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(b) For this sequence we write x(n) = cos(nwo)u(n) = [eJnq+ e-jnq M n ) Therefore, the z-transform is
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with a region of convergence lzl
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1. Combining the two terms together, we have X(z) = 1 - (cos wdz-' I - 2(cos wo)z-I z-2
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Find the z-transform of each of the following sequences. Whenever convenient, use the properties of the z-transform to make the solution easier. (a) x ( n ) = (+)"u(-n)
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(b) x ( n ) = ( i ) " u ( n 2 )
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(a) Using the definition of the z-transform we have
THE 2-TRANSFORM where the sum converges for 1321 < I
121 i
[CHAP. 4
Alternatively, note that the time-reversed sequence y(n) = x(-n) = (;)-"u(n) has a z-transform given by
with a region of convergence given by Izl > 3. Therefore, using the time-reversal property, Y(z) = X(z-I), we obtain the same result. (b) Because x(n) is the sum of two sequences, we will find the z-transform of x(n) by finding the z-transforms of each of these sequences and adding them together. The z-transform of the first sequence may be found easily using the shift property. Specifically, note that because
the z-transform of (;)"u(n
+ 2) is 4.2'
times the z-transform of (a)"u(n), that is.
which has a region of convergence lzl > The second term is a left-sided exponential and has a z-transform that we have seen before, that is,
with a region of convergence Izl < 3. Finally, for the z-transform of .u(n). we have
with a region of convergence
< Izl < 3.
(c) As we saw in Problem 4.3(b), the z-transform of cos(nwo)u(n)is
cos(nwo)u(n) 6 Therefore, using the exponentiation property,
I - (cos w")z-I I -~(COS O&)z-' z-2 '
~zl 1 >
we have with a region of convergence lz 1 > f .
( d ) Writing x(n) as
s ( n ) = a n u ( n ) cr-"u(-n) we may use the linearity and time-reversal properties to write
1 1 X(z) = 1 - ffz-' I - ffz
S(n)
;<1z1<2
which may be simplified to
CHAP. 41
THE Z-TRANSFORM
Without explicitly solving for X(z), find the region of convergence of the z-transform of each of the following sequences:
(a, x ( n ) =
[(i)n( i ) n ] u ( n- 10, +
1 (b) x(n) = 0
-10(n( otherwise
( a ) Because the first sequence is right-sided, the region of convergence is the exterior of a circle. With a pole at z = coming from the term and a pole at z = coming from the term (:)" it follows that the region of convergence must be Izl >
(i)", i.
(b) This sequence is finite in length. Therefore, the region of convergence is at least 0 < Izl < oo. Because x(n) has nonzero values for n < 0 and for n r 0, z = 0 and z = oo are not included within the ROC.
(c) Because this sequence is left-sided, the region of convergence is the interior of a circle. With a pole at z = 2, it follows that the region of convergence is Iz I < 2.
Find the z-transform of the sequence y ( n ) =
xi=-, ( k ) in terms of the z-transform of x ( n ) . x
There are two ways to approach this problem. The first is to note that x(n)may be written in terms of y(n)as follows:
Therefore, if we transform both sides of this equation, and use the shift property of the z-transform, we find
X ( z ) = Y ( z )- z - ' y ( z )
Solving for Y ( z ) ,we find
Thus. which is referred to as the summation property. The second approach is to note that y(n) is the convolution of x ( n )with a unit step,
Therefore, using the convolution theorem, we have
and, with U ( z ) = I / ( I - z - I ) , we obtain the same result as before. For the region of convergence, note that because the ROC of l / ( z )is Iz( > 1, the ROC of Y ( z ) will be at least
R, = R, n ( I Z I > I )
where R, is the ROC of X ( z ) .
Find the z-transform of the sequence y ( n ) where
and y(n) = 0 for n < 0. Assume that la1 < 1.
THE Z-TRANSFORM
[CHAP. 4
For this sequence, we may use a variation of the summation property derived in Prob. 4.6. Specifically, recall that if
then Now consider the two-sided summation,
Y(z)=
X(z) I - z-'
which may be written as
Therefore, if we let
Therefore, we have
Y ( z )=
X I( z ) + X2W - x(0) I - z-'
Finally, with x ( n ) = a " 1 , follows that x l ( n )= x2(n) = a n u ( n ) ,and x ( 0 ) = I. Thus, it
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