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a (z-a)(z-a)
a z-a
-+"-)
z-1/a
1 Z 1 z Y ( z ) = -- - 1 - a 2 Z - 1/a 1 -a2 Z - a and from Table 4-1 we obtain
1 a<lzl<a
which is the same as Eq. (2.137).
4.27. Using t h e z-transform, redo Prob. 2.30.
From Fig. 2-23 and definition ( 4 . 3 )
~ [ n=] ( l , l , l , l ] - X ( z ) h [ n ] = ( l , l , 1)
-H(z)
1 + z - ' +z-2z-3
1+z-' +z-*
CHAP. 41
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
Thus, by the convolution property (4.26) Y ( Z ) = x ( z ) H ( z ) = ( I + Z - ' + Z - ~ + Z - ~ ) + IZ - ' + Z - ~ ) (
~ +4
~ +-
Hence,
h [ n l = {1,2,3,3,2,1)
which is the same result obtained in Prob. 2.30.
4.28. Using the z-transform, redo Prob. 2.32.
Let x[nl and y[nl be the input and output of the system. Then
x [ n ] = u [ n ] -X(z)
2- 1
lzl > 1
Then, by Eq. (4.41)
Using partial-fraction expansion, we have
where Thus,
c, = -
C2 =
1 1-a z H ( z ) = - - -a a Z-a
lzl> a
Taking the inverse z-transform of H(z), we obtain
When n
= 0,
Then
Thus, h [ n ] can be rewritten as
which is the same result obtained in Prob. 2.32.
THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
4.29. The output y [ n ] of a discrete-time LTI system is found to be 2(f )"u[n]when the input x [ n ] is u [ n ] .
( a ) Find the impulse response h [ n ] of the system. ( b ) Find the output y[n] when the input x [ n ] is ( ; ) " u [ n ] .
Hence, the system function H(z) is
Using partial-fraction expansion, we have
where Thus,
Taking the inverse z-transform of H(z), we obtain
h [ n ] = 6 6 [ n ] - 4(;)'u[n]
Then, Again by partial-fraction expansion we have
where Thus,
2(z - 1 )
C, =
c2 =
- 1)
r-L/2
Taking the inverse z-transform of Y(z), we obtain
y [ n ] = [-6(;)'
+ 8($)'lu[n]
CHAP. 41
T H E 2-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
If a discrete-time LTI system is BIBO stable, show that the ROC of its system function
H ( z ) must contain the unit circle, that is, lzl = 1.
A discrete-time LTI system is BIBO stable if and only if its impulse response h[nl is absolutely summable, that is [Eq. (2.49)1,
Now Let z
= ejR
so that lzl = lejRI = 1. Then
Therefore, we see that if the system is stable, then H ( z ) converges for z = ei". That is, for a stable discrete-time LTI system, the ROC of H ( z ) must contain the unit circle lzl = 1.
Using the z-transform, redo Prob. 2.38.
From Prob. 2.38 the impulse response of the system is
Then Since the ROC of H ( z ) is Izl > IaI, z = oo is included. Thus, by the result from Prob. 4.5 we conclude that h[n] is a causal sequence. Thus, the system is causal. ( b ) If l 1 > 1, the ROC of H ( z ) does not contain the unit circle lzl= 1, and hence the system a a will not be stable. If l 1 < 1, the ROC of H ( z ) contains the unit circle lzl = 1, and hence the system will be stable.
A causal discrete-time LTI system is described by
y [ n ] - ;y[n - 11
+ i y [ n - 21 = x [ n ]
(4.88)
where x[n] and y[n] are the input and output of the system, respectively.
( a ) Determine the system function H ( z ) .
( b ) Find the impulse response h[n] of the system. ( c ) Find the step response s[n]of the system.
( a ) Taking the z-transform of Eq. (4.88), we obtain
Y(2) - $ z - ' ~ ( z )+ ; z - ~ Y ( z ) = X ( z )
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
Thus,
( b ) Using partial-fraction expansion, we have
Thus,
Taking the inverse z-transform of H(z), we get
Then
Y(z)
= X ( Z )H ( z ) =
( z - l ) ( z - ; ) ( z - i)
Again using partial-fraction expansion, we have
Thus,
Taking the inverse z-transformation of Y(z), we obtain
CHAP. 41
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
4.33. Using the z-transform, redo Prob. 2.41.
As in Prob. 2.41, from Fig. 2-30 we see that
q [ n ] = 2 q [ n - 11 + x [ n ] ~ [ n l s=[ n l + 3 4 [ n - 11
Taking the z-transform of the above equations, we get
Q ( z )=22-'Q(z) +X ( z ) Y ( z )= Q ( z )+ ~ z - ' Q ( z )
Rearranging, we get
( 1 - 2 2 - ' ) Q ( L )= X ( z )
+ 3 . 2 - ' ) Q ( z )= Y ( z )
from which we obtain
Rewriting Eq. (4.89), we have
(1 - 2 2 - ' ) Y ( 2 ) = ( 1
+3 2 - ' ) x ( ~ )
(4.90)
Y ( z )- 2 2 - ' Y ( z ) = X ( z ) + 32-'X(z)
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