Consider a continuous-time LTI system whose step response is given by in .NET

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2.10. Consider a continuous-time LTI system whose step response is given by
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s ( t )=ebru(t) Determine and sketch the output of this system to the input x ( t ) shown in Fig. 2-15(a). From Fig. 2-15(a) the input x ( t ) can be expressed as x ( t ) = u ( t - 1) - u ( t - 3 ) Since the system is linear and time-invariant, the output y ( t ) is given by y(t)=s(t-1)-s(t-3)
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=e - ( l - l)u(t
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1 ) - e-(+3u(t 3 ) -
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which is sketched in Fig. 2-15(b).
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LINEAR TIME-INVARIANT SYSTEMS
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[CHAP. 2
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Fig. 2-15
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2.11. Consider a continuous-time LTI system described by (see Prob. 1.56)
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Find and sketch the impulse response h ( t ) of the system. ( b ) Is this system causal
Equation (2.72) can be rewritten as
CHAP. 2 1
LINEAR TIME-INVARIANT SYSTEMS
Using Eqs. (2.61) and ( 2 . 9 ) , Eq. (2.73) can be expressed as
Thus, we obtain
h ( t ) = :T[ ~ ( t +
f)- u ( ~ -f)]
- T / 2 < t r T/2 otherwise
( 2.75)
which is sketched in Fig. 2-16. ( b ) From Fig. 2-16 or Eq. (2.75) we see that h ( t ) # 0 for t < 0. Hence, the system is not causal.
Fig. 2-16
2.12. Let y ( t ) be the output of a continuous-time LTI system with input x ( t ) . Find the output of the system if the input is x l ( t ) , where x l ( t ) is the first derivative of x ( t ) .
From Eq. (2.10)
Differentiating both sides of the above convolution integral with respect to
we obtain
which indicates that y l ( t ) is the output of the system when the input is x l ( t ) .
2.13. Verify the B I B 0 stability condition [Eq. (2.21)] for continuous-time LTI systems.
Assume that the input x ( t ) of a continuous-time LTI system is bounded, that is,
Ix(t)ll k,
all t
(2.77)
LINEAR TIME-INVARIANT SYSTEMS
[CHAP.2
Then, using Eq. (2.10),we have
since lx(t - 711 1 k, from Eq. (2.77). Therefore, if the impulse response is absolutely integrable, that is,
then l y ( t )Is k , K = k, and the system is BIBO stable.
2.14. The system shown in Fig. 2-17(a) is formed by connecting two systems in cascade. The impulse responses of the systems are given by h , ( t ) and h 2 ( 0 , respectively, and
h , ( t ) =e-2'u(t) h,(t) = 2e-'u(t)
( a ) Find the impulse response h ( t ) of the overall system shown in Fig. 2-17(b). ( b ) Determine if the overall system is BIBO stable.
Fig. 2-17
( a ) Let w ( t ) be the output of the first system. By Eq. ( 2 . 6 )
~ ) ~ ( t= ) ( t* h , ( t )
(2.78)
Then we have
(2.79) ~ ( t=)w ( t ) * h 2 ( t ) = [ x ( O* h,( t ) ] * h2( t ) But by the associativity property of convolution (2.81, Eq. (2.79)can be rewritten as
~ ( t =)x ( t ) * [ h l ( t ) * h 2 ( t ) ] ~ ( t ) * h ( O =
(2.80) (2.81)
Therefore, the impulse response of the overall system is given by
h ( t ) = h d t )*h2(t) Thus, with the given h ,( t ) and h2(t), we have
CHAP . 21
LINEAR TIME-INVARIANT SYSTEMS
( 6 ) Using the above h( t ), we have
=2(1-+)=I Thus, the system is B I B 0 stable.
EIGENFUNCTIONS OF CONTINUOUS-TIME LTI SYSTEMS
2.15. Consider a continuous-time LTI system with the input-output relation given by
y(t)=
e - ( ' - ' ' x ( r ) dr
(2.82)
Find the impulse response h(t) of this system. ( b ) Show that the complex exponential function e"' is an eigenfunction of the system. ( c ) Find the eigenvalue of the system corresponding to eS' by using the impulse response h(t) obtained in part ( a ) .
From Eq. (2.821, definition (2.1), and Eq. (1.21)we get
h ( f) =
Thus,
/' e - ( 1 - T ) 6 ( 7 )d 7
= e - ( ' - ' ) I 7 =0 -e
( 2.83)
h(t)=e - ' ~ ( 1 )
( 6 ) Let x ( f) = e". Then
Thus, by definition (2.22) e" is the eigenfunction of the system and the associated eigenvalue is
(c) Using Eqs. (2.24) and (2.83),the eigenvalue associated with e"' is given by
which is the same as Eq. (2.85).
2.16. Consider the continuous-time LTI system described by
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
( a ) Find the eigenvalue of the system corresponding to the eigenfunction es'. ( b ) Repeat part ( a ) by using the impulse function h ( t ) of the system. ( a ) Substituting x ( r ) = e" in Eq. (2.861, we obtain
Thus, the eigenvalue of the system corresponding to eS' is
( b ) From Eq. (2.75) in Prob. 2.1 1 we have
h ( t ) = T [u(t+
-u(l -
(:IT
-T/2<tsT/2 otherwise
Using Eq. (2.241, the eigenvalue H(s) corresponding to eS' is given by
which is the same as Eq. (2.87).
2.17. Consider a stable continuous-time LTI system with impulse response h ( t ) that is real and even. Show that cos w t and sin w t are eigenfunctions of this system with the same real eigenvalue.
By setting s = jw in Eqs. (2.23) and (2.241, we see that eJ"' is an eigenfunction of a continuous-time LTI system and the corresponding eigenvalue is
A = H( jw) =
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