 α3 Y (s) = = + + s ( s + 2) 2 s + 2 ( s + 2) 2 s s +1 1 s +1 1 1 α2 = = α3 = = α1 = − s s =−2 2 ( s + 2) 2 s =0 4 4 s +1 α1 α2 α3 −1 / 4 1/ 2 1/ 4Y ( s) = = + + = + + s ( s + 2) 2 s + 2 ( s + 2) 2 s s + 2 ( s + 2) 2 s −1 1 −2t 1 −2t 1 y (t ) = L [Y ( s )] = − e + te + S (t ) 4 2 4
69. 69. Quadratic Factor Example s +1 α α2 α s + α4 Y (s) = = 1 + 2 + 23 s 2 ( s 2 + 4 s + 5) s s s + 4s + 5 s + 1 = α1s ( s 2 + 4 s + 5) + α 2 ( s 2 + 4 s + 5) + (α 3 s + α 4 ) s 2 (α1 + α 3 ) s 3 + (4α1 + α 2 + α 4 ) s 2 + (5α1 + 4α 2 − 1) s + (5α 2 − 1) = 0 s +1 α α2 α s + α4 0.04 0.2 − 0.04s − 0.36Y ( s) = = 1 + 2 + 23 = + 2 + 2 s ( s + 4 s + 5) s s 2 2 s + 4s + 5 s s s + 4s + 5 − 0.04 s − 0.36 − 0.04( s + 2) − 0.28 = + s + 4s + 5 2 ( s + 2) + 1 ( s + 2) 2 + 1 2 y (t ) = L−1[Y ( s )] = 0.04S (t ) + 0.2t − 0.04e −2t cos t − 0.28e −2t sin t
70. 70. Transient Response Example• Component balance dc1 dc1 V = q (ci − c1 ) ⇒ 4 + 2c1 = 2ci c1 (0) = 0 dt dt• Step input 0 t ≤ 0 5 ci (t ) =  ⇒ Ci ( s ) = 5 t > 0 s
71. 71. Transient Response Example cont.• Laplace transform 2 4[ sC1 ( s ) − 0] + 2C1 ( s ) = 2Ci ( s ) ⇒ C1 ( s ) = Ci ( s ) 4s + 2• Substitute input 2 5 5 C1 ( s ) = = 4 s + 2 s s (2 s + 1)• Inverse Laplace transform   −1 c1 (t ) = L  5 s (2 s + 1)  ( = 5 1 − e −t / 2 )  
72. 72. Simulink Solution inlet Inlet 1 2 outlet 4s+2 Step Tank To Workspace 6 >> plot(tout,inlet) Input Output >> hold 5 >> plot(tout,outlet,r)Concentration (kmol/m ) 3 4 3 >> axis([0 15 0 6]) >> ylabel(Concentration 2 (kmol/m^3)) 1 >> xlabel(Time (min)) >> legend(‘Input,Output) 0 0 5 10 15 Time (min)
73. 73. Engineering Circuit AnalysisCh3 Basic RL and RC Circuits 3.1 First-Order RC Circuits 3.2 First-Order RL Circuits 3.3 Examples References: Hayt-Ch5, 6; Gao-Ch5; References
74. 74. Ch3 Basic RL and RC Circuits3.1 First-Order RC CircuitsKey Words: Words Transient Response of RC Circuits, Time constant
75. 75. Ch3 Basic RL and RC Circuits3.1 First-Order RC CircuitsKey Words: Words Transient Response of RC Circuits, Time constant
76. 76. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits• Used for filtering signal by blocking certain frequencies and passing others. e.g. low-pass filter• Any circuit with a single energy storage element, an arbitrary number of sources and an arbitrary number of resistors is a circuit of order 1.• Any voltage or current in such a circuit is the solution to a 1st order differential equation. Ideal Linear Capacitor dq dv i (t ) = =c vc (t +) =v C (t ) dt dt 1 2 Energy stored w = ∫ pdt = ∫ cvdv = cv 2 A capacitor is an energy storage device → memory device.
77. 77. Ch3 Basic RL and RC Circuits3.1 First-Order RC Circuits vr(t) + - R + + vs(t) C vc(t) - - • One capacitor and one resistor • The source and resistor may be equivalent to a circuit with many resistors and sources.
78. 78. Ch3 Basic RL and RC Circuits3.1 First-Order RC CircuitsTransient Response of RC Circuits E − vc ic = Switch is thrown to 1 R KVL around the loop: ic R + vC = E R C dvc C R + vc = E • ο 2 dt t − ο K vC = Ae RC +E 1 E Initial condition vC (0+) =v C (0−) = 0 A = −E t t − − vC = E (1 − e RC ) = E (1 − e τ ) τ = RC dvc E − t Called time constant ic = C = e τ dt R
79. 79. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Time τ = RC Constant5Ω − t dvc E −t / τ R vC = E (1 − e τ ) = e C dt τ 2 dvc E E • ο t =0 = →τ = 10V ο K dt τ dvc 1 t =0∧ E dt 5V R=2k C=0.1µF SEL > > 0V 0s RC 1 ms 2 ms 3 ms 4 ms V( 2 )
80. 80. Ch3 Basic RL and RC Circuits3.1 First-Order RC CircuitsTransient Response of RC Circuits vc + ic R = 0 Switch to 2 dvc ic = C R dt C dv vc + RC c = 0 dt 2 t • ο − K vc = Ae RC ο 1 E Initial condition vC (0+) =v C (0−) = E vc = Ee −t / RC = Ee −t / τ E −t / τ ic = − e R
81. 81. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits Time τ = RC − t − t τ Constant vC (t ) = Ee RC = Ee 5Ω R dvC E C =−Ω dt t =0 τΩ 2 • ο E ο K τ =− 10V 1 dvCIS ∧ E dt t =0 5V R=2k C=0.1µF SEL > > 0V 0s 1 . 0 ms 2 . 0 ms 3 . 0 ms 4 . 0 ms V( 2 ) Ti me
82. 82. Ch3 Basic RL and RC Circuits 3.1 First-Order RC Circuits
83. 83. Ch3 Basic RL and RC Circuits3.2 First-Order RL Circuits Key Words: Words Transient Response of RL Circuits, Time constant
84. 84. Ch3 Basic RL and RC Circuits3.2 First-Order RL Circuits Ideal Linear Inductor t i(t) dψ di (t ) 1 The + v(t ) = =L i (t ) = ∫ v( x)dx rest dt dt L −∞ of L v(t) di the P = iv = Li i L (t +) =i L (t −) - dt circuit 1 Energy stored:(t ) = ∫ pdt = ∫ Lidi = Li 2 (t ) wL 2 • One inductor and one resistor • The source and resistor may be equivalent to a circuit with many resistors and sources.
85. 85. Ch3 Basic RL3.2 First-OrderRC Circuits and RL CircuitsTransient Response of RL Circuits di vL = L Switch to 1 dt R KVL around the loop: iR + vL = E L di 2 E=L + iR • ο dt ο K Initial condition t = 0, i (0 + ) = i (0 − ) = 0 1 E R E − t E → i = (1 − e ) = (1 − e −t /τ ) L R R τ = L/ R → vR = iR = E (1 − e −t /τ ) Called time constant d E  − t  R di E R −Rt vL = L = L ⋅  1 − e L  = L ⋅ ⋅ ⋅ e L = Ee −t /τ dt dt  R      R L
86. 86. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Time constant . i (t) t 0 τ • Indicate how fast i (t) will drop to zero. • It is the amount of time for i (t) to drop to zero if it is dit dropping at the initial rate . dt t =0
87. 87. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits Time constant . i (t) t 0 τ • Indicate how fast i (t) will drop to zero. • It is the amount of time for i (t) to drop to zero if it is dit dropping at the initial rate . dt t =0
88. 88. Ch3 Basic RL and RC Circuits3.2 First-Order RL CircuitsTransient Response of RL Circuits di t′ : 0 → t L + iR = 0 Switch to 2 dt i′ : I 0 → i ( t ) R di R 1 i( t ) t R = − dt ∫I0 i′di′ = ∫0 − L dt ′ L i L 2 • ο R R → i = Ae − t L ln i ′ iI(0t ) = − × ′ t0 t ο K L 1 R E i (t ) R − t ln =− t i (t ) = I 0 e L I0 L E Initial condition t = 0, I 0 = R E − R t E −t / τ i= e L = e R R
89. 89. Ch3 Basic RL and RC Circuits 3.2 First-Order RL Circuits SEL > >Transient Response of RL Circuits Input energy to L 4 . 0 mA R L 2 . 0 mA 2 • ο K ο 0A 1 0s 1 ms 2 ms 3 ms 4 ms E I ( L1) L export its energy , dissipated by R 4 . 0 mA 2 . 0 mA SEL > > 0A 0s 1 ms 2 ms 3 ms 4 ms
90. 90. Ch3 Basic RL and RC Circuits Summary Steady time Initial Value Value (t → constant （t=0） ∞) τ RL Source i0 = 0 iL = E R L/R Circuits (0 state) Source- free E i0 = i=0 L/R (0 input) R Source RC (0 state) v0 = 0 v=E RC Circuits Source- free v0 = E v=0 RC (0 input)
91. 91. Ch3 Basic RL and RC Circuits Summary The Time Constant • For an RC circuit, τ = RC • For an RL circuit, τ = L/R • -1/τ is the initial slope of an exponential with an initial value of 1 • Also, τ is the amount of time necessary for an exponential to decay to 36.7% of its initial value
92. 92. Ch3 Basic RL and RC Circuits Summary• How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types of quantities will remain the same as before the change. – IL(t), inductor current – Vc(t), capacitor voltage• Find these two types of the values before the change and use them as the initial conditions of the circuit after change.
93. 93. Ch3 Basic RLanand RC Circuits 3.3 Examples (Analyzing RC circuit or RL circuit) Method 1 1) Thévenin Equivalent.(Draw out C or L) Simplify the circuit Veq , Req 2) Find Leq(Ceq), and τ = Leq/Req (τ = CeqReq) 3) Substituting Leq(Ceq) and τ to the previous solution of differential equation for RC (RL) circuit .
94. 94. Ch3 Basic RL3.3 Examples RC Circuits or RL circuit) and (Analyzing an RC circuitMethod 2 1) KVL around the loop → the differential equation 2) Find the homogeneous solution. 3) Find the particular solution. 4) The total solution is the sum of the particular and homogeneou solutions.
95. 95. Ch3 Basic RL and RC Circuits 3.3 Examples About Calculation for The Initial Value →i ↓iC ↓iL vC ( 0+ ) = vC ( 0− ) t=0 ( R1 / / R3 ) = 2Ω 2Ω vC ( 0 ) = 8V × = 4V 2Ω + 2Ω i(0+) iC(0+) iL(0+) iL ( 0+ ) = iL ( 0− ) 8V i ( 0) = = 2A + + 2Ω + 2Ω vC(0+)=4V vL(0+) 4Ω _ - iL ( 0 ) = 2A × = 1A ∨ 1A 4Ω + 4Ω
96. 96. Ch3 Basic RL3.3 Examples RC Circuits or RL circuit) and (Analyzing an RC circuitMethod 3 (step-by- tstep) − In general, f (t ) = f (∞) + Ae τ Given f(0+) ， thus A = f(0+) – f(∞) t − f (t ) = f (∞) + [ f (0+ ) − f (∞)]e τ Initial Stead y 1) Draw the circuit for t = 0- and find v(0-) or i(0-) 2) Use the continuity of the capacitor voltage, or inductor current, draw the circuit for t = 0+ to find v(0+) or i(0+) 3) Find v(∞), or i(∞) at steady state 4) Find the time constant τ – For an RC circuit, τ = RC – For an RL circuit, τ = L/R 5) The solution is: f (t ) = f (∞) + [ f (0+ ) − f (∞)]e −t /τ
97. 97. Ch3 Basic RL and RC Circuits 3.3 Examples − − 5Ω 5ΩUS − _ P3.1 vC (0)= 0, Find vC (t) for t ≥ 0. ο i1 6k 5Ω t=0 R1 i2 i3I1 I1 I1 I1 ∨ ∨ Method 3: 5Ω + R2 3k + t E 9V C=1000PF − vc ( t ) = ( ) +  vc ( 0 ) I− ∧ c ( ∞ )  e µU _ vc 1 ∞ -  v  τ pf S E 3K vc ( 0 ) = 0, vc ( ∞ ) = 9V × = 3V βI1 6K + 3K Apply Thevenin theorem : −1  1 1  RTh = + ÷ =2KΩ  6KΩ 3KΩ τ =RThC =2KΩ 1000pF =2 × −6 × 10 s t − vc ( t ) =3 −3e 2× − 10 6 V
98. 98. Ch3 Basic RL and RC Circuits − 3.3 Examples 5Ω 5Ω − − _ C=1000PF P3.2 vC (0)= 0, Find vC (t) for t ≥ 0. 5Ω + -I1 I1 I1 vC 5Ω ο t=0 R1=10k vc ( 0 ) = 0 + + µ U1 - 10KΩ ∧ v 6V vc ( ∞ ) = 6V × IS E = 4.615V _ R2 3k 10KΩ + 3KΩ R1=20k β I1 Apply Thevenin’s theorem : ↑ ↓ −1  1 1  30 RTh =  + ÷ = KΩ  10KΩ 3KΩ  13 30 τ = RThC = KΩ × 1000pF = 2.31× 10−6 s 13 t − vc ( t ) = 4.615 − 4.615e 2.31×10−6 V
99. 99. Differential Equation Solutions of Transient Circuits Dr. Holbert March 3, 2008Lect12 99 EEE 202
100. 100. 1st Order Circuits• Any circuit with a single energy storage element, an arbitrary number of sources, and an arbitrary number of resistors is a circuit of order 1• Any voltage or current in such a circuit is the solution to a 1st order differential equationLect12 EEE 202 100
101. 101. RLC CharacteristicsElement V/I Relation DC Steady-StateResistor vR (t ) = R iR (t ) V=IRCapacitor d vC (t ) I = 0; open iC (t ) = C dtInductor d iL (t ) V = 0; short vL (t ) = L dt ELI and the ICE manLect12 EEE 202 101
102. 102. A First-Order RC Circuit vr(t) + – R + vs(t) + C vc(t) – –• One capacitor and one resistor in series• The source and resistor may be equivalent to a circuit with many resistors and sourcesLect12 EEE 202 102
103. 103. The Differential Equation vr(t) + – R + vs(t) + C – vc(t) –KVL around the loop: vr(t) + vc(t) = vs(t)Lect12 EEE 202 103
104. 104. RC Differential Equation(s) t 1 From KVL: R i (t ) + ∫ i ( x)dx = vs (t ) C −∞Multiply by C; di (t ) dvs (t )take derivative RC + i (t ) = C dt dtMultiply by R; dvr (t ) dvs (t )note vr=R·i RC + vr (t ) = RC dt dt Lect12 EEE 202 104
105. 105. A First-Order RL Circuit + is(t) R L v(t) –• One inductor and one resistor in parallel• The current source and resistor may be equivalent to a circuit with many resistors and sourcesLect12 EEE 202 105
106. 106. The Differential Equations + is(t) R L v(t) –KCL at the top node: t v(t ) 1 + ∫ v( x)dx = is (t ) R L −∞Lect12 EEE 202 106
107. 107. RL Differential Equation(s) t v(t ) 1 From KCL: + ∫ v( x)dx = is (t ) R L −∞Multiply by L; L dv(t ) dis (t )take derivative + v(t ) = L R dt dtLect12 EEE 202 107
108. 108. 1st Order Differential EquationVoltages and currents in a 1st order circuitsatisfy a differential equation of the form dx(t ) + a x(t ) = f (t ) dtwhere f(t) is the forcing function (i.e., theindependent sources driving the circuit)Lect12 EEE 202 108
109. 109. The Time Constant (τ)• The complementary solution for any first order circuit is −t /τ vc (t ) = Ke• For an RC circuit, τ = RC• For an RL circuit, τ = L/R• Where R is the Thevenin equivalent resistanceLect12 EEE 202 109
110. 110. What Does vc(t) Look Like? τ = 10-4Lect12 EEE 202 110
111. 111. Interpretation of τ• The time constant, τ, is the amount of time necessary for an exponential to decay to 36.7% of its initial value• -1/τ is the initial slope of an exponential with an initial value of 1Lect12 EEE 202 111
112. 112. Applications Modeled by a 1st Order RC Circuit• The windings in an electric motor or generator• Computer RAM – A dynamic RAM stores ones as charge on a capacitor – The charge leaks out through transistors modeled by large resistances – The charge must be periodically refreshedLect12 EEE 202 112
113. 113. Important Concepts• The differential equation for the circuit• Forced (particular) and natural (complementary) solutions• Transient and steady-state responses• 1st order circuits: the time constant (τ)• 2nd order circuits: natural frequency (ω0) and the damping ratio (ζ)Lect12 EEE 202 113
114. 114. The Differential Equation• Every voltage and current is the solution to a differential equation• In a circuit of order n, these differential equations have order n• The number and configuration of the energy storage elements determines the order of the circuit• n ≤ number of energy storage elementsLect12 EEE 202 114
115. 115. The Differential Equation• Equations are linear, constant coefficient: d n x(t ) d n −1 x(t ) an n + an −1 n −1 + ... + a0 x(t ) = f (t ) dt dt• The variable x(t) could be voltage or current• The coefficients an through a0 depend on the component values of circuit elements• The function f(t) depends on the circuit elements and on the sources in the circuitLect12 EEE 202 115
116. 116. The Differential Equation• Equations are linear, constant coefficient: d n x(t ) d n −1 x(t ) an n + an −1 n −1 + ... + a0 x(t ) = f (t ) dt dt• The variable x(t) could be voltage or current• The coefficients an through a0 depend on the component values of circuit elements• The function f(t) depends on the circuit elements and on the sources in the circuitLect12 EEE 202 116
117. 117. Building Intuition• Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition to be developed: – Particular and complementary solutions – Effects of initial conditionsLect12 EEE 202 117
118. 118. Differential Equation Solution• The total solution to any differential equation consists of two parts: x(t) = xp(t) + xc(t)• Particular (forced) solution is xp(t) – Response particular to a given source• Complementary (natural) solution is xc(t) – Response common to all sources, that is, due to the “passive” circuit elementsLect12 EEE 202 118
119. 119. Forced (or Particular) Solution• The forced (particular) solution is the solution to the non-homogeneous equation: d n x(t ) d n −1 x(t ) an n + an −1 n −1 + ... + a0 x(t ) = f (t ) dt dt• The particular solution usually has the form of a sum of f(t) and its derivatives – That is, the particular solution looks like the forcing function – If f(t) is constant, then x(t) is constant – If f(t) is sinusoidal, then x(t) is sinusoidalLect12 EEE 202 119
120. 120. Natural/Complementary Solution• The natural (or complementary) solution is the solution to the homogeneous equation: n n −1 d x(t ) d x(t ) an n + an −1 n −1 + ... + a0 x(t ) = 0 dt dt• Different “look” for 1st and 2nd order ODEsLect12 EEE 202 120
121. 121. First-Order Natural Solution• The first-order ODE has a form of dxc (t ) 1 + xc (t ) = 0 dt τ• The natural solution is −t /τ xc (t ) = Ke• Tau (τ) is the time constant • For an RC circuit, τ = RC • For an RL circuit, τ = L/RLect12 EEE 202 121
122. 122. Second-Order Natural Solution• The second-order ODE has a form of d 2 x(t ) dx(t ) 2 + 2ζω 0 + ω 0 x(t ) = 0 2 dt dt• To find the natural solution, we solve the characteristic equation: s + 2ζω 0 s + ω = 0 2 2 0 which has two roots: s1 and s2• The complementary solution is (if we’re lucky) xc (t ) = K1e + K 2 e s1t s2tLect12 EEE 202 122
123. 123. Initial Conditions• The particular and complementary solutions have constants that cannot be determined without knowledge of the initial conditions• The initial conditions are the initial value of the solution and the initial value of one or more of its derivatives• Initial conditions are determined by initial capacitor voltages, initial inductor currents, and initial source valuesLect12 EEE 202 123
124. 124. 2nd Order Circuits• Any circuit with a single capacitor, a single inductor, an arbitrary number of sources, and an arbitrary number of resistors is a circuit of order 2• Any voltage or current in such a circuit is the solution to a 2nd order differential equationLect12 EEE 202 124
125. 125. A 2nd Order RLC Circuit i (t) R vs(t) + C – LThe source and resistor may be equivalentto a circuit with many resistors and sourcesLect12 EEE 202 125
126. 126. The Differential Equation i(t) + vr(t) – R + + C vc(t) – vs(t) vl(t) – – + L KVL around the loop: vr(t) + vc(t) + vl(t) = vs(t)Lect12 EEE 202 126
127. 127. RLC Differential Equation(s)From KVL: t 1 di (t ) R i (t ) + ∫ i ( x)dx + L = vs (t ) C −∞ dtDivide by L, and take the derivative 2 R di (t ) 1 d i (t ) 1 dvs (t ) + i (t ) + 2 = L dt LC dt L dtLect12 EEE 202 127
128. 128. The Differential EquationMost circuits with one capacitor and inductorare not as easy to analyze as the previouscircuit. However, every voltage and currentin such a circuit is the solution to adifferential equation of the following form: 2 d x(t ) dx(t ) 2 + 2ζω 0 + ω0 x(t ) = f (t ) 2 dt dtLect12 EEE 202 128
129. 129. Class Examples• Drill Problems P6-1, P6-2• Suggestion: print out the two-page “First and Second Order Differential Equations” handout from the class webpageLect12 EEE 202 129