Chapter 9
9.2 Solution:
(a). According to eq.(9.3), we will get
X(s)??
???x(t)e?stdt??????e?5tu(t?1)e?stdt??e?(s?5)tdt
1?
e?(s?5)t?e?(s?5)e?(s?5)?? ?
?(s?5)1?(s?5)(s?5)Re{s}>-5
?5tROC:
(b). ? g(t)?AeLT???G(s)?u(?t?t)
0?A(s?5)t0e, Re{s}<-5 s?5? If G(s)?X(s)
then it’s obviously that A=-1, t0??1, Re{s}<-5.
9.5 Solution: (a). 1, 1
?
112s?4?? s?1s?3(s?1)(s?3)? it has a zero in the finite s-plane, that is s??2
And ? because the order of the denominator exceeds the order of the numerator by 1
? X(s) has 1 zero at infinity.
s?1s?11?? 2s?1(s?1)(s?1)s?1 (b). 0, 1
?
? it has no zero in the finite s-plane.
And ? because the order of the denominator exceeds the order of the numerator by 1 ? X(s) has 1 zero at infinity.
(c). 1, 0
s3?1(s?1)(s2?s?1)??s?1 ? 22s?s?1s?s?1? it has a zero in the finite s-plane, that is s?1
And ? because the order of the denominator equals to the order of the numerator ? X(s) has no zero at infinity.
9.7 Solution: ? There are 4 poles in the expression, but only 3 of them have different real part. ? The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across. ? There are 4 signals having the same Laplace transform expression.
9.8 Solution:
? ?
g(t)?e2tx(t)
G(s)?X(s?2) ROC: R(x)+Re{2}
And x(t) have three possible ROC strips: (??,?3),(?3,?1),(?1,??)
? g(t) have three possible ROC strips: (??,?1),(?1,1),(1,??)
IF G(jw)?G(s)|s?jw
Then the ROC of G(s) is (-1,1)
? x(t) is two sides.
9.9 Solution:
It is obtained from the partial-fractional expansion:
X(s)?2(s?2)2(s?2)4?2???,Re{s}??3
s2?7s?12(s?4)(s?3)s?4s?3We can get the inverse Laplace transform from given formula and linear property. x(t)?4e?4tu(t)?2e?3tu(t) 9.10 Solution: (a). H1(s)?
1,.........Re{s}??1
(s?1)(s?3)It’s lowpass.
(b).H2(s)?
s1,.........Re{s}??
2s2?s?1It’s bandpass.
s2,..........Re{s}??1 (c). H3(s)?2s?2s?1 It’s highpass. 9.13 Solution:
? g(t)?x(t)??x(?t),and x(t)??e?tu(t)
The Laplace transform : G(s)?X(s)??X(?s) and X(s)? From the scale property of Laplace transform, X(?s)??s?1,Re{s}??1
,Re{s}?1 ?s?1????(1??)s??(1??)??So G(s)?X(s)??X(?s)?,?1?Re{s}?1 2s?1?s?1s?1sFrom given G(s)?2,?1?Re{s}?1
s?11We can determine : ???1,??
29.21 Solution: (a). ?
?x(t)?e?2tu(t)?e?3tu(t)
X(s)?112s?5??, Re{s}??2 s?2s?3(s?2)(s?3)?
(b). ? (i). ?
x(t)?e?4tu(t)?e?5t(sin5t)u(t)
?
15s2?15s?70, Re{s}??4 X(s)???22s?4(s?5)?5(s?4)(s?5?5j)(s?5?5j)
x(t)??(t)?u(t) X(s)?1??1s??ss
(f). ?
x(t)??(3t)?u(3t)
? X(s)?111s?3???, Re{s}?0 33s/33s
9.22 Solution: (a). ?
X1(s)?11?j/6j/6,Re{s}?0 ???2s?9(s?j3)(s?j3)s?j3s?j3j?j3tj1eu(t)?ej3tu(t)?(sin3t)u(t) 663
?
x1(t)??(b). ?
X2(s)? x2(t)?ss1/21/2,Re{s}?0 ???s2?9(s?j3)(s?j3)s?j3s?j31?j3t1eu(t)?ej3tu(t)?(cos3t)u(t) 22
?
(c) From the property of shifting in the time-domain and (b),we can get
x2(?t)?(cos(?3t))u(?t)? So g(t)??cos(3t)u(?t)??s,Re{s}?0
(?s)2?9s?G(s),Re{s}?0 s2?9 From the property of shifting in the s-domain,we can get X3(s)?s?1Re{s}??1 ?G(s?1,)(s?12)?9and x3(t)?e?tg(t)??e?t(cos3t)u(?t)
9.28. Solution:
(a). All possible ROCs:
(??,?2),(?2,?1),(?1,1),(1,??)
(b). It’s obviouse to see:
(??,?2) unstable & uncausal (?2,?1) (?1,1) (1,??)
unstable & uncausal stable &
uncausal
unstable & causal
9.31. Solution:
d2y(t)dy(t)??2y(t)?x(t) (a). ?
dtdt2
? s2Y(s)?sY(s)?2Y(s)?X(s) ? H(s)?Y(s)111/3?1/3 ?2???X(s)s?s?2(s?2)(s?1)s?2s?1
(b). 1. The system is stable. ? ROC: (-1,2)
2t1?t ? h(t)??13eu(?t)?3eu(t)2. The system is causal.
? ROC: (2,??)
2t1?t ? h(t)?1eu(t)?33eu(t)3. The system is neither stable nor causal
? ROC: (??,?1)
2t1?t ? h(t)??1eu(?t)?33eu(?t)9.32. Solution:
from (1)
? x(t)?e2t, for all t
and x(t) is a eigen function
1? y(t)?H(s)|s?2?e2t?e2t
61? H(s)|s?2?
6from (2)
?
dh(t)?2h(t)?(e?4t)u(t)?bu(t) dt1b? ? sH(s)?2h(s)?s?4s? H(s)?
s?b(s?4)
s(s?4)(s?2) when s?2, h(2)?
2?6b1?
2?6?46? 2?6b?8,b?1