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Taking the inverse Laplace transfer of H ( s ) , we get in VS .NET
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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 CONTINUOUSTIME 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 continuoustime 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 partialfraction expansions, we get
(i) If the system is causal, then h(t) is causal (that is, a rightsided signal) and the ROC of H(s) is Re(s) > 1. Then from Table 31 we get (ii) If the system is stable, then the ROC of H(s) must contain the joaxis. Consequently the ROC of H(s) is  2 < Re(s) < 1. Thus, h(t) is twosided and from Table 31 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 leftsided signal) and from Table 31 we get 331. The feedback interconnection of two causal subsystems with system functions F ( s ) and G ( s ) is depicted in Fig. 313. Find the overall system function H ( s ) for this feedback system. LAPLACE TRANSFORM AND CONTINUOUSTIME LTI SYSTEMS
[CHAP. 3
Fig. 313 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, d41) 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 CONTINUOUSTIME 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) LX(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 righthand 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
e21rl.
LAPLACE TRANSFORM AND CONTINUOUSTIME 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
= e2' a + <R e ( s ) < a  (3.95) X ( t ) = e21'I
for t > 0, which gives
( 2 ) x(t ) = e2' for t < 0. Then x (  t ) = e2' 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.

