The Z-transform and Discrete-time Lti Systems

SIGNALS, SPECTRA, AND SIGNAL PROCESSING

THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS


Objectives:1. To define the z-transform2. To determine the properties of the z-transform3. To describe the methods for inverting the z-

transform of a signal so as to obtain the time-domain representation of the signal

4. To demonstrate the importance of the z-transform in the analysis and characterization of linear time-invariant systems

IntroductionThe z-transform is used to represent digital-time signals or sequences in the z-domain (z is a complex variable).The z-transform converts differential equations into algebraic equations, thereby simplifying the analysis of discrete-time systems.


The z-transformThe z-transform of a discrete-time signal x(n) is

defined as the power series

For convenience, the z-transform is denoted by

whereas the relationship between x(n) and X(z) is indicated by


The region of convergence (ROC) of X(z) is the set of all values of z for which X(z) attains a finite value.Three properties of the ROC:

1. A finite-length sequence has a z-transform with a region of convergence that includes the entire z-plane except, possibly, z = 0 and z = ∞. The point z = ∞ will be included if x(n) = 0 for n < 0, and the point z = 0 will be included if x(n) = 0 for n > 0.


2. A right-sided sequence has a z-transform with a region of convergence that is the exterior of a circle:

3. A left-sided sequence has a z-transform with a region of convergence that is the interior of a circle:


The variable z is generally complex-valued and is expressed in polar form as

Then X(z) can be expressed as


Example 1Determine the z-transforms of the following finite-duration signals.


Solution:


Example 2Determine the z-transform of the signal

Solution:

The z-transform of x(n) is the infinite power series


This is an infinite geometric series. Recall that

Consequently, for |(1/2)z-1| < 1, or equivalently, for |z| > ½, X(z) converges to


Solution:

If |αz-1| < 1 or equivalently |z| > |α|, this power series converges to 1/(1 - αz-1).Thus we have the z-transform pair


The exponential signal x(n) = αnu(n)


The ROC of the z-transform of x(n) = αnu(n)


Properties of the z-Transform1.Linearity

If

then

where a1 and a2 are arbitrary constants.


Example 4Determine the z-transform and the ROC of the signal

Solution:If we define the signals

andthen


Recall that

By setting α = 2 and α = 3, we obtain

Therefore,


2. Time shiftingIf

Then

Special cases:


Example 5Determine the transform of the signal

Solution:

Since x(n) has finite duration, its ROC is the entire z-plane, except z = 0.


3. Scaling in the z-domainIf

Then

For any constant a, real or complex.



Solution:From Table 3.3


4. Time-reversalIf

then


5. Multiplication by n (Differentiation in z)If

then


Example 7Determine the signal x(n) whose z-transform is given by

Solution:By taking the first derivative of X(z), we obtain


Thus

The inverse z-transform of the term in the brackets is (-a)n. The multiplication by z-1 implies a time delay by one sample, which results in (-a)n-1u(n-1). Finally, from the differentiation property we have


6. AccumulationIf

then


7. Convolution of two sequencesIf

then


Computation of the convolution of two signals, using the z-transform, requires the following steps:

1. Compute the z-transforms of the signals to be convolved.


2. Multiply the two z-transforms.

3. Find the inverse z-transforms of X(z).


Example 8Compute the convolution x(n) of the signals


Solution 1:

Multiply X1(z) and X2(z). Thus

or


Solution 2:

Multiply the two signals

or


Rational z-transformMany of the signals of interest in digital signal

processing have z-transforms that are rational functions of z:


If a0 ≠ 0 and b0 ≠ 0, we can avoid the negative powers of z by factoring out the terms b0z-M and a0z-N

as follows:


Since N(z) and D(z) are polynomials in z, they can be expressed in factored form as

Where z1, z2, …, zM values are zeros of a z-transform for which x(z) = 0 and p1, p2, …, pN are poles of a z-transform for which x(z) = ∞.


The Inverse z-TransformTo begin,

Suppose we multiply both sides by zn-1 and integrate both sides over a closed contour within the ROC of X(z). Thus,

Where C denotes the closed contour in the ROC of X(z)


Cauchy integral theorem states that

By applying the theorem to the right hand side of the equation above reduces to 2πjx(n) and hence the desired inversion formula


Three possible approaches of the inverse z-transform:

1. Contour Integration

2. Power Series Expansion

3. Partial Fraction Expansion


Contour IntegrationThis procedure relies on Cauchy's integral theorem, which states that if C is a closed contour that encircles the origin in a counterclockwise direction,

With


Cauchy's integral theorem may be used to show that the coefficients x(n) may be found from X(z) as follows:

where C is a closed contour within the region of convergence of X(z) that encircles the origin in a counterclockwise direction.


Contour integrals of this form may often by evaluated with the help of Cauchy's residue theorem,

If X(z) is a rational function of z with a first-order pole at z = αk,

Contour integration is particularly useful if only a few values of x(n) are needed.


Power Series ExpansionGiven a z-transform X(z) with its corresponding ROC, we expand the X(z) into a power series of the form

which converges in the given ROC


Example 9Determine the inverse z-transform of

when (a) ROC: |z| > 1(b) ROC: |z| < 0.5


IN this case the ROC is the exterior of a circle and the x(n) is a causal signal.

Thus


In this case the ROC is the interior of a circle and the signal x(n) is anticausal.


Thus

Therefore


Partial Fraction ExpansionRecall:

If we assume a0 = 1, then

Note that x(z) is called proper if M < N and aN ≠ 0 and x(z) is called improper if M ≥ N which can be written as the sum of a polynomial and a proper rational function.


Example 10Express the improper rational function

In terms of a polynomial and a proper function.


Solution:First, we should carry out the long division with these two polynomials in reverse order. We stop the division when the order of the remainder becomes z-1. Then we obtain.


Let X(z) be a proper rational function, that is,

where

By multiplying zN both to the numerator and denominator,

which contains only positive powers of z


Since N > M

In performing a partial fraction expansion, we first factor the denominator polynomial into factors that contain the poles p1, p2, …, pN of X(z).


Case 1: Distinct poles

Where: p1, p2, …, pN are the poles and A1, A2, …, AN are the coefficient that need to be determined.


Example 11Determine the partial fraction expansion of the proper function

Solution:First, multiply z2 to both numerator and denominator. Thus


To solve for A1 and A2, multiply the equation by the denominator term (z – 1)(z – 0.5). Thus

Let z = p1 = 1 Let z = p2 = 0.5

Therefore


Case 2: Multiple-order poles

Example 12:Determine the partial fraction expansion of


Solution:First we express the z-transform in terms of positive powers of z

X(z) has a simple pole p1 = -1 and a double pole p2 = p3 = 1.


To determine A1, we multiply both sides of the equation by (z + 1) and evaluate the result at z = -1. Thus


To determine A3, we multiply both sides of the equation by (z - 1)2 and evaluate the result at z = 1. Thus


To determine A2, we differentiate both sides of the equation with respect to z and evaluate the result at z = 1. Thus


The One-Sided z-TransformThe one-sided, or unilateral, z-transform is defined by

The primary use of the one-sided z-transform is to solve linear constant coefficient difference equations that have initial conditions.


Most of the properties of the one-sided z-transform are the same as those for the two-sided z-transform. One that is different, however, is the shift property. Specifically, if x(n) has a one-sided z-transform X1(z), the one-sided z-transform of x(n - 1) is

It is this property that makes the one-sided z-transform useful for solving difference equations with initial conditions.


Example 13Consider the linear constant coefficient difference equation

Let us find the solution to this equation assuming that x(n) = δ(n - 1) with y(-1) = y(-2) = 1.


We begin by noting that if the one-sided z-transform of y(n) is Y1(z), the one-sided z-transform of y(n -2) is

Therefore, taking the z-transform of both sides of the difference equation, we have


where X1(z) = z-1. Substituting for y(-1) and y(-2), and solving for Y1(z), we have

Therefore


The System Function of Discrete-Time LTI A.The System Function

The output y[n] of a discrete-time LTI system equals the convolution of the input x[n] with the impulse response h[n] is

Applying the convolution property of the z-transform, we obtain


The equation can be expressed as

The z-transform H(z) of h[n] is referred to as the system function (or the transfer function) of the system.


B. Characterization of Discrete-Time LTI Systems1. Causality

For a causal discrete-time LTI system, we have

since h[n] is a right-sided signal, the corresponding requirement on H(z) is that the ROC of H(z) must be of the form

the ROC is the exterior of a circle containing all of the poles of H(z) in the z-plane.


if the system is anticausal, that is,

then h[n] is left-sided and the ROC of H(z) must be of the form

the ROC is the interior of a circle containing no poles of H(z) in the z-plane.


2. StabilityA discrete-time LTI system is BIB0 stable if and only if

The corresponding requirement on H(z) is that the ROC of H(z) contains the unit circle (that is, lzl = 1).


3. Causal and Stable SystemsIf the system is both causal and stable, then all of the poles of H(z) must lie inside the unit circle of the z-plane because the ROC is of the form lzl > rmax, and since the unit circle is included in the ROC, we must have rmax < 1.


C. System Function for LTI Systems Described by Linear Constant-Coefficient Difference Equations

The general linear constant-coefficient difference equation

Applying the z-transform and using the time-shift property and the linearity property of the z-transform, we obtain


or

Thus,

Hence, H(z) is always rational.


D. Systems InterconnectionFor two LTI systems (with h1[n] and h2[n], respectively) in cascade, the overall impulse response h[n] is given by

Thus, the corresponding system functions are related by the product


Similarly, the impulse response of a parallel combination of two LTI systems is given by

and


QUESTIONS


The Z-transform and Discrete-time Lti Systems

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