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Unformatted text preview: od (also discussed in Sec. B.52), 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) "coverup" 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 tanJ(~3) 1= tanJ(~). 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 ejJ26.9°
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+B22ABa = 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. ShortCuts
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 righthand 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 ontinuousTime 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
"coverup" p rocedure, as discussed earlier. T he u sual s hortcuts 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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