496 The Laplace Transform in Circuit Analysis Substituting Eq. 13.114 into Eq. 13.96 gives the .,-domain expression for the response: TT/ x A(s cos <f> - a)sm<}>) Y(s) = H(s) -i Y-— 2 ^ . (13.115) We now visualize the partial fraction expansion of Eq. 13.115. The number of terms in the expansion depends on the number of poles of H(s). Because H(s) is not specified beyond being the transfer function of a physically realizable circuit, the expansion of Eq. 13.115 is TO-T^r + T^r- S — )0) S + )0) + 2 terms generated by the poles of H(s). (13.116) In Eq. 13.116, the first two terms result from the complex conjugate poles of the driving source; that is, s 2 + w 2 = (s — jta)(s + joy). However, the terms generated by the poles of H(s) do not contribute to the steady-state response of y(t), because all these poles lie in the left half of the s plane; consequently, the corresponding time-domain terms approach zero as t increases. Thus the first two terms on the right-hand side of Eq. 13.116 determine the steady-state response. The problem is reduced to finding the partial fraction coefficient K\. H(s)A(s cos 4> — a) sin</>) K\ = ; H(jm)A(j(t>eos<f> - fusing) H(ja))A(cos(b + /sin0) 1 , ., = - ^ ^ 2 ' — = -H(j(o)Ae>*. (13.117) In general, H(jo)) is a complex quantity, which we recognize by writing it in polar form; thus H(ja>) = \H(jo))\e j9iM) . (13.118) Note from Eq. 13.118 that both the magnitude, \H(j<o)\, and phase angle, #(&>), of the transfer function vary with the frequency co. When we substi- tute Eq. 13.118 into Eq. 13.117, the expression for K x becomes K, = — \H(j<o)\e mta)+4,] . (13.119) We obtain the steady-state solution for y(t) by inverse-transforming Eq. 13.116 and, in the process, ignoring the terms generated by the poles of H(s). Thus Steady-state sinusoidal response computed using a transfer function • y«(0 = A\H(ja>)\ cos [ajt + <f> + 6»(ft>)], (13.120) which indicates how to use the transfer function to find the steady-state sinusoidal response of a circuit. The amplitude of the response equals the amplitude of the source, A, times the magnitude of the transfer function, \H{j(t))\. The phase angle of the response, <£ + 0(co), equals the phase angle of the source, cj>, plus the phase angle of the transfer function, 0(a)). We evaluate both \H(ja>)\ and 9(to) at the frequency of the source, io. Example 13.4 illustrates how to use the transfer function to find the steady-state sinusoidal response of a circuit. 13.7 The Transfer Function and the Steady-State Sinusoidal Response 497 Example 13.4 Using the Transfer Function to Find the Steady-State Sinusoidal Response The circuit from Example 13.1 is shown in Fig. 13.46. The sinusoidal source voltage is 120 cos (5000/ + 30°) V. Find the steady-state expression for v 0 . The frequency of the voltage source is 5000 rad/s; hence we evaluate H(s) at 7/(/5000): 1000 ft —VYV— o 250 (l 50 mH- IjtF Figure 13.46 A The circuit for Example 13.4. Solution From Example 13.1, H(s) = 1000(.v + 5000) s 2 + 6000s + 25*10* 7/(/5000) 1000(5000 + /5000) -25 * 10 6 + /5000(6000) + 25 X 10 f 1 +/1 1 -/1 V2 /6 6 /-45°. Then, from Eq. 13.120, (120)V2 -—2 cos(5000f + 30° - 45°) o„ 20V2 cos (5000/ - 15°) V. The ability to use the transfer function to calculate the steady-state sinusoidal response of a circuit is important. Note that if we know 77(/a>), we also know 77(^), at least theoretically. In other words, we can reverse the process; instead of using 7/(5) to find 77(ja>), we use H(jco) to find H(s). Once we know H(s), we can find the response to other excitation sources. In this application, we determine H(jo)) experimentally and then construct H{s) from the data. Practically, this experimental approach is not always possible; however, in some cases it does pro- vide a useful method for deriving H(s). In theory, the relationship between H{s) and H(j(x)) provides a link between the time domain and the frequency domain. The transfer function is also a very useful tool in problems concerning the fre- quency response of a circuit, a concept we introduce in the next chapter. t/ASSESSMENT PROBLEMS Objective 4—Know how to use a circuit's transfer function to calculate the circuit's impulse response, unit step response, and steady-state response to sinusoidal input 13.12 The current source in the circuit shown is deliv- ering 10 cos 4/ A. Use the transfer function to compute the steady-state expression for v 0 . Answer: 44.7cos(4/ - 63.43°) V. 13.13 a) For the circuit shown, find the steady-state expression for v 0 when v g = 10cos50,000r V. NOTE: Also try Chapter Problems 13.77 and 13.80. b) Replace the 50 kfl resistor with a variable resistor and compute the value of resistance necessary to cause v 0 to lead v g by 120°. 10 kO 10 kn /vw- Answer: (a) 10 cos (50,000/ + 90°) V; (b) 28,867.51 ft. 498 The Laplace Transform in Circuit Analysis :c, Figure 13.47 • A circuit showing the creation of an impulsive current. Figure 13.48 • The s-domain equivalent circuit for the circuit shown in Fig. 13.47. R 2 <R) Figure 13.49 A The plot of i (t) versus t for two different values of /?. 13.8 The Impulse Function in Circuit Analysis Impulse functions occur in circuit analysis either because of a switching operation or because a circuit is excited by an impulsive source. The Laplace transform can be used to predict the impulsive currents and volt- ages created during switching and the response of a circuit to an impulsive source. We begin our discussion by showing how to create an impulse function with a switching operation. Switching Operations We use two different circuits to illustrate how an impulse function can be created with a switching operation: a capacitor circuit, and a series induc- tor circuit. -£- Capacitor Circuit In the circuit shown in Fig. 13.47, the capacitor Ci is charged to an initial voltage of V () at the time the switch is closed. The initial charge on C 2 is zero. The problem is to find the expression for /'(/) as R —» 0. Figure 13.48 shows the s-domain equivalent circuit. From Fig. 13.48, I = Vjs R + (1/sCj) + (l/sC 2 ) Vo/R s + (\/RC e )' (13.121) where the equivalent capacitance C\C 2 /{C X + C 2 ) is replaced by C e . We inverse-transform Eq. 13.121 by inspection to obtain ^-fe-^^W (13.122) which indicates that as R decreases, the initial current (Vo/R) increases and the time constant (RC e ) decreases. Thus, as R gets smaller, the current starts from a larger initial value and then drops off more rapidly. Figure 13.49 shows these characteristics of/. Apparently i is approaching an impulse function as R approaches zero because the initial value of i is approaching infinity and the duration of i is approaching zero. We still have to determine whether the area under the current function is independent of R. Physically, the total area under the i versus t curve represents the total charge transferred to C 2 after the switch is closed. Thus Area = q V „- K QS-er (13.123) which says that the total charge transferred to C 2 is independent of R and equals V {) C e coulombs. Thus, as R approaches zero, the current approaches an impulse strength V () C e ; that is, I-—v&CAO- (13.124) 13.8 The Impulse Function in Circuit Analysis 499 The physical interpretation of Eq. 13.124 is that when R = 0, a finite amount of charge is transferred to C 2 instantaneously. Making R zero in the circuit shown in Fig. 13.47 shows why we get an instantaneous transfer of charge. With R = 0, we create a contradiction when we close the switch; that is, we apply a voltage across a capacitor that has a zero initial voltage. The only way to have an instantaneous change in capacitor voltage is to have an instantaneous transfer of charge. When the switch is closed, the voltage across C 2 does not jump to V () but to its final value of v 2 C, + C, (13.125) We leave the derivation of Eq. 13.125 to you (see Problem 13.81). If we set R equal to zero at the outset, the Laplace transform analysis will predict the impulsive current response. Thus, (1/Jd) + (l/sC 2 ) C, + C 2 C,V 0 (13.126) In writing Eq. 13.126, we use the capacitor voltages at t = (T. The inverse transform of a constant is the constant times the impulse function; there- fore, from Eq. 13.126, i = C C V 0 8(0- (13.127) The ability of the Laplace transform to predict correctly the occurrence of an impulsive response is one reason why the transform is widely used to analyze the transient behavior of linear lumped-parameter time-invariant circuits. Series Inductor Circuit The circuit shown in Fig. 13.50 illustrates a second switching operation that produces an impulsive response. The problem is to find the time- domain expression for v ( , after the switch has been opened. Note that opening the switch forces an instantaneous change in the current of L 2 , which causes v 0 to contain an impulsive component. Figure 13.51 shows the s-domain equivalent with the switch open. In deriving this circuit, we recognized that the current in the 3 H inductor at t = 0~ is 10 A, and the current in the 2 H inductor at t = 0~ is zero. Using the initial conditions at t = 0" is a direct consequence of our using 0~ as the lower limit on the defining integral of the Laplace transform. We derive the expression for V a from a single node-voltage equation. Summing the currents away from the node between the 15 ft resistor and the 30 V source gives v: 2s + 15 + V a - [(100/5) + 30] 3s + 10 = 0. (13.128) 10 O VW- © 100 V 3H _/-Y-Y"Y-\- Li / = (> sC i5ir 2H T - L 2 Figure 13.50 A A circuit showing the creation of an impulsive Figure 13.51 • The s-domain equivalent circuit for the voltage. circuit shown in Fig. 13.50. Solving for V 0 yields 40(5 + 7.5) 12(5 + 7.5) We anticipate that v 0 will contain an impulse term because the second term on the right-hand side of Eq. 13.129 is an improper rational function. We can express this improper fraction as a constant plus a rational func- tion by simply dividing the denominator into the numerator; that is, 12(5 + 7.5) 30 — •=-*- = 12 + (13.130) s + 5 s + 5 v ' Combining Eq. 13.130 with the partial fraction expansion of the first term on the right-hand side of Eq. 13.129 gives x, 60 20 ,„ 30 V„ = r + 12 + S 5 + 5 5 + 5 „„ 60 10 = 12 + — + -, (13.131) 5 5 + 5 V ' from which v 0 = 125(0 + (60 + 10e~ 5 ')«(0 V. (13.132) Does this solution make sense? Before answering that question, let's first derive the expression for the current when t > 0~. After the switch has been opened, the current in Li is the same as the current in L 2 . If we refer- ence the current clockwise around the mesh, the 5-domain expression is 1 (100/5) + 30 20 55 + 25 s(s + 5) 5 + 5 4 + 5 5 + 5 5 + 5 4 2 = - + r. (13.133 5 5 + 5 ' Inverse-transforming Eq. 13.133 gives i = (4 + 2e~ 5t )u(t) A. (13.134) Before the switch is opened, the current in L 1 is 10 A, and the current in L 2 is 0 A; from Eq. 13.134 we know that at t = Q + , the current in L\ and in L 2 is 6 A.Then, the current in L\ changes instantaneously from 10 to 6 A, while the current in L 2 changes instantaneously from 0 to 6 A. From this value of 6 A, the current decreases exponentially to a final value of 4 A. 13.8 The Impulse Function in Circuit Analysis 501 This final value is easily verified from the circuit; that is, it should equal 100/25, or 4 A. Figure 13.52 shows these characteristics of i { and i 2 . How can we verify that these instantaneous jumps in the inductor cur- rent make sense in terms of the physical behavior of the circuit? First, we note that the switching operation places the two inductors in series. Any impulsive voltage appearing across the 3 H inductor must be exactly bal- anced by an impulsive voltage across the 2 H inductor, because the sum of the impulsive voltages around a closed path must equal zero. Faraday's law states that the induced voltage is proportional to the change in flux linkage (v = dk/dt). Therefore, the change in flux linkage must sum to zero. In other words, the total flux linkage immediately after switching is the same as that before switching. For the circuit here, the flux linkage before switching is A = L x i x + L 2 i 2 = 3(10) + 2(0) - 30 Wb-turns. (13.135) Immediately after switching, it is A = (^ + L 2 )i(0 + ) = 5i(0 + ). (13.136) Combining Eqs. 13.135 and 13.136 gives Z(0 + ) = 30/5 = 6 A. (13.137) Thus the solution for i (Eq. [13.134]) agrees with the principle of the con- servation of flux linkage. We now test the validity of Eq. 13.132. First we check the impulsive term 125(f). The instantaneous jump of i 2 from 0 to 6 A at t = 0 gives rise to an impulse of strength 66(f) in the derivative of i 2 . This impulse gives rise to the 125(f) in the voltage across the 2 H inductor. For f > 0 + , di 2 /dt is -10e~ 5 ' A/s; therefore, the voltage v 0 is Vo = 15(4 + 2e~ 5t ) + 2(-10<T 5 ') / L , i 2 (A) i 2 = i Figure 13.52 A The inductor currents versus t for the circuit shown in Fig. 13.50. (60 + 10e~ 5 ')«(/)V. (13.138) Equation 13.138 agrees with the last two terms on the right-hand side of Eq. 13.132; thus we have confirmed that Eq. 13.132 does make sense in terms of known circuit behavior. We can also check the instantaneous drop from 10 to 6 A in the cur- rent I'I. This drop gives rise to an impulse of -45(f) in the derivative of i h Therefore the voltage across L t contains an impulse of —125(f) at the ori- gin. This impulse exactly balances the impulse across L 2 ; that is, the sum of the impulsive voltages around a closed path equals zero. Impulsive Sources Impulse functions can occur in sources as well as responses; such sources are called impulsive sources. An impulsive source driving a circuit imparts a finite amount of energy into the system instantaneously. A mechanical analogy is striking a bell with an impulsive clapper blow. After the energy has been transferred to the bell, the natural response of the bell deter- mines the tone emitted (that is, the frequency of the resulting sound waves) and the tone's duration. 502 The Laplace Transform in Circuit Analysis V Q 8(t) Figure 13.53 A An RL circuit excited by an impulsive voltage source. In the circuit shown in Fig. 13.53, an impulsive voltage source having a strength of V 0 volt-seconds is applied to a series connection of a resistor and an inductor. When the voltage source is applied, the initial energy in the inductor is zero; therefore the initial current is zero. There is no voltage drop across R, so the impulsive voltage source appears directly across L. An impulsive voltage at the terminals of an inductor establishes an instan- taneous current. The current is jjV 0 8(x)dx. (13.139) Given that the integral of 8(t) over any interval that includes zero is 1, we find that Eq. 13.139 yields .•«n = v ~l A. (13.140) Thus, in an infinitesimal moment, the impulsive voltage source has stored in the inductor. in 2 L (13.141) The current V {) /L now decays to zero in accordance with the natural response of the circuit; that is, l = T e •//T m (13.142) Figure 13.54 • The 5-domain equivalent circuit for the circuit shown in Fig. 13.53. where T = L/R. Remember from Chapter 7 that the natural response is attributable only to passive elements releasing or storing energy, and not to the effects of sources. When a circuit is driven by only an impulsive source, the total response is completely defined by the natural response; the duration of the impulsive source is so infinitesimal that it does not contribute to any forced response. We may also obtain Eq. 13.142 by direct application of the Laplace transform method. Figure 13.54 shows the 5-domain equivalent of the cir- cuit in Fig. 13.53. Hence Vo Vo/L R + sL s + (R/L)' (13.143) ion >vw- 3H _/-Y"VY>_ 505(/) 100 V / = () 15 n 12 H Figure 13.55 A The circuit shown in Fig. 13.50 with an impulsive voltage source added in series with the 100 V source. L -(R/L)t L -t/r U(t). (13.144) Thus the Laplace transform method gives the correct solution for i £: 0 + . Finally, we consider the case in which internally generated impulses and externally applied impulses occur simultaneously. The Laplace trans- form approach automatically ensures the correct solution for t > 0 + if inductor currents and capacitor voltages at t = 0~ are used in constructing the j'-domain equivalent circuit and if externally applied impulses are rep- resented by their transforms. To illustrate, we add an impulsive voltage 13.8 The Impulse Function in Circuit Analysis 503 source of 505(f) in series with the 100 V source to the circuit shown in Fig. 13.50. Figure 13.55 shows the new arrangement. At t = 0", ii(Q~) = 10 A and / 2 (0~) = 0 A. The Laplace transform of 505(0 = 50. If we use these values, the s-domain equivalent circuit is as shown in Fig. 13.56. The expression for I is I = 50 + (100/5) + 30 25 + 5s Figure 13.56 • The s-domain equivalent circuit for the circuit shown in Fig. 13.55. 16 + 20 s + 5 s(s + 5) 16 4 4_ 5 + 5 s 5 + 5 12 4 + - 5 + 5 5 (13.145) from which /(f) = (12e _5/ + 4)M(0 A. (13.146) The expression for V 0 is v /K.OU 32(.y + 7.5) 40(5 + 7.5) V 0 = (15 + 2s) J = - 5 + 5 5(5 + 5) 2.5 \ 60 20 = 32 1 + r )+ 5 + 5) 5 5 + 5 = 32 + 60 60 + —, 5 + 5 5 (13.147) from which v„ = 325(f) + (60e" 5 ' + 60)»(f) V. (13.148) Now we test the results to see whether they make sense. From Eq. 13.146, we see that the current in L x and L 2 is 16 A at f = 0 + . As in the previous case, the switch operation causes /, to decrease instantaneously from 10 to 6 A and, at the same time, causes / 2 to increase from 0 to 6 A. Superimposed on these changes is the establishment of 10 A in L\ and L 2 by the impulsive voltage source; that is, 1 3 + 2 50S(x)dx = 10 A. (13.149) Therefore i { increases suddenly from 10 to 16 A, while / 2 increases sud- denly from 0 to 16 A. The final value of i is 4 A. Figure 13.57 shows i h / 2 , and / graphically. Figure 13.57 • The inductor currents versus t for the circuit shown in Fig. 13.55. We may also find the abrupt changes in i x and / 2 without using super- position. The sum of the impulsive voltages across L x (3 H) and L 2 (2 H) equals 505(f). Thus the change in flux linkage must sum to 50; that is, AA t + AA 2 = 50. (13.150) Because A = Li, we express Eq. 13.150 as 3A/i + 2A/ 2 = 50. (13.151) But because / 5 and / 2 must be equal after the switching takes place, /^(01 + A/i = k(0~) + A/ 2 . (13.152) Then, 10 + All = 0 + A/ 2 . (13.153) Solving Eqs. 13.151 and 13.153 for A/j and A/ 2 yields A/j - 6 A, (13.154) A/ 2 = 16 A. (13.155) These expressions agree with the previous check. Figure 13.57 also indicates that the derivatives of ij and i 2 will contain an impulse at t = 0. Specifically, the derivative of i x will have an impulse of 65(0, and the derivative of / 2 will have an impulse of 165(f). Figure 13.58(a), (b), respectively, illustrate the derivatives of/^ and i 2 . Now let's turn to Eq. 13.148. The impulsive component 325(f) agrees with the impulse 165(f) that characterizes di 2 /dt at the origin. The term (60<T 5 ' + 60) agrees with the fact that for t > 0 + , di v (> = 15/ + 2—. dt We test the impulsive component of di^jdt by noting that it produces an impulsive voltage of (3)65(f), or 185(f), across L\. This voltage, along with 325(f) across L 2 , adds to 505(f). Thus the algebraic sum of the impul- sive voltages around the mesh adds to zero. To summarize, the Laplace transform will correctly predict the creation of impulsive currents and voltages that arise from switching. However, the .y-domain equivalent circuits must be based on initial conditions at t — 0~, that is, on the initial conditions that exist prior to the disturbance caused by the switching. The Laplace transform will correctly predict the response to impulsive driving sources by simply representing these sources in the ,v domain bv their correct transforms. NOTE: Assess your understanding of the impulse function in circuit analysis by trying Chapter Problems 13.87 and 13.88. Practical Perspective Practical Perspective Surge Suppressors As mentioned at the beginning of this chapter, voltage surges can occur in a circuit that is operating in the sinusoidal steady state. Our purpose is to show how the Laplace transform is used to determine the creation of a surge in voltage between the line and neutral conductors of a household circuit when a load is switched off during sinusoidal steady-state operation. Consider the circuit shown in Fig. 13.59, which models a household cir- cuit with three loads, one of which is switched off at time t = 0. To simplify the analysis, we assume that the line-to-neutral voltage, \ ()r is 120 /0° V (rms), a standard household voltage, and that when the load is switched off at t = 0, the value of V^ does not change. After the switch is opened, we can construct the s-domain circuit, as shown in Fig. 13.60. Note that because the phase angle of the voltage across the inductive load is 0°, the initial current through the inductive load is 0. Therefore, only the inductance in the line has a non-zero initial condition, which is modeled in the s-domain circuit as a voltage source with the value L// 0 , as seen in Fig. 13.60. Just before the switch is opened at t = 0, each of the loads has a steady-state sinusoidal voltage with a peak magnitude of 120V2 = 169.7 V. All of the current flowing through the line from the voltage source y g is divided among the three loads. When the switch is opened at t .=? 0, all of the current in the line will flow through the remaining resistive load. This is because the current in the inductive load is 0 at t = 0 and the current in an inductor cannot change instantaneously. Therefore, the voltage drop across the remaining loads can experience a surge as the line current is directed through the resistive load. For example, if the initial current in the line is 25 A (rms) and the impedance of the resistive load is 12 0, the voltage drop across the resistor surges from 169.7 V to (25)( V2 )(12) = 424.3 V when the switch is opened. If the resistive load cannot handle this amount of voltage, it needs to be protected with a surge suppressor such as those shown at the beginning of the chapter. A / = 0 V *(T) h\k*a.Voh\\iX a fcjiife Figure 13.59 • Circuit used to introduce a switching surge voltage. Figure 13.60 A Symbolic s-domain circuit. NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 13.92 and 13.93