Upcoming SlideShare
×

# Circuit analysis ii with matlab

2,474

Published on

Published in: Technology
2 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total Views
2,474
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
439
0
Likes
2
Embeds 0
No embeds

No notes for slide

### Transcript of "Circuit analysis ii with matlab"

1. 1. Orchard Publications, Fremont, Californiawww.orchardpublications.comnum=[0 1100 0]; den=[1 1100 10^5]; w=logspace(0,5,100); bode(num,den,w); gridSteven T. KarrisCircuit Analysis IIwith MATLAB® Applications
2. 2. Circuit Analysis IIwith MATLAB® ApplicationsStudents and working professionals will find CircuitAnalysis II with MATLAB® Applications to be a con-cise and easy-to-learn text. It provides complete,clear, and detailed explanations of advanced electri-cal engineering concepts illustrated with numerouspractical examples.This text includes the following chapters and appendices:• Second Order Circuits • Resonance • Elementary Signals • The Laplace Transformation• The Inverse Laplace Transformation • Circuit Analysis with Laplace Transforms • FrequencyResponse and Bode Plots • Self and Mutual Inductances - Transformers • One and Two PortNetworks • Three Phase Systems • Introduction to MATLAB • Differential Equations • Matricesand Determinants • Constructing Semilog Plots with Microsoft Excel • ScalingEach chapter contains numerous practical applications supplemented with detailed instructionsfor using MATLAB to obtain quick and accurate answers.Steven T. Karris is the president and founder of Orchard Publications. He earned a bachelorsdegree in electrical engineering at Christian Brothers University, Memphis, Tennessee, a mastersdegree in electrical engineering at Florida Institute of Technology, Melbourne, Florida, and hasdone post-master work at the latter. He is a registered professional engineer in California andFlorida. He has over 30 years of professional engineering experience in industry. In addition, hehas over 25 years of teaching experience that he acquired at several educational institutions asan adjunct professor. He is currently with UC Berkeley Extension.Orchard Publications, Fremont, CaliforniaVisit us on the Internetwww.orchardpublications.comor email us: info@orchardpublications.comISBN 0-9709511-9-1\$39.95 U.S.A.
3. 3. Circuit Analysis IIwith MATLAB® ApplicationsSteven T. KarrisOrchard Publicationswww.orchardpublications.com
5. 5. Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsPrefaceThis text is written for use in a second course in circuit analysis. The reader of this book should havethe traditional undergraduate knowledge of an introductory circuit analysis material such as CircuitAnalysis I with MATLAB® Applications by this author. Another prerequisite would be knowledge ofdifferential equations, and in most cases, engineering students at this level have taken all requiredmathematics courses. It encompasses a spectrum of subjects ranging from the most abstract to themost practical, and the material can be covered in one semester or two quarters. Appendix B serves asa review of differential equations with emphasis on engineering related topics and it is recommendedfor readers who may need a review of this subject.There are several textbooks on the subject that have been used for years. The material of this book isnot new, and this author claims no originality of its content. This book was written to fit the needs ofthe average student. Moreover, it is not restricted to computer oriented circuit analysis. While it is truethat there is a great demand for electrical and computer engineers, especially in the internet field, thedemand also exists for power engineers to work in electric utility companies, and facility engineers towork in the industrial areas.Chapter 1 is an introduction to second order circuits and it is essentially a sequel to first order circuitsthat were discussed in the last chapter of as Circuit Analysis I with MATLAB® Applications. Chapter 2is devoted to resonance, and Chapter 3 presents practical methods of expressing signals in terms ofthe elementary functions, i.e., unit step, unit ramp, and unit impulse functions. Thus, any signal can berepresented in the compex frequency domain using the Laplace transformation.Chapters 4 and 5 are introductions to the unilateral Laplace transform and Inverse Laplace transformrespectively, while Chapter 6 presents several examples of analyzing electric circuits using Laplacetransformation methods. Chapter 7 begins with the frequency response concept and Bode magnitudeand frequency plots. Chapter 8 is devoted to transformers with an introduction to self and mutualinductances. Chapter 9 is an introduction to one- and two-terminal devices and presents severalpractical examples. Chapter 10 is an introduction to three-phase circuits.It is not necessary that the reader has previous knowledge of MATLAB®. The material of this textcan be learned without MATLAB. However, this author highly recommends that the reader studiesthis material in conjunction with the inexpensive MATLAB Student Version package that is availableat most college and university bookstores. Appendix A of this text provides a practical introductionto MATLAB. As shown on the front cover of this text the magnitude and phase plots can be easilyobtained with a one line MATLAB code. Moreover, MATLAB will be invaluable in later studies suchas the design of analog and digital filters.
6. 6. Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsAs stated above, Appendix B is a review of differential equations. Appendix C is an introductionto matrices, Appendix D provides instructions on constructing semilog templates to be used withBode plots, and Appendix E discusses scaling methods.In addition to numerous real-world examples, this text contains several exercises at the end ofeach chapter. Detailed solutions of all exercises are provided at the end of each chapter. Therationale is to encourage the reader to solve all exercises and check his effort for correct solutionsand appropriate steps in obtaining the correct solution. And since this text was written to serve asa self-study or supplementary textbook, it provides the reader with a resource to test hisknowledge.The author has accumulated many additional problems for homework assignment and these areavailable to those instructors who adopt this text either as primary or supplementary text, andprefer to assign problems without the solutions. He also has accumulated many sample exams.The author is indebted to the class of the Spring semester of 2001 at San Jose State University,San Jose, California, for providing several of the examples and exercises of this text.Like any other new book, this text may contain some grammar and typographical errors.Accordingly, all feedback for errors, advice, and comments will be most welcomed and greatlyappreciated.Orchard Publicationsinfo@orchardpublications.com
7. 7. Circuit Analysis II with MATLAB Applications iOrchard PublicationsContentsChapter 1Second Order CircuitsThe Response of a Second Order Circuit ....................................................................................1-1The Series RLC Circuit with DC Excitation ...............................................................................1-2Response of Series RLC Circuits with DC Excitation ................................................................1-5Response of Series RLC Circuits with AC Excitation ..............................................................1-11The Parallel GLC Circuit...........................................................................................................1-14Response of Parallel GLC Circuits with DC Excitation............................................................1-16Response of Parallel GLC Circuits with AC Excitation............................................................1-26Other Second Order Circuits .....................................................................................................1-29Summary....................................................................................................................................1-36Exercises....................................................................................................................................1-38Solutions to Exercises................................................................................................................1-40Chapter 2ResonanceSeries Resonance.........................................................................................................................2-1Quality Factor Q0s in Series Resonance.......................................................................................2-4Parallel Resonance.......................................................................................................................2-6Quality Factor Q0p in Parallel Resonance....................................................................................2-9General Definition of Q.............................................................................................................2-10Energy in L and C at Resonance ...............................................................................................2-11Half-Power Frequencies - Bandwidth .......................................................................................2-12A Practical Parallel Resonant Circuit........................................................................................2-16Radio and Television Receivers ................................................................................................2-17Summary....................................................................................................................................2-20Exercises....................................................................................................................................2-22Solutions to Exercises................................................................................................................2-24Chapter 3Elementary SignalsSignals Described in Math Form.................................................................................................3-1The Unit Step Function ...............................................................................................................3-2The Unit Ramp Function...........................................................................................................3-10
8. 8. ii Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsThe Delta Function....................................................................................................................3-12Sampling Property of the Delta Function..................................................................................3-12Sifting Property of the Delta Function ......................................................................................3-13Higher Order Delta Functions ...................................................................................................3-15Summary....................................................................................................................................3-19Exercises....................................................................................................................................3-20Solutions to Exercises................................................................................................................3-21Chapter 4The Laplace TransformationDefinition of the Laplace Transformation...................................................................................4-1Properties of the Laplace Transform ...........................................................................................4-2The Laplace Transform of Common Functions of Time...........................................................4-12The Laplace Transform of Common Waveforms......................................................................4-23Summary....................................................................................................................................4-29Exercises....................................................................................................................................4-34Solutions to Exercises................................................................................................................4-37Chapter 5The Inverse Laplace TransformationThe Inverse Laplace Transform Integral .....................................................................................5-1Partial Fraction Expansion...........................................................................................................5-1Case for m n............................................................................................................................5-13Alternate Method of Partial Fraction Expansion.......................................................................5-15Summary....................................................................................................................................5-18Exercises....................................................................................................................................5-20Solutions to Exercises................................................................................................................5-22Chapter 6Circuit Analysis with Laplace TransformsCircuit Transformation from Time to Complex Frequency ........................................................6-1Complex Impedance Z(s) ............................................................................................................6-8Complex Admittance Y(s).........................................................................................................6-10Transfer Functions.....................................................................................................................6-13Summary....................................................................................................................................6-17Exercises....................................................................................................................................6-18Solutions to Exercises................................................................................................................6-21
9. 9. Circuit Analysis II with MATLAB Applications iiiOrchard PublicationsChapter 7Frequency Response and Bode PlotsDecibel.............................................................................................................................................................7-1Bandwidth and Frequency Response..........................................................................................................7-3Octave and Decade........................................................................................................................................7-4Bode Plot Scales and Asymptotic Approximations ..................................................................................7-5Construction of Bode Plots when the Zeros and Poles are Real............................................................7-6Construction of Bode Plots when the Zeros and Poles are Complex.................................................7-12Corrected Amplitude Plots.........................................................................................................................7-25Summary........................................................................................................................................................7-36Exercises........................................................................................................................................................7-38Solutions to Exercises..................................................................................................................................7-39Chapter 8Self and Mutual Inductances - TransformersSelf-Inductance ............................................................................................................................8-1The Nature of Inductance ............................................................................................................8-1Lenz’s Law ..................................................................................................................................8-3Mutually Coupled Coils...............................................................................................................8-3Establishing Polarity Markings..................................................................................................8-11Energy Stored in a Pair of Mutually Coupled Inductors ...........................................................8-14Circuits with Linear Transformers.............................................................................................8-20Reflected Impedance in Transformers.......................................................................................8-25The Ideal Transformer ...............................................................................................................8-28Impedance Matching..................................................................................................................8-32A Simplified Transformer Equivalent Circuit ...........................................................................8-33Thevenin Equivalent Circuit......................................................................................................8-34Summary....................................................................................................................................8-38Exercises ....................................................................................................................................8-42Solutions to Exercises................................................................................................................8-44Chapter 9One- and Two-port NetworksIntroduction and Definitions .......................................................................................................................9-1One-port Driving-point and Transfer Admittances .................................................................................9-2One-port Driving-point and Transfer Impedances ..................................................................................9-7Two-Port Networks.....................................................................................................................................9-12
10. 10. iv Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsThe y Parameters..........................................................................................................................................9-12The z parameters..........................................................................................................................................9-19The h Parameters.........................................................................................................................................9-24The g Parameters .........................................................................................................................................9-29Reciprocal Two-Port Networks.................................................................................................................9-34Summary........................................................................................................................................................9-38Exercises........................................................................................................................................................9-43Solutions to Exercises .................................................................................................................................9-45Chapter 10Three-Phase SystemsAdvantages of Three-Phase Systems.........................................................................................10-1Three-Phase Connections ..........................................................................................................10-1Transformer Connections in Three-Phase Systems...................................................................10-4Line-to-Line and Line-to-Neutral Voltages and Currents .........................................................10-5Equivalent Y and Loads .......................................................................................................10-10Computation by Reduction to Single Phase ............................................................................10-20Three-Phase Power..................................................................................................................10-21Instantaneous Power in Three-Phase Systems.........................................................................10-23Measuring Three-Phase Power................................................................................................10-27Summary..................................................................................................................................10-30Exercises..................................................................................................................................10-32Solutions to Exercises..............................................................................................................10-33Appendix AIntroduction to MATLAB®MATLAB® and Simulink®.......................................................................................................A-1Command Window.....................................................................................................................A-1Roots of Polynomials..................................................................................................................A-3Polynomial Construction from Known Roots ............................................................................A-4Evaluation of a Polynomial at Specified Values ........................................................................A-6Rational Polynomials..................................................................................................................A-8Using MATLAB to Make Plots................................................................................................A-10Subplots ....................................................................................................................................A-19Multiplication, Division and Exponentiation ...........................................................................A-20Script and Function Files..........................................................................................................A-26Display Formats........................................................................................................................A-31
11. 11. Circuit Analysis II with MATLAB Applications vOrchard PublicationsAppendix BDifferential EquationsSimple Differential Equations ....................................................................................................B-1Classification ..............................................................................................................................B-3Solutions of Ordinary Differential Equations (ODE).................................................................B-6Solution of the Homogeneous ODE ...........................................................................................B-8Using the Method of Undetermined Coefficients for the Forced Response.............................B-10Using the Method of Variation of Parameters for the Forced Response..................................B-20Exercises ...................................................................................................................................B-24Appendix CMatrices and DeterminantsMatrix Definition ........................................................................................................................C-1Matrix Operations.......................................................................................................................C-2Special Forms of Matrices ..........................................................................................................C-5Determinants...............................................................................................................................C-9Minors and Cofactors................................................................................................................C-12Cramer’s Rule...........................................................................................................................C-16Gaussian Elimination Method ..................................................................................................C-19The Adjoint of a Matrix............................................................................................................C-20Singular and Non-Singular Matrices ........................................................................................C-21The Inverse of a Matrix ............................................................................................................C-21Solution of Simultaneous Equations with Matrices..................................................................C-23Exercises ...................................................................................................................................C-30Appendix DConstructing Semilog Plots with Microsoft ExcelThe Excel Spreadsheet Window.................................................................................................D-1Instructions for Constructing Semilog Plots...............................................................................D-2Appendix EScalingMagnitude Scaling ...................................................................................................................... E-1Frequency Scaling....................................................................................................................... E-1Exercises ..................................................................................................................................... E-8Solutions to Exercises................................................................................................................. E-9
12. 12. vi Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsNOTES
13. 13. Circuit Analysis II with MATLAB Applications 1-1Orchard PublicationsChapter 1Second Order Circuitshis chapter discusses the natural, forced and total responses in circuits containing resistors,inductors and capacitors. These circuits are characterized by linear second-order differentialequations whose solutions consist of the natural and the forced responses. We will considerboth DC (constant) and AC (sinusoidal) excitations.1.1 The Response of a Second Order CircuitA circuit containing energy storage devices (inductors and capacitors) is said to be an nth-order cir-cuit, and the differential equation describing the circuit is an nth-order differential equation. For exam-ple, if a circuit contains an inductor and a capacitor, or two capacitors or two inductors, along withother devices such as resistors, it is said to be a second-order circuit and the differential equation thatdescribes it is a second order differential equation. It is possible, however, to describe a circuit havingtwo energy storage devices with a set of two first-order differential equations, a circuit which hasthree energy storage devices with a set of three first-order differential equations and so on. These arecalled state equations*but these will not be discussed here.The response is found from the differential equation describing the circuit, and its solution isobtained as follows:1. We write the differential or integrodifferential (nodal or mesh) equation describing the circuit. Wedifferentiate, if necessary, to eliminate the integral.2. We obtain the forced (steady-state) response. Since the excitation in our work here will be either aconstant (DC) or sinusoidal (AC) in nature, we expect the forced response to have the same formas the excitation. We evaluate the constants of the forced response by substitution of the assumedforced response into the differential equation and equate terms of the left side with the right side.Refer to Appendix B for the general expression of the forced response (particular solution).3. We obtain the general form of the natural response by setting the right side of the differentialequation equal to zero, in other words, solve the homogeneous differential equation using thecharacteristic equation.4. Add the forced and natural responses to form the complete response.5. We evaluate the constants of the complete response from the initial conditions.* State variables and state equations are discussed in Signals and Systems with MATLAB Applications, ISBN 0-9709511-3-2 by this author.Tn
14. 14. Chapter 1 Second Order Circuits1-2 Circuit Analysis II with MATLAB ApplicationsOrchard Publications1.2 The Series RLC Circuit with DC ExcitationLet us consider the series circuit of Figure 1.1 where the initial conditions are ,, and is the unit step function.*We want to find an expression for the currentfor .Figure 1.1. Series RLC CircuitFor this circuit(1.1)and by differentiationTo find the forced response, we must first specify the nature of the excitation , that is, DC or AC.If is DC ( =constant), the right side of (1.1) will be zero and thus the forced response compo-nent . If is AC ( , the right side of (1.1) will be another sinusoid andtherefore . Since in this section we are concerned with DC excitations, the rightside will be zero and thus the total response will be just the natural response.The natural response is found from the homogeneous equation of (1.1), that is,(1.2)The characteristic equation of (1.2) is* The unit step function is discussed in detail in Chapter 3. For our present discussion it will suffice to state thatfor and for .RLC iL 0 I0=vC 0 V0= u0 t i tu0 t 0= t 0 u0 t 1= t 0t 0+vS u0 ti tRLCRi Ldidt-----1C---- i td0tV0+ + + vS= t 0Rdidt----- Ld2idt2------- iC----+ +dvSdt--------= t 0vSvS vSif 0= vS vS V t +cos=if I t +cos=Rdidt----- Ld2idt2------- iC----+ + 0=Ls2Rs1C----+ + 0=
15. 15. 1-3 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsThe Series RLC Circuit with DC Excitationorfrom which(1.3)We will use the following notations:(1.4)where the subscript stands for series circuit. Then, we can express (1.3) as(1.5)or(1.6)Case I: If , the roots and are real, negative, and unequal. This results in the over-damped natural response and has the form(1.7)Case II: If , the roots and are real, negative, and equal. This results in the criticallydamped natural response and has the form(1.8)Case III: If , the roots and are complex conjugates. This is known as the underdampedor oscillatory natural response and has the form(1.9)A typical overdamped response is shown in Figure 1.2 where it is assumed that . This plotwas created with the following MATLAB code:s2 RL---s 1LC-------+ + 0=s1 s2R2L------–R24L2--------1LC-------–=SR2L------=or DampingCoefficient01LC-----------=ResonantFrequencyS S202–=BetaCoefficientnS 02S2–=Damped NaturalFrequencyss1 s2 S– S202– S– S if S202= =s1 s2 S– 02S2– S– nSif 02S2= =S202s1 s2in t k1es1tk2es2t+=S202= s1 s2in t eSt–k1 k2t+=02S2s1 s2in t eS t–k1 nScos t k2 nStsin+ k3eSt–nScos t += =in 0 0=
16. 16. Chapter 1 Second Order Circuits1-4 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationst=0: 0.01: 6; ft=8.4.*(exp( t)-exp( 6.*t)); plot(t,ft); grid; xlabel(t);...ylabel(f(t)); title(Overdamped Response for 4.8.*(exp( t) exp( 6.*t)))Figure 1.2. Typical overdamped responseA typical critically damped response is shown in Figure 1.3 where it is assumed that . Thisplot was created with the following MATLAB code:t=0: 0.01: 6; ft=420.*t.*(exp( 2.45.*t)); plot(t,ft); grid; xlabel(t);...ylabel(f(t)); title(Critically Damped Response for 420.*t.*(exp( 2.45.*t)))Figure 1.3. Typical critically damped responseA typical underdamped response is shown in Figure 1.4 where it is assumed that . Thisplot was created with the following MATLAB code:in 0 0=in 0 0=
17. 17. 1-5 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Series RLC Circuits with DC Excitationt=0: 0.01: 10; ft=210.*sqrt(2).*(exp( 0.5.*t)).*sin(sqrt(2).*t); plot(t,ft); grid; xlabel(t);...ylabel(f(t)); title(Underdamped Response for 210.*sqrt(2).*(exp( 0.5.*t)).*sin(sqrt(2).*t))Figure 1.4. Typical underdamped response1.3 Response of Series RLC Circuits with DC ExcitationDepending on the circuit constants , , and , the total response of a series circuit that isexcited by a DC source, may be overdamped, critically damped, or underdamped. In this section wewill derive the total response of series circuits that are excited by DC sources.Example 1.1For the circuit of Figure 1.5, , , and the resistor represents theresistance of the inductor. Compute and sketch for .Figure 1.5. Circuit for Example 1.1Solution:This circuit can be represented by the integrodifferential equationR L C RLCRLCiL 0 5 A= vC 0 2.5 V= 0.5i t t 0+15u0 t Vi t0.51 mH100 6 mF
18. 18. Chapter 1 Second Order Circuits1-6 Circuit Analysis II with MATLAB ApplicationsOrchard Publications(1.10)Differentiating and noting that the derivatives of the constants and are zero, we obtain thehomogeneous differential equationorand by substitution of the known values , , and(1.11)The roots of the characteristic equation of (1.11) are and . The total responseis just the natural response and for this example it is overdamped. Therefore, from (1.7),(1.12)The constants and can be evaluated from the initial conditions. Thus from the first initial con-dition and (1.12) we get(1.13)We need another equation in order to compute the values of and . With this equation we willmake use of the second initial condition, that is, . Since , we dif-ferentiate (1.12), we evaluate it at , and we equate it with this initial condition. Then,(1.14)Also, at ,Ri Ldidt-----1C---- i td0tvC 0+ + + 15= t 0vC 0 15Rdidt----- Ld2idt2------- iC----+ + 0=d2idt2-------RL---didt----- iLC-------+ + 0=R L Cd2idt2------- 500didt----- 60000i+ + 0=s1 200–= s2 300–=i t in t k1es1tk2es2t+ k1e200– tk2e300– t+= ==k1 k2iL 0 i 0 5 A= =i 0 k1e0k2e0+ 5= =k1 k2+ 5=k1 k2vC 0 2.5 V= iC t i t CdvCdt--------= =t 0+=didt----- 200k– 1e200– t300k2– e300– tand=didt-----t 0+=200k– 1 300– k2=t 0+=Ri 0+Ldidt-----t 0+=vc 0++ + 15=
19. 19. 1-7 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Series RLC Circuits with DC Excitationand solving for we get(1.15)Next, equating (1.14) with (1.15) we get:(1.16)Simultaneous solution of (1.13) and (1.16) yields and . By substitution into (1.12)we find the total response as(1.17)Check with MATLAB:syms t; % Define symbolic variable tR=0.5; L=10^( 3); C=100*10^( 3)/6;% Circuit constantsy0=115*exp( 200*t) 110*exp(-300*t); % Let solution i(t)=y0y1=diff(y0); % Compute the first derivative of y0, i.e., di/dty2=diff(y0,2); % Compute the second derivative of y0, i.e, di2/dt2% Substitute the solution i(t), i.e., equ (1.17)% into differential equation of (1.11) to verify% that correct solution was obtained.% We must also verify that the initial% conditions are satisfiedy=y2+500*y1+60000*y0;i0=115*exp( 200*0) 110*exp( 300*0);vC0= R*i0 L*( 23000*exp( 200*0)+33000*exp( 300*0))+15;fprintf( n);...disp(Solution was entered as y0 = ); disp(y0);...disp(1st derivative of solution is y1 = ); disp(y1);...disp(2nd derivative of solution is y2 = ); disp(y2);...disp(Differential equation is satisfied since y = y2+y1+y0 = ); disp(y);...disp(1st initial condition is satisfied since at t = 0, i0 = ); disp(i0);...disp(2nd initial condition is also satisfied since vC+vL+vR=15 and vC0 = );...disp(vC0);...fprintf( n)Solution was entered as y0 =115*exp(-200*t)-110*exp(-300*t)didt-----t 0+=didt-----t 0+=15 0.5 5– 2.5–103–--------------------------------------- 10000= =200k– 1 300– k2 10000=k– 1 1.5– k2 50=k1 115= k2 110–=i t in t 115e200– t110– e300– t==
20. 20. Chapter 1 Second Order Circuits1-8 Circuit Analysis II with MATLAB ApplicationsOrchard Publications1st derivative of solution is y1 =-23000*exp(-200*t)+33000*exp(-300*t)2nd derivative of solution is y2 =4600000*exp(-200*t)-9900000*exp(-300*t)Differential equation is satisfied since y = y2+y1+y0 = 01st initial condition is satisfied since at t = 0, i0 = 52nd initial condition is also satisfied since vC+vL+vR=15 and vC0= 2.5000We will use the following MATLAB code to sketch .t=0: 0.0001: 0.025; i1=115.*(exp( 200.*t)); i2=110.*(exp( 300.*t)); iT=i1 i2;...plot(t,i1,t,i2,t,iT); grid; xlabel(t); ylabel(i1, i2, iT); title(Response iT for Example 1.1)Figure 1.6. Plot for of Example 1.1In the above example, differentiation eliminated (set equal to zero) the right side of the differentialequation and thus the total response was just the natural response. A different approach however,may not set the right side equal to zero, and therefore the total response will contain both the naturaland forced components. To illustrate, we will use the following approach.The capacitor voltage, for all time , may be expressed as and as before, the circuitcan be represented by the integrodifferential equation(1.18)i ti tt vC t1C---- i td–t=Ri Ldidt----- 1C---- i td–t+ + 15= u0 t
21. 21. 1-9 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Series RLC Circuits with DC Excitationand sincewe rewrite (1.18) as(1.19)We observe that this is a non-homogeneous differential equation whose solution will have both thenatural and the forced response components. Of course, the solution of (1.19) will give us the capaci-tor voltage . This presents no problem since we can obtain the current by differentiation of theexpression for .Substitution of the given values into (1.19) yieldsor(1.20)The characteristic equation of (1.20) is the same as of that of (1.11) and thus the natural response is(1.21)Since the right side of (1.20) is a constant, the forced response will also be a constant and we denote itas . By substitution into (1.20) we getor(1.22)The total solution then is the summation of (1.21) and (1.22), that is,(1.23)As before, the constants and will be evaluated from the initial conditions. First, usingand evaluating (1.23) at , we getori iC CdvCdt--------= =RCdvCdt--------- LCdvC2dt2--------- vC+ + 15= u0 tvC tvC t506------ 103– dvCdt-------- 1 103– 1006---------103– dvC2dt2-------- vC+ + 15= u0 tdvC2dt2-------- 500dvCdt-------- 60000vC+ + 9 105= u0 tvCn t k1es1tk2es2t+ k1e200– tk2e300– t+= =vCf k3=0 0 60000k3+ + 900000=vCf k3 15= =vC t vCn t vCf+= k1e200– tk2e300– t15+ +=k1 k2vC 0 2.5 V= t 0=vC 0 k1e0k2e015+ + 2.5= =
22. 22. Chapter 1 Second Order Circuits1-10 Circuit Analysis II with MATLAB ApplicationsOrchard Publications(1.24)Also,(1.25)Next, we differentiate (1.23), we evaluate it at and equate it with (1.25). Then,(1.26)Equating the right sides of (1.25) and (1.26) we getor(1.27)From (1.24) and (1.27), we get and . By substitution into (1.23), we obtain thetotal solution as(1.28)Check with MATLAB:syms t % Define symbolic variable ty0=22*exp( 300*t) 34.5*exp( 200*t)+15; % The total solution y(t)y1=diff(y0) % The first derivative of y(t)y1 =-6600*exp(-300*t)+6900*exp(-200*t)y2=diff(y0,2) % The second derivative of y(t)y2 =1980000*exp(-300*t)-1380000*exp(-200*t)y=y2+500*y1+60000*y0 % Summation of y and its derivativesy =900000Using the expression for we can find the current as(1.29)k1 k2+ 12.5–=iL iC CdvCdt--------= =dvCdt--------iLC---- anddvCdt--------t 0=iL 0C------------ 51006--------- 103–------------------------- 300= = ==t 0=dvCdt-------- 200k1– e200– t300k2– e300– tanddvCdt--------t 0=200k1– 300k2–==200k1– 300k2– 300=k1– 1.5k2– 1.5=k1 34.5–= k2 22=vC t 22e300– t34.5– e200– t15+ u0 t=vC ti iL= iC CdvCdt--------1006--------- 103–6900e200t–6600– e300t–115e200t–110– e300t–A= == =
23. 23. 1-11 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Series RLC Circuits with AC ExcitationWe observe that (1.29) is the same as (1.17).We will use the following MATLAB code to sketch .t=0: 0.001: 0.03; vc1=22.*(exp( 300.*t)); vc2= 34.5.*(exp( 200.*t)); vc3=15;...vcT=vc1+vc2+vc3; plot(t,vc1,t,vc2,t,vc3,t,vcT); grid; xlabel(t);...ylabel(vc1, vc2, vc3, vcT); title(Response vcT for Example 1.1)Figure 1.7. Plot for of Example 1.11.4 Response of Series RLC Circuits with AC ExcitationThe total response of a series RLC circuit, which is excited by a sinusoidal source, will also consist ofthe natural and forced response components. As we found in the previous section, the naturalresponse can be overdamped, or critically damped, or underdamped. The forced component will be asinusoid of the same frequency as that of the excitation, and since it represents the AC steady-statecondition, we can use phasor analysis to find it. The following example illustrates the procedure.Example 1.2For the circuit of Figure 1.8, , , and the resistor represents theresistance of the inductor. Compute and sketch for .Solution:This circuit is the same as that of Example 1.1 except that the circuit is excited by a sinusoidal source;therefore it can be represented by the integrodifferential equation(1.30)i tvC tiL 0 5 A= vC 0 2.5 V= 0.5i t t 0Ri Ldidt-----1C---- i td0tvC 0+ + + 200 10000tcos= t 0
24. 24. Chapter 1 Second Order Circuits1-12 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsFigure 1.8. Circuit for Example 1.2whose solution consists of the summation of the natural and forced responses. We know its naturalresponse from the previous example. We start with(1.31)where the constants and will be evaluated from the initial conditions after has beenfound. The steady state (or forced) response will have the form in thetime domain ( -domain) and has the form in the frequency domain ( -domain).To find we will use the phasor analysis relation where is the phasor current, isthe phasor voltage, and is the impedance of the phasor circuit which, as we know, is(1.32)The inductive and capacitive reactances areandThen,Also,and this yields . Then, by substitution into (1.32),and thus+200 10000tcos u0 t Vi t0.51 mH100 6 mFi t in t if t+ k1 e200– tk2 e300– tif t+ +==k1 k2 if tif t k3 10 000t +cos=t k3 jif t I V Z= I VZZ R j L 1 C–+ R2L 1 C–2+ L 1 C– R1–tan= =XL L 104103–10= = =XC1C--------1104100 6 103–--------------------------------------------- 6 103–= = =R20.520.25 and L 1 C–210 6 103––299.88= == =L 1 C– R1–tan 10 6 103––0.5------------------------------------1–tan9.9940.5-------------1–tan= =1.52 rads 87.15= =Z 0.25 99.88+o10 87.15o= =IVZ---200 0o10 87.15o--------------------------- 20 87.15–o= = = 20 10000t 87.15–ocos if t=
25. 25. 1-13 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Series RLC Circuits with AC ExcitationThe total solution is(1.33)The constants and are evaluated from the initial conditions. From (1.33) and the first initialcondition we get(1.34)We need another equation in order to compute the values of and . This equation will make useof the second initial condition, that is, . Since , we differentiate(1.33), we evaluate it at , and we equate it with this initial condition. Then,(1.35)and at ,(1.36)Also, atand solving for we get(1.37)Next, equating (1.36) with (1.37) we getor(1.38)i t in t if t+ k1e200– tk2e300– t20 10000t 87.15–ocos+ +==k1 k2iL 0 5 A=i 0 k1e0k2e020 87.15–ocos+ += 5=i 0 k1 k2 20 0.05+ += 5=k1 k2+ 4=k1 k2vC 0 2.5 V= iC t i t CdvCdt--------= =t 0=didt----- 200k– 1e200– t300k2– e300– t2 10510000t 87.15–osin–=t 0=didt-----t 0=200k– 1 300k2– 2 10687.15–osin–= 200k– 1 300k2– 2 105+=t 0+=Ri 0+Ldidt-----t 0+=vc 0++ + 200 0cos 200= =didt-----t 0+=didt-----t 0+=200 0.5 5– 2.5–103–------------------------------------------ 195000= =200k– 1 300– k2 5000–=k1 1.5k2+ 25=
26. 26. Chapter 1 Second Order Circuits1-14 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSimultaneous solution of (1.34) and (1.38) yields and . Then, by substitution into(1.31), the total response is(1.39)The plot is shown in Figure 1.9 and was created with the following MATLAB code:t=0: 0.005: 0.20; i1= 38.*(exp( 200.*t)); i2=42.*(exp( 300.*t));...i3=20.*cos(10000.*t 87.15.*pi./180); iT=i1+i2+i3; plot(t,i1,t,i2,t,i3,t,iT); grid; xlabel(t);...ylabel(i1, i2, i3, iT); title(Response iT for Example 1.2)Figure 1.9. Plot for of Example 1.21.5 The Parallel GLC CircuitConsider the circuit of Figure 1.10 where the initial conditions are , , andis the unit step function. We want to find an expression for the voltage for .Figure 1.10. Parallel RLC circuitFor this circuitk1 38–= k2 42=i t 38– e200– t42e300– t20 10000t 87.15–oAcos+ +=i tiL 0 I0= vC 0 V0=u0 t v t t 0iS u0 tv t G L CiCiLiG
27. 27. 1-15 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsThe Parallel GLC CircuitorBy differentiation,(1.40)To find the forced response, we must first specify the nature of the excitation , that is DC or AC.If is DC ( =constant), the right side of (1.40) will be zero and thus the forced response compo-nent . If is AC ( , the right side of (1.40) will be another sinusoid andtherefore . Since in this section we are concerned with DC excitations, the rightside will be zero and thus the total response will be just the natural response.The natural response is found from the homogeneous equation of (1.40), that is,(1.41)whose characteristic equation isorfrom which(1.42)and with the following notations,(1.43)where the subscript stands for parallel circuit, we can express (1.42) asiG t iL t iC t+ + iS t=Gv1L--- v td0tI0+ + Cdvdt------+ iS= t 0Cdv2dt2-------- Gdvdt------ vL---+ +diSdt-------= t 0iSiS vSvf 0= iS iS I t +cos=vf V t +cos=Cdv2dt2-------- Gdvdt------ vL---+ + 0=Cs2Gs1L---+ + 0=s2 GC----s iLC-------+ + 0=s1 s2G2C-------– G24C2--------- 1LC-------–=PG2C-------=or DampingCoefficient01LC-----------=ResonantFrequencyP P202–=BetaCoefficientnP 02P2–=Damped NaturalFrequencyp
28. 28. Chapter 1 Second Order Circuits1-16 Circuit Analysis II with MATLAB ApplicationsOrchard Publications(1.44)or(1.45)Note: From (1.4) and (1.43) we observe thatAs in a series circuit, the natural response can be overdamped, critically damped, or under-damped.Case I: If , the roots and are real, negative, and unequal. This results in the over-damped natural response and has the form(1.46)Case II: If , the roots and are real, negative, and equal. This results in the criticallydamped natural response and has the form(1.47)Case III: If , the roots and are complex conjugates. This results in the underdampedor oscillatory natural response and has the form(1.48)1.6 Response of Parallel GLC Circuits with DC ExcitationDepending on the circuit constants (or ), , and , the natural response of a parallel cir-cuit may be overdamped, critically damped or underdamped. In this section we will derive the totalresponse of a parallel circuit which is excited by a DC source using the following example.Example 1.3For the circuit of Figure 1.11, and . Compute and sketch for .s1 s2 P– P202– P– P if P202= =s1 s2 P– 02P2– P– nP if 02P2= =S Pvn tP202s1 s2vn t k1es1tk2es2t+=P202= s1 s2vn t ePt–k1 k2t+=02P2s1 s2vn t ePt–k1 nPcos t k2 nPtsin+ k3ePt–nPcos t += =G R L C GLCGLCiL 0 2 A= vC 0 5 V= v t t 0
29. 29. 1-17 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Parallel GLC Circuits with DC ExcitationFigure 1.11. Circuit for Example 1.3Solution:We could write the integrodifferential equation that describes the given circuit, differentiate, and findthe roots of the characteristic equation from the homogeneous differential equation as we did in theprevious section. However, we will skip these steps and start with(1.49)and when steady-state conditions have been reached we will have ,and .To find out whether the natural response is overdamped, critically damped, or oscillatory, we need tocompute the values of and using (1.43) and the values of and using (1.44) or (1.45).Then will use (1.46), or (1.47), or (1.48) as appropriate. For this example,orandThenor and . Therefore, the natural response is overdamped and from (1.46) we get(1.50)and the constants and will be evaluated from the initial conditions.From the initial condition and (1.50) we get10u0 t Av t32 1 640 FiCiLiR10 Hv t vf t vn t+=v vL L di dt 0= = = vf 0=v t vn t=P 0 s1 s2PG2C-------12RC-----------12 32 1 640------------------------------------- 10= = = =P2100=02 1LC-------110 1 640---------------------------- 64= = =s1 s2 P– P202– 10– 6= =s1 4–= s2 16–=v t vn t k1es1tk2es2t+ k1e4– tk2e16– t+= ==k1 k2vC 0 v 0 5 V= =
30. 30. Chapter 1 Second Order Circuits1-18 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationsor(1.51)The second equation that is needed for the computation of the values of and is found fromother initial condition, that is, . Since , we differentiate (1.50),evaluate it at , and we equate it with this initial condition.Then,(1.52)Also, atand solving for we get(1.53)Next, equating (1.52) with (1.53) we getor(1.54)Simultaneous solution of (1.51) and (1.54) yields , , and by substitutioninto (1.50) we get the total response as(1.55)Check with MATLAB:syms t % Define symbolic variable ty0=291*exp( 4*t)/6 261*exp( 16*t)/6; % Let solution v(t) = y0y1=diff(y0) % Compute and display first derivativey1 =-194*exp(-4*t)+696*exp(-16*t)v 0 k1e0k2e0+ 5= =k1 k2+ 5=k1 k2iL 0 2 A= iC t CdvCdt-------- Cdvdt------= =t 0+=dvdt------ 4k– 1e4– t16k2– e16– tand=dvdt------t 0+=4k– 1 16– k2=t 0+=1R---v 0+iL 0+Cdvdt------t 0+=+ + 10=dvdt------t 0+=dvdt------t 0+=10 5 32– 2–1 640------------------------------- 502= =4k– 1 16– k2 502=2k– 1 8– k2 251=k1 291 6= k2 261– 6=v t vn t2916---------e4– t 2616---------– e16– t==16--- 291e4– t261– e16– tV=
31. 31. 1-19 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Parallel GLC Circuits with DC Excitationy2=diff(y0,2) % Compute and display second derivativey2 =776*exp(-4*t)-11136*exp(-16*t)y=y2/640+y1/32+y0/10 % Verify that (1.40) is satisfiedy =0The plot is shown in Figure 1.12 where we have used the following MATLAB code:t=0: 0.01: 1; v1=(291./6).*(exp( 4.*t)); v2= (261./6).*(exp( 16.*t));...vT=v1+v2; plot(t,v1,t,v2,t,vT); grid; xlabel(t);...ylabel(v1, v2, vT); title(Response vT for Example 1.3)Figure 1.12. Plot for of Example 1.3From the plot of Figure 1.12, we observe that attains its maximum value somewhere in the inter-val and sec., and the maximum voltage is approximately . If we desire to compute pre-cisely the maximum voltage and the exact time it occurs, we can find the derivative of (1.55), set itequal to zero, and solve for . Thus,(1.56)Division of (1.56) by yieldsorv tv t0.10 0.12 24 Vtdvdt------t 0=1164e4– t– 4176e16– t+ 0= =e16t–1164e12t– 4176+ 0=
32. 32. Chapter 1 Second Order Circuits1-20 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsorandBy substitution into (1.55)(1.57)A useful quantity, especially in electronic circuit analysis, is the settling time, denoted as , and it isdefined as the time required for the voltage to drop to of its maximum value. Therefore, is anindication of the time it takes for to damp-out, meaning to decrease the amplitude of toapproximately zero. For this example, , and we can find by substitutioninto (1.55). Then,(1.58)and we need to solve for the time . To simplify the computation, we neglect the second term insidethe parentheses of (1.58) since this component of the voltage damps out much faster than the othercomponent. This expression then simplifies tooror(1.59)Example 1.4For the circuit of Figure 1.13, and , and the resistor is to be adjusted sothat the natural response will be critically damped. Compute and sketch for .e12t 34897---------=12t34897---------ln= 1.2775=t tmax1.277512---------------- 0.106 s= = =vmax16--- 291e4– x0.106261– e16– x0.10623.76 V= =tS1% tSv t v t0.01 23.76 0.2376 V= tS0.01vmax 0.237616--- 291e4t–261e16t––= =t0.237616--- 291e4– ts=4– tS 0.005ln 5.32–= =tS 1.33 s=iL 0 2 A= vC 0 5 V=v t t 0
33. 33. 1-21 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Parallel GLC Circuits with DC ExcitationFigure 1.13. Circuit for Example 1.4Solution:Since the natural response is to be critically damped, we must have because the L and C val-ues are the same as in the previous example. Please refer to (1.43). We must also haveoror and thus . The natural response will have the form(1.60)Using the initial condition and evaluating (1.60) at , we getor(1.61)and (1.60) simplifies to(1.62)As before, we need to compute the derivative in order to apply the second initial condition andfind the value of the constant .We obtain the derivative using MATLAB as follows:syms t k2; v0=exp( 8*t)*(5+k2*t); v1=diff(v0); % v1 is 1st derivative of v0v1 =-8*exp(-8*t)*(5+k2*t)+exp(-8*t)*k210u0 t Av t1 640 FiCiLiR10 H0264=PG2C-------12RC----------- 0= = =1LC------- 8= =1R--- 82640--------- 140------= =R 40= s1 s2 P– 8–= = =v t vn t= ePt–k1 k2t+ or v t vn t= e8t–k1 k2t+==vC 0 5 V= t 0=v 0 e0k1 k20+= 5=k1 5=v t e8t–5 k2t+=dv dtk2
34. 34. Chapter 1 Second Order Circuits1-22 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsThen,and(1.63)Also, or and(1.64)or(1.65)Equating (1.63) with (1.65) and solving for we getor(1.66)and by substitution into (1.62), we obtain the total solution as(1.67)Check with MATLAB:syms t; y0=exp( 8*t)*(5+5080*t); y1=diff(y0)% Compute 1st derivativey1 =-8*exp(-8*t)*(5+5080*t)+5080*exp(-8*t)y2=diff(y0,2) % Compute 2nd derivativey2 =64*exp(-8*t)*(5+5080*t)-81280*exp(-8*t)y=y2/640+y1/40+y0/10 % Verify differential equation, see (1.40)y =0The plot is shown in Figure 1.14 where we have used the following MATLAB code:dvdt------ 8e–8t–5 k2t+= k2e8t–+dvdt------t 0=40–= k2+iC Cdvdt------=dvdt------iCC----=dvdt------t 0+=iC 0+C---------------IS iR 0+– iL 0+–C-------------------------------------------= =dvdt------t 0=IS vC 0 R– iL 0–C-------------------------------------------------10 5 40– 2–1 640-------------------------------7.8751 640---------------- 5040= = = =k240– k2+ 5040=k2 5080=v t e8t–5 5080t+ V=
35. 35. 1-23 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Parallel GLC Circuits with DC Excitationt=0: 0.01: 1; vt=exp( 8.*t).*(5+5080.*t); plot(t,vt); grid; xlabel(t);...ylabel(vt); title(Response vt for Example 1.4)Figure 1.14. Plot for of Example 1.4By inspection of (1.67), we see that at , and thus the second initial condition is sat-isfied. We can verify that the first initial condition is also satisfied by differentiation of (1.67). We canalso show that approaches zero as approaches infinity with L’Hôpital’s rule as follows:(1.68)Example 1.5For the circuit of Figure 1.15, and . Compute and sketch for .Figure 1.15. Circuit for Example 1.5Solution:This is the same circuit as the that of the two previous examples except that the resistance has beenincreased to . For this example,v tt 0= v t 5 V=v t tv ttlim e8t–5 5080t+tlim 5 5080t+e8t----------------------------tlim d 5 5080t+ dtd e8tdt----------------------------------------tlim 50808e8t------------tlim 0= = = ==iL 0 2 A= vC 0 5 V= v t t 010u0 t Av t50 1 640 FiCiLiR10 H50
36. 36. Chapter 1 Second Order Circuits1-24 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationsorand as before,Also, . Therefore, the natural response is underdamped with natural frequencySince , the total response is just the natural response. Then, from (1.48),(1.69)and the constants and will be evaluated from the initial conditions.From the initial condition and (1.69) we getor(1.70)To evaluate the constants and we differentiate (1.69), we evaluate it at , we write the equa-tion which describes the circuit at , and we equate these two expressions. Using MATLAB weget:syms t k phi; y0=k*exp( 6.4*t)*cos(4.8*t+phi); y1=diff(y0)y1 =-32/5*k*exp(-32/5*t)*cos(24/5*t+phi)-24/5*k*exp(-32/5*t)*sin(24/5*t+phi)pretty(y1)- 32/5 k exp(- 32/5 t) cos(24/5 t + phi)- 24/5 k exp(- 32/5 t) sin(24/5 t + phi)Thus,(1.71)andPG2C-------12RC-----------12 50 1 640------------------------------------- 6.4= = = =P240.96=02 1LC-------110 1 640---------------------------- 64= = =02P2nP 02P2– 64 40.96– 23.04 4.8= = = =vf 0=v t vn t kePt–nPt +cos== ke6.4t–4.8t +cos=kvC 0 v 0 5 V= =v 0 ke00 +cos 5= =kcos 5=k t 0=t 0+=dvdt------ 6.4ke6.4t–– 4.8t +cos 4.8ke6.4t–4.8t +sin–=
37. 37. 1-25 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Parallel GLC Circuits with DC ExcitationBy substitution of (1.70), the above expression simplifies to(1.72)Also, or andor(1.73)Equating (1.72) with (1.73) we getor(1.74)The phase angle can be found by dividing (1.74) by (1.70). Then,orThe value of the constant is found from (1.70) asorand by substitution into (1.69), the total solution is(1.75)The plot is shown in Figure 1.16 where we have used the following MATLAB code:dvdt------t 0=6.4k– cos 4.8ksin–=dvdt------t 0=32– 4.8ksin–=iC Cdvdt------= dvdt------iCC----=dvdt------t 0+=iC 0+C---------------IS iR 0+– iL 0+–C-------------------------------------------= =dvdt------t 0=IS vC 0 R– iL 0–C------------------------------------------------- 10 5 50– 2–1 640------------------------------- 7.9 640 5056= = = =32– 4.8ksin– 5056=ksin 1060–=ksinkcos--------------- tan1060–5--------------- 212–= = =212–1–tan 1.566 rads– 89.73 deg–= = =kk 1.566–cos 5=k 51.566–cos------------------------------ 1042= =v t 1042e6.4t–4.8t 89.73–cos=
38. 38. Chapter 1 Second Order Circuits1-26 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationst=0: 0.005: 1.5; vt=10.42.*exp( 6.4.*t).*cos(4.8.*t 89.73.*pi./180);...plot(t,vt); grid; xlabel(t); ylabel(vt); title(Response v(t) for Example 1.5)Figure 1.16. Plot for of Example 1.5We can also use a spreadsheet to plot (1.75). From the columns of that spreadsheet we can read thefollowing maximum and minimum values and the times these occur.Alternately, we can find the maxima and minima by differentiating the response of (1.75) and settingit equal to zero.1.7 Response of Parallel GLC Circuits with AC ExcitationThe total response of a parallel GLC (or RLC) circuit that is excited by a sinusoidal source also con-sists of the natural and forced response components. The natural response will be overdamped, criti-cally damped, or underdamped. The forced component will be a sinusoid of the same frequency asthat of the excitation, and since it represents the AC steady-state condition, we can use phasor analy-sis to find the forced response. We will derive the total response of a parallel GLC (or RLC) circuitwhich is excited by an AC source with the following example.Example 1.6For the circuit of Figure 1.17, and . Compute and sketch for .t (sec) v (V)Maximum 0.13 266.71Minimum 0.79 4.05v tiL 0 2 A= vC 0 5 V= v t t 0
39. 39. 1-27 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsResponse of Parallel GLC Circuits with AC ExcitationFigure 1.17. Circuit for Example 1.6Solution:This is the same circuit as the previous example where the DC source has been replaced by an ACsource. The total response will consist of the natural response which we already know from theprevious example, and the forced response which is the AC steady-state response, will be foundby phasor analysis.The -domain to -domain transformation yieldsThe admittance iswhereand thusNow, we find the phasor voltage asand -domain to -domain transformation yieldsThe total solution is(1.76)iS 20 6400t 90+ usin 0 t A=v t50 1 640 FiCiLiR10 HiSvn tvf tt jis t 20 6400t 90+sin 20 6400t Icos 20 0= = =YY G j C 1L-------–+ G2C 1L-------–2+ C 1L-------– G1–tan= =G1R---=150------= C 64001640--------- 10 and1L------- 16400 10------------------------164000---------------= = = =,Y150------210164000---------------–2+ 10164000---------------–150------1–tan 10 89.72= =VVIY---20 010 89.72--------------------------- 2 89.72–= = =j tV 2 89.72– vf t 2 6400t 89.72–cos= =v t vn t vf t+ ke6.4t–4.8t +cos= = 2 6400t 89.72–cos+
40. 40. Chapter 1 Second Order Circuits1-28 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsNow, we need to evaluate the constants and .With the initial condition (1.76) becomesor(1.77)To make use of the second initial condition, we differentiate (1.76) using MATLAB as follows, andthen we evaluate it at .syms t k phi; y0=k*exp( 6.4*t)*cos(4.8*t+phi)+2*cos(6400*t 1.5688);y1=diff(y0); % Differentiate v(t) of (1.76)y1 =-32/5*k*exp(-32/5*t)*cos(24/5*t+phi)-24/5*k*exp(-32/5*t)*sin(24/5*t+phi)-12800*sin(6400*t-1961/1250)orand(1.78)With (1.77) we get(1.79)Also, or andor(1.80)Equating (1.79) with (1.80) and solving for we getkvC 0 5 V=v 0 vC 0 ke0cos= = 2 89.72–cos+ 5=kcos 5t 0=dvdt------ 6.4ke6.4t–– 4.8t +cos 4.8ke6.4t–4.8t +sin– 12800 6400t 1.5688–sin–=dvdt------t 0=6.4k– cos 4.8ksin– 12800 1.5688–sin– 6.4k– cos 4.8ksin– 12800+= =dvdt------t 0=32– 4.8ksin– 12800+= 4.8ksin– 12832+iC Cdvdt------= dvdt------iCC----=dvdt------t 0+=iC 0+C---------------iS 0+iR 0+– iL 0+–C-----------------------------------------------------= =dvdt------t 0=iS 0+vC 0 R– iL 0–C----------------------------------------------------------- 20 5 50– 2–1 640------------------------------- 11456= = =k4.8ksin– 12832+ 11456=
41. 41. 1-29 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsOther Second Order Circuitsor(1.81)Then with (1.77) and (1.81),orThe value of the constant is found from (1.77), that is,By substitution into (1.76), we obtain the total solution as(1.82)With MATLAB we get the plot shown in Figure 1.18.Figure 1.18. Plot for of Example 1.61.8 Other Second Order CircuitsSecond order circuits are not restricted to series and parallel circuits. Other second ordercircuits include amplifiers and filters. It is beyond the scope of this text to analyze such circuits indetail. In this section we will use the following example to illustrate the transient analysis of a secondorder active low-pass filter.ksin 287=ksinkcos--------------- tan2875--------- 57.4= = =1.53 rad 89= =kk 5 89cos 279.4= =v t 279.4e6.4t–4.8t 89+cos= 2 6400t 89.72–cos+v tRLC GLC
42. 42. Chapter 1 Second Order Circuits1-30 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsExample 1.7The circuit of Figure 1.19 a known as a Multiple Feed Back (MFB) active low-pass filter. For this cir-cuit, the initial conditions are . Compute and sketch for .Figure 1.19. Circuit for Example 1.7Solution:At node :(1.83)At node :(1.84)We observe that (virtual ground).Collecting like terms and rearranging (1.83) and (1.84) we get(1.85)and(1.86)Differentiation of (1.86) yields(1.87)vC1 vC2 0= = vout t t 0vin vout40 K200 K50 K25 nF10 nFR2R1C2C1R3v1 v2++vin t 6.25 6280tu0 tcos=v1v1 vin–R1----------------- C1dv1dt--------v1 vout–R2--------------------v1 v2–R3---------------+ + + 0 t 0=v2v2 v1–R3--------------- C2dvoutdt-------------=v2 0=1R1-----1R2-----1R3-----+ + v1 C1dv1dt--------1R2-----vout–+1R1-----vin=v1 R3C2dvoutdt-----------–=dv1dt-------- R3C2dv2outdt2-------------–=
43. 43. 1-31 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsOther Second Order Circuitsand by substitution of given numerical values into (1.85) through (1.87), we getor(1.88)(1.89)(1.90)Next, substitution of (1.89) and (1.90) into (1.88) yields(1.91)orDivision by yieldsor(1.92)We use MATLAB to find the roots of the characteristic equation of (1.92).syms s; y0=solve(s^2+2*10^3*s+2*10^6)y0 =[ -1000+1000*i][ -1000-1000*i]that is,12 105------------------14 104------------------15 104------------------+ + v1 25 109– dv1dt--------14 104------------------vout–+12 105------------------vin=0.05 103–v1 25 109– dv1dt-------- 0.25 104–vout–+ 0.5 105–vin=v1 5 104–dvoutdt-------------–=dv1dt-------- 5 104– d2voutdt2---------------–=0.05 103–5 104–dvoutdt-------------– 25 109–5 104––d2voutdt2---------------0.25 104–vout–+0.5 105–vin=125– 1013– d2voutdt2--------------- 0.25 107–dvoutdt-------------– 0.25 104–vout– 104–vin=125– 1013–d2voutdt2---------------- 2 103dvoutdt------------- 2 106vout+ + 1.6 105– vin=d2voutdt2---------------- 2 103dvoutdt------------- 2 106vout+ + 1066280tcos–=
44. 44. Chapter 1 Second Order Circuits1-32 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsWe cannot classify the given circuit as series or parallel and therefore, we should not use the dampingratio or . Instead, for the natural response we will use the general expression(1.93)whereTherefore, the natural response is oscillatory and has the form(1.94)Since the right side of (1.92) is a sinusoid, the forced response has the form(1.95)Of course, for the derivation of the forced response we could use phasor analysis but we must firstderive an expression for the impedance or admittance because the expressions we’ve used earlier arevalid for series and parallel circuits only.The coefficients and will be found by substitution of (1.95) into (1.92) and then by equatinglike terms. Using MATLAB we get:syms t k3 k4; y0=k3*cos(6280*t)+k4*sin(6280*t); y1=diff(y0)y1 =-6280*k3*sin(6280*t)+6280*k4*cos(6280*t)y2=diff(y0,2)y2 =-39438400*k3*cos(6280*t)-39438400*k4*sin(6280*t)y=y2+2*10^3*y1+2*10^6*y0y =-37438400*k3*cos(6280*t)-37438400*k4*sin(6280*t)-12560000*k3*sin(6280*t)+12560000*k4*cos(6280*t)Equating like terms with (1.92) we get(1.96)Simultaneous solution of the equations of (1.96) is done with MATLAB.s1 s2, – j 1000– j1000 1000 1– j1= = =S P vn tvn t Aes1t= Bes2t+ et–k1 tcos k2 tsin+=s1 s2, – j 1000– j1000= =vn t e1000t–k1 1000tcos k2 1000tsin+=vf t k3 6280tcos k4 6280tsin+=k3 k437438400 k3– 12560000 k4+ 6280tcos 106– 6280tcos=12560000 k3– 37438400 k4– 6280tsin 0=
45. 45. 1-33 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsOther Second Order Circuitssyms k3 k4; eq1= 37438400*k3+12560000*k4+10^6;...eq2= 12560000*k3-37438400*k4+0; y=solve(eq1,eq2)y =k3: [1x1 sym]k4: [1x1 sym]y.k3ans =0.0240y.k4ans =-0.0081that is, and . Then, by substitution into (1.95)(1.97)The total response is(1.98)We will use the initial conditions to evaluate and . We observe thatand at relation (1.98) becomesor and thus (1.98) simplifies to(1.99)To evaluate the constant , we make use of the initial condition . We observe thatand by KCL at node we have:ork3 0.024= k4 0.008–=vf t 0.024 6280tcos 0.008– 6280tsin=vout t vn t vf t+ e1000t–k1 1000tcos k2 1000tsin+0.024 6280tcos 0.008– 6280tsin+= =vC1 vC2 0= = k1 k2 vC2 vout=t 0=vout 0 e0k1 0cos 0+ 0.024 0cos 0–+ 0= =k1 0.024–=vout t e1000t–0.024– 1000tcos k2 1000tsin+0.024 6280tcos 0.008– 6280tsin+=k2 vC1 0 0=vC1 v1= v1v1 v2–R3--------------- C2dvoutdt-------------+ 0=v1 0–5 104----------------- 108––=dvoutdt-------------
46. 46. Chapter 1 Second Order Circuits1-34 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationsorand since , it follows that(1.100)The last step in finding the constant is to differentiate (1.99), evaluate it at , and equate itwith (1.100). This is done with MATLAB as follows:y0=exp( 1000*t)*( 0.024*cos(1000*t)+k2*sin(1000*t))...+0.024*cos(6280*t) 0.008*sin(6280*t);y1=diff(y0)y1 =-1000*exp(-1000*t)*(-3/125*cos(1000*t)+k2*sin(1000*t))+exp(-1000*t)*(24*sin(1000*t)+1000*k2*cos(1000*t))-3768/25*sin(6280*t)-1256/25*cos(6280*t)orand(1.101)Simplifying and equating (1.100) with (1.101) we getorand by substitution into (1.99),(1.102)v1 5 104––=dvoutdt-------------vC1 0 v1 0 0= =dvoutdt-------------t 0=0=k2 t 0=dvoutdt------------- 1000e1000t– 3–125--------- 1000t k2 1000tsin+cos–e1000t–24 1000t 1000k2 1000tcos+sin376825------------ 6280tsin–125625------------ 6280tcos–+=dvoutdt-------------t 0=10003–125---------– 1000k2125625------------–+=1000k2 26.24– 0=k2 0.026=vout t e1000t–0.024– 1000tcos 0.026 1000tsin+0.024 6280tcos 0.008– 6280tsin+=
47. 47. 1-35 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsOther Second Order CircuitsWe use Excel to sketch . In Column A we enter several values of time and in Column B. The plot is shown in Figure 1.20.Figure 1.20. Plot for Example 1.7vout t tvout t-0.03-0.02-0.010.000.010.020.030.000 0.002 0.004 0.006 0.008Time (s)Voltage(V)
48. 48. Chapter 1 Second Order Circuits1-36 Circuit Analysis II with MATLAB ApplicationsOrchard Publications1.9 SummaryCircuits that contain energy storing devices can be described by integrodifferential equations andupon differentiation can be simplified to differential equations with constant coefficients.A second order circuit contains two energy storing devices. Thus, an RLC circuit is a second ordercircuit.The total response is the summation of the natural and forced responses.If the differential equation describing a series RLC circuit that is excited by a constant (DC) volt-age source is written in terms of the current, the forced response is zero and thus the totalresponse is just the natural response.If the differential equation describing a parallel RLC circuit that is excited by a constant (DC) cur-rent source is written in terms of the voltage, the forced response is zero and thus the totalresponse is just the natural response.If a circuit is excited by a sinusoidal (AC) source, the forced response is never zero.The natural response of a second order circuit may be overdamped, critically damped, or under-damped depending on the values of the circuit constants.For a series RLC circuit, the roots and are found fromorwhereIf , the roots and are real, negative, and unequal. This results in the overdamped nat-ural response and has the formIf , the roots and are real, negative, and equal. This results in the critically dampednatural response and has the forms1 s2s1 s2 S– S202– S– S if S202= =s1 s2 S– 02S2– S– nSif 02S2= =SR2L------= 01LC-----------= S S202–= nS 02S2–=S202s1 s2in t k1es1tk2es2t+=S202= s1 s2in t eSt–k1 k2t+=
49. 49. 1-37 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSummaryIf , the roots and are complex conjugates. This is known as the underdamped oroscillatory natural response and has the formFor a parallel GLC circuit, the roots and are found fromorwhereIf , the roots and are real, negative, and unequal. This results in the overdamped nat-ural response and has the formIf , the roots and are real, negative, and equal. This results in the critically dampednatural response and has the formIf , the roots and are complex conjugates. This results in the underdamped or oscil-latory natural response and has the formIf a second order circuit is neither series nor parallel, the natural response if found fromorordepending on the roots of the characteristic equation being real and unequal, real and equal, orcomplex conjugates respectively.02S2s1 s2in t eS t–k1 nScos t k2 nStsin+ k3eSt–nScos t += =s1 s2s1 s2 P– P202– P– P if P202= =s1 s2 P– 02P2– P– nP if 02P2= =PG2C-------= 01LC-----------= P P202–= nP 02P2–=P202s1 s2vn t k1es1tk2es2t+=P202= s1 s2vn t ePt–k1 k2t+=02P2s1 s2vn t ePt–k1 nPcos t k2 nPtsin+ k3ePt–nPcos t += =yn k1es1t= k2es2t+yn k1 k2 t+ es1t=yn et–k3 tcos k4 sin t+ et–k5 t +cos==
50. 50. Chapter 1 Second Order Circuits1-38 Circuit Analysis II with MATLAB ApplicationsOrchard Publications1.10 Exercises1. For the circuit of Figure 1.21, it is known that and . Compute and sketchand for .Figure 1.21. Circuit for Exercise 12. For the circuit of Figure 1.22, it is known that and . Compute and sketchand for .Figure 1.22. Circuit for Exercise 23. In the circuit of Figure 1.23, the switch has been closed for a very long time and opens at. Compute for .Figure 1.23. Circuit for Exercise 3vC 0 0= iL 0 0=vC t iL t t 0+100u0 t V10 0.2 H8 mFiL t+vC tvC 0 0= iL 0 0=vC t iL t t 0+100u0 t V4 5 H21.83 mFiL t+vC tSt 0= vC t t 0++20 H100100 V400vC t1 120 FSt 0=
51. 51. 1-39 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsExercises4. In the circuit of Figure 1.24, the switch has been closed for a very long time and opens at .Compute for .Figure 1.24. Circuit for Exercise 45. In the circuit of Figure 1.25, the switch has been in position for closed for a very long timeand it is placed in position at . Find the value of that will cause the circuit to becomecritically damped and then compute and forFigure 1.25. Circuit for Exercise 56. In the circuit of Figure 1.26, the switch has been closed for a very long time and opens at .Compute for .Figure 1.26. Circuit for Exercise 6S t 0=vC t t 0++20 H100400vC t1 120 FSt 0=vSvS 100 tcos u0 t V=S AB t 0= RvC t iL t t 0+ +12 V32RABS63 HvC t1 12 FiL tt 0=S t 0=vAB t t 0+12 V42 H1 4 FBAS t 0=2
52. 52. Chapter 1 Second Order Circuits1-40 Circuit Analysis II with MATLAB ApplicationsOrchard Publications1.11 Solutions to ExercisesDear Reader:The remaining pages on this chapter contain the solutions to the exercises.You must, for your benefit, make an honest effort to find the solutions to the exercises without firstlooking at the solutions that follow. It is recommended that first you go through and work out thoseyou feel that you know. For the exercises that you are uncertain, review this chapter and try again.Refer to the solutions as a last resort and rework those exercises at a later date.You should follow this practice with the rest of the exercises of this book.
53. 53. 1-41 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to Exercises1.and since , the above becomesFrom the characteristic equationwe get (critical damping) andThe total solution is(1)With the first initial condition the above expression becomes orand by substitution into (1) we get(2)To evaluate we make use of the second initial condition and since , and, we differentiate (2) using the following MATLAB code:+100u0 t V10 0.2 H8 mFiL t+vC tiRi Ldidt----- vC+ + 100= t 0i iC CdvCdt--------= =RCdvCdt-------- LCd2vCdt2---------- vC+ + 100=d2vCdt2----------RL---dvCdt-------- 1LC-------vC+ +100LC---------=d2vCdt2----------100.2-------dvCdt-------- 10.2 8 103–--------------------------------- vC+ + 1000.2 8 103–---------------------------------=d2vCdt2---------- 50dvCdt-------- 625 vC+ + 62500=s250s 625+ + 0=s1 s2 25–= = S R 2L 25= =vC t vCf vCn+ 100 eS t–k1 k2t++ 100 e25 t–k1 k2t++= = =vC 0 0= 0 100 e0k1 0++=k1 100–=vC t 100 e25 t–k2t 100–+=k2 iL 0 0= iL iC=i iC C dvC dt= =
54. 54. Chapter 1 Second Order Circuits1-42 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationssyms t k2v0=100+exp( 25*t)*(k2*t 100); v1=diff(v0)v1 =-25*exp(-25*t)*(k2*t-100)+exp(-25*t)*k2Thus,and(3)Also, and at(4)From (3) and (4) or and by substitution into (2)(5)We find by differentiating (5) and multiplication by . Using MATLAB we get:syms tC=8*10^( 3);i0=C*(100-exp( 25*t)*(100+2500*t)); iL=diff(i0)iL =1/5*exp(-25*t)*(100+2500*t)-20*exp(-25*t)Thus,The plots for and are shown on the next page.dvCdt-------- k2 e25t–25e25t–k2 t 100––=dvCdt--------t 0=k2 2500+=dvCdt--------iCC----iLC----= = t 0=dvCdt--------t 0=iL 0C--------------- 0= =k2 2500+ 0= k2 2500–=vC t 100 e25t–– 2500t 100+=iL t iC t= CiL t iC t 0.2e25t–100 2500t+ 20e25t––= =vC t iL t
55. 55. 1-43 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to Exercises2.The general form of the differential equation that describes this circuit is same as in Exercise 1,that is,From the characteristic equation and the MATLAB code belowt vC(t) iL(t)0.000 0 00.005 0.7191 2.2060.010 2.6499 3.8940.015 5.4977 5.1550.020 9.0204 6.0650.025 13.02 6.6910.030 17.336 7.0850.035 21.838 7.2950.040 26.424 7.3580.045 31.011 7.3050.050 35.536 7.1630.055 39.951 6.9530.060 44.217 6.6940.065 48.311 6.40.070 52.212 6.0820.075 55.91 5.7510.080 59.399 5.4130.085 62.677 5.0760.090 65.745 4.7430.095 68.608 4.4180.100 71.27 4.1040.105 73.741 3.8030.110 76.027 3.5160.115 78.139 3.244vC(t)0204060801000.0 0.1 0.2 0.3 0.4 0.5TimeVoltsiL(t)024680.0 0.1 0.2 0.3 0.4 0.5TimeAmps+100u0 t V4 5 H21.83 mFiL t+vC td2vCdt2----------RL---dvCdt--------1LC-------vC+ +100LC---------= t 0d2vCdt2---------- 0.8dvCdt-------- 9.16vC+ + 916=s20.8s 9.16+ + 0=
56. 56. Chapter 1 Second Order Circuits1-44 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationss=[1 0.8 9.16]; roots(s)ans =-0.4000 + 3.0000i-0.4000 - 3.0000iwe find that and . Therefore, the total solution iswhereandThus,(1)and with the initial condition we get or(2)To evaluate and we differentiate (1) with MATLAB and evaluate it at .syms t k phi; v0=100+k*exp( 0.4*t)*cos(3*t+phi); v1=diff(v0)v1 =-2/5*k*exp(-2/5*t)*cos(3*t+phi)-3*k*exp(-2/5*t)*sin(3*t+phi)Thus,and with (2)(3)Also, and at(4)s1 0.4– j3+= s2 0.4– j3–=vC t vCf vCn+ 100 keSt–nSt +cos+= =S R 2L 0.4= =nS 02S2– 1 LC R24L2– 9.16 0.16– 3= = = =vC t 100 ke0.4t–3t +cos+=vC 0 0= 0 100 k 0 +cos+=kcos 100–=k t 0=dvCdt-------- 0.4k– e0.4t–3t +cos 3ke0.4t–3 t +sin–=dvCdt--------t 0=0.4k– cos 3ksin–=dvCdt--------t 0=40 3ksin–=dvCdt--------iCC----iLC----= = t 0=dvCdt--------t 0=iL 0C--------------- 0= =
57. 57. 1-45 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to ExercisesFrom (3) and (4)(5)and from (2) and (5)The value of can be found from either (2) or (5). From (2)and by substitution into (1)(6)Since , we use MATLAB to differentiate (6).syms t; vC=100 100.8*exp( 0.4*t)*cos(3*t 0.1326); C=0.02183; iL=C*diff(vC)iL =137529/156250*exp(-2/5*t)*cos(3*t-663/5000)+412587/62500*exp(-2/5*t)*sin(3*t-663/5000)137529/156250, 412587/62500ans =0.8802ans =6.6014The plots for and are shown on the next page.3ksin 40=3ksinkcos-----------------40100–------------=3tan 0.4–=0.4– 31–tan 0.1326 rad– 7.6–= = =kk 0.1236–cos 100–=k100–0.1236–cos--------------------------------- 100.8–= =vC t 100 100.8– e0.4t–3t 7.6–cos=iL t iC t C dvC dt= =iL t 0.88e0.4t–3t 7.6– 6.6e0.4t–3t 7.6–sin+cos=vC t iL t
58. 58. Chapter 1 Second Order Circuits1-46 Circuit Analysis II with MATLAB ApplicationsOrchard Publications3.At the circuit is as shown below.At this time the inductor behaves as a short and the capacitor as an open. Then,and this establishes the first initial condition as . Also,and this establishes the first initial condition as .For the circuit is as shown below.t vC(t) iL(t)0.000 -0.014 -0.0020.010 0.0313 0.1980.020 0.1677 0.3950.030 0.394 0.5910.040 0.7094 0.7840.050 1.1129 0.9750.060 1.6034 1.1640.070 2.1798 1.350.080 2.8407 1.5340.090 3.5851 1.7140.100 4.4115 1.8920.110 5.3185 2.0660.120 6.3046 2.2380.130 7.3684 2.4050.140 8.5082 2.570.150 9.7224 2.730.160 11.009 2.887vC(t)-50501500 3 6 9 12iL(t)-10100 3 6 9 12t 0=++20 H100100 V400vC 01 120 FiL 0iL 0 100 100 400+ I0 0.2 A= = =I0 0.2 A=vC 0 v400 400 iL 0 400 0.2 V0 80 V= = = = =V0 80 V=t 0
59. 59. 1-47 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to ExercisesThe general form of the differential equation that describes this circuit is same as in Exercise 1,that is,From the characteristic equation we find that and and the totalresponse for the capacitor voltage is(1)Using the initial condition we getor(2)Differentiation of (1) and evaluation at yields(3)Also, and at(4)Equating (3) and (4) we get(5)and simultaneous solution of (2) and (5) yields and .++20 H100100 VvC t1 120 Fd2vCdt2----------RL---dvCdt--------1LC-------vC+ +100LC---------= t 0d2vCdt2---------- 5dvCdt-------- 6vC+ + 600=s25s 6+ + 0= s1 2–= s2 3–=vC t vCf vCn+ 100 k1es1tk2es2t+ + 100 k1e2t–k2e3t–+ += = =V0 80 V=vC 0 V0 80 V 100 k1e0k2e0+ += ==k1 k2+ 20–=t 0=dvCdt--------t 0=2k1– 3k2–=dvCdt--------iCC----iLC----= = t 0=dvCdt--------t 0=iL 0C--------------- 0.21 120---------------- 24= = =2k1– 3k2– 24=k1 36–= k2 16=
60. 60. Chapter 1 Second Order Circuits1-48 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsBy substitution into (1) we find the total solution as4.This is the same circuit as in Exercise 3 where the DC voltage source has been replaced by an ACsource that is being applied at . No initial conditions were given so we will assume thatand . Also, the circuit constants are the same and thus the naturalresponse has the form .We will find the forced (steady-state) response using phasor circuit analysis where ,, , and . The phasor circuit is shown below.Using the voltage division expression we getand in the -domain . Therefore, the total response is(1)Using the initial condition and (1) we getvC t vCf vCn+ 100 36– e2t–16e3t–+= =++20 H100400vC t1 120 FSt 0=vSvS 100 tcos u0 t V=t 0+=iL 0 0= vC 0 0=vCn k1e2t–k2e3t–+=1=j L j20= j– C j120–= 100 t 100 0cos++j20100VSVS 100 0 V=VCj– 120VCj120–100 j20 j120–+----------------------------------------100 0j120–100 j100+--------------------------100 0120 90 100 0–100 2 45---------------------------------------------------- 60 2 135–= = ==t vCf 60 2 t 135–cos=vC t 60 2 t 135–cos k1e2t–k2e3t–+ +=vC 0 0=
61. 61. 1-49 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to Exercisesand since the above expression reduces to(2)Differentiating (1) we getand(3)Also, and at(4)Equating (3) and (4) we get(5)Simultaneous solution of (2) and (5) yields and . Then, by substitution into (1)5.We must first find the value of before we can establish initial conditions for and. The condition for critical damping is wherevC 0 0 60 2 135–cos k1 k2+ += =135–cos 2– 2=k1 k2+ 60=dvCdt-------- 60 2 t 45+sin 2k– 1e2t–3k2– e3t–+=dvCdt--------t 0=60 2 45sin 2k– 1 3k2–=dvCdt--------t 0=60 2k– 1 3k2–=dvCdt--------iCC----iLC----= = t 0=dvCdt--------t 0=iL 0C--------------- 0= =2k1 3k2+ 60=k1 120= k2 60–=vC t 60 2 t 135–cos 120e2t–60– e3t–+=+ +12 V32RABS63 HvC t1 12 FiL tt 0=R iL 0 0=vC 0 0= P202– 0= P G 2C 1 2RC= =
62. 62. Chapter 1 Second Order Circuits1-50 Circuit Analysis II with MATLAB ApplicationsOrchard Publicationsand . Then, where . Therefore,, or , or , or and thus .At the circuit is as shown below.From the circuit aboveandAt the circuit is as shown below.Since the circuit is critically damped, the solution has the formwhere and thus(1)021 LC= P2 12R 1 12---------------------------202= = 13 1 12----------------------= R R 2+=122 R 2+---------------------24=6R 2+-------------24= R 2+236 4 9= = R 2+ 3= R 1=t 0=+ +12 V3 16vC 0iL 0v6+vC 0 v663 1 6+ +--------------------- 12 7.2 V= = =iL 0v66---------7.26------- 1.2 A= = =t 0+=+63 HvC t1 12 FiL t12iC tiR tvC t eP t–k1 k2t+=P12 1 2+ 1 12--------------------------------------- 2= =vC t e2 t–k1 k2t+=
63. 63. 1-51 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to ExercisesWith the initial condition relation (1) becomes orand (1) simplifies to(2)Differentiating (2) we getand(3)Also, and at(4)because at the capacitor is an open circuit.Equating (3) and (4) we get or and by substitution into (2)We find from or whereand . Then,6.At the circuit is as shown below where , , and thusthe initial conditions have been established.vC 0 7.2 V= 7.2 e0k1 0+= k1 7.2 V=vC t e2 t–7.2 k2t+=dvCdt-------- k2 e2– t2e2 t–7.2 k2t+–=dvCdt--------t 0=k2 2 7.2 0+– k2 14.4–= =dvCdt--------iCC----= t 0=dvCdt--------t 0=iC 0C------------0C---- 0= = =t 0=k2 14.4– 0= k2 14.4=vC t e2 t–7.2 14.4t+ 7.2e2t–2t 1+= =iL t iR t iC t iL t+ + 0= iL t iC t iR t––= iC t C dvC dt=iR t vR t 1 2+ vC t 3= =iL t112------ 14.4e2t–2t 1+ 14.4e2t–+––7.23-------e2t–2t 1+– 2.4e2t–t 1+–= =t 0= iL 0 12 2 6 A= = vC 0 12 V=+12 V42 H1 4 FBA2+vC 0iL 0
64. 64. Chapter 1 Second Order Circuits1-52 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsFor the circuit is as shown below.For this circuitand with the above relation can be written asThe characteristic equation of the last expression above yields and and thus(1)With the initial condition and (1) we get(2)Differentiating (1) we getand(3)Also, and att 042 H 1 4 FBA2+vC tiL tR1 R2LCR1 R2+ iL vC LdiLdt-------+ + 0=iL iC C dvC dt= =R1 R2+ CdvCdt-------- LCd2vCdt2---------- vC+ + 0=d2vCdt2----------R1 R2+L----------------------dvCdt--------1LC-------vC+ + 0=d2vCdt2---------- 3dvCdt-------- 2vC+ + 0=s1 1–= s2 2–=vC t k1et–k2 e2t–+=vC 0 12 V=k1 k2+ 12=dvCdt-------- k– 1et–2k2– e2t–=dvCdt--------t 0=k– 1 2k2–=dvCdt--------iCC----iLC----= = t 0=
65. 65. 1-53 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsSolutions to Exercises(4)From (3) and (4)(5)and from (2) and (5) and . By substitution into (1) we getThen,dvCdt--------t 0=iL 0C------------61 4---------- 24= = =k– 1 2k2– 24=k1 48= k2 36–=vC t 48et–36– e2t–=vAB vL t vC t– LdiLdt------- vC t– LCd2iCdt2---------- vC t–= = =0.5d2dt2------- 48et–36– e2t–48et–36– e2t––=0.5 48et–144– e2t–48et–36– e2t––=24– et–108– e2t–24 et–4.5e2t–+–==
66. 66. Chapter 1 Second Order Circuits1-54 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsNOTES
67. 67. Circuit Analysis II with MATLAB Applications 2-1Orchard PublicationsChapter 2Resonancehis chapter defines series and parallel resonance. The quality factor is then defined in termsof the series and parallel resonant frequencies. The half-power frequencies and bandwidth arealso defined in terms of the resonant frequency.2.1 Series ResonanceConsider phasor series circuit of Figure 2.1.Figure 2.1. Series RLC phasor circuitThe impedance is(2.1)or(2.2)Therefore, the magnitude and phase angle of the impedance are:(2.3)and(2.4)The components of are shown on the plot of Figure 2.2.TQRLCVSIR1 j Cj LZImpedance ZPhasor VoltagePhasor Current---------------------------------------VSI------ R j L1j C----------+ + R j L1C--------–+= = = = =Z R2L 1 C–2+ L 1 C– R1–tan=Z R2L 1 C–2+=Z L 1 C– R1–tan=Z
68. 68. Chapter 2 Resonance2-2 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsFigure 2.2. The components of in a series RLC circuitThe frequency at which the capacitive reactance and the inductive reactanceare equal is called the resonant frequency. The resonant frequency is denoted as or and thesecan be expressed in terms of the inductance and capacitance by equating the reactances, that is,(2.5)and(2.6)We observe that at resonance where denotes the impedance value at resonance, and. In our subsequent discussion the subscript zero will be used to indicate that the circuit vari-ables are at resonance.Example 2.1For the circuit shown in Figure 2.3, compute , , , , , and . Then, draw a phasordiagram showing , , and .Series Resonance CurvesRadian FrequencyMagnitudeofImpedance01 / C|Z|RLL 1 / C0ZXC 1 C= XL L=0 f0L C0L 10C----------=02 1LC-------=01LC-----------=f012 LC------------------=Z0 R= Z0Z 0=I0 0 C VR0 VL0 VC0VR0 VL0 VC0
69. 69. Circuit Analysis II with MATLAB Applications 2-3Orchard PublicationsSeries ResonanceFigure 2.3. Circuit for Example 2.1Solution:At resonance,and thusThen,Sinceit follows thatTherefore,orNow,andThe phasor diagram showing , , and is shown in Figure 2.4.VSIL=0.2 mHCj– XC120 0 VR 1.2= jXL j10=jXL jXC–=Z0 R 1.2= =I0120 V1.2-------------- 100 A= =XL0 0L 10= =010L------100.2 103–------------------------ 50000 rad s= = =XC0 XL0 1010C----------= = =C110 50000--------------------------- 2 F= =VR0 RI0 1.2 100 120 V= = =VL0 0LI0 50000 0.2 103–100 1000= = =VC010C----------I0150000 2 106–----------------------------------------- 100 1000 V= = =VR0 VL0 VC0
70. 70. Chapter 2 Resonance2-4 Circuit Analysis II with MATLAB ApplicationsOrchard PublicationsFigure 2.4. Phasor diagram for Example 2.1Figure 2.4 reveals that and these voltages are much higher than the appliedvoltage of . This illustrates the useful property of resonant circuits to develop high voltagesacross capacitors and inductors.2.2 Quality Factor Q0s in Series ResonanceThe quality factor *is an important parameter in resonant circuits. Its definition is derived from thefollowing relations:At resonance,andThen(2.7)and(2.8)At series resonance the left sides of (2.7) and (2.8) are equal and therefore,* We denote the quality factor for series resonant circuits as , and the quality factor for parallel resonant cir-cuits as .VR0 = 120 V|VL0| = 1000 V|VC0| = 1000 VVL0 VC0 1000 V= =120 VQ0SQ0P0L 10C----------=I0VSR---------=VL0 0LI0 0LVSR--------- 0LR--------- VS= = =VC010C----------I010C----------VSR---------10RC-------------- VS= = =0LR---------10RC--------------=
71. 71. Circuit Analysis II with MATLAB Applications 2-5Orchard PublicationsQuality Factor Q0s in Series ResonanceThen, by definition(2.9)In a practical circuit, the resistance in the definition of above, represents the resistance of theinductor and thus the quality factor is a measure of the energy storage property of the inductancein relation to the energy dissipation property of the resistance of that inductance.In terms of , the magnitude of the voltages across the inductor and capacitor are(2.10)and therefore, we say that there is a “resonant” rise in the voltage across the reactive devices and it isequal to the times the applied voltage. Thus in Example 2.1,The quality factor is also a measure of frequency selectivity. Thus, we say that a circuit with a highhas a high selectivity, whereas a low circuit has low selectivity. The high frequency selectivity ismore desirable in parallel circuits as we will see in the next section.Figure 2.5 shows the relative response versus for , and where we observe thathighest provides the best frequency selectivity, i.e., higher rejection of signal components outsidethe bandwidth which is the difference in the frequencies.Figure 2.5. Selectivity curves with , andQ0S0LR---------10RC--------------= =Quality Factor at Series ResonanceR Q0SQ0SL RQ0SVL0 VC0 Q0S VS= =Q0SQ0SVL0VS------------VC0VS------------1000120------------253------= = = =QQ QQ 25 50= 100QBW 2 1–= 3 dBSelectivity Curves for Different Qs0.00.20.40.60.81.01.2r/sRelativeResponse(gain)1 0 2Q=25Q=50Q=100Q 25 50= 100