Hi v _\ 13.1 Circuit Elements in the s Domain p. 468 13.2 Circuit Analysis in the s Domain p. 470 13.3 Applications p. 472 13.4 The Transfer Function p. 484 13.5 The Transfer Function in Partial Fraction Expansions p. 486 13.6 The Transfer Function and the Convolution Integral p. 489 13.7 The Transfer Function and the Steady-State Sinusoidal Response p. 495 13.8 The Impulse Function in Circuit Analysis p. 498 1 Be able to transform a circuit into the s domain using Laplace transforms; be sure you understand how to represent the initial conditions on energy-storage elements in the 5 domain. 2 Know how to analyze a circuit in the s-domain and be able to transform an s-domain solution back to the time domain. 3 Understand the definition and significance of the transfer function and be able to calculate the transfer function for a circuit using s-domain techniques. 4 Know how to use a circuit's transfer function to calculate the circuit's unit impulse response, its unit step response, and its steady-state response to a sinusoidal input. The Laplace Transform in Circuit Analysis The Laplace transform has two characteristics that make it an attractive tool in circuit analysis. First, it transforms a set of linear constant-coefficient differential equations into a set of linear polynomial equations, which are easier to manipulate. Second, it automatically introduces into the polynomial equations the initial values of the current and voltage variables. Thus, initial condi- tions are an inherent part of the transform process. (This con- trasts with the classical approach to the solution of differential equations, in which initial conditions are considered when the unknown coefficients are evaluated.) We begin this chapter by showing how we can skip the step of writing time-domain integrodifferential equations and transform- ing them into the s domain. In Section 13.1, we'll develop the s-domain circuit models for resistors, inductors, and capacitors so that we can write s-domain equations for all circuits directly. Section 13.2 reviews Ohm's and Kirchhoff's laws in the context of the s domain. After establishing these fundamentals, we apply the Laplace transform method to a variety of circuit problems in Section 13.3. Analytical and simplification techniques first introduced with resistive circuits—such as mesh-current and node-voltage methods and source transformations—can be used in the s domain as well. After solving for the circuit response in the s domain, we inverse transform back to the time domain, using partial fraction expansion (as demonstrated in the preceding chapter). As before, checking the final time-domain equations in terms of the initial conditions and final values is an important step in the solution process. The s-domain descriptions of circuit input and output lead us, in Section 13.4, to the concept of the transfer function. The trans- fer function for a particular circuit is the ratio of the Laplace transform of its output to the Laplace transform of its input. In Chapters 14 and 15, we'll examine the design uses of the transfer function, but here we focus on its use as an analytical tool. We continue this chapter with a look at the role of partial fraction 466 Practical Perspective Surge Suppressors With the advent of home-based personal computers, modems, fax machines, and other sensitive electronic equipment, it is necessary to provide protection from voltage surges that can occur in a household circuit due to switching. A commer- cially available surge suppressor is shown in the accompany- ing figure. How can flipping a switch to turn on a light or turn off a hair dryer cause a voltage surge? At the end of this chapter, we will answer that question using Laplace transform tech- niques to analyze a circuit. We will illustrate how a voltage surge can be created by switching off a resistive load in a cir- cuit operating in the sinusoidal steady state. 467 468 The Laplace Transform in Circuit Analysis expansion (Section 13.5) and the convolution integral (Section 13.6) in employing the transfer function in circuit analysis. We conclude with a look at the impulse function in circuit analysis. 13.1 Circuit Elements in the s Domain The procedure for developing an ^-domain equivalent circuit for each cir- cuit element is simple. First, we write the time-domain equation that relates the terminal voltage to the terminal current. Next, we take the Laplace transform of the time-domain equation. This step generates an algebraic relationship between the s-domain current and voltage. Note that the dimension of a transformed voltage is volt-seconds, and the dimension of a transformed current is ampere-seconds. A voltage-to-current ratio in the s domain carries the dimension of volts per ampere. An impedance in the s domain is measured in ohms, and an admittance is measured in Siemens. Finally, we construct a circuit model that satisfies the relation- ship between the ^-domain current and voltage. We use the passive sign convention in all the derivations. A Resistor in the s Domain We begin with the resistance element. From Ohm's law, v = RL Because R is a constant, the Laplace transform of Eq. 13.1 is (13.1) V = RI, (13.2) 4 + V < 4 1 » :K\I 1 a + T vi • *!• b b (a) (b) Figure 13.1 A The resistance element, (a) Time domain, (b) Frequency domain. where V = %{v\ and J = .£{*}. Equation 13.2 states that the s-domain equivalent circuit of a resistor is simply a resistance of R ohms that carries a current of I ampere-seconds and has a terminal voltage of V volt-seconds. Figure 13.1 shows the time- and frequency-domain circuits of the resistor. Note that going from the time domain to the frequency domain does not change the resistance element. vL iW' An Inductor in the s Domain Figure 13.2 shows an inductor carrying an initial current of /Q amperes. The time-domain equation that relates the terminal voltage to the termi- nal current is Figure 13.2 A An inductor of L henrys carrying an initial current of / 0 amperes. v = L di dt (13.3) The Laplace transform of Eq. 13.3 gives V = L[sl - /(0 - )] = sLl - LI ( (13.4) 13.1 Circuit Elements in the s Domain 469 Two different circuit configurations satisfy Eq. 13.4. The first consists of an impedance of sL ohms in series with an independent voltage source of L/ ( ) volt-seconds, as shown in Fig. 13.3. Note that the polarity marks on the voltage source L/ () agree with the minus sign in Eq. 13.4. Note also that LI () carries its own algebraic sign; that is, if the initial value of i is opposite to the reference direction for i, then / ( , has a negative value. The second .y-domain equivalent circuit that satisfies Eq. 13.4 consists of an impedance of sL ohms in parallel with an independent current source of I {) /s ampere-seconds, as shown in Fig. 13.4. We can derive the alternative equivalent circuit shown in Fig. 13.4 in several ways. One way is simply to solve Eq. 13.4 for the current I and then construct the circuit to satisfy the resulting equation. Thus / — — T . sL sL s (13.5) Two other ways are: (1) find the Norton equivalent of the circuit shown in Fig. 13.3 and (2) start with the inductor current as a function of the induc- tor voltage and then find the Laplace transform of the resulting integral equation. We leave these two approaches to Problems 13.1 and 13.2. If the initial energy stored in the inductor is zero, that is, if / u = 0, the y-domain equivalent circuit of the inductor reduces to an inductor with an impedance of sL ohms. Figure 13.5 shows this circuit. A Capacitor in the s Domain An initially charged capacitor also has two s-domain equivalent circuits. Figure 13.6 shows a capacitor initially charged to V {) volts. The terminal current is Figure 13.3 • The series equivalent circuit for an inductor of L henrys carrying an initial current of /„ amperes. Figure 13.4 • The parallel equivalent circuit for an inductor of L henrys carrying an initial current of / 0 amperes. / = C dv dt (13.6) VisLU Transforming Eq. 13.6 yields or I = C[sV - v(0~)] I = sCV - CV (h (13.7) which indicates that the .v-domain current I is the sum of two branch cur- rents. One branch consists of an admittance of sC Siemens, and the second branch consists of an independent current source of CVQ ampere-seconds. Figure 13.7 shows this parallel equivalent circuit. We derive the series equivalent circuit for the charged capacitor by solving Eq. 13.7 for V: Figure 13.5 • The s-domain circuit for an inductor when the initial current is zero. + 4 vCT- + Figure 13.6 • A capacitor of C farads initially charged to V 0 volts. V a»? (13.8) Figure 13.8 shows the circuit that satisfies Eq. 13.8. In the equivalent circuits shown in Figs. 13.7 and 13.8, V () carries its own algebraic sign. In other words, if the polarity of V 0 is opposite to the reference polarity for v, V {) is a negative quantity. If the initial voltage on 470 The Laplace Transform in Circuit Analysis l/.vC Figure 13.7 A The parallel equivalent circuit for a capacitor initially charged to V 0 volts. + fa 1/sC V 0 /s Figure 13.8 • The series equivalent circuit for a capacitor initially charged to V 0 volts. /sC Figure 13.9 • The s-domain circuit for a capacitor when the initial voltage is zero. the capacitor is zero, both equivalent circuits reduce to an impedance of \/sC ohms, as shown in Fig. 13.9. In this chapter, an important first problem-solving step will be to choose between the parallel or series equivalents when inductors and capacitors are present. With a little forethought and some experience, the correct choice will often be quite evident. The equivalent circuits are sum- marized in Table 13.1. TABLE 13.1 Summary of the s-Domain Equivalent Circuits TIME DOMAIN FREQUENCY DOMAIN »i V*R b v = Ri H fa + i = C dvjdu sL] V (|)/„A -*b V = sLl - Ll {) *b v At I = + sL s i/vc: / = sCV - CV {) 13.2 Circuit Analysis in the s Domain Before illustrating how to use the s-domain equivalent circuits in analysis, we need to lay some groundwork. First, we know that if no energy is stored in the inductor or capacitor, the relationship between the terminal voltage and current for each passive element takes the form: Ohm's law in the s-domain • V = ZI, (13.9) where Z refers to the s-domain impedance of the element. Thus a resistor has an impedance of R ohms, an inductor has an impedance of sL ohms. 13.2 Circuit Analysis in the 5 Domain 471 and a capacitor has an impedance of 1/sC ohms. The relationship con- tained in Eq. 13.9 is also contained in Figs. 13.1(b), 13.5, and 13.9. Equation 13.9 is sometimes referred to as Ohm's law for the s domain. The reciprocal of the impedance is admittance. Therefore, the s domain admittance of a resistor is \/R Siemens, an inductor has an admittance of 1/sL Siemens, and a capacitor has an admittance of sC Siemens. The rules for combining impedances and admittances in the s domain are the same as those for frequency-domain circuits.Thus series-parallel sim- plifications and A-to-Y conversions also are applicable to .v-domain analysis. In addition, Kirchhoff s laws apply to s-domain currents and voltages. Their applicability stems from the operational transform stating that the Laplace transform of a sum of time-domain functions is the sum of the transforms of the individual functions (see Table 12.2). Because the alge- braic sum of the currents at a node is zero in the time domain, the alge- braic sum of the transformed currents is also zero. A similar statement holds for the algebraic sum of the transformed voltages around a closed path. The s-domain version of Kirchhoff s laws is alg2/ = 0, (13.10) alg 2> = 0. (13.11) Because the voltage and current at the terminals of a passive element are related by an algebraic equation and because Kirchhoff s laws still hold, all the techniques of circuit analysis developed for pure resistive networks may be used in s-domain analysis. Thus node voltages, mesh currents, source transformations, and Thevenin-Norton equivalents are all valid techniques, even when energy is stored initially in the inductors and capacitors. Initially stored energy requires that we modify Eq. 13.9 by simply adding independent sources either in series or parallel with the element impedances. The addition of these sources is governed by Kirchhoff s laws. /ASSESSMENT PROBLEMS Objective 1—Be able to transform a circuit into the s domain using Laplace transforms 13.1 A 500 O, resistor, a 16 mH inductor, and a 25 11F capacitor are connected in parallel. a) Express the admittance of this parallel com- bination of elements as a rational function of s. b) Compute the numerical values of the zeros and poles. Answer: (a) 25 X 10"V + 80,000.? + 25 x 1(f)/s; (b) -z { = -40,000 - /30,000; - Z2 = -40,000 + /30,000; p x = 0. 13.2 The parallel circuit in Assessment Problem 13.1 is placed in series with a 2000 fl resistor. a) Express the impedance of this series combi- nation as a rational function of s. b) Compute the numerical values of the zeros and poles. Answer: (a) 2000(5 + 50,000) 2 /C* 2 + 80,0005 + 25 X 10 8 ); (b) -z x = -z 2 = -50,000; -p 1 = -40,000 - /30,000, -p 2 = -40,000 + 730,000. NOTE: Also try Chapter Problems 13.4 and 13.6. 472 The Laplace Transform in Circuit Analysis 13.3 Applications We now illustrate how to use the Laplace transform to determine the tran- sient behavior of several linear lumped-parameter circuits. We start by ana- lyzing familiar circuits from Chapters 7 and 8 because they represent a simple starting place and because they show that the Laplace transform approach yields the same results. In all the examples, the ease of manipulat- ing algebraic equations instead of differential equations should be apparent. Figure 13.10 • The capacitor discharge circuit. + / RkV Figure 13.11 A An s-domain equivalent circuit for the circuit shown in Fig. 13.10. The Natural Response of an RC Circuit We first revisit the natural response of an RC circuit (Fig. 13.10) via Laplace transform techniques. (You may want to review the classical analysis of this same circuit in Section 7.2). The capacitor is initially charged to V 0 volts, and we are interested in the time-domain expressions for i and v. We start by finding i. In transfer- ring the circuit in Fig. 13.10 to the s domain, we have a choice of two equiv- alent circuits for the charged capacitor. Because we are interested in the current, the series-equivalent circuit is more attractive; it results in a single- mesh circuit in the frequency domain. Thus we construct the .s-domain cir- cuit shown in Fig. 13.11. Summing the voltages around the mesh generates the expression s sC (13.12) Solving Eq. 13.12 for /yields / = CV 0 V {) /R RCs + 1 s + (1/RC)' (13.13) Figure 13.12 A An s-domain equivalent circuit for the circuit shown in Fig. 13.10. Note that the expression for I is a proper rational function of s and can be inverse-transformed by inspection: (13.14) which is equivalent to the expression for the current derived by the classi- cal methods discussed in Chapter 7. In that chapter, the current is given by Eq. 7.26, where T is used in place of RC. After we have found /, the easiest way to determine v is simply to apply Ohm's law; that is, from the circuit, V = Ri = V {) e~' /RC u(t). (13.15) We now illustrate a way to find v from the circuit without first finding /'. In this alternative approach, we return to the original circuit of Fig. 13.10 and transfer it to the 5 domain using the parallel equivalent circuit for the charged capacitor. Using the parallel equivalent circuit is attractive now because we can describe the resulting circuit in terms of a single node volt- age. Figure 13.12 shows the new s-domain equivalent circuit. The node-voltage equation that describes the new circuit is ^ + sCV = CV Q . (13.16) Solving Eq. 13.16 for V gives 13.3 Applications 473 V Vn (13.17) s + {i/Rcy Inverse-transforming Eq. 13.17 leads to the same expression for v given by Eq. 13.15, namely, v = V {) er ! l RC = V 0 e^ T u(t). (13.18) Our purpose in deriving by direct use of the transform method is to show that the choice of which 5-domain equivalent circuit to use is influ- enced by which response signal is of interest. /ASSESSMENT PROBLEM Objective 2—Know how to analyze a circuit in the s domain and be able to transform an s domain solution to the time domain 13.3 The switch in the circuit shown has been in position a for a long time. At t = 0, the switch is thrown to position b. a) Find I, V h and V 2 as rational functions of s. b) Find the time-domain expressions for i, v h and v 2 . Answer: (a) I = 0.02/(5 + 1250), V x = 80/(5 + 1250), V 2 = 20/(5 + 1250); NOTE: Also try Chapter Problems 13.9 and 13.12. (b) i = 20<T 1250f M(0 mA, V] = 80e -125( V0 V, v 2 = 20e- u50t u(t) V. io kn \ 100V A (j\ 0.2 M F; 0.8 JUF; f = 0 + + ||5ka The Step Response of a Parallel Circuit Next we analyze the parallel RLC circuit, shown in Fig. 13.13, that we first analyzed in Example 8.7. The problem is to find the expression for i L after the constant current source is switched across the parallel elements. The initial energy stored in the circuit is zero. As before, we begin by constructing the 5-domain equivalent circuit shown in Fig. 13.14. Note how easily an independent source can be trans- formed from the time domain to the frequency domain. We transform the source to the s domain simply by determining the Laplace transform of its time-domain function. Here, opening the switch results in a step change in the current applied to the circuit. Therefore the 5-domain current source is !£{I\} C u(t)}, or lfe/$. To find I L , we first solve for V and then use I, = sL (13.19) to establish the 5-domain expression for 1 L . Summing the currents away from the top node generates the expression sCV + — H—- = — R sL 5 Solving Eq. 13.20 for V gives V = UJC 5 2 + {i/RC)s + (1/LC) (13.20) (13.21) R h\L 625 Ct 25mH| Figure 13.13 • The step response of a parallel RLC circuit. Figure 13.14 • The s-domain equivalent circuit for the circuit shown in Fig. 13.13. Substituting Eq. 13.21 into Eq. 13.19 gives hJLC L s[s 2 + (1/RQs + (1/LC)]' Substituting the numerical values of R, L, C, and / dc into Eq. 13.22 yields 384 X 10 5 h = — > 5T« (13.23) i'(.? 2 + 64,000.v + 16 X 10 8 ) Before expanding Eq. 13.23 into a sum of partial fractions, we factor the quadratic term in the denominator: 384 X 10 5 L " s(s + 32,000 - /24,000)(5 + 32,000 + /24,000) * (13 ' 24) Now, we can test the ,v-domain expression for I L by checking to see whether the final-value theorem predicts the correct value for i L at t = co. All the poles of I L , except for the first-order pole at the origin, lie in the left half of the s plane, so the theorem is applicable. We know from the behavior of the circuit that after the switch has been open for a long time, the inductor will short-circuit the current source.Therefore, the final value of i L must be 24 m A. The limit of si L as s —> 0 is ,. T 384 X 10 5 n , A hm sir = r = 24 mA. (13.25) *-o 16 X 10 8 (Currents in the .v domain carry the dimension of ampere-seconds, so the dimension of sI L will be amperes.) Thus our s-domain expression checks out. We now proceed with the partial fraction expansion of Eq. 13.24: s s + 32,000 - /24,000 Kl + s + 32,000 + /24,000' (13 ' 26) The partial fraction coefficients are 384 X 10 3 ,„ , K ] = jT- = 24 X 10" 3 , (13.27) 16 x 10 8 384 X 10 5 A.' (-32,000 + /24,000)(/48,000) = 20 X 10" 3 /126.87°. (13.28) Substituting the numerical values of K x and K 2 into Eq. 13.26 and inverse- transforming the resulting expression yields i L = [24 + 40e" 320(M) 'cos(24,000f + 126.87° )]«(f)mA. (13.29) The answer given by Eq. 13.29 is equivalent to the answer given for Example 8.7 because 40cos(24,000f + 126.87°) = -24 cos 24,000/ - 32 sin 24,000/. If we weren't using a previous solution as a check, we would test Eq. 13.29 to make sure that / L (0) satisfied the given initial conditions and i L (oo) satisfied the known behavior of the circuit. 13.3 Applications 475 /ASSESSMENT PROBLEM Objective 2—Know how to analyze a circuit in the s domain and be able to transform an s domain solution to the time domain 13.4 The energy stored in the circuit shown is zero Answer: (a) I = 40/(5 2 + 1.2s + 1); at the time when the switch is closed. (b) , = (5Qe -o.6r sin Q8t)m A; a) Find the 5-domain expression for I. ( c ) v = 160^/(^ 2 + 1.2s + 1); (d) v = [200£f a6 'cos(0.8r + 36.87°)]u(0 V. b) Find the time-domain expression for i when t > 0. V 4.8 II 4 H c) Find the s-domain expression for V. ^J?\ ' = () d) Find the time-domain expression for v when t > 0. NOTE: Also try Chapter Problems 13.10 and 13.21. The Transient Response of a Parallel RLC Circuit Another example of using the Laplace transform to find the transient behav- ior of a circuit arises from replacing the dc current source in the circuit shown in Fig. 13.13 with a sinusoidal current source. The new current source is if> = / m costof A, (13.30) where /„, = 24 mA and o> = 40,000 rad/s. As before, we assume that the initial energy stored in the circuit is zero. The .v-domain expression for the source current is Sim I g = 2 2 . (13.31) h r + w 2 The voltage across the parallel elements is (I s /C)s s 2 + (l/RC)s + (1/L Substituting Eq. 13.31 into Eq. 13.32 results in (IJC)s 2 L + (1/RQs from which V__ (IJLC)s sL (s 2 + a> 2 )[ s 2 + (1/RQs + {1/LC)] Substituting the numerical values of / m , w, R, L, and C into Eq. 13.34 gives 384 X 10 5 s (s z + 16 X 10 8 )(s 2 + 64,000^ + We now write the denominator in factored form: -vw- + v — V = 2 , - /r J^ -777777- ( 13 - 32 ) y - 7X7 2u.2 • 7717777, . 777777;, (13.33) J L = — == 7TT-^2 . ,+ ,— 7777- (13.34) h = ~h T7^ 5T- (13.35) :0.25 F 384 X 10 5 5 '" ~ (5 - j(o)(s + j(o)(s + a- //3)(5 + a+Wy