barcode generator for ssrs For example, if H ( z ) is a rational function of z as given in Eq. (5.2).the inverse system is in Software

Encoder USS Code 128 in Software For example, if H ( z ) is a rational function of z as given in Eq. (5.2).the inverse system is

For example, if H ( z ) is a rational function of z as given in Eq. (5.2).the inverse system is
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Thus, the poles of H ( z ) become the zeros of G(z), and the zeros of H ( z ) become the poles of G(z). The region of convergence that is associated with the inverse system is determined by the requirement that H ( z ) and G ( z ) have overlapping regions of convergence.'
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EXAMPLE 5.2.1
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the inverse system is
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There are two possible regions of convergence for ,q(n). The ti rst is IzJ > f ,and the second is lzI i$. Because lzl < f does not overlap the region of convergence for H(z), the only possibility for the inverse system is l z l > {. In this case, the unit sample response is ~ ( n = () u(n) - 0 . 8 ( ~ ) " - ' u ( n- I ) ) 1"
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'If this were not the case, H ( r ) G ( z ) would not be the identity system. because the region of convergence would be empty.
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CHAP. 51
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TRANSFORM ANALYSIS OF SYSTEMS
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which is stable and causal. However, suppose that
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H (z) =
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0.5 - Z - ' I - 0.82-'
Izl > 0.8
In this case, the inverse system is
where the region of convergence may be either (zI > 2 or IzI c 2. Because both regions of convergence overlap the region of convergence of H ( z ) , both are valid inverse systems. The first, which has a region of convergence li( > 2, has a unit sample response ~ ( n = 2(2)"u(n) - 1,6(2)"-'u(n -- I) ) and is causal but unstable. The second, with a region of convergence lzl < 2, has a unit sample response
and is stable but noncausal.
5.2.3 Unit Sample Response for Rational System Functions
A linear shift-invariant system with a rational system function may be written in factored form as follows:
k Assuming only first-order poles, with a # fraction expansion as follows:
pr for all k and I, if p
> q , H ( z ) may be expanded using a partial
If the system is causal, the unit sample response is
When p I the partial fraction expansion has the form q,
and, if the system is causal, the unit sample response becomes
If p = 0,H ( z ) has only zeros,
and h(n) is finite in length with
TRANSFORM ANALYSIS OF SYSTEMS
[CHAP. 5
These systems are called finite-length impulse response (FIR) filters. If p r 0, H(z) is infinite in length, and these systems are called infinite-length impulse response (IIR) filters. If h(n) is real, H (z) = H *(z*), and the complex poles and zeros of H (z) occur in complex-conjugate pairs. is l For example, if ak = rkeJWAa complex-valued pole, a; = r k e p J W h i l also be a pole. This symmetry implies that the complex terms in Eq. (5.5) may be combined to form terms of the form
5.2.4 Frequency Response for Rational System Functions
The frequency response of a linear shift-invariant system may be found from the system function by evaluating H(z) on the unit circle. For a rational function of z, the frequency response may be found geometrically from the poles and zeros of H(z). With H(z) written in factored form as in Eq. (5.4), the frequency response is
Because the magnitude of the frequency response is
IH(ejW)lis IAl times the product of the terms I I - bke-jwl divided by the product of the terms 11 - ake-JWI. Each term in the numerator 11 - ,#ke-j"l = lejw - BkI is the length of the vector from the zero at z = Bk to the unit circle at z = ej" (labeled vl in Fig. 5-2). Similarly, each term in the denominator
11 - uke-jol = lej" - a k I
is the length of the vector from the pole at z = a k to the unit circle at z = ejw (labeled v2 in Fig. 5-2). When a pole is close to the unit circle, a k = rkejwk with r k = 1, the magnitude of the frequency response becomes large for w = w because the length of the vector from the pole to the unit circle becomes small. Similarly, if there is k with r k 1, the magnitude of the frequency response becomes small a zero close to the unit circle, Bk = rkeJWk k ~ ~ for w = w (if the zero is on the unit circle at z = ej"', ~ ( e j = 0). ) The analysis for the phase is similar. Assuming that A is a positive real number, the phase corresponding to the frequency response H (el") given by Eq. (5.7) is
Thus, & ( a ) is the sum of the phases associated with the terms (1 - Bke-jw),minus the sum of the phases of the terms (1 - ake-Jw). Because . . 1 - pke-jo = e-Jw(eJw - B k
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