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I) will be causal if h(n) is causal. Finally, because H(z) is a rational function of z,
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Therefore, if the poles of H(z) are inside the unit circle, the poles of G(z) are inside the unit circle, and g(n) is stable.
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(c) With G(z) = ~ ( z - ' ) note that if H(z) has a pole at := zo, G(:) will have a pole at z = l / z ~ .Therefore, ,
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all of the poles of G(z) will be outside the unit circle. and g(n) cannot be both stable and causal. Because the replacement of z with z-' corresponds to a time reversal,
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g(n) is noncausal. Furthermore, because time-reversing a sequence does not affect its absolute summability,
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the region of convergence for G ( z ) will include the unit circle. Thus, g(n) is stable.
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(d) With G(;) = H(-z), note that replacing z with -z corresponds
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modulating h(n) by (- I)":
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Therefore, if h(n) is causal and stable, so is g(t7).
Find the inverse system of
The inverse system is
TRANSFORM ANALYSIS OF SYSTEMS
[CHAP. 5
and there are two possible regions of convergence: Izl > 2 and lzl < 2. Because both of these overlap with the region of convergence for H(z), both are valid inverse systems. For J z J 2, the unit sample response is >
which is a causal but unstable system. For (zl < 2, the unit sample response is ~ ( n= -2"u(-n )
+ 0.6(2)"-'u(-n)
which is stable but noncausal. Note. however, that the system
is both stable and causal, and the magnitude of the frequency response is the same as that of the inverse system. Therefore, this system is realizable, and the system that is the cascade of H ( z ) with G ( z )has a frequency response with a magnitude of I .
Let H ( z ) be a stable and causal filter with a system function
(a) Make a pole-zero plot of the system function, and use geometric arguments to show that if r the system is a notch filter.
(h) At what frequency does I H ( e I w ) l reach its maximum value
and a pair of complex poles just inside ( a ) This system has a pair of complex zeros on the unit circle at z = A pole-zero diagram for H ( z ) is shown in the figure below. the unit circle at z =
The first thing to note is that, due to the zeros that are on the unit circle, the frequency response goes to zero at w = i~wo. The second thing to observe is that, as we move away from the unit circle zeros, the lengths of the vectors from the poles to the unit circle approach the lengths of the vectors from the zeros to the unit circle. Furthermore, the closer r is to I , the more rapidly the lengths of these vectors become the same. Therefore, if r 1, H(e1") is a notchfilter, with a frequency response that is approximately constant except within a narrow band of frequencies around w = fwo, where the frequency response goes to zero.
(17)
The magnitude of the frequency response increases monotonically as we more away from the unit circle zeros. Therefore, I H(P]")Iwill reach its maximum value either at w = 0 or w = n. Because the frequency response at w = 0 is
(I - e J w ) ( l - C J W ) 2 - 2coswo = A 1 -r e -r e I r Z - 2r cos w0
CHAP. 51
TRANSFORM ANALYSIS OF SYSTEMS and the frequency response at w = n is
I H(ejo)l will reach its maximum value at w = 0 if n/2 < wo w=1~ifO<w~<~/2.
A signal y ( n ) contains a primary signal, x ( n ) , plus two echos:
~r, and
it will reach its maximum value at
Find a realizable flter that will recover x ( n ) from y ( n ) .
Because Y ( z ) is related to X ( z ) as follows:
the inverse filter is
We must check. however, to see whether or not this filter is realizable First. note that we may write G ( z ) as
G ( z )= F (znd )
where The poles of F ( z ) are at
F ( z )=
I I + +-I + iz-1
z = -;(I
&j h )
which are inside the unit circle at a radius of r = 0.5. Therefore, the poles of G ( z ) are inside the unit circle, at a radius of r' = (0.5)-"d, and G ( z )is realizable.
A causal system with areal-valued unit sample response has a frequency response with areal part given by
HR(eJ") 1 =
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