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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online