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LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
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Re(s) < a
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A finite-duration signal x ( t ) is defined as
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t,I t I t ,
otherwise
where I , and I, are finite values. Show that if X ( s ) converges for at least one value of s , then the ROC of X ( s ) is the entire s-plane.
Assume that X(s) converges at s
= a,;then
by Eq. (3.3)
Since (u, 0,) > 0, e-(ul-"~)l is a decaying exponential. Then over the interval where the maximum value of this exponential is e-("l-"o)'l, and we can write
x(t)
+ 0,
Thus, X(s) converges for Re(s) = a, > u,,.By a similar argument, if
a, < u,,
then (3.57)
dl < e
( w ~ - ~ ~ ) l ~
l''l~(r)le-~~'<m dt
and again X(s) converges for Re(s) = u, <u,. Thus, the ROC of X(s) includes the entire s-plane.
~ ( t=) O I t l T
otherwise
Find the Laplace transform of x ( t ) .
By Eq. (3.3)
s + a
[ 1 -e-(s+u)T~
( 3.58)
Since x(f is a finite-duration signal, the ROC of X(s) is the entire s-plane. Note that from Eq. (3.58) it appears that X(s) does not converge at s = -a. But this is not the case. Setting
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
s = -a in the integral in Eq. (3.581, we have
The same result can be obtained b applying L'Hospitalls rule to Eq. (3.58). y
Show that if x ( t ) is a right-sided signal and X(s)converges for some value of s, then the R O C of X ( s ) is of the form
equals the maximum real part of any of the poles of X ( s ) . where amax
Consider a right-sided signal x(t) so that
and X(s) converges for Re(s) = a,. Then
Thus, X(s) converges for Re(s) = a , and the ROC of X(s) is of the form Re($) > a ( , . Since the ROC of X(s) cannot include any poles of X(s), we conclude that it is of the form Re( s > ~ m a x where a , ,equals the maximum real part of any of the poles of X(s). ,,,
Find the Laplace transform X ( s ) and sketch the pole-zero plot with the ROC for the following signals x( t ):
( a ) x ( t )=e-"u(t) +eP3'u(t) ( b ) x ( t ) =e-"u(t) + e2'u(-t) (c) x ( t ) = e 2 ' u ( t )+ e - 3 ' u ( - t )
(a) From Table 3-1
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
Fig. 3-11
We see that the ROCs in Eqs. (3.59) and (3.60) overlap, and thus,
From Eq. (3.61) we see that X(s) has one zero at s = - and two poles at s s = - 3 and that the ROC is Re(s) > - 2, as sketched in Fig. 3-1 l(a). (b) From Table 3-1
- 2 and
We see that the ROCs in Eqs. (3.62) and (3.63) overlap, and thus,
From Eq. (3.64) we see that X(s) has no zeros and two poles at s = 2 and s that the ROC is - 3 < Re(s) < 2, as sketched in Fig. 3-1 l(b). From Table 3-1
- 3 and
Re(s) < - 3 (3.66) s +3 We see that the ROCs in Eqs. (3.65) and (3.66) do not overlap and that there is no common ROC; thus, x(t) has no transform X(s).
e - 3 r ~ -(t )
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Find X(s) and sketch the zero-pole plot and the ROC for a > 0 and a < 0.
The signal x ( t ) is sketched in Figs. 3-12(a) and ( b ) for both a > 0 and a < 0. Since x ( t ) is a two-sided signal, we can express it as
x ( t ) =e-"u(t) +ea'u(-r) (3.67)
Note that x ( t ) is continuous at
t =0
and x(O-) =x(O) = x ( O + ) = 1 . From Table 3-1
earu( - t )
Re(s) < a
(3.69)
Fig. 3-12
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
If a > 0, we see that the ROCs in Eqs. (3.68)and (3.69) overlap, and thus, 1 1 - 2a x ( s )= -- -= - a < Re(s) < a s+a s-a sZ-aZ From Eq. (3.70)we see that X ( s ) has no zeros and two poles at s = a and s = - a and that the ROC is - a < Re(s) < a , as sketched in Fig. 3-12(c).If a < 0, we see that the ROCs in Eqs. (3.68) and (3.69) do not overlap and that there is no common ROC; thus, X ( I ) has no transform X ( s ) . PROPERTIES OF THE LAPLACE TRANSFORM
Verify the time-shifting property (3.161, that is,
By definition ( 3 . 3 )
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