the z-transform and discrete-time lti systems
TRANSCRIPT
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SIGNALS, SPECTRA, AND SIGNAL PROCESSING
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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
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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.
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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
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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.
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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:
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The variable z is generally complex-valued and is expressed in polar form as
Then X(z) can be expressed as
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Example 1Determine the z-transforms of the following finite-duration signals.
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Solution:
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Example 2Determine the z-transform of the signal
Solution:
The z-transform of x(n) is the infinite power series
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This is an infinite geometric series. Recall that
Consequently, for |(1/2)z-1| < 1, or equivalently, for |z| > ½, X(z) converges to
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Example 3Determine the z-transform of the signal
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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
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The exponential signal x(n) = αnu(n)
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The ROC of the z-transform of x(n) = αnu(n)
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Properties of the z-Transform1.Linearity
If
then
where a1 and a2 are arbitrary constants.
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Example 4Determine the z-transform and the ROC of the signal
Solution:If we define the signals
andthen
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Recall that
By setting α = 2 and α = 3, we obtain
Therefore,
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2. Time shiftingIf
Then
Special cases:
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Example 5Determine the transform of the signal
Solution:
Since x(n) has finite duration, its ROC is the entire z-plane, except z = 0.
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3. Scaling in the z-domainIf
Then
For any constant a, real or complex.
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Example 6Determine the z-transform of the signal
Solution:From Table 3.3
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4. Time-reversalIf
then
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5. Multiplication by n (Differentiation in z)If
then
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Example 7Determine the signal x(n) whose z-transform is given by
Solution:By taking the first derivative of X(z), we obtain
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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
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6. AccumulationIf
then
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7. Convolution of two sequencesIf
then
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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.
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2. Multiply the two z-transforms.
3. Find the inverse z-transforms of X(z).
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Example 8Compute the convolution x(n) of the signals
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Solution 1:
Multiply X1(z) and X2(z). Thus
or
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Solution 2:
Multiply the two signals
or
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Rational z-transformMany of the signals of interest in digital signal
processing have z-transforms that are rational functions of z:
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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:
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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) = ∞.
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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)
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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
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Three possible approaches of the inverse z-transform:
1. Contour Integration
2. Power Series Expansion
3. Partial Fraction Expansion
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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
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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.
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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.
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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
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Example 9Determine the inverse z-transform of
when (a) ROC: |z| > 1(b) ROC: |z| < 0.5
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IN this case the ROC is the exterior of a circle and the x(n) is a causal signal.
Thus
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In this case the ROC is the interior of a circle and the signal x(n) is anticausal.
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Thus
Therefore
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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.
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Example 10Express the improper rational function
In terms of a polynomial and a proper function.
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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.
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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
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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).
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Case 1: Distinct poles
Where: p1, p2, …, pN are the poles and A1, A2, …, AN are the coefficient that need to be determined.
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Example 11Determine the partial fraction expansion of the proper function
Solution:First, multiply z2 to both numerator and denominator. Thus
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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
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Case 2: Multiple-order poles
Example 12:Determine the partial fraction expansion of
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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.
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To determine A1, we multiply both sides of the equation by (z + 1) and evaluate the result at z = -1. Thus
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To determine A3, we multiply both sides of the equation by (z - 1)2 and evaluate the result at z = 1. Thus
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To determine A2, we differentiate both sides of the equation with respect to z and evaluate the result at z = 1. Thus
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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.
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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.
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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.
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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
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where X1(z) = z-1. Substituting for y(-1) and y(-2), and solving for Y1(z), we have
Therefore
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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
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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.
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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.
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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.
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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).
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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.
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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
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or
Thus,
Hence, H(z) is always rational.
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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
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Similarly, the impulse response of a parallel combination of two LTI systems is given by
and
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QUESTIONS
THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS