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with the same ROC as for X ( s ) itself. Hence, where R and R' are the ROCs before and after the time-shift operation.
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Verify the time-scaling property (3.181, that is,
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= a t with a > 0, we have I w ( x a )= x(r)e-('/")'dr = - X a -, a
RP=aR
Note that because of the scaling s / a in the transform, the ROC of X ( s / a ) is aR. With a < 0, we have
CHAP. 31
LAF'LACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Thus, combining the two results for a > 0 and a < 0, we can write these relationships as
Find the Laplace transform and the associated ROC for each of the following signals:
( a ) x ( t ) = S(t - t o ) ( b ) x ( t ) = u(t - to)
( c ) ~ ( t=)e - " [ u ( t ) - u ( t - 5 ) ]
(dl x(t)=
S(t - k T )
( e ) x ( t ) = S(at
+ b ) , a , b real constants
S ( I - I,,)
( a ) Using Eqs. (3.13) and (3.161, we obtain e-s'fl
all s
( b ) Using Eqs. (3.14) and (3.16), we obtain
Rewriting x ( l ) as
Then, from Table 3-1 and using Eq. (3.161, we obtain
Using Eqs. (3.71) and (1.99), we obtain
~ ( s= ) C e-.~'T=
k -0
C (e-sT)li = 1 - 1 s T e
Re(s) > 0
(3.73)
( e ) Let
f ( 0= s ( a t )
Then from Eqs. (3.13) and (3.18)we have
f ( t ) = S(a1)
F(s)= la l
all s
Now Using Eqs. (3.16) and (3.741, we obtain
1 X ( s ) = e s b / a ~ (= ) S -esh/"
la l
all s
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
3.10. Verify the time differentiation property (3.20),that is,
From Eq. (3.24) the inverse Laplace transform is given by
Differentiating both sides of the above expression with respect to t, we obtain
Comparing Eq. ( 3 . 7 7 ) with Eq. (3.76), we conclude that h ( t ) / d t is the inverse Laplace transform of sX(s). Thus,
R'3R
Note that the associated ROC is unchanged unless a pole-zero cancellation exists at s = 0.
3.11. Verify the differentiation in s property (3.21), that is,
From definition (3.3)
Ij I
Differentiating both sides of the above expression with respect to s , we have
Thus, we conclude that
3.12. Verify the integration property (3.22), that is,
Then
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Applying the differentiation property (3.20), we obtain X(s) =sF(s) Thus,
The form of the ROC R' follows from the possible introduction of an additional pole at s = 0 by the multiplying by l/s. Using the various Laplace transform properties, derive the Laplace transforms of the following signals from the Laplace transform of u(t).
( a ) 6(t)
tu(t) (e) te-"'u(t) (g) e-"'cos w,tu(t)
( b ) 6'(t) ( d ) e-"'u(t)
(f) coso,tu(t)
From Eq. (3.14)we have
1 u(t) H S
for Re( s) > 0
From Eq. ( 1.30) we have
Thus, using the time-differentiation property (3.20), we obtain S(t) H S 1
all s
( b ) Again applying the time-differentiation property (3.20) to the result from part (a), we obtain
a'(!)
( 3.78)
Using the differentiation in s property (3.211, we obtain
Using the shifting in the s-domain property (3.17), we have e-a'u(t)
Re(s) > -a
From the result from part (c) and using the differentiation in s property (3.21), we obtain
( f ) From Euler's formula we can write
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
Using the linearity property (3.15) and the shifting in the s-domain property (3.17), we obtain
Applying the shifting in the s-domain property (3.17) to the result from part (f),we obtain
3.14. Verify the convolution property (3.23), that is,
y(t) =x,(t) * x2(t) = Then, by definition (3.3)
x,(r)x2(t - r ) d r
Noting that the bracketed term in the last expression is the Laplace transform of the shifted signal x2(t - 71, by Eq. (3.16) we have
with an ROC that contains the intersection of the ROC of X,(s) and X2(s). If a zero of one transform cancels a pole of the other, the ROC of Y(s) may be larger. Thus, we conclude that
3.15. Using the convolution property (3.23), verify Eq. (3.22), that is,
We can write [Eq. (2.601, Prob. 2.21
From Eq. (3.14)
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
and thus, from the convolution property (3.23) we obtain
x ( t ) * ~ ( t ) -X(S)
with the ROC that includes the intersection of the ROC of X(s) and the ROC of the Laplace transform of u( t 1. Thus,
INVERSE LAPLACE TRANSFORM 3.16. Find the inverse Laplace transform of the following X(s):
( a ) X ( s >=
-, Re(s) >
s+l 1
(b) X ( s ) = - Re(s) < - 1 , s+l
( c ) X(s) = - Re(s) > 0 , s2+4 s+l (dl X(s) = , Re(s) > - 1 ( s + 1)'+4
From Table 3-1 we obtain
( b ) From Table 3-1 we obtain
~ ( t = -e-'u(-t) )
From Table 3-1 we obtain
x(t)
( d ) From Table 3-1 we obtain
= cos2tu(t)
~ ( t = e-'cos2tu(t) )
3.17. Find the inverse Laplace transform of the following X(s):
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
Expanding by partial fractions, we have
Using Eq. (3.30), we obtain
Hence,
(a) The ROC of X(s) is Re(s) > - 1. Thus, x(t) is a right-sided signal and from Table 3-1 we obtain
x(t)
= eP'u(t)
+ e - 3 ' ~ ( t ) (e-' + e - 3 ' ) ~ ( t ) =
( b ) The ROC of X(s) is Re(s) < -3. Thus, x(t) is a left-sided signal and from Table 3-1 we obtain
x(t)
-e-'u(-t)
- eC3'u(
- ( e - ' +e-3')u( -1)
(c) The ROC of X(s) is - 3 < Re(s) < - 1. Thus, x(t) is a double-sided signal and from Table 3-1 we obtain
3.18. Find the inverse Laplace transform of
We can write
Then
where
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Thus,
The ROC of X(s) is Re(s) > 0.Thus, x ( t ) is a right-sided signal and from Table 3-1 we obtain
into the above expression, after simple computations we obtain
Alternate Solution: We can write X(s) as
As before, by Eq. (3.30) we obtain
Then we have
Thus,
Then from Table 3-1 we obtain
3.19. Find the inverse Laplace transform of
We see that X(s) has one simple pole at s = - 3 and one multiple pole at s multiplicity 2. Then by Eqs. (3.29) and (3.31)we have
-5 with
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
By Eqs. (3.30) and (3.32) we have
Hence,
The ROC of X(s) is R d s ) > -3. Thus, x(t) is a right-sided signal and from Table 3-1 we obtain
Note that there is a simpler way of finding A , without resorting to differentiation. This is shown as follows: First find c , and A, according to the regular procedure. Then substituting the values of c , and A, into Eq. (3.84), we obtain
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