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THE ZTRANSFORM AND DISCRETETIME LTI SYSTEMS in VS .NET
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THE ZTRANSFORMAND DISCRETETIME LTI SYSTEMS
Verify Eq. (4.211, that is, By definition (4.3) A pole (or zero) at z = zk in X(z) moves to z = zoz,, and the ROC expands or contracts by the factor Izol. Thus, we have 4.10. Find the ztransform and the associated ROC for each of the following sequences: (a) From Eq. (4.15) S[n] all z
Applying the timeshifting property (4.181, we obtain
( b ) From Eq. (4.16) IZI> I
Again by the timeshifting property (4.18) we obtain
From Eqs. (4.8) and (4.10) anu[n] w By Eq. (4.20) we obtain an+ 'u[n
za z za
Izl> la1
+ I] lal< lzl < m
(4.73) ( d l From Eq. (4.16) THE ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
[CHAP. 4
By the timereversal property (4.23) we obtain
( e ) From Eqs. (4.8) and (4.10) anu[n] Izl> la1
Again by the timereversal property (4.23) we obtain
4.11. Verify the multiplication by n (or differentiation in z ) property (4.24), that is, From definition (4.3) Differentiating both sides with respect to
we have
Thus, we conclude that
4.12. Find the ztransform of each of the following sequences: (a) x [ n ] = n a n u [ n ] ( b ) x [ n ] = nan l u [ n ] ( a ) From Eqs. (4.8) and (4.10) anu[n] o za
I z I > la1
Using the multiplication by n property (4.24), we get
CHAP. 41
T H E ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
( b ) Differentiating Eq. (4.76)with respect to a, we have
Note that dividing both sides of Eq. (4.77)by a , we obtain Eq. (4.78). 4.13. Verify the convolution property (4.26), that is, By definition (2.35) Thus, by definition (4.3) Noting that the term in parentheses in the last expression is the ztransform of the shifted signal x 2 [ n  k ] , then by the timeshifting property (4.18)we have with an ROC that contains the intersection of the ROC of X , ( z ) and X , ( z ) . If a zero of one transform cancels a pole of the other, the ROC of Y ( z )may be larger. Thus, we conclude that 4.14. Verify the accumulation property (4.25), that is, From Eq. (2.40)we have
Thus, using Eq. (4.16) and the convolution property (4.26),we obtain
with the ROC that includes the intersection of the ROC of X ( z ) and the ROC of the ztransform of u [ n ] .Thus, THE ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
[CHAP. 4
INVERSE ZTRANSFORM
4.15. Find the inverse ztransform of
X ( z ) = z 2 ( l  i2')(1 2')(I
+227 3 +zI
0 < lzl< 00
(4.79) Multiplying out the factors of Eq. (4.79),we can express X ( z ) as X ( Z )= z 2 + t z Then, by definition ( 4 . 3 ) X ( z ) = x [  2 ] z 2 + x [  1 ] z + x [ o ]+ x [ 1 ] z  ' and we get
x [ n ] = { ...,O , l , $ ,  5 , 1 , 0 ,... } 4.16. Using the power series expansion technique, find the inverse ztransform of the following X ( 2): ( a ) Since the ROC is ( z ( >la(, that is, the exterior of a circle, x [ n ] is a rightsided sequence. Thus, we must divide to obtain a series in the power of z  ' . Carrying out the long division, we obtain Thus, 1  az' and so by definition ( 4 . 3 ) we have
x[n]=O x[O]=1 n<O x[l]=a x[2]=a2 X(z)= = 1+ a ~  ' + a ~ z  ~ +
Thus, we obtain
x [ n ]= anu[n] ( 6 ) Since the ROC is lzl < lal, that is, the interior of a circle, x[n] is a leftsided sequence. Thus, we must divide so as to obtain a series in the power of z as follows, Multiplying both the numerator and denominator of X ( z ) by z , we have z X(z) CHAP. 41
THE zTRANSFORM AND DISCRETETIME LTI SYSTEMS
and carrying out the long division, we obtain a'z  a  2 z 2  a  3 z 3  Thus,

