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EXAMPLE 4.4.1
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Suppose that a sequence x ( n ) has a z-transform
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with a region of convergence l z l z the form
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f . Because p = q
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= 2, and the two poles are simple, the partial fraction expansion has
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The constant C is found by long division:
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THE 2-TRANSFORM
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[CHAP. 4
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Therefore, C = 2 and we may write X(z) as follows:
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Next, for the coefficients A , and Az we have
and Thus, the complete partial fraction expansion becomes
Finally, because the region of convergence is the exterior of the circle Izl >
i, x(n) is the right-sided sequence
4.4.2 Power Series
The z-transform is a power series expansion,
where the sequence values x ( n ) are the coefficients of z-" in the expansion. Therefore, if we can find the power the series expansion for X(z), sequence values x ( n ) may be found by simply picking off the coefficients of z-".
EXAMPLE 4.4.2
Consider the z-transform X(:) = log(l The power series expansion of this function is
+ a:-')
Izl > la1
Therefore, the sequence x(n) having this z-transform is
CHAP. 41
THE z-TRANSFORM
4.4.3 Contour Integration
Another approach that may be used to find the inverse z-transform of X(z) is to use contour integration. This procedure relies on Cauchy's integral theorem, which states that if C is a closed contour that encircles the origin in a counterclockwise direction,
With
X(z) = n=-w
x(n)zPn
Cauchy's integral theorem may be used to show that the coefficients .x(n) may be found from X(z) as follows:
where C is a closed contour within the region of convergenceof X(z) that encircles the origin in acounterclockwise direction. Contour integrals of this form may often by evaluated with the help of Cauchy's residue theorem, x ( z ) z n ldz =
[residues of x ( z ) z n ' a t the poles inside C]
If X(z) is a rational function of z with a first-order pole at z = ak, ~es[x(z)z"-l at z = a k ] = [(I - cmz-l)~(z)zn-']z=ak Contour integration is particularly useful if only a few values of x(n) are needed.
4.5 THE ONE-SIDED Z-TRANSFORM
The z-transform defined in Sec. 4.2 is the two-sided, or bilateral, z-transform. The one-sided, or unilateral, z-transform is defined by
The primary use of the one-sided z-transform is to solve linear constant coefficient difference equations that have initial conditions. Most of the properties of the one-sided z-transform are the same as those for the two-sided z-transform. One that is different, however, is the shift property. Specifically, if x(n) has a one-sided z-transform X 1 ( ~ the one-sided z-transform of x(n - 1) is ),
It is this property that makes the one-sided z-transform useful for solving difference equations with initial conditions.
EXAMPLE 4.5.1
Consider the linear constant coefficient difference equation
Let us find the solution to this equation assuming that x ( n ) = S(n - I ) with y ( - I ) = y(-2) = 1 . We begin by noting that if the one-sided z-transform of y(n) is Y l ( z ) , one-sided z-transform of y(n the
- 2 ) is
THE z-TRANSFORM
[CHAP. 4
Therefore, taking the z-transform of both sides of the difference equation, we have
YI = 0.25[y(-2) (z)
+ y(-1)z-I + Z - ~ Y ~ ( +)XI(z) Z ]
where X 1 (z) = z-' . Substituting for y ( - 1) and y(-2), and solving for Yl we have (z),
To find y(n), note that Yl may be expanded as follow^:^ (z)
Therefore.
Solved Problems
Computing z-'hansforms
The z-transform of a sequence x ( n ) is
If the region of convergence includes the unit circle, find the DTFT of x ( n ) at w = n.
If X(z) is the z-transform of x(n), and the unit circle is within the region of convergence, the DTFT of x(n) may be found by evaluating X(z) around the unit circle: X(eJW) x(z)J;_-, = Therefore. the DTFT at o = 7r is
and we have
Find the z-transform of each of the following sequences: (a) x ( n ) = 3S(n)+ S(n - 2 ) + S(n
(b) x ( n ) = u ( n ) - u ( n - 1 0 )
(a) Because this sequence is finite in length, the z-transform is a polynomial,
and the region of convergence is 0 < Izl < m. Note that because x(n) has nonzero values for n < 0, the region of convergence does not include IzI = co,and because x(n) has nonzero values for n 0, the region of convergence does not include the point z = 0.
2 ~ e the discussion in Sec. 4.4.1 on partial fraction expansions. e
CHAP. 41 ( h ) For this sequence,
THE Z-TRANSFORM
which converges for all lzl > 0. Note that the roots of the numerator are solutions to the equation
z10 =
These roots are
= ej2nkl10
k = 0 , 1 , ..., 9
which are 10 equally spaced points around the unit circle. Thus, the pole at z = 1 in the denominator of X(z) is canceled by the zero at z = 1 in the numerator, and the z-transform may also be expressed in the form
Find the z-transform of each of the following sequences:
(a) x ( n ) = 2"u(n) 3 ( ; ) " u ( n )
(b) x ( n ) = cos(noo)u(n).
(a) Because x(n) is a sum of two sequences of the form anu(n), using the linearity property of the z-transform, and
the z-transform pair ffnu(n)
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