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the poles and zeros of H ( 2 ) occur in conjugate pairs.
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It follows from the symmetry property of the z-transform that if h ( n )is real. H ( z ) = Ht(z*).Therefore,
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Without evaluating the inverse z-transform, determine which of the following z-transforms could be the system function of a causul hut not necessurily stable discrete-time linear shift-invariant system:
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(Z - I l3 (b) X(2) = - -
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A causal sequence is one that is equal to zero for n < 0. Therefore, the z-transform of a causal sequence may be written as a one-sided summation:
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TRANSFORM ANALYSIS OF SYSTEMS
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[CHAP. 5
What distinguishes the z-transform of a causal signal from one that is not is the fact that X(z) does not contain any positive powers of z. Consequently, if we let lzl -t m, X(z) + x(O), which is a statement of the initial value theorem. It follows, therefore, that if .r(n) is causal, this limit must be finite. For noncausal signals, on the other hand, this limit will tend to infinity, because the z-transform will contain positive powers of z. For example, the sequence x(n) = u ( n I) has a z-transform
lim X(z) = m
:I-CU
Thus, a z-transform may be the system function of a causal system only if
Of the transforms listed, only ( a ) and ( c )have a finite limit as Izl + rn and, therefore, are the only ones that could be the z-transform of a causal signal.
The result of a particular computer-aided filter design is the following causal second-order filter:
H (z) =
1 - 2z-I
+ 22-' + zp2 +1.33~-~
Show that this filter is unstable, and find a causal and stable filter that has the same magnitude response
as H(z).
This filter is clearly not stable, because the coefficient for z-' in the denominator, which is the product of the roots of H(z), is greater than 1. Specifically, if the poles of H(z) are crl and crz, then a l . a = 1.33, and this implies that 2 at least one of the roots is outside the unit circle. Because the discriminant of the polynomial is negative,
the roots are complex with crl = reJ%nd cr2 = re-IH where ,- = and % = c o s p ' ( l / m ) . Recall that if we form a new system function given by Hf(z)= H(z)G,,(z), where Gap(z)is an allpass filter of the form
I ~ ' ( e j " ) l= I H (ejW)l.Therefore, if
Gap(z)=
I - 2:-I 1 .33zr2 1.33 - 22-I + z r 2
the effect of Gap(:) is to replace the pair of complex poles in H(z) that are outside the unit circle with a complex pole pair inside the unit circle at the reciprocal locations while preserving the magnitude response. Thus. a stable filter that has a frequency response with the same magnitude as H(eJW) the following: is H '(z) =
1 2z-I + z 1.33 - 22-1 + zp2
The system function of a discrete-time linear shift-invariant system is H (z). Assume that H ( z ) is a rational
function of z and that H ( z ) is causal and stable. Determine which of the following systems are stable and which are causal:
(a) G ( z ) = H(z)H*(z*)
d (b) G ( z ) = H1(z) where H1(z) = -[ H ( z ) ]
CHAP. 51
TRANSFORM ANALYSIS OF SYSTEMS
(c) G(z) = ~ ( z - I )
(d) G ( z ) = H(-Z)
With H(z) a rational function of z , if h(n) is stable and causal, the poles of H(z) (if any) are inside the unit circle, and the region of convergence is the extenor of a circle and includes the unit circle.
( a ) If H(z) is the z-transform of h(n), then H*(z*)is the z-transfonn of h*(n), and the region of convergence is the same as that for H(z). Because the region of convergence of G(:) = H(z)H*(i*)includes the regions of convergence of H(z) and H*(z*),the region of convergence of G(z) will be the exterior of a circle and include the unit circle. Therefore, g(n) is stable and causal.
( h ) Recall that if H(z) is the z-transform of h(n),
Therefore, delaying the sequence nh(n) by I yields the following z-transform pair:
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