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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
For example, if H ( z ) is a rational function of z as given in Eq. (5.2).the inverse system is Read Code 128A In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128C Creator In None Using Barcode drawer for Software Control to generate, create USS Code 128 image in Software applications. 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.' Recognize Code 128C In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Code 128B Printer In Visual C# Using Barcode encoder for .NET Control to generate, create Code128 image in .NET applications. EXAMPLE 5.2.1
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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 shiftinvariant system with a rational system function may be written in factored form as follows: k Assuming only firstorder 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 finitelength impulse response (FIR) filters. If p r 0, H(z) is infinite in length, and these systems are called infinitelength impulse response (IIR) filters. If h(n) is real, H (z) = H *(z*), and the complex poles and zeros of H (z) occur in complexconjugate pairs. is l For example, if ak = rkeJWAa complexvalued 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 shiftinvariant 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  bkejwl divided by the product of the terms 11  akeJWI. Each term in the numerator 11  ,#kej"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. 52). Similarly, each term in the denominator 11  ukejol = 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. 52). 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  Bkejw),minus the sum of the phases of the terms (1  akeJw). Because . . 1  pkejo = eJw(eJw  B k

