486 The Laplace Transform in Circuit Analysis ASSESSMENT PROBLEM Objective 3—Understand the definition and significance of the transfer function; be able to derive a transfer function 13.9 a) Derive the numerical expression for the transfer function V 0 fl g for the circuit shown. b) Give the numerical value of each pole and zero of H(s). NOTE: Also try Chapter Problem 13.50. Answer: (a) H(s) = 10(5 + 2)/(s 2 + 2s + 10); (b) -/;, = -1 + /3, -p 2 = -1 - /3, -z = -2. 2a 1H 0.1 F The Location of Poles and Zeros of H(s) For linear lumped-parameter circuits, H(s) is always a rational function of s. Complex poles and zeros always appear in conjugate pairs. The poles of H(s) must lie in the left half of the s plane if the response to a bounded source (one whose values lie within some finite bounds) is to be bounded. The zeros of H(s) may lie in either the right half or the left half of the s plane. With these general characteristics in mind, we next discuss the role that H(s) plays in determining the response function. We begin with the partial fraction expansion technique for finding y(t). 13.5 The Transfer Function in Partial Fraction Expansions From Eq. 13.93 we can write the circuit output as the product of the trans- fer function and the driving function: Y(s) = H(s)X(s). (13.96) We have already noted that H(s) is a rational function of s. Reference to Table 13.1 shows that X(s) also is a rational function of s for the excitation functions of most interest in circuit analysis. Expanding the right-hand side of Eq. 13.96 into a sum of partial frac- tions produces a term for each pole of H(s) and X(s). Remember from Chapter 12 that poles are the roots of the denominator polynomial; zeros are the roots of the numerator polynomial. The terms generated by the poles of H(s) give rise to the transient component of the total response, whereas the terms generated by the poles of X(s) give rise to the steady- state component of the response. By steady-state response, we mean the response that exists after the transient components have become negligi- ble. Example 13.2 illustrates these general observations. 13.5 The Transfer Function in Partial Fraction Expansions 487 Example 13.2 Analyzing the Transfer Function of a Circuit The circuit in Example 13.1 (Fig. 13.31) is driven by a voltage source whose voltage increases linearly with lime, namely, v g = 50tu(t). a) Use the transfer function to find v a . b) Identify the transient component of the response. c) Identify the steady-slate component of the response. d) Sketch v a versus / for 0 < t =£ 1.5 ms. Solution a) From Example 13.1, 10()0(5 + 5000) H(S) = -: ' s 2 + 6000^ + 25 X 10 6 The transform of the driving voltage is 50/.v~; therefore, the .v-domain expression for the out- put voltage is K, = 1000(5 + 5000) 50 (s 2 + 6000s + 25 x 10 6 ) .v 2 * The partial fraction expansion of V a is _ K { V ° ~ s + 3000 - /4000 K i K 2 K 3 + s + 3000 + /4000 I 2 V' We evaluate the coefficients K^ K 2 , and /t 3 by using the techniques described in Section 12.7: /<, = 5V5 x l(r 4 /79.70°; K\ = 5V5 x IQ- 4 /-79.70°, K 2 = 10, The time-domain expression for v a is v a = [10V5 X 10"" 4 e~ 3,)()()/ cos (4000/ + 79.70°) + 10? - 4 X 10~ 4 }u(t) V. b) The transient component of v a is 10V5 X 10-V 3000 'cos (4000/ + 79.70°). Note that this term is generated by the poles (-3000 4- /4000) and (-3000 - /4000) of the transfer function. c) The steady-state component of the response is (10/ - 4 X 10" 4 )z<(/). These two terms are generated by the second- order pole (K/s 2 ) of the driving voltage. d) Figure 13.33 shows a sketch of v <7 versus /. Note that the deviation from the steady-state solution 10,000/ - 0.4 mV is imperceptible after approxi- mately 1 ms. MmV) 16 14- 12 8h 6 4 - 2 - (10.()00/- 0.4) mV, V 0.2 0.4 0.6 0.8 1.0 1.2 1.4 t (ms) Figure 13.33 A The graph of i\, versus t for Example 13.2. 488 The Laplace Transform in Circuit Analysis /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 13.10 Find (a) the unit step and (b) the unit impulse response of the circuit shown in Assessment Problem 13.9. Answer: (a) [2 + (10/3)e" f cos (3/ - 126.87°)]w(0 V; (b) 10.54*? - 'cos (3/ - 18.43>(0 V. NOTE: Also try Chapter Problems 13.79(a) and (b). 13.11 The unit impulse response of a circuit is v () (t) = 10,000e _70 'cos(240r + 6) V, where tan 6 = ^. a) Find the transfer function of the circuit. b) Find the unit step response of the circuit. Answer: (a) 9600s/(s 2 + 140s + 62,500); (b)40<T 70 'sin240r V. Observations on the Use of H(s) in Circuit Analysis Example 13.2 clearly shows how the transfer function H(s) relates to the response of a circuit through a partial fraction expansion. However, the example raises questions about the practicality of driving a circuit with an increasing ramp voltage that generates an increasing ramp response. Eventually the circuit components will fail under the stress of excessive voltage, and when that happens our linear model is no longer valid. The ramp response is of interest in practical applications where the ramp function increases to a maximum value over a finite time interval. If the time taken to reach this maximum value is long compared with the time constants of the circuit, the solution assuming an unbounded ramp is valid for this finite time interval. We make two additional observations regarding Eq. 13.96. First, let's look at the response of the circuit due to a delayed input. If the input is delayed by a seconds, £{x(t - a)u(t - a)} = e- ai X(s), and, from Eq. 13.96, the response becomes Y(s) = H(s)X(s)e- as . Uy(t) = %- l {H(s)X(s)}, then, from Eq. 13.97, y(t - a)u{t - a) = ^T 1 {B{s)X{s)e- as }. (13.97) (13.98) Therefore, delaying the input by a seconds simply delays the response function by a seconds. A circuit that exhibits this characteristic is said to be time invariant. Second, if a unit impulse source drives the circuit, the response of the circuit equals the inverse transform of the transfer function. Thus if x(t) = 5(0, then X{s) = 1 and Y(s) = Ms). (13.99) 13.6 The Transfer Function and the Convolution Integral 489 Hence, from Eq. 13.99, y(t) = //(/), (13.100) where the inverse transform of the transfer function equals the unit impulse response of the circuit. Note that this is also the natural response of the circuit because the application of an impulsive source is equivalent to instantaneously storing energy in the circuit (see Section 13.8). The subsequent release of this stored energy gives rise to the natural response (see Problem 13.90). Actually, the unit impulse response of a circuit, h(t), contains enough information to compute the response to any source that drives the circuit. The convolution integral is used to extract the response of a circuit to an arbitrary source as demonstrated in the next section. 13.6 The Transfer Function and the Convolution Integral The convolution integral relates the output y(t) of a linear time-invariant circuit to the input x{t) of the circuit and the circuit's impulse response h(t). The integral relationship can be expressed in two ways: y(t) = / h(X)x(t - X)dX = / h(t - X)x(X)dX. (13.101) We are interested in the convolution integral for several reasons. First, it allows us to work entirely in the time domain. Doing so may be beneficial in situations where x(t) and h{t) are known only through experimental data. In such cases, the transform method may be awkward or even impossible, as it would require us to compute the Laplace trans- form of experimental data. Second, the convolution integral introduces the concepts of memory and the weighting function into analysis. We will show how the concept of memory enables us to look at the impulse response (or the weighting function) h{t) and predict, to some degree, how closely the output waveform replicates the input waveform. Finally, the convolution integral provides a formal procedure for finding the *(0 *j K f ) ")'(*) inverse transform of products of Laplace transforms. We based the derivation of Eq. 13.101 on the assumption that the cir- cuit is linear and time invariant. Because the circuit is linear, the principle Fi 9 ure 13 * 34 A A block diagram of a general dKufL of superposition is valid, and because it is time invariant, the amount of the response delay is exactly the same as that of the input delay. Now con- sider Fig. 13.34, in which the block containing h{t) represents any linear time-invariant circuit whose impulse response is known, x(t) represents the excitation signal and y(t) represents the desired output signal. We assume that x(t) is the general excitation signal shown in Fig. 13.35(a). For convenience we also assume that x(t) = 0 for t < (T. Once you see the derivation of the convolution integral assuming x{t) = 0 for t < 0~, the extension of the integral to include excitation functions that exist over all time becomes apparent. Note also that we permit a discontinuity in x(t) at the origin, that is, a jump between 0~ and0 + . 490 The Laplace Transform in Circuit Analysis x(t) x(t) *(X () ) (a) liYX V V(X;?) x 0 \j x 2 x? . • • K • (b) x(X () ) AX. Xo X ! ^ r ~ -*> 2 x 7 * 3 9> 7 s X •i (c) Figure 13.35 • The excitation signal of x(t). (a) A general excitation signal, (b) Approximating x(t) with a series of pulses, (c) Approximating x(t) with a series of impulses. h(t) 0 y(') (a) ^ , N \^ Approximation of v(f) (b) Figure 13.36 A The approximation of y(f). (a) The impulse response of the box shown in Fig. 13.34. (b) Summing the impulse responses. Now we approximate x(t) by a series of rectangular pulses of uni- form width AA, as shown in Fig. 13.35(b). Thus x(t) = x 0 (t) + Xl (t) + ••• + x t {t) + (13.102) where x^t) is a rectangular pulse that equals x(A,) between A/ and A (+1 and is zero elsewhere. Note that the /th pulse can be expressed in terms of step functions; that is, Xi(t) = x(Xi){u(t - A/) - u[t - (A, + AA)]}. The next step in the approximation of x(t) is to make AA small enough that the /th component can be approximated by an impulse func- tion of strength x(Aj)AA. Figure 13.35(c) shows the impulse representa- tion, with the strength of each impulse shown in brackets beside each arrow. The impulse representation of x(t) is x(t) = x(A 0 )AA5(f - An) + x(X^)AXS(t - A,) + + jc(A f )AA5(/ - A/) + • • • (13.103) Now when x{t) is represented by a series of impulse functions (which occur at equally spaced intervals of time, that is, at An, Aj, A 2 , ), the response function y{t) consists of the sum of a series of uniformly delayed impulse responses. The strength of each response depends on the strength of the impulse driving the circuit. For example, let's assume that the unit impulse response of the circuit contained in the box in Fig. 13.34 is the exponential decay function shown in Fig. 13.36(a). Then the approximation of y{t) is the sum of the impulse responses shown in Fig. 13.36(b). Analytically, the expression for y(t) is y(t) = x(X {) )AXh(t - A () ) + x(M)AXk(t - X{) + x(X 2 )AXh(t - A 2 ) + + x(A, : )AA/j(/ - A/) + • (13.104) As AA^-0, the summation in Eq. 13.104 approaches a continuous integration, or oo r°° ^x(Ai)h(t- A;) A A -> I x{X)h{t - X)dX. (13. i=0 J0 105) Therefore, y{t) = I x(X)h(t - X)dX. (13.106) If x{t) exists over all time, then the lower limit on Eq. 13.106 becomes -oo; thus, in general, y{t) x{X)h{t - X)dX, (13.107) 13.6 The Transfer Function and the Convolution Integral 491 which is the second form of the convolution integral given in Eq. 13.101. We derive the first form of the integral from Eq. 13.107 by making a change in the variable of integration. We let u = t — A, and then we note that du = — dX, u — -co when A = oo, and u = + oo when A = -co. Now we can write Eq. 13.107 as m x(t - u)h(u)(-du), or y(t) = / x{t - it)h(ii)(du). (13.108) But because ti is just a symbol of integration, Eq. 13.108 is equivalent to the first form of the convolution integral, Eq. 13.101. The integral relationship between y(t),h(t), and x(t), expressed in Eq. 13.101, often is written in a shorthand notation: y(t) = h(t)*x{t)^x(t)*h(t), (13.109) where the asterisk, signifies the integral relationship between h(t) and x(t). Thus h(t) * x(t) is read as "//(f) is convolved with *(/)" and implies that h(t)*x(t) = J h(X)x(t - X)dX, whereas x(t) * //(/) is read as "x(t) is convolved with //(0" and implies that x(\)h(t - X)dX. The integrals in Eq. 13.101 give the most general relationship for the convolution of two functions. However, in our applications of the convolu- tion integral, we can change the lower limit to zero and the upper limit to /•.Then we can write Eq. 13.101 as y(t) = / h(X)x(t - X)dX = / x(X)h(t - X)dX. (13.110) Jo Jo We change the limits for two reasons. First, for physically realizable circuits, h(t) is zero for t < 0. In other words, there can be no impulse response before an impulse is applied. Second, we start measuring time at the instant the excitation x{t) is turned on; therefore x{t) = 0 for t < 0". A graphic interpretation of the convolution integrals contained in Eq. 13.110 is important in the use of the integral as a computational tool. We begin with the first integral. For purposes of discussion, we assume that the impulse response of our circuit is the exponential decay function shown in Fig. 13.37(a) and that the excitation function has the waveform shown in Fig. 13.37(b). In each of these plots, we replaced t with A, the symbol of integration. Replacing A with -A simply folds the excitation function over the vertical axis, and replacing -A with t — X slides the folded function to the right. See Figures 13.37(c) and (d).This h{\) A 0 (a) x{\) M 0 (b) -v(-X) M - -Ti -Ti 0 (c) .v(/ - \) M t—Tl t (d) h(k)x( MA 0 t-\) y(t) = Area t - T] t (e) Figure 13.37 A A graphic interpretation of the convolution integral f' ] h(X)x{t - X)d\. (a) The impulse response, (b) The excitation function, (c) The folded excitation function, (d) The folded excitation function displaced t units, (e) The product h(X)x(t - A). folding operation gives rise to the term convolution. At any specified value of f, the response function y(t) is the area under the product func- tion /j(A).v(f - A), as shown in Fig. 13.37(e). It should be apparent from this plot why the lower limit on the convolution integral is zero and the upper limit is t. For A < 0, the product h{k)x(t - A) is zero because h{k) is zero. For A > r, the product h(X)x{t — A) is zero because x(t - A) is zero. Figure 13.38 shows the second form of the convolution integral. Note that the product function in Fig. 13.38(e) confirms the use of zero for the lower limit and t for the upper limit. Example 13.3 illustrates how to use the convolution integral, in conjunction with the unit impulse response, to find the response of a circuit. (b) /2(- 0 -x) A (c) h(t - X) A 0 — ^> t (d) h(t - X)x(X) MA y— > ^ 1 X i >'(/) = Area X u (e) Figure 13.38 • A graphic interpretation of the convolu- tion integral J {) h(t - \)x{X)d\. (a) The impulse response, (b) The excitation function, (c) The folded impulse response, (d) The folded impulse response displaced / units, (e) The product h(t - A)x(A). 13.6 The Transfer Function and the Convolution Integral 493 Example 13.3 Using the Convolution Integral to Find an Output Signal The excitation voltage v- t for the circuit shown in Fig. 13.39(a) is shown in Fig. 13.39(b). a) Use the convolution integral to find v 0 . b) Plot v a over the range of 0 ^ t ^ 15 s. 1 H v, 20 V (a) 5 10 (b) f(s) Figure 13.39 A The circuit and excitation voltage for Example 13.3. (a) The circuit, (b) The excitation voltage. Impulse response Figure 13.40 • The impulse response and the folded excitation function for Example 13.3. Solution a) The first step in using the convolution integral is to find the unit impulse response of the circuit. We obtain the expression for V () from the s-domain equivalent of the circuit in Fig. 13.39(a): K = V, s + 1 (1)- When v t is a unit impulse function 5(f), v„ = h(t) = e~'ii(t), from which //(A) = e' A u(A) Using the first form of the convolution integral in Eq. 13.110, we construct the impulse response and folded excitation function shown in Fig. 13.40, which are helpful in selecting the lim- its on the convolution integral. Sliding the folded excitation function to the right requires breaking the integration into three intervals: 0 < t < 5; 5 < t < 10; and 10 < t < oo. The breaks in the excitation function at 0,5, and 10 s dictate these break points. Figure 13.41 shows the positioning of the folded excitation for each of these intervals. The analytical expression for Vi in the time interval 0 < / < 5 is Vt = 4/, 0 < t < 5 s. /i(\) 1.0 0 (t - 10) [t - 5) 0 t 5 ^ (t - \) 20 5ssr«10 [t - 10) 0 (/ - 5) 5 t 10 v t (t - X) 20 10 =£ /=¾ x 0(/-10)5(/-5)10 / Figure 13.41 A The displacement of ?;,(/ - A) for three different time intervals. 494 The Laplace Transform in Circuit Analysis Hence, the analytical expression for the folded excitation function in the interval t — 5 =£ A < f is Vi(t - A) = 4(r - A), t - 5 < A < t. We can now set up the three integral expressions for v a . For 0 ^ t < 5 s: MV) V n = 4(t - \)e~ x dX For 5 < t = 4(e~' + f - 1)V. 10 s, v„ = 20e~ x d\ + / 4(r - A)e _A dA Jf-5 = 4(5 + e~ l - e' { '~ 5) ) V. And for 10 < t < oo s. = 4(eT' - e" ( '- 5) + 5e~ {t ~ m ) V ,-A 20t> _A rfA + /4(7- \)e A dX 10 if-5 Figure 13.42 A The voltage response versus time for Example 13.3. b) We have computed v 0 for 1 s intervals of time, using the appropriate equation. The results arc tabulated in Table 13.2 and shown graphically in Fig. 13.42. NOTE: Assess your understanding of convolution by trying Chapter Problems 13.62 and 13.63. TABLE 13.2 t 1 2 3 4 5 6 7 8 Numerical Values of v 0 (t) *0 1.47 4.54 8.20 12.07 16.03 18.54 19.56 19.80 t 9 10 11 12 13 14 15 v 0 19.93 19.97 7.35 2.70 0.99 0.37 0.13 v,(t-k) Future (will happen) Past (has happened) (t - 10) (/ Figure 13.43 • The past, present, and future values of the excitation function. The Concepts of Memory and the Weighting Function We mentioned at the beginning of this section that the convolution inte- gral introduces the concepts of memory and the weighting function into circuit analysis. The graphic interpretation of the convolution integral is the easiest way to begin to grasp these concepts. We can view the folding and sliding of the excitation function on a timescale characterized as past, present, and future. The vertical axis, over which the excitation function x(t) is folded, represents the present value; past values of x(t) lie to the right of the vertical axis, and future values lie to the left. Figure 13.43 shows this description of x(t). For illustrative purposes, we used the exci- tation function from Example 13.3. When we combine the past, present, and future views of x(t — r) with the impulse response of the circuit, we see that the impulse response weights x(t) according to present and past values. For example, Fig. 13.41 shows that the impulse response in Example 13.3 gives less weight to past values of x(t) than to the present value of x(t). In other words, the circuit retains less and less about past input values. Therefore, in Fig. 13.42, v a quickly approaches zero when the present value of the input is zero (that is, when t > 10 s). In other words, because the present value of the input receives more weight than the past values, the output quickly approaches the present value of the input. 13.7 The Transfer Function and the Steady-State Sinusoidal Response 495 The multiplication of x{t — A) by /z(A) gives rise to the practice of referring to the impulse response as the circuit weighting function. The weighting function, in turn, determines how much memory the circuit has. Memory is the extent to which the circuit's response matches its input. For example, if the impulse response, or weighting function, is flat, as shown in Fig. 13.44(a), it gives equal weight to all values of x(t), past and present. Such a circuit has a perfect memory. However, if the impulse response is an impulse function, as shown in Fig. 13.44(b), it gives no weight to past values of x(t). Such a circuit has no memory. Thus the more memory a cir- cuit has, the more distortion there is between the waveform of the excita- tion function and the waveform of the response function. We can show this relationship by assuming that the circuit has no memory, that is, h(t) — A8(t), and then noting from the convolution integral that /i(0 y(0 h(X)x(t - X)d\ A8(X)x(t - \)dX = Ax(t). (13.111) Equation 13.111 shows that, if the circuit has no memory, the output is a scaled replica of the input. The circuit shown in Example 13.3 illustrates the distortion between input and output for a circuit that has some memory. This distortion is clear when we plot the input and output waveforms on the same graph, as in Fig. 13.45. 13.7 The Transfer Function and the Steady-State Sinusoidal Response Once we have computed a circuit's transfer function, we no longer need to perforin a separate phasor analysis of the circuit to determine its steady- state response. Instead, we use the transfer function to relate the steady- state response to the excitation source. First we assume that 1.0 (a) h(t) 1.0 (b) Figure 13.44 • Weighting functions, (a) Perfect mem- ory, (b) No memory. 10 12 14 Figure 13.45 • The input and output waveforms for Example 13.3. x(t) = A cos (<ot + (f>), (13.112) and then we use Eq. 13.96 to find the steady-state solution of y(t). To find the Laplace transform of x(t), we first write x(t) as x(t) = A cos lot cos 4> - A sin wt sin (/>, (13.113) from which X{s) (A cos 4>)s (A sin (f>)(o $ + co s + or Ais cos 4> - OJ sin 4>) s 2 + o> 2 ~' (13.114) . -z = -2. 2a 1H 0.1 F The Location of Poles and Zeros of H(s) For linear lumped-parameter circuits, H(s) is always a rational function of s. Complex poles and zeros always appear in conjugate. X)dX. (13.110) Jo Jo We change the limits for two reasons. First, for physically realizable circuits, h(t) is zero for t < 0. In other words, there can be no impulse response before an