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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 zTRANSFORM AND DISCRETETIME 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 ztransform 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 leftsided 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 rightsided sequence. Thus, we must divide so as to obtain a series in power of z as follows: Thus, CHAP. 41
THE zTRANSFORM AND DISCRETETIME LTI SYSTEMS
and so by definition (4.3) we obtain x[n] { O , T , ~ , E.) ,. 1 3 7 4.19. Using partialfraction expansion, redo Prob. 4.18.
X ( z ) = 2 ~ 2  3 ~ + 2 ( z  l ) ( z  + ) 1 Using partialfraction expansion, we have
1 lzl< 2
where and we get
Since the ROC of X(z) is lzl < i, x[n] is a leftsided sequence, and from Table 41 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 rightsided sequence, and from Table 41 we get x[n] which gives = u[n]  (i)'u[n]  (i)']u[n] 4.20. Find the inverse ztransform of
Using partialfraction expansion, we have
where
THE zTRANSFORM AND DISCRETETIME 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 rightsided sequence, and from Table 41 we get
4.21. Find the inverse ttransform 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 leftsided sequence, and from Table 41 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 zTRANSFORM AND DISCRETETIME LTI SYSTEMS
4.22. Find the inverse ztransform of
X( 2) can be rewritten as lzl > 2 Since the ROC is J z > 2, x[n] is a rightsided sequence, and from Table 41 we have J Using the timeshifting property (4.181, we have 2nLu[n l ] ~ z  1 Thus, we conclude that x[n] = 3(2)"'u[n Z
 11 4.23. Find the inverse ztransform of
X(2)= 2 + z  + 324 ~ z 2 + 4z + 3
= (2.7I
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 timeshifting 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 rightsided sequence, and from Table 41 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 2TRANSFORM AND DISCRETETIME LTI SYSTEMS
[CHAP.4
4.24. Find the inverse ztransform 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 timeshifting 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 ztransform, 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 41 h[n]=anu[n] H(z)=Za
lzl> l 1 a
Then, by Eq. (4.40) Using partialfraction expansion, we have
where Thus, CHAP. 41
THE zTRANSFORM AND DISCRETETIME LTI SYSTEMS
Taking the inverse ztransform of Y(z), we get
which is the same as Eq. (2.134). Using the ztransform, 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 41 ~ [ n= a n u [ n ]H X ( Z ) = ] a Izl> l 1
Then Using partialfraction expansion, we have
where Thus, C, =  zP,=, C2 = z  a ~ = p
P ap
y(z)  a  p 2(Y ap 2p
> max(a,p ) and which is the same as Eq. (2.135). When a = P , Using partialfraction expansion, we have
where and
T H E ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
[CHAP. 4
Setting z = 0 in the above expression, we have
Thus, Y(z) and from Table 41 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 41 and Eq. (4.75) (za)

