446 Introduction to the Laplace Transform or Then K 3 is 96(-3)(4) (-8)(-2) K? = -72. (12.47) 96(5 + 5)(5 + 12) s(s + 8) K 3 = 48. (12.48) From Eq. 12.45 and the K values obtained, 96(5 + 5)(5 + 12) s(s + 8)(5 + 6) 120 + 48 72 5 + 6 5 + 8 [12.49) At this point, testing the result to protect against computational errors is a good idea. As we already mentioned, a partial fraction expansion creates an identity; thus both sides of Eq. 12.49 must be the same for all s values. The choice of test values is completely open; hence we choose values that are easy to verify. For example, in Eq. 12.49, testing at either -5 or -12 is attractive because in both cases the left-hand side reduces to zero. Choosing -5 yields 120 48 72 •24 + 48 - 24 = 0, whereas testing -12 gives 120 48 72 n „ ,„ n + —: = -10 - 8 + 18 = 0. 12 -6 -4 Now confident that the numerical values of the various K$ are correct, we proceed to find the inverse transform: %- ,)96(5 + 5)(5 + 12) s(s + 8)(5 + 6) (120 + 48e" 6 ' - 72eT 8 ')"(0- (12.50) i/ASSESSMENT PROBLEMS Objective 2—Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace transform table 12.3 Find/(0 if 65 2 + 265 + 26 ( ' V) ~ (5+ 1)(5 + 2)(5 + 3)* Answer: f(t) = (3e' 1 + 2e' 2 ' + e~^)u(t). 12A Find/(f) if 75 2 + 635 + 134 (5) ~ (5 + 3)(5 + 4)(5+ 5)* Answer: /(0 = (4<T 3 ' + 6e~ 41 - 3e" 5/ )»(0- NOTE: Also try Chapter Problems 12.40(a) and (b). Partial Fraction Expansion: Distinct Complex Roots of D(s) The only difference between finding the coefficients associated with dis- tinct complex roots and finding those associated with distinct real roots is that the algebra in the former involves complex numbers. We illustrate by expanding the rational function: F(s) = 100(5 + 3) (s + 6)(5 2 + 6s + 25) (12.51) We begin by noting that F(s) is a proper rational function. Next we must find the roots of the quadratic term s 2 + 6s + 25: s 2 + 6s + 25 = (s + 3 - j4)(s + 3 + /4). (12.52) With the denominator in factored form, we proceed as before: 100(5 + 3) = (s + 6)(s 2 + 6s + 25) ~~ /v, + K? + K, s + 6 s + 3 - /4 .v + 3 + /4 To find K u K 2 , and K 3 , we use the same process as before: K, = 100(5 + 3) s 2 + 6s + 25 ,= -6 100(-3) 25 -12, (12.53) (12.54) K 7 = 100(5 + 3) (s + 6)(5 + 3 + /4) 100(/4) ,=-3+/ ~ (3 + /4)(/8) = 6-/8= l0e~ j5Uy , (12.55) K, = 100(5 + 3) (5 + 6)(5 + 3 - /4) 100(-/4) -3-,4 ~ (3 - /4)(-/8) Then = 6 + /8= 10e' 5313 100(5 + 3) 12 10/-53.13° (5 + 6)(5 2 + 65 + 25) 5 + 6 5 + 3 - /4 (12.56) + 10/53.13° s + 3 + /4 (12.57) Again, we need to make some observations. First, in physically realiz- able circuits, complex roots always appear in conjugate pairs. Second, the coefficients associated with these conjugate pairs are themselves conju- gates. Note, for example, that /<C 3 (Eq. 12.56) is the conjugate of K 2 (Eq. 12.55). Thus for complex conjugate roots, you actually need to calcu- late only half the coefficients. Before inverse-transforming Eq. 12.57, we check the partial fraction expansion numerically. Testing at —3 is attractive because the left-hand side reduces to zero at this value: -12 10/-53.13° 10/53.13° /4 /4 = -4 + 2.5 /36.87° + 2.5 /-36.87° = -4 + 2.0 + /1.5 + 2.0 - /1.5 = 0. We now proceed to in verse-transform Eq. 12.57: J 100(5 + 3) 1 = & + ^-/53.13^-(3-/4), \{s + 6)(5 2 + 6^ + 25)] v + 10e' 53 - ,3 V (3+ ' 4) ')«(*)• (12.58) In general, having the function in the time domain contain imaginary com- ponents is undesirable. Fortunately, because the terms involving imaginary components always come in conjugate pairs, we can eliminate the imagi- nary components simply by adding the pairs: 10e -j53A3 e -O-j4) t + 10e /53.13 V (3 + /4)r = 10e V X4 ' _53LV) + e**- 5 " 3 '*) = 20cT 3 'cos(4r - 53.13°), (12.59) which enables us to simplify Eq. 12.58: 100(5 + 3) (5 + 6)(5^ + 65 + 25) = [-12fT 6 ' + 20e" 3 'cos(4r - 53.13°)]u(f). (12.60) Because distinct complex roots appear frequently in lumped-parameter linear circuit analysis, we need to summarize these results with a new transform pair. Whenever D(s) contains distinct complex roots—that is, factors of the form (5 + a - //3)(5 + a + //3)-a pair of terms of the form K K* , + TZ (12.61) 5 + a - //3 5 + a + //3 appears in the partial fraction expansion, where the partial fraction coeffi- cient is, in general, a complex number. In polar form, K = \K\e' e = 1*1/0°, (12.62) 12.7 Inverse Transforms where \K\ denotes the magnitude of the complex coefficient.Then K* = \K\e~ j0 = \K\/-0°. (12.63) The complex conjugate pair in Eq. 12.61 always inverse-transforms as K K* 1 sr 1 - + s + a - jB s + a + jB J = 2|K|<r°"cos(j3f + 0). (12.64) In applying Eq. 12.64 it is important to note that K is defined as the coeffi- cient associated with the denominator term s + a — jB, and K" is defined as the coefficient associated with the denominator s + a + yj8. /"ASSESSMENT PROBLE Objective 2—Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace transform table 12.5 Find /(f) if F(s) = 1Q (* 2 + 119 ) (s + 5)(s 2 + 10s + 169)' NOTE: Also try Chapter Problems 12.40(c) and (d). Answer: /(f) = (lOe - * - 8.33c" 5 ' sin 12f)«(f). Partial Fraction Expansion: Repeated Real Roots of D(s) To find the coefficients associated with the terms generated by a multiple root of multiplicity r, we multiply both sides of the identity by the multiple root raised to its rth power. We find the K appearing over the factor raised to the rth power by evaluating both sides of the identity at the multiple root. To find the remaining (r — 1) coefficients, we differentiate both sides of the identity (r — 1) times. At the end of each differentiation, we evaluate both sides of the identity at the multiple root. The right-hand side is always the desired K, and the left-hand side is always its numerical value. For example. 100(5 + 25) K l + Ki K, K A s(s + 5f s (s + 5) 3 (s + 5) 2 We find K x as previously described; that is, S + 5 K: = 100(5' + 25) (s + 5) 3 s=() 100(25) 125 20. (12.65) (12.66) To find Ko, we multiply both sides by (s + 5) J and then evaluate both sides at -5: 100(5 + 25) K.is + 5) 3 + K 2 + K 3 (s + 5)U- 5 + K 4 (s + 5) 2 100(20) s—5 (12.67) (-5) Ki X 0 + K 2 + K 3 X 0 + K 4 X 0 = Ko = -400. (12.68) To find K?, we first must multiply both sides of Eq. 12.65 by (s + 5) 3 . Next we differentiate both sides once with respect to s and then evaluate at s = -5: d ds 10Q(s + 25) s L_ _ d v =_ 5 ds + Ki(s + 5) 3 s &**" s=-5 + -[K 3 (s + 5)L = _ 5 + — [K 4 (s + 5) 2 ], = _ 5 , (12.69) 100 s - (s + 25) = K 3 = -100. (12.70) To find K 4 we first multiply both sides of Eq. 12.65 by (s + 5) 3 . Next we differentiate both sides twice with respect to s and then evaluate both sides at s = —5. After simplifying the first derivative, the second deriva- tive becomes n d 100— as 25] s- _ ,,=-5 ds \s + 5f{2s-S) .v=-5 + 0 + J^3l*=-5 + j- s l 2K *( S + 5 )]*=-5< or -40 = 2K 4 . (12.71) Solving Eq. 12.71 for K 4 gives K 4 = -20. (12.72) Then IOOQT + 25) _ 20 400 100 20 s(s + 5) 3 " .v ' (.v + 5) 3 (s + 5) 2 s + 5' (12.73) At this point we can check our expansion by testing both sides of Eq. 12.73 at s = -25. Noting both sides of Eq. 12.73 equal zero when s = —25 gives us confidence in the correctness of the partial fraction expansion. The inverse transform of Eq. 12.73 yields /100(5 + 25) 1 s(s + 5) 3 = [20 - 200t 2 e~ 5t - lOOte'* 1 - 20e~ 5 ']w(f). (12.74) 12.7 Inverse Transforms 451 I/'ASSESSMENT PROBLEM Objective 2—Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace transform table 12.6 Find/(0 if (4s 2 + Is + 1) s(s + 1) NOTE: Also try Chapter Problems 12.41(a), (b), and (d). Answer: fit) = (1 + 2te~' + 3<T r )"(0- Partial Fraction Expansion: Repeated Complex Roots of D(s) We handle repeated complex roots in the same way that we did repeated real roots; the only difference is that the algebra involves complex num- bers. Recall that complex roots always appear in conjugate pairs and that the coefficients associated with a conjugate pair are also conjugates, so that only half the Ks need to be evaluated. For example, F(s) = 768 (s 2 + 65 + 25) : ' After factoring the denominator polynomial, we write 768 F(s) (s + 3 - jA)\s + 3 + /4) 2 (12.75) K> + K, (s + 3 - /4) 2 s + 3 - /4 + K\ + K\ (s + 3 + /4) 2 s + 3 + /4 (12.76) Now we need to evaluate only K { and K 2 , because K\ and K\ are conju- gate values. The value of K { is K, = 768 (s + 3 + /4) 2 768 s=-3+j4 (;8) : -12. The value of K 2 is Ki = els 768 (5 + 3 + /4) 2(768) s=-3+/4 (s + 3 + /4) 3 2(768) " (/8) 3 = -/3 = 3/-90°. s=—3+/4 (12.77) (12.78) From Eqs. 12.77 and 12.78, K\ = -12, (12.79) K 2 = /3 = 3 /90° . (12.80) We now group the partial fraction expansion by conjugate terms to obtain F(s) = 12 -12 .(.V + 3-/4) 2 (. + 3+,4) 2 3 /~9(T 3 /90° s + 3 - /4 s + 3 + /4 (12.81) We now write the inverse transform of F(s): f(t) = [-24f£T 3f eos4f + 6c~ 3 'cos(4/ - 90°)]u(f). (12.82) Note that if F(s) has a real root a of multiplicity /• in its denominator, the term in a partial fraction expansion is of the form A' (s + a) r The inverse transform of this term is We-* (s + ayj (r-1)! «(')• (12.83) If F(s) has a complex root of a + //3 of multiplicity r in its denominator, the term in partial fraction expansion is the conjugate pair K + K (s + a- jp) r (s + a+ j{3) r The inverse transform of this pair is K K* 2T 1 (s + a - y/3)' (.v + a + j(3) r 2\K\t r 1 (r - 1)! e~°" co$((3t + 6) u{t). (12.84) Equations 12.83 and 12.84 are the key to being able to inverse-transform any partial fraction expansion by inspection. One further note regarding these two equations: In most circuit analysis problems, r is seldom greater than 2. Therefore, the inverse transform of a rational function can be han- dled with four transform pairs. Table 12.3 lists these pairs. 12.7 Inverse Transforms TABLE 12.3 Four Useful Transform Pairs Pair Number 1 2 3 4 Nature of Roots Distinct real Repeated real Distinct complex Repeated complex F(s) f{() K s <* s + a K + a) 2 K + a - K 7/3 + - s i + K a K* (s + a - jpf (s + a + jpy Ke- at u{t) Kte-"'u(t) 2|/<|e _a 'cos(/3r + d)u(t) 2t\K\e- al cos ((3t + $)u(t) Note: In pairs 1 and 2, K is a real quantity, whereas in pairs 3 and 4. K is the complex quantity | K | /0. ^/ASSESSMENT PROBLEM Objective 2—Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace transform table 12.7 Find f(t) if Answer: f(t) = (-20te~ 21 cos t + 20e" 2f sin t)u(t). F{s) = 40 W (s 2 + As + 5) 2 NOTE: Also try Chapter Problem 12.41(e). Partial Fraction Expansion: Improper Rational Functions We conclude the discussion of partial fraction expansions by returning to an observation made at the beginning of this section, namely, that improper rational functions pose no serious problem in finding inverse transforms. An improper rational function can always be expanded into a polynomial plus a proper rational function. The polynomial is then inverse-transformed into impulse functions and derivatives of impulse functions. The proper rational function is inverse-transformed by the techniques outlined in this section. To illustrate the procedure, we use the function ,, s 4 + 13.v 3 + 66^ 2 + 200* + 300 s + 9s + 20 Dividing the denominator into the numerator until the remainder is a proper rational function gives •» . ^ 305 + 100 F(s) = s 2 + 4.v + 10 + (12.86) s + 9s + 20 where the term (30s + 100)/(s 2 + 9s + 20) is the remainder. Next we expand the proper rational function into a sum of partial fractions: 30* + 100 305 + 100 -20 50 + (12.87) s 2 + 95 + 20 (s + 4)(5 + 5) 5 + 4 5 + 5 454 Introduction to the Laplace Transform Substituting Eq. 12.87 into Eq. 12.86 yields 7 ,« 20 50 F(s) = s 2 + 4s + 10 - - + .v + 4 s + 5" Now we can inverse-transform Eq. 12.88 by inspection. Hence ,, , d 2 8(t) d8(t) - (2()e- 41 - 5Qe~ 5t )u(t). (12.88) (12.89) ^ASSESSMENT PROBLEMS Objective 2—Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace transform table 12.8 Find/(0 if (5s 2 + 29s + 32) F(s) = (s + 2)(5 + 4) Answer: /(0 = 58(0 ~ (3e" 2/ - 2<T 4/ >/(0- NOTE: Also try Chapter Problem 12.42(c). 12.9 Find/(0 if (25 3 + 85 2 + 25 - 4) F(s) (s 2 + 5s + 4) Answer: /(0 = 2 £/5(Q -4/. - 25(0 + 4e~ 4t u(t) 12.8 Poles and Zeros of F(s) The rational function of Eq. 12.42 also may be expressed as the ratio of two factored polynomials. In other words, we may write F(s) as F(s) K(s + Zi)(s + z 2 ) {s + z a ) (12.90) (s + p t )(s + p 2 ) • • • (s + p M ) ' where K is the constant ajb m . For example, we may also write the function 85 2 + 1205 + 400 as F(s) F(s) 2s A + 205 3 + 705 2 + 1005 + 48 8(5 2 + 155 + 50) 2(5 4 + 105 3 + 355 2 + 505 + 24) 4(5 + 5)(5 + 10) (5 + 1)(5 + 2)(5 + 3)(5 +4)' (12.91) The roots of the denominator polynomial, that is, -pi, -p 2 , ~p$, , -/?„,, are called the poles of F(s); they are the values of s at which F(s) becomes infinitely large. In the function described by Eq. 12.91, the poles of F(s) are -1, -2, -3, and -4. The roots of the numerator polynomial, that is, — Z[, ~z 2 , ~z^ , -z„, are called the zeros of F(s); they are the values of s at which F(s) becomes zero. In the function described by Eq. 12.91, the zeros of F(s) are -5 and -10. 12.9 Initial- and Final-Value Theorems 455 In what follows, you may find that being able to visualize the poles and zeros of F(s) as points on a complex s plane is helpful. A complex plane is needed because the roots of the polynomials may be complex. In the complex s plane, we use the horizontal axis to plot the real values of s and the vertical axis to plot the imaginary values of .v. As an example of plotting the poles and zeros of F(s), consider the function F(s) 10(.y + 5)(s + 3 - j4)(s + 3 + /4) s(s + 10)(s + 6 - J8){s + 6 + /8) (12.92) The poles of F(s) are at 0, -10,-6 4- /8, and -6 — y'8. The zeros are at -5, -3 + /4, and -3 — ;4. Figure 12.17 shows the poles and zeros plotted on the s plane, where X ? s represent poles and O's represent zeros. Note that the poles and zeros for Eq. 12.90 are located in the finite s plane. F(s) can also have either an rth-order pole or an rth-order zero at infinity. For example, the function described by Eq. 12.91 has a second- order zero at infinity, because for large values of s the function reduces to 4/s 2 , and F(s) = 0 when .v = oo. In this text, we are interested in the poles and zeros located in the finite s plane. Therefore, when we refer to the poles and zeros of a rational function of s, we are referring to the finite poles and zeros. s plane -6+ /8 >f I I I I I I I I Ml I I -10 r° i i e— -3-/4 -3+;4_5 p__ I <ti' i \y\ 11 —5 -6-/8 X- h Figure 12.17 A Plotting poles and zeros on the s plane. 12.9 Initial- and Final-Value Theorems The initial- and final-value theorems are useful because they enable us to determine from F(s) the behavior of /(0 at 0 and oo. Hence we can check the initial and final values of /(/) to see if they conform with known circuit behavior, before actually finding the inverse transform of F(s). The initial-value theorem states that lim /(0 = lim sF(s), (12.93) ^1 Initial value theorem and the final-value theorem states that lim /(0 = lim sF(s). (12.94) ^ Final value theorem The initial-value theorem is based on the assumption that /(0 contains no impulse functions. In Eq. 12.94, we must add the restriction that the theo- rem is valid only if the poles of F(s), except for a first-order pole at the origin, lie in the left half of the s plane. To prove Eq. 12.93, we start with the operational transform of the first derivative: Now we take the limit as s —» oo: ^/ dt e' sr dt. (12.95) lim [sF(s) - /(0 - )] = lim • s 'dt. (12.96) . + 3 + /4 (12.57) Again, we need to make some observations. First, in physically realiz- able circuits, complex roots always appear in conjugate pairs. Second, the coefficients associated with