Signal Processing and Linear Systems-B.P.Lathi copy

# 20 cannot be satisfied b ut t he t ransform of 1 0

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Unformatted text preview: le 6 .1 6.1 T he Laplace T ransform 373 Checking t he answer A S hort T able o f ( Unilateral) L aplace T ransforms I t is easy to make a mistake in partial fraction computations. Fortunately i t is simple to check the answer by recognizing t hat F (s) a nd its partial fractions must be equal for every value of s if t he p artial fractions are correct. Let us verify this assertion in Eq. (6.25a) for some convenient value, say s = 1. S ubstitution o f s = 1 in Eq. (6.25a) yieldst o tt) 1 2 u (t) 3 t nu(t) 1 6 We can now be sure of our answer with a high margin of confidence. Using P air 5 (Table 6.1) in Eq. (6.25a), we o btain t u(t) 4 5 n! s n+l f (t) 3 =.c- 1 (_4_ + _3_) = (4e8 +2 2t s -3 + 3 e3t) u (t) (6.25b) ( b) 1 e Atu(t) 4 6 =3-2 s 2 2 28 +5 = 28 +5 8 + 38 + 2 ( 8 + 1)(8 + 2) Observe t hat F (s) is an improper function with m = n. In such a case we c an express F (s) as a sum o f t he coefficient bn ( the coefficient of t he highest power in t he n umerator) plus partial fractions corresponding to t he poles of F(8) (see Sec. B.5-4). In t he p resent case bn = 2. T herefore F (s)= s -).. 2 6 7 t neAtu(t) 8a 1 (s - )..)2 t eAtu(t) cos b tu(t) n! k1 S S2 where 8b 9a s in b tu(t) 2 S2+5 + 2) s +a e - at c os b tu(t) a nd 28 k 2= 9b e - at s in b tu(t) l Oa r e- at c os (bt + Ii) u (t) ( rcos Ii)s lOb r e- at c os (bt + Ii) u (t) O.5re)9 + b2 (s + a)2 s +a- + ( arcos Ii - jb 2 +5 ( s+I)111 8 +2 I = 8 =-1 I 8 =-2 ~ -1 + 2 =7 8+5 = - 2+1 = -13 Therefore + 2 as + 82 k + -2 - F (s) = 2 + - s+1 + b2 (a 2 + 7 brsin Ii) b2 ) F (s) = 2 + - s+1 From Table 6.1, Pairs 1 a nd 5, we o btain O .5re- j9 f (t) = 2o(t) + 8 + a + jb + (7e- t - 1 - -3 - s +2 13e- 2t ) u (t) (6.26) (e) l Oc r e- at cos (bt _j T- A s+B + Ii) u (t) A 'ctB'-2ABa , c a2 8 Ii = t an-1 2 + 2as + c A a-B A~ b=~ lOd [A cos bt + B~ Aa sin bt] u(t) b = vc - a e - at 2 A s+B 82 + 2as + c + 34) + lOs + 34) 6 (s + 34) s (s + 5 - j3)(s + 5 + j 3) 6 (s F (s) = s(s2 k1 =- s + k2 s +5- k 2· j3 + ---::----:::s + 5 + j3 t Because F (s) = 0 0 a t i ts poles, we should avoid t he p ole values ( -2 a nd 3 in t he p resent case) for checking. I t is p ossible t hat t he a nswers may check e ven if p artial f ractions a re wrong. T his s ituation c an o ccur w hen two o r m ore errors cancel t heir effects. B ut t he c hances o f t his p roblem arising for r andomly s elected values o f s a re e xtremely small. 374 6 Continu~us-Time S ystem A nalysis Using t he L aplace T ransform N ote t hat t he c oefficients (k2 a nd k2 *) o fthe c onjugate t erms m ust a lso b e c onjugate (see Sec. B .5). Now kJ = 6(8 + 34) 1 (8 2 + 108 + 34) I _ 6 x 34 - _C-..-_~I + 5 + j 3) 34 . =0 - 6 - 6.1 T he L aplace T ransform Alternative Method Using Quadratic Factors T he above procedure involves considerable manipulation of complex numbers. As indicated by P air 10c (Table 6.1), t he inverse transform of quadratic terms (with complex conjugate poles) can be found directly without having t o find first-order partial fractions. We now show t he a lternative meth...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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