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Taking the limit of X(z) as z + oo, we see that X(z) + 1. Because the limit exists, x(n) is causal, and x(0) = 1.
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Find the value of x ( 0 ) for the sequence that has a z-transform
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Because the region of convergence of X(z) is the exterior of a circle, x(n) is right-sided. However, if we write X(z) in terms of positive powers of z,
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CHAP. 41 4
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THE z-TRANSFORM
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we see that X(z) -+ oo as Izl -+ oo.Therefore, x(n) is not causal. However, because x(n) is right-sided, it may be delayed so that it is causal. Specifically, if we delay x(n) by 1 to form the sequence y(n) = x(n - I),
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;) (z2 - ;)
m. Thus, y(n) is causal, and we conclude that y(0) = x(-I) = I . Because
X(z) - x(-l)z is the z-transform of a causal sequence, and it follows from the initial value theorem that
With
we have
x(0) = lim [X(z) - x(- I )Z1 =
Izlbm
Generalize the initial value theorem to find the value of a causal sequence x ( n ) at n = 1 , and find x ( 1 ) when
If x(n) is causal, X(z) = x(0)
+ x(1)z-I + , ~ ( 2 ) z -+ ~
Therefore, note that if we subtract x(0) from X(z),
Multiplying both sides of this equation by z, we have
If we let z + m, we obtain the value for x(l), x ( l ) = lirn (z[X(z) - x(O)]J
Izl-rm
For the given z-transform we see that x(0) = lim X(z) = $
(21-+m
Therefore,
x(1) = lim (z[X(z) - x(0)I) =
Izl+m
THE z-TRANSFORM
[CHAP. 4
Let x ( n ) be a left-sided sequence that is equal to zero for n > 0 . If
find x(0).
For a left-sided sequence that is zero for n > 0, the z-transform is
Therefore, it follows that x(0) = lim X(z)
For the given z-transform, we see that 32-' 2zp2 32 2 = lim =2 x(0) = lim X(z) = lim i-ro 2-0 3 - z-' + z-2 2-0 3z2 - z 1
If x(n) is real and even with a rational z-transform, show that
and describe what constraints this places on the poles and zeros of X ( z ) .
If x(n) is even, x(n) = x(-n) Therefore, it follows immediately from the time-reversal property that
If X(z) has a zero at z = zO, X(z0) = 0 then
x (z,')
which implies that X(z) will also have a zero at z = 1/20, The same holds true for poles. That is, if there is a pole at zO,there must also be a pole at z = I /zo.
Use the derivative property to find the z-transform of the following sequences:
(a) x ( n ) = n(;)"u(n - 2 )
1 (b) x(n) = ; ( - 2 ) - " 4 - n
- 1)
(a) The derivative property states that if X(z) is the z-transform of x(n),
If we let x(n) = nw(n), where w(n) = (;)"u(n from the delay property and the z-transform pair
- 2) = f (i)
- 2)
CHAP. 41 it follows that
THE 2-TRANSFORM
Therefore, using the derivative property, we have the z-transform of x(n),
(b) Evaluating the z-transform of this sequence directly is difficult due to the factor of n-I. However, if we define a new sequence, y(n), as follows,
y(n) = nx(n) = (-2)-"4-n the z-transform of y(n) is easily determined to be
Y (z) = - l + ;z-'
II < z
Noting the relationship between x(n) and y(n), we can apply the derivative property to set up a differential equation for X(z),
The solution to this differential equation is
X(z)= log (z + f )
and the region of convergence is Iz 1 <
Up-sampling is an operation that stretches a sequence in time by inserting zeros between the sequence values. For example, up-sampling a sequence x ( n ) by a factor of L results in the sequence
y(n> =
otherwise
Express the z-transform of y(n) in terms of the z-transform of x(n).
Because y(n) isequal to zero for all n signal is
+ kL, with y(n) equal tox(n/L) forn = kL, the z-transform of the up-sampled
If X(z) converges for cu < Izl < /3, Y (z) will converge for cu < lzlL < p, or
a'/'. <
(zI < pIIL
42 .2
Find the z-transform of the sequence
an/10 x(n) = n = 0, 10,20, . . .
else
where l 1 -= 1. a
THE z-TRANSFORM
[CHAP. 4
We recognizex(n) as an exponential sequence that has been up-sampled by a factor of 10 (see Prob. 4.21). Therefore, because
the z-transform of x(n) is
Inverse z-'Ransforms
Find the inverse of each of the following z-transforms:
Because X(z) is a finite-order polynomial, x(n) is a finite-length sequence. Therefore, x(n) is the coefficient that multiplies z-" in X(z). Thus, x(0) = 4 and x(2) = x(-2) = 3. This z-transform is a sum of two first-order rational functions of z. Because the region of convergence of X(z) is the exterior of a circle, x(n) is a right-sided sequence. Using the z-transform pair for a right-sided exponential, we may invert X(z) easily as follows:
Here we have a rational function of z with a denominator that is a quadratic in z. Before we can find the inverse z-transform, we need to factor the denominator and perform a partial fraction expansion:
Because x(n) is right-sided, the inverse z-transform is
One way to invert this z-transform is to perform a partial fraction expansion. With X(z) =
I -(I-z-)(~-z-~ (I-
the constants A , B I, and B2 are as follows:
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