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THE z-TRANSFORM we may use the up-sampling property (Prob. 4.21) to obtain
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[CHAP. 4
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Therefore, we have x(n) =
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n = 0 , 10,20. otherwise
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In many cases one is interested in computing the inverse z-transform of a rational function
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Because a partial fraction expansion requires knowledge of the roots of A(z), if the order of the denominator is large, finding the roots may be difficult. Although a partial fraction expansion would give a closed-form solution for x(n) for all n. if one only wants to plot x(n) for a limited range of values for n, a closed-form expression is not required. Given that x ( n ) = 0 for n c 0, find a recursion that generates x ( n ) for n 5 0.
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If we consider x(n) to be the unit sample response of a linear shift-invariant system, we may straightforwardly specify the filter in terms of a linear constant coefficient difference equation. This leads to a recursively computable difference equation for x(n). Specifically, note that because
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we may express this in the time domain as follows:
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Writing out this convolution explicitly, we have
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Bringing the first term out of the summation and dividing by a(0)gives
Therefore, given that x(n) = 0 for n < 0, this recursion allows us to compute s ( n )for all n 2 0 . For example,
Note that h(n) = 0 for n > q. Thus. for n z q, the recursion simplifies to
CHAP. 41
THE z-TRANSFORM
One-sided z-'hansforms
Find the one-sided z-transform of the following sequences:
(a) x ( n ) = ( { ) " u ( n 3 ) (b) x ( n ) = S(n - 5 ) S(n) + 2 " - ' 4 - n )
In the following, let x+(n) denote the sequence that is formed fromx(n) by setting x ( n )equal to zero for n < 0, that is,
( a ) Because x.,(n)
= ( ; l n u ( n ) ,the one-sided z-transform of x ( n ) is
(b) For this sequence, because
x+(n) = S(n - 5 )
then
X l ( z ) = z-%
+ S(n) + T 1 6 ( n ) I + $ = 1.5 + z 4
Let XI(z) be the one-sided z-transform of x ( n ) . Find the one-sided z-transform of y ( n ) = x ( n
The one-sided 2-transform of x ( n ) is
+ 1).
If x(n) is advanced in time by one, y ( n ) = x(n
+ I), the one-sided z-transform of y(n) is
Therefore. Comparing this to X I (z), we see that
Y I ( z )= x(1)
+ x(2)z-' + x(3)z -Z + .
Y I ( z )= z[X1(z) - x(0)l
Consider the LCCDE
y(n)-iy(n-2)=S(n) n 2 O
Find a set of initial conditions on y(n) for n < 0 so that y ( n ) = 0 for n 2 0 .
The one-sided z-transform of the LCCDE is
Solving for Yl(z),we have
Y I ( z )=
+ S [ , Y ( - ~ ) + y(1 - iz-2
I)z- ll
In order for y ( n ) to be equal to zero for n
0, Y l ( z )must be equal to zero. This will be the case when
THE z-TRANSFORM
[CHAP. 4
Consider a system described by the difference equation
Find the response of this system to the input
with initial conditions y(- 1) = 0.75 and y(-2) = 0.25.
This is the same problem as Prob. 1.37. Whereas this difference equation was solved in Chap. 1 by finding the particular and homogeneous solutions, here we will use the one-sided z-transform. First, we take the one-sided z-transform of each term in the difference equation
Substituting the given values for the initial conditions, we have Y(z) = z - ' Y ( z ) +
z - ~ Y ( z)
$2-'
+ ;x(~)+;Z-'~(Z)
Collecting all of the terms that contain Y (z) onto the left side of the equation gives
Because x(n) = (i)"u(n).
which gives Y(z) =
- az-1 1 - 2-1 + z-2
;+ + (1 - iz-l)(l fz-I + z-2) - z-1
Expanding the second term using a partial fraction expansion, we have
Y (z) = - 2L z-- I 2
;+ ;z-'
1 - 2-1
+ z-2
Therefore, the solution is
A digital filter that is implemented on a DSP chip is described by the linear constant coefficient difference equation
y(n) = y(n - I) - ky(n - 2) + x ( n ) In evaluating the performance of the filter, the unit sample response is measured (i.e., the response y(n) to the input x(n) = S(n) is determined). The internal storage registers on the chip, however, are not set to zero prior to applying the input. Therefore, the output of the filter contains the effect of the initial conditions, which are ( - I ) = -1 and y(-2)= 1
CHAP. 41
THE Z-TRANSFORM
Determine the response of the filter for all n p 0 and compare it with the zero state response (i.e., the output with y(.-I) = y(-2) = 0). Here we want to solve a difference equation that has initial conditions. Using the one-sided z-transform, we have
With X(z) = I and the given initial conditions, this becomes
Solving for Y (2). we find
Performing a partial fraction expansion gives
Thus, with an inverse z-transform we have y(n) =
[-;(a)''
+ :(;)"]u(n)
The zero state response, on the other hand, is simply the unit sample response of the filter. With
it follows that
Applications
There are two kinds of particles inside a nuclear reactor. Every second, an cr particle will split into eight B particles and a B particle will split into an a! particle and two / particles. If there is a single cr particle in the reactor at time t = 0, how may particles are there altogether at time t = 1 00 In this problern we need to begin by writing down, in mathematical terms, what is happening within the reactor. Let a ( n ) be the number of a particles in the reactor at time n, and let B(n) be the number of /3 particles. Because there are eight B particles created from each a particle and two from each ~9particle, we have
Also, because one a particle is created from each B particle,
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