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

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

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: 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) "cover-up" 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"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 "cover-up" p rocedure, as discussed earlier. T he u sual s hort-cuts a re t hen u sed t o d etermine t h...
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