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Continuous-time signals and systems are commonly analyzed using the Fourier trans-form and the Laplace transtrans-form (will be introduced in Chapter 6). For discrete-time systems,the transform corresponding to the Laplace transform is the z-transform. The z-transform yields a frequency-domain description of discrete-time signals and systems, and provides a powerful tool in the design and implementation of digital filters. In this section,we will introduce the z-transform,discuss some important properties,and show its importance in the analysis of linear time-invariant (LTI) systems.

4.2.1 Definitions and Basic Properties

The z-transform (ZT) of a digital signal, x…n†, 1 < n < 1,is defined as the power series

X…z† ˆ X¹

nˆ 1

x…n†z ⁿ, …4:2:1†

where X…z† represents the z-transform of x…n†. The variable z is a complex variable,and can be expressed in polar form as

z ˆ re^jy, …4:2:2†

where r is the magnitude (radius) of z,and y is the angle of z. When r ˆ 1, jzj ˆ 1 is called the unit circle on the z-plane. Since the z-transform involves an infinite power series,it exists only for those values of z where the power series defined in (4.2.1)

converges. The region on the complex z-plane in which the power series converges is called the region of convergence (ROC).

As discussed in Section 3.1,the signal x…n† encountered in most practical applications is causal. For this type of signal,the two-sided z-transform defined in (4.2.1) becomes a one-sided z-transform expressed as

X…z† ˆX¹

nˆ0

x…n†z ⁿ: …4:2:3†

Clearly if x…n† is causal,the one-sided and two-sided z-transforms are equivalent.

Example 4.5: Consider the exponential function x…n† ˆ aⁿu…n†:

The z-transform can be computed as

X…z† ˆ X¹

nˆ 1

aⁿz ⁿu…n† ˆX¹

nˆ0

…az ¹†ⁿ:

Using the infinite geometric series given in Appendix A.2,we have

X…z† ˆ 1

1 az ¹ if jaz ¹j < 1:

The equivalent condition for convergence (or ROC) is jzj > jaj:

Thus we obtain X…z† as

X…z† ˆ z

z a, jzj > jaj:

There is a zero at the origin z ˆ 0 and a pole at z ˆ a. The ROC and the pole±zero plot are illustrated in Figure 4.2 for 0 < a < 1,where `' marks the position of the pole and `o' denotes the position of the zero. The ROC is the region outside the circle with radius a. Therefore the ROC is always bounded by a circle since the convergence condition is on the magnitude jzj. A causal signal is characterized by an ROC that is outside the maximum pole circle and does not contain any pole.

The properties of the z-transform are extremely useful for the analysis of discrete-time LTI systems. These properties are summarized as follows:

1. Linearity (superposition). The z-transform is a linear transformation. Therefore the z-transform of the sum of two sequences is the sum of the z-transforms of the individual sequences. That is,

134 FREQUENCY ANALYSIS

|z| = a

|z| = 1

Re[z]

Im[z]

Figure 4.2 Pole,zero,and ROC (shaded area) on the z-plane

ZT‰a1x1…n† ‡ a2x2…n†Š ˆ a1ZT‰x1…n†Š ‡ a2ZT‰x2…n†Š

ˆ a₁X₁…z† ‡ a₂X₂…z†, …4:2:4†

2. where a1 and a2 are constants,and X1…z† and X2…z† are the z-transforms of the signals x1…n† and x2…n†,respectively. This linearity property can be generalized for an arbitrary number of signals.

2. Time shifting. The z-transform of the shifted (delayed) signal y…n† ˆ x…n k† is

Y…z† ˆ ZT‰x…n k†Š ˆ z ^kX…z†, …4:2:5†

2. where the minus sign corresponds to a delay of k samples. This delay property states that the effect of delaying a signal by k samples is equivalent to multiplying its z-transform by a factor of z ^k. For example,ZT‰x…n 1†Š ˆ z ¹X…z†. Thus the unit delay z ¹in the z-domain corresponds to a time shift of one sampling period in the time domain.

3. Convolution. Consider the signal

x…n† ˆ x₁…n†  x₂…n†, …4:2:6†

2. where  denotes the linear convolution introduced in Chapter 3,we have

X…z† ˆ X1…z†X2…z†: …4:2:7†

2. Therefore the z-transform converts the convolution of two time-domain signals to the multiplication of their corresponding z-transforms.

Some of the commonly used signals and their z-transforms are summarized in Table 4.3.

Table 4.3 Some common z-transform pairs

The inverse z-transform can be expressed as

x…n† ˆ ZT ¹‰X…z†Š ˆ 1 2pj

‡

CX…z†z^{n 1}dz, …4:2:8†

where C denotes the closed contour in the ROC of X…z† taken in a counterclockwise direction. Several methods are available for finding the inverse z-transform. We will discuss the three most commonly used methods ± long division,partial-fraction expan-sion,and residue method.

Given the z-transform X…z† of a causal sequence,it can be expanded into an infinite series in z ¹or z by long division. To use the long-division method,we express X…z† as the ratio of two polynomials such as

X…z† ˆB…z†

where A…z† and B…z† are expressed in either descending powers of z,or ascending powers of z ¹. Dividing B…z† by A…z† obtains a series of negative powers of z if a positive-time sequence is indicated by the ROC. If a negative-time function is indicated,we express X…z† as a series of positive powers of z. The method will not work for a sequence defined

136 FREQUENCY ANALYSIS

in both positive and negative time. In addition,it is difficult to obtain a closed-form solution of the time-domain signal x…n† via the long-division method.

The long-division method can be performed recursively. That is,

x…n† ˆ b_n Xⁿ

mˆ1

x…n m†a_m

" #,

a₀, n ˆ 1,2, . . . …4:2:10†

where

x…0† ˆ b0=a0: …4:2:11†

This recursive equation can be implemented on a computer to obtain x…n†.

Example 4.6: Given

X…z† ˆ 1 ‡ 2z ¹‡ z ² 1 z ¹‡ 0:3561z ² using the recursive equation given in (4.2.10),we have

x…0† ˆ b₀=a₀ ˆ 1,

x…1† ˆ ‰b1 x…0†a1Š=a0ˆ 3,

x…2† ˆ ‰b2 x…1†a1 x…0†a2Š=a0 ˆ 3:6439, . . .

This yields the time domain signal x…n† ˆ f1,3,3:6439, . . .g obtained from long division.

The partial-fraction-expansion method factors the denominator of X…z† if it is not already in factored form,then expands X…z† into a sum of simple partial fractions.

The inverse z-transform of each partial fraction is obtained from the z-transform tables such as Table 4.3,and then added to give the overall inverse z-transform. In many practical cases,the z-transform is given as a ratio of polynomials in z or z ¹ as shown in (4.2.9). If the poles of X…z† are of first order and M ˆ L 1,then X…z† can be expanded as

X…z† ˆ c0‡X^{L 1}

lˆ1

cl

1 p_lz ¹ˆ c0‡X^{L 1}

lˆ1

clz

z p_l, …4:2:12†

where p_l are the distinct poles of X…z† and c_l are the partial-fraction coefficients. The coefficient c_lassociated with the pole p_lmay be obtained with

c_lˆX…z†

z …z p_l† zˆpl

: …4:2:13†

If the order of the numerator B(z) is less than that of the denominator A(z) in (4.2.9), that is L 1 < M,then c₀ ˆ 0. If L 1 > M,then X(z) must be reduced first in order to make L 1  M by long division with the numerator and denominator polynomials written in descending power of z ¹.

Example 4.7: For the z-transform

X…z† ˆ z ¹

1 0:25z ¹ 0:375z ² we can first express X(z) in positive powers of z,expressed as

X…z† ˆ z

z² 0:25z 0:375ˆ z

…z 0:75†…z ‡ 0:5†ˆ c1z

z 0:75‡ c2z z ‡ 0:5: The two coefficients are obtained by (4.2.13) as follows:

c1ˆX…z†

The overall inverse z-transform x(n) is the sum of the two inverse z-transforms.

From entry 3 of Table 4.3,we obtain

x…n† ˆ 0:8‰…0:75†ⁿ … 0:5†ⁿŠ , n  0:

The MATLAB function residuez finds the residues,poles and direct terms of the partial-fraction expansion of B…z†=A…z† given in (4.2.9). Assuming that the numerator and denominator polynomials are in ascending powers of z ¹,the function

[c, p, g]= residuez(b, a);

finds the partial-fraction expansion coefficients, cl,and the poles,pl,in the returned vectors c and p,respectively. The vector g contains the direct (or polynomial) terms of the rational function in z ¹if L 1  M. The vectors b and a represent the coefficients of polynomials B(z) and A(z),respectively.

If X(z) contains one or more multiple-order poles,the partial-fraction expansion must include extra terms of the form P_m

jˆ1 gj

…z pl†^j for an mth order pole at z ˆ p_l. The coefficients g_j may be obtained with

138 FREQUENCY ANALYSIS

g_jˆ 1

Example 4.8: Consider the function

X…z† ˆ z²‡ z …z 1†²: We first express X(z) as

X…z† ˆ g₁

The residue method is based on Cauchy's integral theorem expressed as 1

Thus the inversion integral in (4.2.8) can be easily evaluated using Cauchy's residue theorem expressed as

The residue of X…z†z^{n 1}at a given pole at z ˆ p_l can be calculated using the formula

where m is the order of the pole at z ˆ pl. For a simple pole,Equation (4.2.17) reduces to Rzˆpl ˆ …z pl†X z†z^{n 1}

zˆpl: …4:2:18†

Example 4.9: Given the following z-transform function:

X…z† ˆ 1

For the case that n  1,the residue theorem is applied to obtain x…n† ˆ R_zˆ1‡ R_zˆ0:5

We have discussed three methods for obtaining the inverse z-transform. A limitation of the long-division method is that it does not lead to a closed-form solution. However, it is simple and lends itself to software implementation. Because of its recursive nature, care should be taken to minimize possible accumulation of numerical errors when the number of data points in the inverse z-transform is large. Both the partial-fraction-expansion and the residue methods lead to closed-form solutions. The main disadvan-tage with both methods is the need to factor the denominator polynomial,which is done by finding the poles of X(z). If the order of X(z) is high,finding the poles of X(z) may be

140 FREQUENCY ANALYSIS

a difficult task. Both methods may also involve high-order differentiation if X(z) contains multiple-order poles. The partial-fraction-expansion method is useful in gen-erating the coefficients of parallel structures for digital filters. Another application of z-transforms and inverse z-z-transforms is to solve linear difference equations with constant coefficients.

In document Presentación del Secretario General de Sanidad y Consumo (página 132-137)