For the system described by Eq. (2.55)the impulse response h[n] is given by in .NET framework

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For the system described by Eq. (2.55)the impulse response h[n] is given by
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Note that the impulse response for this system has finite terms; that is, it is nonzero for only a finite time duration. Because of this property, the system specified by Eq. (2.55) is known as a finite impulse response (FIR) system. On the other hand, a system whose impulse response is nonzero for an infinite time duration is said to be an infinite impulse response (IIR) system. Examples of finding impulse responses are given in Probs. 2.44 and 2.45. In Chap. 4, we will find the impulse response by using transform techniques.
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RESPONSES OF A CONTINUOUS-TIME LTI SYSTEM AND CONVOLUTION
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Verify Eqs. ( 2 . 7 )and ( 2 . 8 ) ,that is,
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(a) x(t)*h(t)=h(t)*x(t) ( b ) ( x ( t ) *h , ( t ) J * , ( t ) = x ( t ) * ( h , ( t ) h,(t)J h *
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( a ) By definition (2.6)
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By changing the variable t - T
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= A,
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we have
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CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
( b ) Let x ( t ) * h , ( t ) =f , ( t ) and h , ( t ) * h 2 ( t ) =f2(t). Then
Substituting A = a - T and interchanging the order of integration, we have
Now, since
we have
Thus,
Show that
x(t)
* u(t - to)=
I-to
( a ) By definition ( 2 . 6 ) and Eq. (1.22) we have
( b ) By Eqs. ( 2 . 7 ) and (1.22)we have
By Eqs. ( 2 . 6 ) and (1.19) we have
since u(t - 7 ) =
7<t 7>t'
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
( d l In a similar manner, we have
x ( t )* u(t - t,,) =
x ( r ) u ( t- 7 - t o ) dr
Let y ( r ) = x ( r ) * h ( t 1. Then show that
By Eq. ( 2 . 6 )we have
and Let r
A . Then
T =A
+ t , and Eq. (2.63b)becomes
Comparing Eqs. ( 2 . 6 3 ~and (2.63~1, see that replacing ) we Thus, we conclude that obtain Eq. ( 2 . 6 3 ~ ) .
in Eq. ( 2 . 6 3 ~by r - r )
, - r,,
The input x ( t ) and the impulse response h ( t ) of a continuous time LTI system are given by
(a) Compute the output y ( t ) by Eq. ( 2 . 6 ) . ( b ) Compute the output y ( t ) by Eq. (2.10).
( a ) By Eq. ( 2 . 6 )
Functions X ( T ) and h(t - r ) are shown in Fig. 2-4(a)for t < 0 and t > 0. From Fig. 2-4(a) we see that for t < 0, x ( r ) and h(t - T ) do not overlap, while for t > 0, they overlap from T = 0 to T = I . Hence, for t < 0, y ( t ) = 0. For t > 0, we have
Thus, we can write the output y ( t ) as
CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
( b ) By Eq. (2.10)
Functions h ( r ) and x(t - 7)are shown in Fig. 2-4(b) for t < 0 and t > 0. Again from Fig. 2-4(b) we see that for t < 0, h(7)and x(t - 7 ) do not overlap, while for t > 0, they overlap from 7 = 0 to r = t . Hence, for t < 0, y ( t ) = 0. For t > 0, we have
Thus, we can write the output y ( t ) as
which is the same as Eq. (2.64).
F g 2-4 i.
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
Compute the output y(t for a continuous-time LTI system whose impulse response h ( t ) and the input x ( 0 are given by
By Eq. ( 2 . 6 )
y ) ( t )* h )=
1 x ( r ) h ( t -r ) dr
Functions ~ ( 7and h(t - 7 ) are shown in Fig. 2-5(a) for t < 0 and t > 0. From Fig. 2-5(a) we ) see that for t < 0, X ( T ) and h(t - 7 ) overlap from 7 = - w to 7 = t , while for t > 0, they overlap from 7 = -01 to 7 = 0. Hence, for t < 0, we have
y(r)
e'Te-'V-T)
- rn
= e - a ' G 2 a ' d r = -eat 2a
(2.66~)
For t > 0, we have
Fig. 2-5
CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
) Combining Eqs. ( 2 . 6 6 ~and (2.6681, we can write y ( t ) as
y(t)
1 -e-uIrl
CY>O
which is shown in Fig. 2 - S b ) .
Evaluate y ( t ) = x ( t ) * h ( t ), where x ( t ) and h ( t ) are shown in Fig. 2-6, ( a ) by an analytical technique, and ( b ) by a graphical method.
Fig. 2-6
( a ) We first express x(t and h ( t ) in functional form:
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