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CHAP. 41
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THE z-TRANSFORM Inverse transforming each term, we have x(n) = a[(-1)"
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+ 1 + 2(n .+ I)]u(n)
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Another way to invert this z-transform is to note that x(n) is the convolution of the two sequences. x(n) = x ~ ( n* xz(n) ) where xl (n) = u(n) and x2(n) is a step function that is up-sampled by a factor of 2. Because xl(n) * x2(n) = { I , 1 , 2 , 2 , 3 , 3 , 4 , 4 , . . .) we have the same result as before.
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Find the inverse z-transform of the second-order system
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Here we have a second-order pole at z = f . The partial fraction expansion for X(z) is
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The constant A is
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and the constant A2 is
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Therefore,
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Find the inverse of each of the following z-transforms:
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(a) X z = log (1 - iz-') ()
Izl >
(b) X(z) = ellZ, with x ( n ) a right-sided sequence
(a) There are several ways to solve this problem. One is to look up or compute the power series expansion of the log function. Another way is to differentiate X(z). Specifically, because
if we multiply both sides of this equation by (-z), we have
THE z-TRANSFORM Note that the region of convergence for X(z) is Izl > as it is for X(z), the inverse z-transform of Y (z) is
[CHAP. 4
i. Because the region of convergence for Y (z) is the same
y(n) = -($)"u(n - I ) Now, from the derivative property, y(n) = nx(n), and i t follows that x(n) = - i ( i ) " u ( n
- I)
(6) For this z-transform, we could determine the inverse by tinding the power series expansion of X(z). However,
another approach is to do what we did in part (a) and take the derivative. Differentiating X(z), we find d -X(z) dz Multiplying both sides by (-z), we have
= -z-~x(z)
and taking the inverse z-transform gives nx(n) = x(n - 1) which is a recursion for x(n). To solve this recursion, we need an initial condition. Because x(n) is a right-sided sequence, we may use the initial value theorem to find x(0). Specifically, x(0) = lim
l:+m
X (z) = I
Thus. the recursion that we want to solve is
with x(0) = 1. The solution for n > 0 is
and we have
42 .6
Find the inverse z-transform of X(z) = sin z.
To find the inverse z-transform of X(z) = sin z, we expand X(z) in a Taylor series about z = 0 as follows:
Because
~ ( z= )
) x(n)z-" :
)I=-N
we may associate the coefficients in the Taylor series expansion with the sequence values x(n). Thus, we have x(n) = (- 1)" I (2lnl+ I)! n = -1,-3,
-5,...
CHAP. 41
THE Z-TRANSFORM
Evaluate the following integral:
where the contour of integration C is the unit circle. Recall that for a sequence x(n) that has a z-transform X(z), the sequence may be recovered using contour integration as follows:
Therefore, the integral that is to be evaluated corresponds to the value of the sequence x(n) at n = 4 that has a z-transform
x (z) =
+ 2z-I
- 2-2
- +z-1)(1 - +-I)
Thus, we may find x(n) using a partial fraction expansion of X(z) and then evaluate the sequence at n = 4. With this approach, however, we are finding the values of x(n) for all n. Alternatively, we could perform long division would then be the value of and divide the numerator of X(z) by the denominator. The coefficient multiplying z - ~ x(n) at n = 4, and the value of the integral. However, because we are only interested in the value of the sequence at n = 4, the easiest approach is to evaluate the integral directly using the Cauchy integral theorem. The value of the integral is equal to the sum of the residues of the poles of x(z)z3 inside the unit circle. Because
has poles at z =
and z =
Therefore, we have
Find the inverse z-transform of
Note that the denominator of X(z) is a tenth-order polynomial. Although the roots may be found easily, performing a partial fraction expansion would be time consuming. For this problem, it is much better to exploit the properties of the z-transform. Note, for example, that X(z) = Y (zJO) where I Y(z) = 1 - culoz-l
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