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with a region of convergence Izl > I .
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Let x ( n ) be a finite-length sequence that is nonzero only for 0 5 n 5 N - 1, and consider the one-sided periodic sequence, y(n), that is formed by periodically extending x(n) as follows:
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Express the z-transform of y(n) in terms of X(z) and find the region of convergence of Y(z).
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The one-sided periodic sequence y(n) may be written as the convolution of x ( n ) with the pulse train
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CHAP. 41 In other words,
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THE z-TRANSFORM
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Therefore, the z-transform of y(n) is the product of the z-transforms of x(n) and pN(n). Because pN(n)is a sum of shifted unit samples, and because the z-transform of S(n - kN) is equal to z - ' ~ , the z-transform of pN(n)is
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Thus, the z-transform of the one-sided periodic sequence y(n) is
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Because x(n) is finite in length and zero for n c 0, the region of convergence for X(z) is J z J> 0. Therefore, the region of convergence of Y (z) is I z 1 > 1.
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Consider the sequence shown in the figure below.
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The sequence repeats periodically with a period N = 4 for n 2 0 and is zero for n < 0. Find the z-transform of this sequence along with its region of convergence.
This is a problem that may be solved easily using the property derived in Prob. 4.8. Because
where N = 4 and w(n) = S(n then and we have
- I) + 2S(n - 2) + S(n - 3)
W(z) = z-l[l
+ 2z-' + z - ~ ]
Because x(nj is right-sided and X(z) has four poles at lzl = I, the region of convergence is I z l > 1.
Properties
Use the z-transform to perform the convolution of the following two sequences:
THE z-TRANSFORM The convolution theorem for z-transforms states that if y(n) = h(n) H(z)X (z). With
[CHAP. 4
* x(n), the z-transform of y(n) is Y(z)
it follows that Y(z) = H(z)X(z) = ( I Multiplying these two polynomials, we have
+ iz-' + : Z - ~ ) ( I+ z-I + 4z-')
By inspection, we then have for the sequence y(n),
Evaluate the convolution of the two sequences
h ( n ) = (OS)"u(n) and x ( n ) = 3"u(-n)
To evaluate this convolution, we will use the convolution property of the z-transform. The z-transform of h(n) is
and the z-transform of x(n) may be found from the time-reversal and shift properties, or directly as follows:
Therefore, the z-transform of the convolution, y(n) = x(n) * h(n), is
The region of convergence is the intersection of the regions JzJ> and lzl < 3, which is inverse z-transform, we perform a partial fraction expansion of Y (z),
< lzl < 3. To find the
Y (z) = ----- + 1 - I,-I
B 1 - 32-1
and Therefore, it follows that
B = [(I - 3z-')Y (z)],,~ = - 6 5
Y(.) = ($)(;)"u(n)
+ (;)3nu(-n
CHAP. 41
THE Z-TRANSFORM
Let x(n) be an absolutely summable sequence,
with a rational z-transform. If X(z) has a pole at z = and limlzl+m X(z) = 1, what can be said about the extent of x(n) (i.e., finite-in-length, right-sided, etc.)
Because x(n) is an absolutely summable sequence, the ROC of X ( z ) includes the unit circle, Izl = I. With a pole at z = f , the region of convergence will either be an annulus of the form r- < lzl < r,, or it will be the exterior of a circle, r - < Izl. However, because X (z) converges as Izl -+ m, the region of convergence will be the exterior of a circle, and it follows that x(n) is right-sided (infinite in length) with x(n) = 0 for n < 0.
Find the z-transform of x(n) = lnl(;)lnl.
Using the derivative property and the z-transform pair
it follows that the z-transform of w(n) = n(;)"u(n) is
Because x(n) may be written as
x(n) = \nl(f)"' = n(f)"u(n) - n ( f ) zd-n)
using linearity and the time-reversal property, we have
which has a region of convergence
< Izl < 2.
Let y(n) be a sequence that is generated from a sequence x(n) as follows:
(a) Show that y(n) satisfies the time-varying difference equation
and show that -z2 dX(z) Y (z) = - Z - I dz where X(z) and Y (z) are the z-transforms of x(n) and y(n), respectively. (b) Use this property to find the z-transform of
THE Z-TRANSFORM
[CHAP. 4
(a) From the definition of y(n), we see that
and it follows immediately that y(n)- y(n
- 1) = nx(n)
From this difference equation, we may take the z-transform of both sides. Because
then
Y(z) - z-'Y (z) = -z
dX(z) dz
(b) To find the z-transform of the given sequence, note that
where Because the z-transform of x(n) is
x(n) = (f)"u(n)
1 X(z) = 1 - {z-I
then
-z2 dX(z) -z2 -:z 1 -2 Y(z) = --2 z - 1 dz 2 - 1 (1 - iZ-l)
3 (1 - ; z - ~ ) 2 ( l- z - ~ )
Because x(n) is right-sided, then the region of convergence is the exterior of a circle. Having poles at z = 1 and z = it follows that the region of convergence is Izl > I.
Find the value of x(0) for the sequence that has a z-transform X(z) =
1 1 - az-'
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