Taking the inverse z-transform of Eq. (4.90) and using the time-shifting property (4.181, we obtain in VS .NET

Encoding QR Code in VS .NET Taking the inverse z-transform of Eq. (4.90) and using the time-shifting property (4.181, we obtain

Taking the inverse z-transform of Eq. (4.90) and using the time-shifting property (4.181, we obtain
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y [ n ] - 2 y [ n - 1 1 = x [ n ]+ 3 x [ n - 1 1
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which is the same as Eq. (2.148).
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4.34. Consider the discrete-time system shown in Fig. 4-8. For what values of k is the system BIB0 stable
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THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
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[CHAP. 4
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From Fig. 4-8 we see that
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Taking the z-transform of the above equations, we obtain
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Y(z) = Q ( z ) Rearranging, we have
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+ z-'Q(z)
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from which we obtain
which shows that the system has one zero at z = - k / 3 and one pole at z = k / 2 and that the ROC is J z l > lk/2l. Thus, as shown in Prob. 4.30, the system will be BIB0 stable if the ROC contains the unit circle, lzl= 1. Hence the system is stable only if IkJ< 2.
UNILATERAL Z-TRANSFORM
4.35. Find the unilateral z-transform of the following x[n]:
( a ) x[n] = a n u [ n ]
( b ) x[n] = a n 'u[n
+ 11
= X(z)
(a) Since x[nl = 0 for n < 0, X,(z)
and from Example 4.1 we have
( b ) By definition (4.49) we have
1-az-'
Izl > la1
Note that in this case x[n] is not a causal sequence; hence X,(z) Prob. 4.101.
+ X(z) [see Eq. (4.73) in
CHAP. 41
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
4.36. Verify Eqs. (4.50) and (4.51), that is, for m 2 0,
( a ) x [ n - m ] -z-"x,(z)
+ Z - ~ + ' X [ - 11 + Z-""
'x[-2]
+x[-m]
( a ) By definition (4.49) with m 2 0 and using the change in variable k = n
- m, we have
( b ) With m 1 0 8,{x[n
+m ] )= C x[n +m]z-" = C x [ k ] ~ - ( ~ - ~ )
n =O
4.37. Using the unilateral z-transform, redo Prob. 2.42.
The system is described by
~ [ n- ay[n - 11 = x [ n ] ]
with y [ - 11 = y - and x[n 1 = Kbnu[n].Let
~ [ " *l Y I ( z )
Then from Eq. (4.50)
11 ~ [ -n - z - ~ Y , ( z ) + y [ - I ] = Z - ' & ( Z ) + y - I
From Table 4-1 we have
x [ n ] -X,(Z) =KZ
Izl> lbl
Taking the unilateral z-transform of Eq. (4.931, we obtain
&(z)
- a { z - ' ~ ( z + y - J = KZ ) z-b
THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
Thus, y(z)
= ay-\
-+ K
z (z-a)(z-b)
Using partial-fraction expansion, we obtain Y,(z) =ay-,-
z z-a
z z - ~ ( ~ z - b z-a
Taking the inverse z-transform of Y,(z), we get b Y [ n ] = a y - ,anu[n] + K-bnu[n] b-a
- K-anu[n]
which is the same as Eq. (2.158).
4.38. For each of the following difference equations and associated input and initial conditions, determine the output y[n I: ( a ) y[nl - f y [ n - 11 = x [ n l , with x [ n ] = (:In, y[- 11 = 1 ( b ) 3 y [ n ] - 4 y [ n - 11 + y[n - 21 = A n ] , with x[n] = (i)", - 11 = 1, y[- 21 = 2 y[
Taking the unilateral z-transform of the given difference equation, we get Y,(z) - +{z-'Y,(z) + y [ - I]} = X , ( z ) Substituting y[- 11 = 1 and X,(z) into the above expression, we get
Thus,
Hence, y [ n ) = 7(;)"+' - 2 ( f ) n (b) x[n] ++ X,(z)
n 2 -1
z - - 7
Taking the unilateral z-transform of the given difference equation, we obtain
2Y, 3Y,(z) - 4{z- ' 5( z ) + Y [ - I ] } + { ~ - ( z ) + z - ' y [ - 11 + ~ [ - 2 ] }= X I ( Z )
CHAP. 41
THE Z-TRANSFORMAND DISCRETE-TIME LTI SYSTEMS
Substituting y[- 1 = 1, y[-2]= 2, and X,(z) into the above expression, we get 1
Thus,
Hence,
y[n]=$
- (f)"+ +(f)"
n r -2
439. Let x [ n ] be a causal sequence and
x b l -X(z)
Show that
x[O] = lim X(z)
Equation (4.94) is called the initial value theorem for the z-transform.
Since x [ n ] = 0 for n < 0, we have
As z
+ oa,
0 for n > 0. Thus, we get
lim X ( z ) = x[O]
2--.m
4.40. Let x [ n ] be a causal sequence and Show that if X(z) is a rational function with all its poles strictly inside the unit circle except possibly for a first-order pole at z = 1, then
N-tm
lim x [ N ] = lim (1 -2-')X(z)
(4.95)
Equation (4.95) is called the final value theorem for the z-transform.
From the time-shifting property (4.19) we have The left-hand side of Eq.(4.96) can be written as
T H E Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
If we now let z
---, 1,
then from Eq. (4.96) we have
lim ( I - 2 - ' ) X ( z )
z-+ 1
N-+m n = O
( x [ n ]- x [ n - I]}
lim x [ N ]
Supplementary Problems
Find the z-transform of the following x [ n ] :
(a) (b) (c) (d) Am.
x [ n l = ( ; , I , - $1 x [ n ]= 2 S [ n + 21 - 3S[n - 21 x [ n ]= 3 ( - f)"u[n] - 2(3)"u[-n - I ] x [ n l = 3 ( i ) " u [ n l - 2(a)"u[-n - 11 (a) X(z)= f
+ z-'
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