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

# 63 answer a l e 2s for all s b l e 2s e

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Unformatted text preview: od (also discussed in Sec. B.5-2), which uses this observation. For this purpose we shall express F (s) as F (s) = = 2 9+j3 = - 3+j4 6 (s + 34) + lOs + 34) =~+ S (S2 - 3 - j5 . =-5+j3 375 S As + B s2 + lOs + 34 We have already determined t hat k l = 6 by t he (Heaviside) &quot;cover-up&quot; method. Therefore T herefore 6 (s + 34) 6 As + B S(s2 + lOs + 34) = :; + 82 + lOs + 34 k2* = - 3 - j4 T o u se P air l Ob ( Table 6 .1), we n eed t o e xpress k2 a nd k2* i n p olar f orm. -3 + j4 = ( \1&quot;3 2 + 42 ) Clearing t he fractions by multiplying b oth sides by s(s2 + lOs + 34) yields e j t an-'(4/-3) = 5e jtan -'(4/-3) 6 (s + 34) O bserve t hat tan-J(~3) 1= tan-J(-~). T his f act is e vident i n F ig. 6.4. R emember t hat e lectronic c alculators c an give answers o nly for t he a ngles i n t he f irst a nd t he f ourth q uadrant. F or t his r eason i t is i mportant t o p lot t he p oint (e.g. - 3 + j 4) i n t he c omplex p lane, a s d epicted i n Fig. 6.4, a nd d etermine t he a ngle. F or f urther d iscussion o f t his t opic, see E xample B .l. F rom F ig. 6 .4, we observe t hat k2 = -3 + j4 = 5 ejJ26.9° = 6 (s2 + 108 + 34) = (6 + + 8(A8+B) A )S2 + (60 + B )s + 204 Now, e quating t he coefficients of 82 a nd s on b oth sides yields o = (6 + A ) 6 = 60 + B and 6 =- A = - 6 =- B = -54 - 68 - 54 F (8) = :; + 82 + 108 + 34 s o t hat We now use Pairs 2 a nd lOc t o find t he inverse Laplace transform. The parameters for P air 10c are A = - 6, B = - 54, a = 5, c = 34, a nd b = vic - a 2 = 3, a nd T herefore 6 F (8) 5 ejJ26.9° 5 e-jJ26.9° r = -; + 8 + 5 - j3 + 8 + 5 + j 3 F rom T able 6 .1 ( Pairs 2 a nd l Ob), we o btain = V A 2 c+B2-2ABa = 10 , . / c -a 2 Therefore j (t) = [6 + 1 0e- 5t cos (3t + 126.9°)] u (t) f (t) (6.27) = [6 + B = t an- 1 ~ A V c -a 2 = 126.9° lOe- 5t cos (3t + 126.9°)]u(t) which agrees with t he previous result. Short-Cuts T he p artial fractions with quadratic terms also can be obtained by using short cuts. We have F (8) = 6 (8+34) =~+ A s+B S (8 2 + lOs + 34) 8 82 + 108 + 34 We can determine A by eliminating B o n the right-hand side. This step can b e accomplished by multiplying b oth sides of t he above equation by s a nd t hen l etting s - + 0 0. This procedure yields 0 =6+A 3 -j4 F ig. 6 .4 t an- 1 4 ( -3 ) , Han- (~). 1 Therefore =- A=-6 6 (s + 34) = ~+ - 68 + B 8(8 2 +108+34) 8 8 2 +10s+34 To find B , we let 8 take on any convenient value, say 8 = 1, in this equation to obtain 376 6 C ontinuous-Time S ystem A nalysis U sing t he L aplace T ransform 210 _ 6 B - 6 45 - + 45 M ultiplying b oth sides of t his e quation by 45 yields 210 = 270 + B - 6 ~ 6.1 T he L aplace T ransform E quating coefficients o f 8 3 a nd 377 8 2 o = (2 + a2) F8 _ 8 8+10 ( )- (8+1)(8+2)3 8 = 30 + 3 al = ~ + _ _ _o_ + _ _ _l_ + ~ a a 8 +1 ( 8+2)3 ( S+2)2 8 +2 10 = 22 I =2 d[ a2 = ds (8 I n t his m ethod, t he s impler coefficients k l a nd aD a re d etermined b y t he Heaviside &quot;cover-up&quot; p rocedure, as discussed earlier. T he u sual s hort-cuts a re t hen u sed t o d etermine t h...
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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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