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Unformatted text preview: 9.9. Using partial fraction expansion
4 2 3+4 3+3. X(s) = Taking the inverse Laplace transform, 30,) m tie—“um  2e“3‘u(t). 9.29. (3.) Using Table 9.2, we obtain : “R > —1
Ms) 3+1, e{s}
and 1 c
= R > ——2
III(3) 8+2, e{s} (1)) Since y(t) = $05} * Mt), we may use the convolution property to obtain 1 The ROG of Y(s) is 72:2{3} > ~1. (c) Performing partial fraction expansion on Y(s), we obtain 1 1
3+1 3+2' Taking the inverse Laplace transform, we get Y(s) 2 y(t) = e—tu(t) — e‘2tu(t). (d) Explicit convolution of z(t) and h(t) gives us y(t) = fie h('r)x(t —— ﬂair 00 00
2 / e‘gTe"(t"T}u(t—'r)dv
0 H a
(ft/ (Td'r for t > 0
0 [6“ w 6‘2‘}u(t). H (C) Us ‘3 15(4) Figure 89.37 9.40. Taking the unilateral Laplace transform of both sides of the given differential equation, we
get 3339(8) H 3211(0“)  Sy'(0")  ENG") + 63231(3)  63940") —6y(0—) + 11sy{s) — 11y(0‘) + 6343) = “ml {3940—1) (a) For the zero state response, assume that all the initial conditions are zero. Furthermore,
from the given :r(t) we may determine 1
(15(8) 2 m, R€{S} > —4. From eq. (89.401), we get 1 r3 2
. r + +11 +6 3 .
32(9)“; 63 s ] 3+4 Therefore,
1 (s+4)(s3 +632+113+6)' ' Taking the inverse unilateral Laplace transform of the partial fraction expansion of the
above equation, we get y(t) = ée”‘u(t) — éeﬂuuﬁ) + 32(3) = ée‘ztuﬁ) — —e—3tu[t). (b) For the zero—input response, we assume that X(.9) : 0. Assuming that the initial
conditions are as given, we obtain from (89.404) 32+53+6 1 s3+632~§~lls+6 = 3+1‘
Taking the inverse unilateral Laplace transform of the above equation, we get y(t) = e“u(t). (c) The total response is the sum of the zerostate and zeroinput responses. 1 1 1
y(t) = gee—tum — E‘s—“1413 + ifs”um — 56—311136). NS) = ...
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 Spring '06
 V."Ragu"Balakrishnan

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