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

B5 4 in t he p resent case bn 2 t herefore f s s

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Unformatted text preview: e r emaining coefficients. Using t he values k l = 2 a nd aD = 6, d etermined earlier by t he Heaviside m ethod, we h ave 88 + 10 + 1)[ 88 + 10 2 6 al a2 (8 + 1)(8 + 2)3 = 8 + 1 + (8 + 2)3 + (8 + 2)2 + 8 + 2 ~ {:822 [~]} 8=-2 = - 2 T herefore 2 6 2 T here a re two unknowns, a l a nd a2. I f we mUltiply b oth sides by we e liminate a l. T his p rocedure yields o = 2 + a2 2 F(8) = 8 + 1 + (s + 2)3 - (s + 2)2 - 8 + 2 T herefore a nd j (t) = [2e- + (3t t 2 + 2al + 4a2 Another Alternative: A Hybrid o f Heaviside and Short-Cuts . =-1 8s + 10 = '7(8-+--::-:1)--- al = { + 8a2 S ubstitution o f a l = a2 = - 2, o btained e arlier, satisfies these equations. T his s tep a ssures t he c orrectness o f o ur answers. w here (see Sec. B.5-2) aD = -2 We c an s top h ere if we wish, since t he t wo desired coefficients a l a nd a2 have already been found. However, e quating t he coefficients of 8 1 a nd 8 0 serves as a check o n o ur answers. T his s tep yields ( d) 1 (8 + 2)3 ~ a2 0 = 12 + a l + 5a2 = 2 + a l ~ a l = - 2 B = - 54 a d eduction w hich agrees with t he r esults we f ound earlier. 8 8+10 o n b oth sides, we o btain - 2t - 2)e- 2t u(t) I n t his m ethod, t he s impler coefficients k l a nd aD a re d etermined by t he Heaviside "cover-up" p rocedure, as discussed earlier. To d etermine t he remaining coefficients, we use t he c learing-fraction m ethod (see Sec. B.5-3). Using t he values k l = 2 a nd a D = 6 o btained e arlier by t he Heaviside "cover-up" m ethod, we have 8 8+10 = _ 2_+ _ _ _+_a_I_+~ 6 ( s+I)(8+2)3 8 +1 ( 8+2)3 ( s+2)2 8 +2 tWe could have cleared fractions without finding k l and aD. This alternative, however, proves more laborious because it increases the number of unknowns to 4. By predetermining k l and ao, we reduce the unknowns t o 2. Moreover, this method provides a convenient check on t he solution. This hybrid procedure achieves the best of both methods. 8 - + 0 0, a2 = - 2 T here is now only one unknown, a l. T his value c an b e d etermined r eadily b y s etting e qual t o a ny c onvenient value, say 8 = O. T his s tep yields o 8 C omputer E xample C 6.1 F ind t he inverse laplace t ransform o f t he following functions using p artial f raction expansion method: We now clear fractions by multiplying b oth sides o f t he e quation by (s + 1)(s + 2)3. T his p rocedure y ieldst 88 + 10 = 2 (s + 2)3 + 6(s + 1) + a l(8 + 1)(8 + 2) + a2(8 + 1)(8 + 2)2 3 = ( 2 + a2)8 + (12 + a l + 5a2)s2 + (30 + 3 al + 8a2)8 + (22 + 2al + 4a2) a nd t hen let 88 + 10 2 6 al 2 (8 + 1)(8 + 2)3 = 8 + 1 + (8 + 2)3 + (s + 2)2 - 8 + 2 (6.28) Alternative Method: A Hybrid o f Heaviside and Clearing Fractions ~ 8 2 28 + 5 88 2 + 218 + 19 b 28 2 + 78 + 4 (a) 82 + 38 + 2 ( ) (8 + 1)(8 + 2)2 (c) (8 + 2)(8 2 + 8 + 7) ( a) n um=[2 0 5]; d en=[l 3 2]; [ r,p,k]=residue(num,den) r - -13, 7 P - 2,-1 k•2 6 Continuous-Time System Analysis Using t he Laplace Transform 378 t!. Thus, F (s) = - 13 s +2 + _ 7_ and s +1 f (t) = ( _13e- 2t + 7 e- t )u(t) + 26(t) ( b) n...
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