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Thus, this system is nonlinear if y(0) = yo # 0. ( b ) If y(0) = 0, the system is linear. This is shown as follows. Let x J t ) and x,(t) be two input signals and let y ,(t) and y,(r) be the corresponding outputs. That is,
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with the auxiliary conditions
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=~ 2 ( 0= 0 )
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Consider x(r) =a,x,(t) +a,x*(t) where a , and a, are any complex numbers. Multiplying Eq. (2.111) by a , and Eq. (2.112) by a and adding, we see that )'(r) = a I ~ l ( t +azy,(t) ) satisfies the differential equation dy(t) +ay(t) = x ( I ) dr and also, from Eq. (2.113)
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= ~ ( 0 = a,y,(O) + ~ , Y , ( O ) 0 )
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Therefore, y(t) is the output corresponding to x(t), and thus the system is linear.
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CHAP. 2 1
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LINEAR TIME-INVARIANT SYSTEMS
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2.22. Consider the system in Prob. 2.20. Show that the initial rest condition y ( 0 ) = 0 also implies that the system is time-invariant.
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Let y , ( t ) be the response to an input x , ( t ) and
xI(t) =0 t10 (2.114)
Then and
Y do) = 0
(2.116)
Now, let y 2 ( t ) be the response to the shifted input x2(t ) = x ,(t - T ) . From Eq. (2.114) we have
x2(t)= 0 t s r (2.117)
Then y , (
must satisfy
and Now, from Eq. (2.115) we have
~ ~ ( = 01 7
If we let y 2 ( t ) = y l ( t - T), then by Eq. (2.116) we have
yz(7) = Y , ( T- 7 ) = ~ 1 ( 0= 0 )
Thus, Eqs. (2.118) and (2.119) are satisfied and we conclude that the system is time-invariant.
2.23. Consider the system in Prob. 2.20. Find the impulse response h ( r ) of the system.
The impulse response h ( t ) should satisfy the differential equation
The homogeneous solution h h ( t ) to Eq. (2.120) satisfies
To obtain h h ( t ) we assume
h h ( t ) = ceS'
Substituting this into Eq. (2.121) gives
sces' + aces' = ( s + a)ces' = 0
from which we have s = - a and
h h ( t ) = ce-"'u(t)
We predict that the particular solution h p ( t ) is zero since h p ( t ) cannot contain N t ) . Otherwise, h ( t ) would have a derivative of S ( t ) that is not part of the right-hand side of Eq. (2.120). Thus,
h ( t ) =ce-"'u(t) (2.123)
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
To find the constant c , substituting Eq. (2.123) into Eq. (2.120), we obtain
Using Eqs. (1.25)and (1.301, the above equation becomes
so that c = 1. Thus, the impulse response is given by
h ( t ) =e-"'u(t)
2-24 Consider the system in Prob. 2.20 with y(0) = 0.
( a ) Find the step response
h ( t 1.
s(t)
of the system without using the impulse response
( b ) Find the step response d r ) with the impulse response h ( t ) obtained in Prob. 2.23. (c) Find the impulse response h(r from s ( t ).
( a ) In Prob. 2.20
Setting K = 1, b = 0, we obtain x ( t ) = u ( t ) and then y ( t ) = s ( t 1. Thus, setting K b = 0, and y ( 0 ) = y o = 0 in Eq. (2.109),we obtain the step response
( b ) Using Eqs. ( 2 . 1 2 )and (2.124) in Prob. 2.23, the step response s ( t ) is given by
which is the same as Eq. (2.125). (c) Using Eqs. (2.13) and (2.125),the impulse response h ( t ) is given by
Using Eqs. (1.25)and (1.30), we have
Thus, which is the same as Eq. (1.124).
h ( t ) =e-"'u(t)
CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
2.25. Consider the system described by
yf(t)
+2y(t)=x(t) +xl(t)
( 2.127)
Find the impulse response h ( t ) of the system.
The impulse response h ( t ) should satisfy the differential equation
h'(t)+ 2h(t) =6 ( r )+ S'(t)
The homogeneous solution h,(t) to Eq. (2.127) is [see Prob. 2.23 and Eq. (2.12211
h h ( t )= ~ ~ e - ~ ' u ( t )
Assuming the particular solution h J t ) of the form the general solution is
h ( t ) = C ~ ~ - ~ ' U t 6I( ) ) c( t
(2.128)
The delta function 6 ( t ) must be present so that h l ( t )contributes S1(t)to the left-hand side of Eq. (1.127).Substituting Eq. (2.128) into Eq. (2.1271, we obtain
=6(t)
+6'(t)
Again, using Eqs. (1.25) and (1.301, we have
( c , + 2 c 2 )6 ( t ) + c 2 S 1 ( t = 6 ( t ) + 6 ' ( t ) )
Equating coefficients of 6 ( t ) and 6'(t), we obtain from which we have c , = - 1 and c,
1. Substituting these values in Eq. (2.128).we obtain
RESPONSES OF A DISCRETE-TIME LTI SYSTEM AND CONVOLUTION
2.26. Verify Eqs. ( 2 . 3 6 ) and (2.37), that is,
( a ) By definition (2.35)
By changing the variable n - k = m, we have
( b ) Let x[nl* h,[nl =fJn1 and h,[nI* h2[nl = f 2 [ n l .Then
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
Substituting r = m - k and interchanging the order of summation, we have
Now, since
we have
2.27. Show that
( a ) x [ n ]* 6 [ n ]= x [ n ]
( b ) x [ n ] * 6 [ n- n , ] = x [ n - n o ]
(el x [ n ] * u [ n ] =