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Setting s
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0 on both sides of the above expression, we have
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from which we obtain A ,
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3.20. Find the inverse Laplace transform of the following X(s): 2s+ 1 ( a ) X(s) = - Re(s) > -2 , s+2
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X(S)=
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2s + 1 2(s 2) - 3 3 =2- X ( s ) = -s +2 s 2 s+2 Since the ROC of X(s) is Re(s) > -2, x ( t ) is a right-sided signal and from Table 3-1 we obtain
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s3 + 2s' + 6 , Re(s) > 0 s z + 3s
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( b ) Performing long division, we have
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where
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Hence,
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The ROC of X(s) is Re(s) > - 1. Thus, x ( r ) is a right-sided signal and from Table 3-1 we obtain
Proceeding similarly, we obtain
where
Hence,
The ROC of X(s) is Re(s) > 0.Thus, x(t) is a right-sided signal and from Table 3-1 and Eq. (3.78) we obtain
Note that all X(s) in this problem are improper fractions and that x(t) contains S(t) or its derivatives.
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
3.21. Find the inverse Laplace transform of 2 + 2se-" X(s) =
We see that X ( s ) is a sum
+ 4eP4' s2+ 4 s + 3
Re(s) > - 1
where X I ( $ )=
2 + 4s
4 XAs)
X2(s) = st
+ 4s + 3
s 2 + 4s + 3
x l ( f) - X d s ) x2(t) # X Z ( S )
~ 3 ( f- X 3 ( s )
then by the linearity property (3.15) and the time-shifting property (3.16) we obtain ~ ( t= )x l ( t ) + x 2 ( t - 2 ) + x 3 ( t - 4 ) Next, using partial-fraction expansions and from Table 3-1, we obtain (3.85)
3.22. Using the differentiation in s property (3.211, find the inverse Laplace transform of
We have
and from Eq. (3.9) we have e-"u(t)
Re(s) > -a
Thus, using the differentiation in s property (3.21), we obtain
X ( I ) = te-"'u(t)
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
SYSTEM FUNCTION
3.23. Find the system function H(s) the impulse response h ( t ) of the RC circuit in Fig. and 1-32 (Prob. 1.32).
( a ) Let
In this case, the RC circuit is described by [Eq. (1.105)]
Taking the Laplace transform of the above equation, we obtain
Hence, by Eq. (3.37) the system function H(s) is
Since the system is causal, taking the inverse Laplace transform of H(s), the impulse response h(t ) is
( b ) Let
In this case, the RC circuit is described by [Eq. (1.107)l
Taking the Laplace transform of the above equation, we have
( s + &)Y(S)
1 Rsx(s)
Hence, the system function H(s) is
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
In this case, the system function H(s) is an improper fraction and can be rewritten as
Since the system is causal, taking the inverse Laplace transform of H(s), the impulse response h(t ) is
Note that we obtained different system functions depending on the different sets of input and output.
3.24. Using the Laplace transform, redo Prob. 2.5.
From Prob. 2.5 we have h(t) Using Table 3-1, we have H(s) X(s) Thus,
= e-"'u(t)
~ ( t =)e a ' u ( - t )
Re(s) > -a Re(s) < a
-s-a
and from Table 3-1 (or Prob. 3.6) the output is
which is the same as Eq. (2.67).
3.25. The output y ( t ) of a continuous-time LTI system is found to be 2e-3'u(t) when the input x ( t is u(t ).
(a) Find the impulse response h ( t ) of the system. (6) Find the output y ( t ) when the input x ( t ) is e-'u(f).
x(f) = u(t), y(t) = 2e-3'u(t) Taking the Laplace transforms of x(t) and we obtain
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Hence, the system function H(s) is
Rewriting H(s) as
and taking the inverse Laplace transform of H(s), we have
Note that h ( t ) is equal to the derivative of 2 e - " d l ) which is the step response s(r) of the system [see Eq. (2.1311.
x(t) =e-'dt)
Re(s)> - 1
Thus,
Using partial-fraction expansions, we get
Taking the inverse Laplace transform of Y(s), we obtain
y(t)
(-e-'
+ 3e-")u(r)
3.26. If a continuous-time LTI system is B I B O stable, then show that the ROC of its system function H ( s ) must contain the imaginary axis, that is, s = jo.
A continuous-time LTI system is BIBO stable if and only if its impulse response h ( t ) is absolutely integrable, that is [Eq. (2.2111,
By Eq. (3.3)
Let s = jw. Then
Therefore, we see that if the system is stable, then H(s) converges for s = jo. That is, for a stable continuous-time LTI system, the ROC of H(s) must contain the imaginary axis s = j w .
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
3 2 . Using the Laplace transfer, redo Prob. 2.14. .7
( a ) Using Eqs. ( 3 . 3 6 ) and ( 3 . 4 0 , we have Y ( s )=X(s)H,(s)H,(s) = X ( s ) H ( s )
where H ( s ) = H , ( s ) H , ( s ) is the system function of the overall system. Now from Table 3-1 we have
Hence,
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