Electric Circuits, 9th Edition P51 docx

476 The Laplace Transform in Circuit Analysis where co = 40,000, a = 32,000, and (3 = 24,000. We can't test the final value of i L with the final-value theorem because l L has a pair of poles on the imaginary axis; that is, poles at ±/4 X 10 4 . Thus we must first find i L and then check the validity of the expression from known circuit behavior. When we expand Eq. 13.36 into a sum of partial fractions, we generate the equation /, = K: + Kl + K, s -/40,000 s + /40,000 s + 32,000 - /24,000 + K\ s + 32,000 + /24,000 The numerical values of the coefficients K\ and K 2 are 384 X 10 5 (/40,000) (13.37) *1 = 7T (/80,000)(32,000 + /16,000)(32,000 + /64,000) = 7.5 X 10~ 3 /-90°, (13.38) K, = 384 X 10 5 ( -32,000 + /24,000) (-32,000 - /16,000)(-32,000 + /64,000)(/48,000) 12.5 X 10~ 3 /90°. (13.39) Substituting the numerical values from Eqs. 13.38 and 13.39 into Eq. 13.37 and inverse-transforming the resulting expression yields i L = [15 cos (40,000r - 90°) + 25e -32.000? cos(24,000r + 90°)] mA, = (15 sin 40,000r - 2Se~ nmt sin 24,000/)tt(0 mA. (13.40) We now test Eq. 13.40 to see whether it makes sense in terms of the given initial conditions and the known circuit behavior after the switch has been open for a long time. For t = 0, Eq. 13.40 predicts zero initial current, which agrees with the initial energy of zero in the circuit. Equation 13.40 also predicts a steady-state current of i La = 15sin40,000rmA, which can be verified by the phasor method (Chapter 9). (13.41) Figure 13.15 • A multiple-mesh RL circuit. The Step Response of a Multiple Mesh Circuit Until now, we avoided circuits that required two or more node-voltage or 48 ft mesh-current equations, because the techniques for solving simultaneous differential equations are beyond the scope of this text. However, using Laplace techniques, we can solve a problem like the one posed by the multiple-mesh circuit in Fig. 13.15. 13.3 Applications 477 Here we want to find the branch currents /j and i 2 that arise when the 336 V dc voltage source is applied suddenly to the circuit. The initial energy stored in the circuit is zero. Figure 13.16 shows the .v-domain equiv- alent circuit of Fig. 13.15. The two mesh-current equations are 336 = (42 + 8.45)/] - 42/ 2 , 0 = -42/j + (90 + 10s)I 2 . Using Cramer's method to solve for l x and I 2 , we obtain A = 42 + 8.4s - 42 - 42 90 + 10s (13.42) (13.43) 48 a Figure 13.16 • The s-domain equivalent circuit for the circuit shown in Fig. 13.15. = 84(5 2 + 14.9 + 24) = 84( s + 2)(5 + 12), (13.44) A', 336/5 - 42 0 90 + 105 3360(5 + 9) (13.45) No = 42 + 8.45 336/5 - 42 0 14,112 (13.46) Based on Eqs. 13.44-13.46, /> = /Vi _ 40(5 + 9) A " s(s + 2)(5 + 12)' (13.47) A = N; 168 A 5(5 + 2)(5 + 12)' Expanding I] and / 2 into a sum of partial fractions gives 14 1 15 5 5 + 2 5 + 12' (13.48) (13.49) 8.4 1.4 + 5 5 + 2 5+12 (13.50) 478 The Laplace Transform in Circuit Analysis We obtain the expressions tor i { and i 2 by inverse-transforming Eqs. 13.49 and 13.50, respectively: I, = (15 - Ue~ 2t - e~ ]2t )u(t) A, (13.51) / 2 = (7 - 8.4e" z ' + \Ae~ l *)u(t) A. (13.52) Next we test the solutions to see whether they make sense in terms of the circuit. Because no energy is stored in the circuit at the instant the switch is closed, both /'i(0~) and /2(0 - ) must be zero. The solutions agree with these initial values. After the switch has been closed for a long time, the two induc- tors appear as short circuits. Therefore, the final values of i { and i 2 are , x 336(90) /(00) = *—'- = 15 A, 1V ) 42(48) (13.53) 15(42) / 2 (oo)=^ = 7A. (13.54) One final test involves the numerical values of the exponents and calcu- lating the voltage drop across the 42 11 resistor by three different methods. From the circuit, the voltage across the 42 Q resistor (positive at the top) is dii d'h v = 42(/, - h) = 336 - 8.4—- = 48/, + 10- 2 dt dt (13.55) /'ASSESSMENT PROBLEM You should verify that regardless of which form of Eq. 13.55 is used, the voltage is v = (336 - 235.2e~ 2r - 100.80e" 12r )«(0 V. We are thus confident that the solutions for i x and i 2 are correct. 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 13.5 The dc current and voltage sources are applied simultaneously to the circuit shown. No energy is stored in the circuit at the instant of application. a) Derive the 5-domain expressions for V\ and V 2 . b) For t > 0, derive the time-domain expres- sions for V\ and v 2 . c) Calculate ^(0+) and v 2 (0 + ). d) Compute the steady-state values of V\ and Vi. Answer: (a) V l = [5(s + 3)]/[s(s + 0.5)(5 + 2)], V 2 = [2.5(s 2 + 6)]/[s(s + 0.5)(5 + 2)]; (b) v, = (15 50 -0.5/ + f<r 2 >(0 v, v 2 = 05 - l -fe-°- 51 + ?e*)«(0 V; (c) u,(0 + ) = 0, v 2 (0 + ) = 2.5 V; (d) vi = v 2 = 15 V. f>A <Z IF: 1H _/-WY-\_ + v\ 3 n f v- 15a 15VI NOTE: Also try Chapter Problems 13.22 and 13.29. 13.3 Applications 479 The Use of Thevenin's Equivalent In this section we show how to use Thevenin's equivalent in the s domain. Figure 13.17 shows the circuit to be analyzed. The problem is to find the capacitor current that results from closing the switch.The energy stored in the circuit prior to closing is zero. To find i c , we first construct the 5-domain equivalent circuit and then find the Thevenin equivalent of this circuit with respect to the terminals of the capacitor. Figure 13.18 shows the 5-domain circuit. The Thevenin voltage is the open-circuit voltage across terminals a, b. Under open-circuit conditions, there is no voltage across the 60 Q resis- tor. Hence VTI,= {480/s)(0.002s) 20 + 0.002* 480 5 + 10 4 ' (13.56) The Thevenin impedance seen from terminals a and b equals the 60 ft resistor in series with the parallel combination of the 20 fl resistor and the 2 mH inductor. Thus Z Th = 60 + 0.0025(20) 80(5 + 7500) 20 + 0.0025 .v + KV (13.57) Using the Thevenin equivalent, we reduce the circuit shown in Fig. 13.18 to the one shown in Fig. 13.19. It indicates that the capacitor current I c equals the Thevenin voltage divided by the total series impedance. Thus, Ir = 480/(5 + 10 4 ) [80(5 + 7500)/(5 + 10 4 )] + [(2 X 10 5 )/5] We simplify Eq. 13.58 to /c = 65 6.v 5 2 + 10,0005 + 25 X 10 6 (s + 5000) 2 A partial fraction expansion of Eq. 13.59 generates Ir = -30,000 + (5 + 5000)' the inverse transform of which is 5 + 5000' i c = (-30,000f«T 50,M) ' + 6e- 5m ')u(t) A. (13.58) (13.59) (13.60) (13.61) We now test Eq. 13.61 to see whether it makes sense in terms of known circuit behavior. From Eq. 13.61, / c (0) = 6 A. (13.62) This result agrees with the initial current in the capacitor, as calculated from the circuit in Fig. 13.17. The initial inductor current is zero and the initial capacitor voltage is zero, so the initial capacitor current is 480/80, or 6 A. The final value of the current is zero, which also agrees with Eq. 13.61. Note also from this equation that the current reverses sign when t exceeds 6/30.000, or 200 /AS. The fact that ic reverses sign makes sense because, when the switch first closes, the capacitor begins to charge. Eventually this charge is reduced to zero because the inductor is a short circuit at t = co. The sign reversal of i c reflects the charging and discharging of the capacitor. Let's assume that the voltage drop across the capacitor v c is also of inter- est. Once we know i c . we find v c by integration in the time domain; that is, v c = 2 x 10 5 /(6- 30, -5(K«h 0Q0x)e~^ m,x dx. (13.63) 20 fi 60 a a 'VW- Figure 13.17 A A circuit to be analyzed using Thevenin's equivalent in the s domain. 20 O 60 0 a -^vw •- N4 |^ 10.002 s V c /c t -±z2x 10 5 Figure 13.18 • The 5-domain model of the circuit shown in Fig. 13.17. v c i c ;^2xio 5 Figure 13.19 • A simplified version of the circuit shown in Fig. 13.18, using a Thevenin equivalent. 480 The Laplace Transform in Circuit Analysis Although the integration called for in Eq. 13.63 is not difficult, we may avoid it altogether by first finding the s-domain expression for V c and then finding v c by an inverse transform. Thus V c = -pic sC 2 X 10 5 6.v s (s + 5000) 2 from which 12 X 10 5 (s + 5000)^ v c = 12 x io 5 rtr 50f)0 'u(0- (13.64) (13.65) You should verify that Eq. 13.65 is consistent with Eq. 13.63 and that it also supports the observations made with regard to the behavior of i c (see Problem 13.33). 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 13.6 The initial charge on the capacitor in the circuit shown is zero. a) Find the 5-domain Thevenin equivalent cir- cuit with respect to terminals a and b. b) Find the s-domain expression for the current that the circuit delivers to a load consisting of a 1 H inductor in series with a 2 ft resistor. Answer: (a) V Th = V &h = [20(5 + 2A))/[s(s + 2)), Z Ttl = 5(5 + 2.8)/(s + 2); NOTE: Also try Chapter Problem 13.34. (b) J ab = [20(5 + 2A)]/[s(s + 3)(5 + 6)]. 0.5 F A Circuit with Mutual Inductance The next example illustrates how to use the Laplace transform to analyze the transient response of a circuit that contains mutual inductance. Figure 13.20 shows the circuit. The make-before-break switch has been in position a for a long time. At t = 0, the switch moves instantaneously to position b. The problem is to derive the time-domain expression for i 2 . We begin by redrawing the circuit in Fig. 13.20, with the switch in position b and the magnetically coupled coils replaced with a T-equivalent circuit. 1 Figure 13.21 shows the new circuit. 9fi © 3 a -'VW- / = 0 2H 2D, ^vw- 60 V 2H 8H Figure 13.20 • A circuit containing magnetically coupled coils. h ion Sec Appendix C. 13.3 Applications 481 We now transform this circuit to the s domain. In so doing, we note that fl( °" )= f = 5A ' WO = 0. (13.66) (13.67) Because we plan to use mesh analysis in the s domain, we use the series equivalent circuit for an inductor carrying an initial current. Figure 13.22 shows the s-domain circuit. Note that there is only one independent volt- age source. This source appears in the vertical leg of the tee to account for the initial value of the current in the 2 H inductor of /,(0 - ) + / 2 (0~), or 5 A. The branch carrying /, has no voltage source because L t - M = 0. The two i'-domain mesh equations that describe the circuit in Fig. 13.22 are (3 + 2s)I } + 2sl 2 = 10 2sT t + (12 + 85)/ 2 = 10. Solving for / 2 yields /2 = 2.5 (s + l)(j + 3) Expanding Eq. 13.70 into a sum of partial fractions generates 1.25 1.25 /7 = s + 1 s + 3 Then, i 2 = (1.25e"' - 1.25e" 3/ )w(0 A. (13.68) (13.69) (13.70) (13.71) (13.72) Equation 13.72 reveals that i 2 increases from zero to a peak value of 481.13 mA in 549.31 ms after the switch is moved to position b. Thereafter, / 2 decreases exponentially toward zero. Figure 13.23 shows a plot of i 2 ver- sus t. This response makes sense in terms of the known physical behavior of the magnetically coupled coils. A current can exist in the L 2 inductor only if there is a time-varying current in the L, inductor. As i\ decreases from its initial value of 5 A, / 2 increases from zero and then approaches zero as i, approaches zero. (L, - M) (L 2 - M) 3H OH 6H 2ti (MH2H 10 a Figure 13.21 A The circuit shown in Fig. 13.20, with the magnetically coupled coils replaced by a T-equivalent circuit. Figure 13.22 A The s-domain equivalent circuit for the circuit shown in Fig. 13.21. i 2 (mA) 481.13 549.31 /(ms) Figure 13.23 A The plot of i 2 versus t for the circuit shown in Fig. 13.20. /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 13.7 a) Verify from Eq. 13.72 that / 2 reaches a peak value of 481.13 mA at t = 549.31 ms. b) Find i h for t > 0, for the circuit shown in Fig. 13.20. c) Compute di x fdt when i 2 is at its peak value. d) Express i 2 as a function of dii/dt when i 2 is at its peak value. e) Use the results obtained in (c) and (d) to calculate the peak value of i 2 . Answer: (a) di 2 /dt = 0 when t = |ln3 (s); (b) /, = 2.5(6'"' + e" 3 >(0 A; (c) -2.89 A/s; (d) / 2 = -{MdiJdt)/\2\ (e) 481.13 mA. NOTE: Also try Chapter Problems 13.39 and 13.40. 482 The Laplace Transform in Circuit Analysis + y Vj < Ri Figure 13.24 • A circuit showing the use of super- position in s-domain analysis. Figure 13.25 • The s-domain equivalent for the circuit of Fig. 13.24. Figure 13.26 • The circuit shown in Fig. 13.25 with V^ acting alone. The Use of Superposition Because we are analyzing linear lumped-parameter circuits, we can use superposition to divide the response into components that can be identi- fied with particular sources and initial conditions. Distinguishing these components is critical to being able to use the transfer function, which we introduce in the next section. Figure 13.24 shows our illustrative circuit. We assume that at the instant when the two sources are applied to the circuit, the inductor is car- rying an initial current of p amperes and that the capacitor is carrying an initial voltage of y volts. The desired response of the circuit is the voltage across the resistor R 2 , labeled v 2 . Figure 13.25 shows the s-domain equivalent circuit. We opted for the parallel equivalents for L and C because we anticipated solving for V 2 using the node-voltage method. To find V 2 by superposition, we calculate the component of V 2 result- ing from each source acting alone, and then we sum the components. We begin with V g acting alone. Opening each of the three current sources deactivates them. Figure 13.26 shows the resulting circuit. We added the node voltage V{ to aid the analysis. The primes on V t and V 2 indicate that they are the components of Vj and V 2 attributable to V g acting alone. The two equations that describe the circuit in Fig. 13.26 are J_ Ri l sL V* + — + sC \V\ - sCV' 2 = -P-, fli \-jf + sC (13.73) (13.74) For convenience, we introduce the notation Y » = i + i + sC (13.75) Y l2 = sC; (13.76) K 22 = 4- + ^. K 2 (13.77) Substituting Eqs. 13.75-13.77 into Eqs. 13.73 and 13.74 gives Y n V\ + Y n V 2 = V g /R h (13.78) Y„V\ + YnV'r, = 0. 12" 1 (13.79) Solving Eqs. 13.78 and 13.79 for V' 2 gives V'o ^11¾ _ *12 (13.80) With the current source I g acting alone, the circuit shown in Fig. 13.25 reduces to the one shown in Fig. 13.27. Here, V'{ and V 2 are the compo- nents of V x and V 2 resulting from L. If we use the notation introduced in 13.3 Applications 483 Eqs. 13.75-13.77, the two node-voltage equations that describe the circuit in Fig. 13.27 are Y u V"i + YnV'i = 0 (13.81) and 1/sC -1(- sLAV'[ V'^R, Y l2 Vl + Y 22 V'i = I r Solving Eqs. 13.81 and 13.82 for V'i yields V'i = ^-j-/ Ml'22 — '12 Figure 13.27 • The circuit shown in Fig. 13.25, with (13.82) j g acting alone. (13.83) To find the component of V 2 resulting from the initial energy stored in the inductor {V%), we must solve the circuit shown in Fig. 13.28, where Fi 9U"-e 13.28 • The circuit shown in Fig. 13.25, with the energized inductor acting alone. YnV'{' + Y l2 V 2 " = -p/s, (13.84) Y 12 Vf + Y 22 V'{ = 0. Til us V'i' = Yjs Ml*22 -yl V- * 12 (13.85) (13.86) From the circuit shown in Fig. 13.29, we find the component of V 2 iy'i') resulting from the initial energy stored in the capacitor. The node-voltage equations describing this circuit are YuVT + Y l2 V? = yC, K„Vr + Yy>V7 = -yC. 22^ 2 Solving for Vf yields Y\\Y 2 2 Yyi The expression for V 2 is V 2 =V' 2 + V'i + V'i' + V'i" (13.87) (13.88) (13.89) Figure 13.29 A The circuit shown in Fig. 13.25, with the energized capacitor acting alone. -(Yn/Ri) Y\\Y 22 - Y\\ *i + — — 7 .? Y\]Y 22 - Yyi + Y ]2 /s Y]\Y 22 Y\ 2 -C(Y U + Y n ) p ^ T~ y- Y\\Y 22 — Y i2 (13.90) We can find V 2 without using superposition by solving the two node- voltage equations that describe the circuit shown in Fig. 13.25. Thus 484 The Laplace Transform in Circuit Analysis (13.91) Y X2 V X + Y 22 V 2 = I - yC. (13.92) You should verify in Problem 13.43 that the solution of Eqs. 13.91 and 13.92 for V 2 gives the same result as Eq. 13.90. •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 13.8 The energy stored in the circuit shown is zero at the instant the two sources are turned on. a) Find the component of v for t > 0 owing to the voltage source. b) Find the component of v for t > 0 owing to the current source. c) Find the expression for v when t > 0. NOTE: Also try Chapter Problem 13.42. Answer: (a) [(100/3)e" 2 ' - (100/3)<T 8r jw(0 V; (b) [(50/3)<T 2 ' - (50/3)e-*')u(t) V; (c) [50<T 2 ' - 50e" 8 ']«(0 V. 2fl 20u(t)f + 1.25IH » :50mF(t ^ 5u{t) 13.4 The Transfer Function The transfer function is defined as the s-domain ratio of the Laplace trans- form of the output (response) to the Laplace transform of the input (source). In computing the transfer function, we restrict our attention to circuits where all initial conditions are zero. If a circuit has multiple inde- pendent sources, we can find the transfer function for each source and use superposition to find the response to all sources. The transfer function is Definition of a transfer function • H(s) = Y(s) X(s)' (13.93) sL \/sC Figure 13.30 • A series RLC circuit. where Y(s) is the Laplace transform of the output signal, and X(s) is the Laplace transform of the input signal. Note that the transfer function depends on what is defined as the output signal. Consider, for example, the series circuit shown in Fig. 13.30. If the current is defined as the response signal of the circuit, H(s) = 1 sC ]_ __ V~R + sL+ 1/sC ~ S 2 LC + RCs + 1 (13.94) In deriving Eq. 13.94, we recognized that /corresponds to the output Y(s) and V g corresponds to the input X(s). 13.4 The Transfer Function 485 If the voltage across the capacitor is defined as the output signal of the circuit shown in Fig. 13.30, the transfer function is H(s) = V 1/sC 1 R + sL + 1/sC s 2 LC + RCs + 1 (13.95) Thus, because circuits may have multiple sources and because the definition of the output signal of interest can vary, a single circuit can generate many transfer functions. Remember that when multiple sources are involved, no single transfer function can represent the total output—transfer functions associated with each source must be combined using superposition to yield the total response. Example 13.1 illustrates the computation of a transfer function for known numerical values of R, L, and C. Example 13.1 Deriving the Transfer Function of a Circuit The voltage source v„ drives the circuit shown in Fig. 13.31. The response signal is the voltage across the capacitor, v (> . a) Calculate the numerical expression for the trans- fer function. b) Calculate the numerical values for the poles and zeros of the transfer function. 1000 a AM, 250fi 50 mH + 1 ju.F v t> Figure 13.31 A The circuit for Example 13.1. Solution a) The first step in finding the transfer function is to construct the 5-domain equivalent circuit, as shown in Fig. 13.32. By definition, the transfer function is the ratio of V 0 /V s , which can be com- puted from a single node-voltage equation. Summing the currents away from the upper node generates V - V <> .? 1000 + V' V„s 250 + 0.055 10 6 = 0. ooo n AM-— s Figure 13.32 • The s-domain equivalent circuit for the circuit shown in Fig. 13.31. Solving for V () yields 1000(5 + 5()00)¾ Vo = 5 2 + 60005 + 25 X 10 6 ' Hence the transfer function is s 1000(5 + 5000) " 5 2 + 60005 + 25 X 10 6 " b) The poles of H{s) are the roots of the denomina- tor polynomial. Therefore -p ] = -3000 - y'4000, -p 2 = -3000 + /4000. The zeros of H(s) are the roots of the numera- tor polynomial; thus H(s) has a zero at -Zi = -5000. . acting alone. The Use of Superposition Because we are analyzing linear lumped-parameter circuits, we can use superposition to divide the response into components that can be identi- fied