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y [ n ] = a y [ n - 1 1 + x [ n ] =an(acu K ) =an+'cu + anK
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Similarly, we can also determine y [ n ] for n < 0 by rearranging Eq. (2.149) as
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y[-11 = a
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y[-n]
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+ 1 1 - x [ - n + 1 1 ) =a-"+'a
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Combining Eqs. (2.163) and (2.169, we obtain
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y [ n ] = a n + ' a+ K a n u [ n ]
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
2.44. Consider the discrete-time system in Prob. 2.43 for an initially at rest condition.
( a ) Find in impulse response h [ n ] of the system. ( b ) Find the step response s [ n ] of the system. (c) Find the impulse response h [ n ] from the result of part ( b ) .
Setting K
1 and y [ - 11 = a
in Eq. (2.166), we obtain
Setting K
1, b = 1 , and y [ - 1]
=y - , = 0
in Eq. (2.161), we obtain
From Eqs. (2.41) and (2.168) the impulse response h [ n ] is given by
When n = 0.
When n r 1,
Thus, which is the same as Eq. (2.167).
h [ n ]= anu[n]
2.45. Find the impulse response h [ n ] for each of the causal LTI discrete-time systems satisfying the following difference equations and indicate whether each system is a FIR or an IIR system.
( a ) y [ n ] = x [ n ] - 2 x [ n - 21 + x [ n - 31 ( b ) y [ n ] + 2 y [ n - 11 = x [ n ] + x [ n - 11 (c) y [ n ] - t y [ n - 21 = 2 x [ n ] - x [ n - 21
By definition (2.56)
Since h [ n ] has only four terms, the system is a FIR system.
CHAP. 2 1
LINEAR TIME-INVARIANT SYSTEMS
( b ) h[nl = -2h[n - 11 + 6 [ n ]+ 6 [ n- 11 Since the system is causal, h[- 1 ] = 0. Then h[O]= -2h[ - 1 ] + 6 [ 0 ]+ 6[ - 1 1 = S [ O ] = 1 h [ l ]= -2h[O] + 6 [ 1 ]+ S [ O ] = -2
+ 1 = -1
=2 =
h [ 2 ]= - 2 h [ l ] + 6 [ 2 ]+ S [ l ]= - 2 ( - 1 ) h [ 3 ]= -2h[2] + 6 [ 3 ]+ 6 [ 2 ]= - 2 ( 2 )
Hence,
h [ n ]= 6 [ n ]+ ( - 1 ) " 2 " - ' u [ n - 1 1
Since h[nl has infinite terms, the system is an IIR system. 1 (c) h[nl = i h [ n - 2 + 26[n]- 6[n - 21 Since the system is causal, h [ - 2 = h[- 11 = 0. Then 1
Hence,
h [ n ]= 2 6 [ n ]
Since h[nl has only one term, the system is a FIR system.
Supplementary Problems
Compute the convolution y(t ) = x ( t (a)
X(I) =
* h(t ) of the
-a < t < a 1 otherwise , h ( t )= O<rsT otherwise ' 2a - It1 It( < 2a 111 L 2a
following pair of signals: -a < t l a otherwise O<r52T otherwise
Am. ( a ) Y O ) =
r<O O<t_<T T<rs2T 2T<rs3T 3T<t
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
Compute the convolution sum y [ n ] = x [ n l * h[nl of the following pairs of sequences:
( a ) y [ n ]=
ns0 n>O
Show that if y ( t ) = x ( t ) * h ( t ) , then
yl(r)
' ( r h(r) = x ( t ) * hl(t) *)
Hint: Differentiate Eqs. (2.6) and (2.10) with respect to
Show that
~ ( t* S'(1) = x t ( t ) ) Hint: Use the result from Prob. 2.48 and Eq. (2.58).
Let y [ n ] = x [ n ] *h [ n ] .Then show that
x [ n - n , ] *h [ n - n , ] = y [ n - n , - n , ] Hint: See Prob. 2.3.
Show that
for an arbitrary starting point no.
Hint: See Probs. 2.31 and 2.8.
The step response s ( t ) of a continuous-time LTI system is given by
Find the impulse response h(r) of the system.
h ( t ) = S ( t ) - w,[sin w , , f l u ( t )
CHAP. 2 1
LINEAR TIME-INVARIANT SYSTEMS
The system shown in Fig. 2-31 is formed by connection two systems in parallel. The impulse responses of the systems are given by
)= e 2 ( )
h,(t) = 2ee'u(t)
( a ) Find the impulse response h ( t ) of the overall system. ( b ) Is the overall system stable Ans. (a) h ( t ) = ( e - 2 ' + 2e-')u(t) ( b ) Yes
Consider an integrator whose input x ( t ) and output y ( t ) are related by
( a ) Find the impulse response h ( t ) of the integrator. ( b ) Is the integrator stable Ans. ( a ) h ( t ) = u ( t ) ( b ) No
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