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4.17. Find the inverse z-transform of the following X(z):
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(a) The power series expansion for log(1 - r) is given by
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X(z)= log
-az-,)
-lo@
-afl)
Izl> lal
Since the ROC is lzl> lal, that is, laz-'I< 1 , by Eq. (4.80), X(z) has the power sel expansion
from which we can identify x[n] as
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
Since the ROC is J z J lal, that is, la-'zl < 1, by Eq. (4.801, X(z) has the power series < expansion
from which we can identify x [ n ] as
4.18. Using the power series expansion technique, find the inverse z-transform of the following X( z): Z 1 ( a ) X(z)= IzI< 2 z 2 - 3z + 1 2
(a) Since the ROC is (zl < i, x [ n ] is a left-sided sequence. Thus, we must divide to obtain a series in power of z. Carrying out the long division, we obtain z + 3 z 2 + 7z3 + 15z4+ - . .
15z4 Thus, and so by definition (4.3) we obtain
( b ) Since the ROC is lzl> 1, x [ n ] is a right-sided sequence. Thus, we must divide so as to obtain a series in power of z- as follows:
Thus,
CHAP. 41
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
and so by definition (4.3) we obtain x[n]
{ O , T , ~ , E-.) ,.
1 3 7
4.19. Using partial-fraction expansion, redo Prob. 4.18.
X ( z ) = 2 ~ 2 - 3 ~ +- 2 ( z - l ) ( z - + ) 1 Using partial-fraction expansion, we have
1 lzl< 2
where and we get
Since the ROC of X(z) is lzl < i, x[n] is a left-sided sequence, and from Table 4-1 we get x[n] which gives x[n] = (..., 15,7,3,l,O)
-u[-n
- I] +
- I ] = [(i)n I]u[-n -
- 11
Since the ROC of X(z) is lzl> 1, x[n] is a right-sided sequence, and from Table 4-1 we get x[n] which gives
= u[n]
- (i)'u[n]
- (i)']u[n]
4.20. Find the inverse z-transform of
Using partial-fraction expansion, we have
where
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
Substituting these values into Eq. (4,831, we have
Setting z = 0 in the above expression, we have
Thus,
Since the ROC is lzl> 2, x [ n ] is a right-sided sequence, and from Table 4-1 we get
4.21. Find the inverse t-transform of
Note that X(Z) an improper rational function; thus, by long division, we have is
Let Then where
Thus.
3 X(z)=2z+--2
2 - 1
+ -- - 2 22
- 1)
Since the ROC of X ( z ) is ( z( < 1, x [ n ] is a left-sided sequence, and from Table 4-1 we get
x [ n ] = 2 S [ n + 11
+ i S [ n ]+ u [ - n - 11 - $2"u[-n = 2 S [ n + 11 + $ [ n ] + (1 - 2 " - ' ) u [ - n - 11
CHAP. 41
T H E z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
4.22. Find the inverse z-transform of
X( 2) can be rewritten as lzl > 2 Since the ROC is J z > 2, x[n] is a right-sided sequence, and from Table 4-1 we have J
Using the time-shifting property (4.181, we have 2n-Lu[n l ] ~ z - 1 Thus, we conclude that x[n]
= 3(2)"-'u[n Z
- 11
4.23. Find the inverse z-transform of
X(2)=
2 + z - + 32-4 ~ z 2 + 4z + 3
= (2.7-I
11 2
We see that X ( z ) can be written as X(z) where Thus, if then by the linearity property (4.17) and the time-shifting property (4.18), we get
+ z - ~+ ~ z - ~ ) x , ( z )
where Then 1 z 1 z lz1 > 0 2z+1 2z+3 Since the ROC of X,(z) is Izl > 0, x,[n] is a right-sided sequence, and from Table 4-1 we get X,(z)
= - -- - -
x1[n] = ;[(-I)" - (-3)"]u[n] Thus, from Eq. (4.84) we get x[n]
[ ( - I ) " - ' - (-3)"-']u[n - I ] + f [ ( - I ) " - ' - (-3)"-']u[n - 3 1
+ f [( -I)"-'
- (-3)"-']u[n
- 51
THE 2-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP.4
4.24. Find the inverse z-transform of
X(2)=
From Eq. (4.78) (Prob. 4.12)
(1 a
( z -a )
l z l > lal
Now, from Eq. (4.85)
and applying the time-shifting property (4.20) to Eq. (4.86), we get
~ [ n=]( n + l ) a n u [ n+ 1 ] = ( n + l ) a n u [ n ]
since x [ - 1 ] = 0 a t
n = -1.
SYSTEM FUNCTION
4.25. Using the z-transform, redo Prob. 2.28.
From Prob. 2.28, x [ n ] and h [ n ] are given by
x[n]=u[n] h [ n ]=anu[n]
O<a<1
From Table 4-1
h[n]=anu[n]
H(z)=Z-a
lzl> l 1 a
Then, by Eq. (4.40)
Using partial-fraction expansion, we have
where Thus,
CHAP. 41
THE z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
Taking the inverse z-transform of Y(z), we get
which is the same as Eq. (2.134).
Using the z-transform, redo Prob. 2.29.
(a) From Prob. 2.29(a), x [ n ] and h [ n ] are given by
x [ n ] =anu[n] h[n] =Pnu[n]
From Table 4-1
~ [ n= a n u [ n ]H X ( Z ) = ]
a Izl> l 1
Then Using partial-fraction expansion, we have
where Thus,
C, = -
z-P,=,
C2 =
z - a ~ = p
P a-p
y(z)
-- a - p 2-(Y a-p 2-p
> max(a,p )
and which is the same as Eq. (2.135). When a = P ,
Using partial-fraction expansion, we have
where and
T H E Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
Setting z = 0 in the above expression, we have
Thus, Y(z) and from Table 4-1 we get Thus, we obtain the same results as Eq. (2.135). ( b ) From Prob. 2.29(b), x [ n ] and h[nl are given by From Table 4-1 and Eq. (4.75)
(z-a)
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