Taking the inverse Laplace transfer of H ( s ) , we get in VS .NET

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Taking the inverse Laplace transfer of H ( s ) , we get
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h ( t )=2(ep' -e-2')u(t)
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Since the ROC of H ( s ) , Re(s) >
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1, contains the jo-axis, the overall system is stable.
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3 2 . Using the Laplace transform, redo Prob. 2.23. .8
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The system is described by
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Taking the Laplace transform of the above equation, we obtain
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s Y ( s ) + a Y ( s ) =X ( s )
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( s + a ) Y ( s ) =X ( s )
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Hence, the system function H ( s ) is
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Assuming the system is causal and taking the inverse Laplace transform of H ( s ) , the impulse response h ( t ) is
h ( t ) =e-"'u(t)
which is the same as Eq. (2.124).
3 2 . Using the Laplace transform, redo Prob. 2.25. .9
The system is described by
yf(t)+2y(t) = x ( t ) +xl(t)
Taking the Laplace transform of the above equation, we get
sY(s) + 2Y(s) =X ( s )
+s X ( s )
( s + 2 ) Y ( s )= ( s + l ) X ( s )
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Hence, the system function H(s) is
Assuming the system is causal and taking the inverse Laplace transform of H(s), the impulse response h(t ) is
3.30. Consider a continuous-time LTI system for which the input x ( t ) and output y ( t ) are related by
~ " ( 1 +yl(t) - 2y(t) =x(t) )
( a ) Find the system function H(s).
(3.86)
( b ) Determine the impulse response h ( t ) for each of the following three cases: (i)
the system is causal, (ii) the system is stable, (iii) the system is neither causal nor stable.
Taking the Laplace transform of Eq. (3.86),we have s 2 ~ ( s+ sY(s) ) or Hence, the system function H(s) is
- 2Y(s) = X ( s )
( s 2+ s - ~ ) Y ( s ) X ( s ) =
( b ) Using partial-fraction expansions, we get
(i) If the system is causal, then h(t) is causal (that is, a right-sided signal) and the ROC of H(s) is Re(s) > 1. Then from Table 3-1 we get
(ii)
If the system is stable, then the ROC of H(s) must contain the jo-axis. Consequently the ROC of H(s) is - 2 < Re(s) < 1. Thus, h(t) is two-sided and from Table 3-1 we get
(iii) If the system is neither causal nor stable, then the ROC of H(s) is Re(s) < -2. Then h(r) is noncausal (that is, a left-sided signal) and from Table 3-1 we get
331. The feedback interconnection of two causal subsystems with system functions F ( s ) and G ( s ) is depicted in Fig. 3-13. Find the overall system function H ( s ) for this feedback system.
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
Fig. 3-13 Feedback system.
Let Then,
Y ( s ) = E ( s )F ( s ) R(s)= Y(s)G(s) (3.87) (3.88)
Since
e ( t )= x ( t )+ r ( t )
we have
E ( s ) =X ( s ) + R ( s ) (3.89)
Substituting Eq. (3.88) into Eq. (3.89) and then substituting the result into Eq. (3.87), we obtain
Y ( s ) = [ X ( s )+ Y ( s ) G ( s ) l F ( s )
[ l - ~ ( s ) G ( s~ ]( s =) F ( s )X ( s ) )
Thus, the overall system function is
UNILATERAL LAPLACE TRANSFORM
3.32. Verify Eqs. ( 3 . 4 4 ) and ( 3 . 4 5 ) ,that is,
d-41) -H s X I ( s )- x ( O - )
( a ) Using Eq. (3.43)and integrating by parts, we obtain
Thus, we have
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
(b) Applying the above property to signal xt(t) = du(t)/dt, we obtain
Note that Eq. (3.46) can be obtained by continued application of the above procedure.
3.33. Verify Eqs. (3.47) and (3.481, that is, 1
(a) L-X(T) d~
-X,(S)
Let Then Now if
dl)++G,(s)
then by Eq. (3.44) X1(s) =sG1(s) -g(O-) =sGI(s) Thus,
We can write
Note that the first term on the right-hand side is a constant. Thus, taking the unilateral Laplace transform of the above equation and using Eq. (3.47), we get
3.34. ( a ) Show that the bilateral Laplace transform of x ( t ) can be computed from two unilateral Laplace transforms.
(b) Using the result obtained in part ( a ) , find the bilateral Laplace transform of
e-21rl.
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
( a ) The bilateral Laplace transform of x ( t ) defined in Eq. (3.3) can be expressed as
Now Next. let
/ : x ( t ) e p " dl
=X,(s)
R e ( s ) > o+
(3.92)
Then
~ , ~ x ( - ~ ) e = d r~ ( - r ) e - ' ~ ' ~ = d t; ( - s ) " 'X
R e ( s ) < o-
(3.94)
Thus, substituting Eqs. (3.92)and (-3.94)into Eq. (3.91),we obtain
X(s) =X,(s) +X,( - s )
( 1 ) x(t
= e-2'
a + <R e ( s ) < a -
(3.95)
X ( t ) = e-21'I
for t > 0, which gives
( 2 ) x(t ) = e2' for t < 0. Then x ( - t ) = e-2' for t > 0, which gives
Thus,
( 3 ) According to Eq. (3.95),we have
which is equal to Eq. (3.701, with a = 2, in Prob. 3.6.
3.35. Show that
( a ) ~ ( 0 ' )= lim s X , ( s )
(3.97) (3.98)
( b ) lim x ( t ) = lirn s X , ( s )
s -0
Equation ( 3 . 9 7 ) is called the initial value theorem, while Eq. (3.98)is called the final calue theorem for the unilateral Laplace transform.
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