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Unformatted text preview: A Figure 59.21 (c) The total response is the sum of the zero—state and zeroinput responses. This is
y(t) = 26—‘u(t) — e’2tu(t). 9.21. The pole zero plots for all the subparts are shown in Figure 89.21.
(a) The Laplace transform of x(t) is 00
X(s) = / (e‘2t+e'3t)e_”dt
o = [—eWﬂ/(s + 2m? + [—e“3+3)‘/(s + 3>II3° l 1 2s+5 s+2+s+3=82+5s+6 The region of convergence (ROC) is ’Re{s} > —2. (b) Using an approach similar to that shown in part (a), we have
e'4tu(t) (—> 3—, Re{s} > —4. Also, and . 1
e—5te—15tu(t) (5, s + 5 +175, Re{s} > —5.
Hour this we obtain
_ . 1 _ ‘ . L 5
5t __ 5t 51 —5t — 5t ____
e Sln(5t)U(t) — 517p: 6" ‘6 ‘3 J Mt) ‘ a (5+5)2 +25’ where Re{s} > —5. Therefore, 2
_4t _5t . L .5 +155 + 70
t ——————— R —5.
e u( )+e sm(5t)u(t)(—) s3+1432+903+100’ e{s}>
(c) The Laplace transform of :1;(t) is
0
X(s) = / (em + €3t)e_’tdt
—oo
= he‘s—”Vs ~ 2)]:‘100 + [—e—<s3>‘/(s — 3)] ‘loo
1 1 25 —— 5
= + : ____._._
5—2 5—3 32—55+6
The region of convergence (R00) is 'Re{s} < 2.
((1) Using an approach along the lines of part (a), we obtain
*2: E 1
—— 'R —2. 8921—1
6 u(t)(—> 5+2, e{s}> ( )
Using an approach along the lines of part (c), we obtain
e2‘u(t) A s i 2, 1245} < 2. (5921—2)
Horn these we obtain
2
e—2ltl = e‘2tu(t) + e2‘u(—t) (i; 52 j 4, —2 < ’Re{s} < 2.
Using the differentiation in the s—domain property, we obtain
d 2.5 252 + 8
t‘zltl 5 —— =——— —2 R . 2.
e (———> (18 32—4 (32—4)2’ < e{e}<
(e) Using the differentiation in the sdomain property on eq. (39.211), we get
(1 l 1
r” t ‘— = , ’R >~2.
e u( ) H d3 [8 +2] (3 + 2)2 e{s}
Using the differentiation in the s—domain property on eq. (89.212), we get
_ _ _ = — 2.
te u( t) (—) d3 [5 _2] (3—2)2, Re{s} <
Therefore,
1; —4s Itle‘ml = te‘2tu(t) + —teztu(—t) (—2 —2 < Re{s} < 2. (s + 2)2(s — 2)2 ’ 9.22. (a) From Table 9.2, we have :t;(t) = ésin(3t)u(t). (b) From Table 9.2 we know that L s
t t +—+ 0.
cos(3 )u( ) 52 + 9, ’Re{s} >
Using the time scaling property, we obtain
1; s
cos(3t)u(—t) +—% _32 + 9, Re{s} < 0. Therefore, the inverse Laplace transform of X (s) is
:r,(t) = —cos(3t)u(——t).
(c) From Table 9.2 we know that — 1
Ct cos(3t)u(t) (—1.1) m, R€{S} > 1.
Using the time scaling property, we obtain
_t c s + 1
e cos(3t)u(—t) 4—) —m, ’Re{s} < —1. Therefore, the inverse Laplace transform of X (s) is
z(t) = —e_‘cos(3t)u(—t). ((1) Using partial fraction expansion on X (s), we obtain 2 1
X(s)—3+4s+3' Horn the given ROC, we know that 2(t) must be a twosided signal. Therefore,
(C(t) = 2e’4tu(t) + e‘3tu(—t). (8) Using partial fraction expansion on X (s), we obtain 2 l
X(s)—3+3_s+2' From the given ROC, we know that :1:(t) must be a twosided signal. Therefore, a:(t) = 2e‘3tu(t) + e‘2tu(—t). (f) We may rewrite X(s) as X“) = ”ST—3%?
_ 1+ 3.5
' (s — 1/2)2 + («i/2)?
= 1+3 5—1/2 3/2 (s — 1/2)? + («i/2)? + (s — 1/2)? + («i/2V Using Table 9.2, we obtain
$(t) = 6(t) + 3et/2 cos(\/§t/2)u(t) + «ﬁat/2 sin(\/§t/2)u(t). 334 (g) We may rewrite X(s) as 35
X = 1 — .
(s) (s + 1)2
From Table 9.2, we know that
1; l
tu(t) <——> 3—2, Re{s} > 0.
Using the shifting property, we obtain
—t C 1
e tu(t) 6—) (s + D2, ’Re{s} > —1. Using the differentiation property, di:[e'tu(t)] = etuu) — te‘um «i» 8
(s+1)2’ Re{s} > —1. Therefore,
:1:(t) = 6(t) — 3e“tu(t) — 3te‘tu(t). 9.23. The four polezero plots shown may have the following possible ROCs: oPlot (a): Re{s} < —2 or —2 < Re{s} < 2 or Re{s} > 2.
oPlot (b): 'Re{s} < —2 or ’Re{s} > —2. oPlot (c): Re{s} < 2 or ’Re{s} > 2. oPlot (d): Entire splane. Also, suppose that the signal z(t) has a Laplace transform X (s) with R00 R.
(1) We know from Table 9.1 that e‘3ta;(t) ti) X(s + 3). The ROC R1 of this new Laplace transform is R shifted by 3 to the left. If .7:(t)e~3t is
absolutely integrable, then R1 must include the jw axis. oFor plot (a), this is possible only if R was Re{s} > 2. oFor plot (b), this is possible only if R was Re{s} > —2. oFor plot (c), this is possible only if R was Re{s} > 2. oFor plot (d), R is the entire splane. (2) We know from Table 9.2 that 1
3+1 e‘tu(t) +5) , ’Re{s} > —1. Also, from Table 9.1 we obtain
—t L X(s)
:1:(t) * [e u(t)] (~—> 3 +1, If e“u(t) * 22(t) is absolutely integrable, then R2 must include the jwaxis. R2 = Rn [Re{s} > —1]. oFor plot (a), this is possible only if R was —2 < ’Re{s} < 2.
oFor plot (b), this is possible only if R was Re{s} > —2.
oFor plot (c), this is possible only if R was 'Re{s} < 2.
oFor plot (d), R is the entire s—plane.
(3) If :c(t) = 0 for t > 1, then the signal is a left—sided signal or a ﬁniteduration signal.
oFor plot (a), this is possible only if R was Re{s} < —2.
oFor plot. (b), this is possible only if R was Re{s} < ~2.
0For plot (c), this is possible only if R was Re{s} < 2.
oFor plot (d), R is the entire s—plane.
(4) If $(t) = 0 for t < —1, then the signal is a right—sided signal or a ﬁnite—duration signal
oFor plot (a), this is possible only if R was Re{s} > 2.
oFor plot (b), this is possible only if R was Re{s} > —2.
0For plot (c), this is possible only if R was Re{s} > 2.
0For plot (d), R is the entire s—plane. (0‘ Cb) "Mb
N
“Al““4..—
_No (A. w 4. N. N
(C) (d)
(e) (1})
I430) “'10, wihn
with w
w
—NO 040 £0 9.27. From clues 1 and 2, we know that X (s) is of the form X‘3)=rs:2><T+T> is —1 + j. Since (C(t) is real, the Furthermore, we are given that one of the poles of X (s)
a:1—jandb=1+) poles of X (3) must occur in conjugate reciprocal pairs. Therefore,
and A
(s) (s+l—])(s+1+j) From clue 5, we know that X (0) = 8. Therefore, we may deduce that A = 16 and . From the pole locations we know that there are two Let R denote the ROC of X (s)
—1 or Re{s} > 1. We will now use clue possible choices of R. R may either be Re{s} <
4 to pick one. Note that y(t) = 6mm“) (—5 Y(s) = X(s — 2). The ROC of Y(s) is R shifted by 2 t0 the right. Since it is given that y(t) is not absolutely
integrable, the ROC of Y(s) should not include the jwaxis. This is possible only of R is Re{s} > ——1. ...
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 Fall '08
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