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THE -TRANSFORM AND ITS APPLICATION TO THE ANALYSIS OF LTI SYSTEMS

3.0. Introduction

Transform techniques are an important tool in the analysis of signals and linear time-invariant (LTI)

systems. This chapter introduces thez-transform, develop its properties and demonstrate its importance inthe analysis and characterization of linear time-invariant systems.Thez-transform plays the same role in the analysis of discrete-time signals and LTI systems as theLaplace transform does in the analysis of continuous-time signals and LTI systems. Many properties of thez-transform can be applied to simplify the analysis of LTI systems, and the idea of poles and zeros provideanother perspective in characterizing the response of the system.3.1. Thez-TransformThe direct -transform. The direct

z-transform of a discrete-time signal

xn is defined as the power

series

Xz xnz (3.1)wherez is a complex variable ( re). The directz-transform transforms the time-domain signalxn intoits complex-plane representationXz. The inverse procedure, that is, obtainingxnfromXz, is calledthe inversez-transform. Thez-transform of a signalxn is denoted by

Xz Zxn (3.2)whereas the relationship betweenxn andXz is indicated by

xn Xz (3.3)Since thez-transform is an infinite power series, it exists only for those values ofz for which the seriesconverges. The region of convergence (ROC) ofXz is the set of all values ofz for whichXz attains afinite value. Thus, anytime that the

z-transform of a function is cited, its ROC must also be cited.

Example 3.1 Determine thez-transform of the following finite-duration signals.(a)xn 1 2 5 7 0 1 (b)xn 1 2 5 7 0 1

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(c)xn 0 0 1 2 5 7 0 1 (d)xn n (e)

xn n k,

k 0

(f)xn n k,k 0 Answer: (a)Xz 1 2z 5z 7z z; ROC: the entirez-plane exceptz 0.(b)Xz z 2z 5 7z z; ROC: the entirez-plane exceptz 0 andz .(c)Xz z 2z 5z 7z z; ROC: the entirez-plane exceptz 0.(d)

Xz 1; ROC: the entire

z-plane.

(e)Xz z; ROC: the entirez-plane the entirez-plane exceptz 0.(f)Xz z; ROC: the entirez-plane the entirez-plane exceptz .Example 3.2 Determine the closed-form expression of thez-transform of the signalxn 12un Answer: Xz ;ROC:|z| The complex variablez expressed in polar form isz re, wherer |z|(magnitude ofz) and (angle or argument ofz). Then Equation 3.1 can be expressed as

Xz | xnre (3.4)In the ROC ofXz, |Xz | . But,

|Xz | xnre xnre |xnr| (3.5)

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Example 3.3 Determine thez-transform of the signals(a)xn aun (b)xn au n 1 Answers: (a)Xz ; ROC:|z| |a| (b)

Xz ; ROC:

|z| |a|

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The previous example raises two very important issues with regards to thez-transform:

The closed-form expression of thez-transform alone does not uniquely determine the signal.Specifying the ROC will remove the ambiguity. A discrete-time signalxn

is uniquely determined

by itsz-transformXz and the region of convergence ofXz. The ROC of a causal signal is the exterior of a circle of some radiusr while the ROC of an anti-causal signal is the interior of a circle of some radiusr.Example 3.4 Determine thez-transform of the signalxn aun bu n 1 Answer: If|b| |a|,X z does not exist because the ROCs of the first and second terms of the signal do notoverlap. If

|b| |a|,

Xz exists and is equal to

Xz b aa b z abz with ROC ofX z is|a| |z| |b|.

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To summarize, the ROC of thez-transform of a signal depends whether the signal has a finite or infiniteduration, and on whether it is causal, anti-causal or two-sided. These are summarized in Table 3.1 below.It should be noted that thez-transform defined by 3.1 is referred to as the two-sided or bilateralz-transform. The one-sided or unilateralz-transform is given by

Xz xnz (3.7)Note that when the signalxn is causal, the two-sided and the one-sidedz-transform are equal. In anyother case, they are different.

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The inverse -transform. The procedure for transforming from thez-domain to the time-domain is calledthe inversez-transform. An inversion formula for obtainingxn fromXz can be derived using theCauchy integral theorem, which is an important theorem in the theory of complex variables.To findxnfromXz, it can be shown that

xn 1j2 XzzdzC

(3.8)

where is any contour that encloses the origin within the ROC ofXz and taken in counterclockwisedirection. Although this integral provides the desired inversion formula, this shall not be used directly in theevaluation of inversez-transform. A lookup table will be developed where the forward and inversez-transform will be evaluated.

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3.2. Properties ofz-TransformLinearity. Ifxn Xzandxn Xz, then

xn axn axn Xz aXz aXz (3.9)or thez-transform of the linear combination ofxn is the linear combination of their transforms.

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Example 3.5 Determine thez-transform and the ROC of the signalxn 32 43un Answer:

Xz 31 2z 41 3z with ROC|z| 3.Example 3.6 Determine thez-transform of the signals(a)xn cosnu n (b)xn sinnun Answers: (a)Xz ; ROC:|z| 1 (b)Xz ; ROC:|z| 1 Time shifting.If xn X z, then

xn k

zXz (3.10)

or when shifting the signalxn byk number of samples, itsz-transform is multiplied by the factorz.The ROC ofzXz is the same as that ofXz except forz 0 ifk 0 andz ifk 0.Example 3.7 Ifxn 1 2 5 7 0 1, using itsz-transform, determine thez-transform of(a)x xn 2 (b)x xn 2 Answers: (a)Xz z 2z 5 7z z; ROC: the entirez-plane exceptz 0 andz .(b)Xz z 2z 5z 7z z; ROC: the entirez-plane exceptz 0.

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Example 3.8 Determine the transform of the signal

Answer:

Scaling in the -domain.Ifxn Xz, ROC:r |z| r, thenaxn Xaz; ROC:|a|r |z| |a|r (3.11)

for any constanta, real or complex. To better understand the meaning and implications of the scalingproperty, let a re and z re and let w az. Thus, Zxn Xz andZaxn Xw. It can be seen thatw az 1rr e (3.12)

Thus, when a signalxn is multiplied by an exponential (real or complex) this causes the signal to bescaled in z-domain, that is it shrinks when|a| r 1 or expands when |a| r 1 with acorresponding rotation (if 2k) of thez-plane. This is illustrated in Figure 3.7.

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Example 3.9 Determine thez-transforms of the signals(a)xn acosnun (b)

xn asinnun

Answers:

(a)Xz ; ROC:|z| |a| (b)Xz ; ROC:|z| |a| Time reversal.If xn Xz, ROCr |z| r, then

x n Xz ; ROC |z| (3.13)Example 3.10 Determine thez-transform of the signal xn un Answer: Xz ;ROC|z| 1 Differentiation in the -domain.Ifxn Xz, thennxn zdXzdz (3.14)Note that both transforms have the same ROC.

Example 3.11 Determine thez-transform of the signal

xn naun

Answer:

Xz ; ROC|z| |a|

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Example 3.12 Determine the signalxn whosez-transform is given byXz log1 az and ROC

|z| |a|

Answer:

xn 1an un 1 Convolution of two sequences .If xn Xzandxn Xz, then

xn xn xn Xz XzXz (3.15)The ROC ofXz is at least the intersection that forXzandXz. Note that convolving signals in time-domain causes theirz-transforms to be multiplied.Example 3.13 Compute the convolutionxn of the signals

Answer:

The convolution property is one of the most powerful properties of thez-transform because it converts theconvolution of two signals (time-domain) to multiplication of their transforms using the following steps:1. Compute thez-transform of the signals to be convolved.2. Multiply the two transforms.3. Find the inverse of the product in (2).

This procedure is, in many cases, computationally easier than the direct evaluation of the convolutionsummation.

Initial value theorem. Ifxn is causal, thenx0 limXz (3.16)

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Drill Problems 3.1 1. Determine thez-transform of the following signals:

2. Determine thez-transform of the following signals.

3. Determine thez-transform of the following signals and sketch the ROC .

4. Determine the

z-transform of the following signals

5. Express the

z-transform of

yn xk in terms ofXz.

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6. Compute the convolution of the following signals by means of thez-transform.

3.3. Rationalz-TransformsAn important family ofz-transforms are those for whichXz is a rational function, that is, a ratio of twopolynomials inz (orz).Poles and Zeros . The zeros of az-transformXz are the values ofz for whichXz 0. The poles of az-transform are the values ofz for whichXz . IfXz is a rational function, then

Xz NzDz b bz bMz Ma az az bzM az (3.17)Ifa 0 andb 0, negative powers ofz can be avoided by factoring out the termsbz M andaz,thus,

Xz NzDz bz Maz zM zM z z

(3.18)

SinceNz andDz are polynomials inz, they can be expressed in factored form asXz NzDz ba zM z zz z z zMz pz p z p

Xz GzM z zM z p (3.19)

whereG called the system gain. Thus,Xz has finite zeros atz z,z, ,zM (the roots of thenumerator polynomial),N finite poles atz p,p, ,p (the roots of the denominator polynomial) and|N| zeros (ifN ) or poles (ifN ) at the originz 0. Poles and zeros may also occur atz . A zero exists atz ifX 0 and a pole exists atz ifX . If we count thepoles and zeros at zero and infinity, there will be exactly an equal number of poles as zeros.

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Xz can be represented graphically by a pole-zero plot (or pattern) in the complex plane, which shows thelocation of poles by crosses ( and the location of zeros by circles (). The multiplicity of multiple-orderpoles or zeros is indicated by a number close to the corresponding cross or circle.

Example 3.14 Determine the pole zero plot for the signal xn aun fora 0.Answer:

Example 3.15 Determine the pole-zero plot for the signal

Answer:

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Example 3.16 Determine thez-transform and the signal that corresponds to the to the pole-zero plot below:

Answer:

Xz G 1 rzcos1 2rzcos rz with ROC|z| r. The signal corresponding to this isxn G rcosn un As was shown before, thez-transformXz is a complex function of the complex variablez Rzj Im z. Its magnitude,|Xz| is a real and positive function ofz. Since,z represents a point in thecomplex plane,

|Xz| is a two-dimensional function and describes a surface. For the transform

Xz z z1 1.2732z 0.81z the plot of its magnitude|Xz | is shown below.

. .

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and forz e,

Pole Location and Time-Domain Behavior for Causal Signals . The characteristic behavior of causalsignals depends whether the poles of the transform are contained in the region|z| 1, or in the region|z| 1 or on the circle|z| 1. Since the circle has a radius of1, it is called the unit circle.If a real signal has az-transform with one pole, this pole has to be real. The only such signal is the realexponential

xn aun Xz 11 az whose ROC is|z| |a| having one zero atz 0 and one pole atp a on the real axis. Figure 3.10summarizes the time-domain behavior of a single real pole causal signal as a function of the pole locationwith respect to the unit circle.

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A causal real signal with a double real pole has the form

xn naun and its behavior is summarized in figure 3.11.

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A causal signal with a complex conjugate pair of poles results in an exponentially weighted sinusoidalsignal, as shown in the Figure 3.12.

Two pairs of complex conjugate poles on the unit circle results in the following time-domain behavior.

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From the foregoing discussions, it can be observed that:

A causal signal decays if its poles (whether single or double real, or complex conjugates) is insidethe unit circle, has a bounded value when the poles are on the unit circle except when there aredouble poles in the unit circle, and grows when its poles are outside the unit circle.

When the poles on the unit circle have multiplicity greater than one, the resulting signal is one thatgrows over time.

For the complex conjugate poles, the distance of the poles from the origin dictates the envelope ofthe sinusoidal signal and their angle with the real positive axis dictates the relative frequency of thesignal.

A signal with a pole or complex conjugate poles near the origin decays more rapidly than the oneassociated with a pole near and inside the unit circle.

It should be remembered that everything that has been said about causal signals applies as well to causalLTI systems since their impulse response is a causal signal. Hence if a pole of a system is outside the unitcircle, the impulse response of the system becomes unbounded and consequently, the system is unstable.

System Function of an LTI System. The output of a relaxed, LTI system to an input sequencexn canbe obtained by computing the convolution sum ofxn and the unit sample response of the system. Theconvolution property allowed the expression of this relationship inz-domain asYz HzXz (3.20)whereYz is thez-transform of the output sequenceyn,X z is thez-transform of the input sequencexn

and

Hz is the

z-transform of the unit sample response

hn. Thus, if

hn is the time-domain

characterization of the system, itsz-domain characterization isHz which is called the system function orthe transfer function.If the system is described by a difference equation

yn ayn k bxn kM (3.21)the system function can be determined by taking the

z-transform of 3.21 and solving for the ratio

Hz , thusHz YzXz bkzk 01 akzNk 1 (3.22)

Thus, an LTI system described by a constant coefficient difference equation has a rational system function.

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From 3.22, ifa 0 for1 k N, this will be reduced toHz 1zM bzM

M (3.23)

which has zeros and poles at the origin. This results to an impulse response that is finite in duration.A systemdescribed by this system function is called all-zero system or an FIR system. On the other hand,from 3.22, ifb 0 for1 k , the system function reduces to

Hz bz az (3.24)with

a 1. This reduces to an impulse response that is of infinite length because of the presence of non-

zero poles. A system described by this system function is called an all-pole system or an IIR system.

Example 3.17 Determine the system function and the unit sample response of the system described by the differenceequation

yn 12yn 12xn Answers: The system function is

Hz 21 z

with the impulse response as

hn 212un

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Drill Problems 3.2Determine thez-transform of the following signals and sketch the pole-zero patterns.1. xn 1 nun 2.

xn a aun,

a is real

3. xn Arcosn un,0 r 1 4. xn n nun 1 5. xn un un 10 3.4. Inversion ofz-TransformThere are three methods that are often used for the evaluation of the inversez-transform: Direct evaluation by contour integration.

Expansion into a series of terms, in the variablesz andz. Partial fraction expansion and table look-up.In this section, the last two methods will be discussed.

Inverse -transform by Power Series Expansion. The basic idea is this: Given az-transformXz withits corresponding ROC, the functionXz can be expanded into a power series of the formXz cz (3.25)

which converges in the given ROC. Then, by uniqueness of thez-transform,xn c for alln. WhenXz is rational, the expansion can be performed by long division.Example 3.18 Determine the inversez-transform of

Xz 11 1.5z 0.5z

when the ROC is (a)|z| 1 and (b)|z| 0.5.Answers: (a)

xn 1,32,74,158 ,3116,

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(b)

xn 62,30,14,6,2,0, Note that long division will not provide a closed-form solution of the inverse

z-transform except if the

resulting pattern is simple enough to infer the general termxn. Hence this method is used only if onewished to determine the values of the first few samples of the signal.Example 3.19 Determine the inversez-transform of Xz log1 az using power series expansion.Answer:

xn 1an un 1

Expansion of irrational functions into power series can be obtained from tables.

Inverse -transform by Partial Fraction Expansion . In the table lookup method, the functionXz will befirst expressed as a linear combinationXz aXz aXz aXz aXz (3.26)

If such decomposition is possible, then

xn can be found using the linearity property as

xn axn axn axn axn (3.27)This approach is particularly useful ifXz is a rational function. Assume thata 1, then the generalform of the rationalz-transform as

Xz NzDz b bz bMz M1 az az (3.28)A rational function of the form 3.28 is called proper ifa 0 and N. Otherwise, the function is calledimproper (when N). An improper rational function can always be expressed as a sum of a polynomialand a proper rational function.

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Example 3.20 Express the improper rational function

Xz 1 3z z z1

z z

Answer:

Xz 1 2z z1 z z By inspection, the inverse of the polynomial can be found easily. Thus, the expansion of the proper rationalfunction to its partial fraction components will be given emphasis. For the proper rational function

Xz NzDz b bz bMzM

1 az az (3.28)wherea 1 and , negative powers ofz must be eliminated first by multiplying thru numeratorand denominators byz, making 3.28 as

Xz bz bz bMz Mz az a (3.29)and then dividing both sides by

z to make 3.29 a proper rational function again, thus

Xzz bz bz bMz Mz az a (3.30)Factor the denominator of 3.30 into linear factors. The factors will contain the polesp,p, p ofXz.Two cases will be specified.Distinct poles. Suppose that the polesp,p, p are all different. Then the expansion will be of theform

Xzz Az p Az p Az p (3.31)The coefficientsA,A, ,A can be determined using the formula

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A z pXzz (3.32)for

k 1,2,3,,N. Of course, the old trick will still work, but this method is better when evaluating partial

fraction expansion of functions with complex poles.

Example 3.21 Determine the partial fraction expansion of the proper function

Xz 11 1.5z 0.5z Answer: Xz

z 2z 1

1z 0.5

Example 3.22 Determine the partial fraction expansion

Xz 1 z1 z 0.5z Answer:

Xzz jz j

jz j Note that complex conjugate poles result in complex conjugate coefficients in the partial fraction expansion.

Multiple Poles. IfXz has a pole of multiplicity, that is it contains in its denominator the factorz p, the expansion becomesXzz Az p Az p Az p (3.33)The method of determining the coefficients

A,

A, ,

A is illustrated in the following example.

Example 3.23 Determine the partial fraction expansion of

Xz 11 z1 z

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Answer: Xzz 3/4z 1 1/2z 1 1/4z 1 Inversion of the partial fraction expansion. When the poles are real and distinct, the partial fraction

Xzz Az p Az p Az p (3.31)becomes

Xz A1 pz A1 pz A1 pz (3.34)Thus, by table look-up, each term of 3.34 has an inversexn pun when the ROC ofXz is theexterior of the circle whose radius isp (causal signal) andxn pu n 1when the ROC ofXz is the interior of the circle whose radius isp (anticausal signal). Thus, a real, single pole generatesan exponential signal.If the poles are distinct but complex conjugates, the coefficients of their partial fraction expansion are alsoconjugates. Thus ifp is a pole, then its conjugatep is also a pole, and the coefficients of their partialfractions areA andA, which are also complex conjugates. Letting

A |A|e

and

p re then the inverse of

Xz A1 pz

A1 pz

will be

xn 2|A|rcosn un (3.35)Thus, each pair of complex conjugate poles results in a causal sinusoidal signal component with anexponential envelope (a damped sinusoid). The distancer of the pole from the origin determines the

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exponential weighting (growing whenr 1, decaying whenr 1, and constant whenr 1). Theangle of the poles with respect to the positive real axis provides the frequency of the sinusoidal signal. Thezeros, or equivalently, the numerator of the rational transform, affect only indirectly the amplitude and thephase ofxn thruA.In the case of multiple poles, either real or complex, the inverse transform of terms of the formA/z p is required. In the case of the double pole, the transform pair

Z pz1 pz npun (3.36)can be used, provided that the ROC is|z| |p|.Example 3.24 Determine the inverse transform of

Xz 11 1.5z 0.5z when (a) ROC:|z| 1; (b) ROC:|z| 0.5; (c) ROC:0.5 |z| 1 Answers: a. xn 2 0.5 un b. xn 2 0.5u n 1 c.

xn 21u n 1 0.5un

Example 3.25 Determine the causal signal whosez-transform is

Xz 1 z1 z 0.5z Answer:

xn 10 1 2

cosn4 1.249 un

Example 3.26 Determine the causal signalxn having thez-transform

Xz 11 z1 z

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Answer:

xn 14 1 34 n2 un Drill Problems 3.3 Determine the causal signalxn if itsz-transformXz is given by:1. Xz 2. Xz 3. Xz 4.

Xz

5. Xz . 3.5. The One-Sidedz-TransformThe two-sidedz-transform requires that the corresponding signals be specified for the entire time range n . This requirement prevents its use for the evaluation of nonrelaxed systems. Recall thatthese systems are described by difference equations with nonzero initial conditions. Thus, the one-sidedz-transform allows the solution of difference equations with initial conditions.Definition. The one-sided or unilateralz-transform of the signalxn is defined by

Xz xnz (3.37)The one-sidedz-transform differs from the two-sided transform in the lower limit of the summation, which isalways zero, whether or not the signalxn is zero forn 0 (is causal). Due to the choice of lower limit,the one-sidedz-transform has the following characteristics:

It does not contain information about the signalxn for negative values ofn. It is unique only for causal signals, because only these signals are zero forn 0. The one-sidedz-transformXz ofxn is identical to the two-sidedz-transform of the signalxnun. Sincexnun is causal, the ROC of its transform, and hence the ROC ofXz isalways the exterior of a circle. Thus, with the one-sidedz-transform, it is not necessary to refer totheir ROC.

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Example 3.27 Determine the one-sidedz-transform of the following signals (note that these signals are also given inExample 3.1)(a)xn 1 2 5 7 0 1 (b)xn 1 2 5 7 0 1 (c)xn 0 0 1 2 5 7 0 1 (d)xn n (e)xn n k,k 0 (f)xn n k,k 0 Answers:(a)Xz 1 2z 5z 7z z (b)Xz 5 7z z (c)Xz z 2z 5z 7z z (d)Xz 1;(e)Xz z (f)Xz 0 Note that for noncausal signals,X is not unique. Also for anticausal signals,X is always zero.Almost all properties of two-sidedz-transform apply to one-sidedz-transform with the exception of shiftingproperty.Shifting Property of the One-Sided -transform: Case 1 Time Delay. Ifxn Xz, then

xn k

z Xz xnz (3.38)

withk 0. In casexn is causal, thenxn k zXz (3.39)

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Example 3.28 Determine the one-sidedz-transform of the signals(a)xn aun (b)

xn xn 2 where

xn a.

Answers: (a)Xz (b)Xz az a Shifting Property of the One-Sided -transform: Case 2 Time Advance. Ifxn Xz, then

xn k zXz xnz (3.40)fork 0.Example 3.29 Withxn a, find the one-sidedz-transform ofxn xn k.Answer:

Xz z1 az z az Final Value Theorem. Ifxn Xz, thenlimxn limz 1Xz (3.41)The limit of 3.41 exists if the ROC of

z 1Xz includes the unit circle.

Solution of Difference Equations. The one-sidedz-transform is a very efficient tool for the solution ofdifference equations with nonzero initial conditions. It achieves that by reducing the difference equationrelating the two time-domain signals to an equivalent algebraic equation relating their one-sidedz-transforms. This equation can be easily solved to obtain the transform of the desired signal. The signal inthe time-domain is obtained by inverting the resultingz-transform.

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Example 3.30 Determine a closed-form expression for thenth term of the Fibonacci sequence.Answer:

yn 1 512 1 5 1 5un Example 3.31 Determine the step response of the systemyn ayn 1 xn with 1 1 when the initial condition isy 1 1.Answer:

yn 11 a1 a un Drill Problems 3.4 Use the one-sidedz-transform to determineyn,n 0 in the following cases:1. yn yn 1yn 2 0;y 1 y2 1 2. yn 1.5yn 1 0.5yn 2 0;y 1 1,y 2 0 3. yn yn 1 xn;xn un;y 1 1 4. yn yn 2 xn;xn un;y 1 0,y 2 1 3.6. Analysis of LTI Systems in thez-DomainIn this section, the method by which linear, time-invariant systems are analyzed usingz-transform isdiscussed. In particular, the focus is given on the important class of pole-zero systems represented bylinear constant coefficient difference equations with arbitrary initial conditions.

Response of Systems with Rational System Function . If a system is described by the differenceequation

yn ayn k bxn kM (3.21)

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whose system function is defined as

Hz YzXz bkzk 01 akzNk 1 BzAz (3.22)where the system functionHz will be represented by the rational functionBz/Az. Thus,Bz contains the zeros ofHz andAz contains the poles ofHz. If there is an input signalxn to thesystem, the transform of the output can be computed as the product

Yz HzXz (3.20)whereYz is the transform of the outputyn,Hz is the system function and the transform of theimpulse responsehn andXz is the transform of the inputxn. Since the inputXz will also be arational function, it can be expressed in the form

z NzQz (3.42)If the system is initially relaxed, thez-transform of the output will have the form

Yz HzXz BzNzAzQz (3.43)This system will have system poles due toAz as p,p, ,p and input poles due toQz as q,q,,qL. We assume p q for allk 1,2,,N and m 1,2,, and that there are no polescoincident with the zeros ofBz andNz. The partial fraction expansion of 3.43 yields

Yz A1 pz Q1 qzL

(3.44)

The outputyn which is the inverse ofYz will beyn Apun QqunL (3.45)

It can be observed that the output response has two parts:

The response due to the system polesp,p, ,p, which is called the natural response. The response due to the input polesq,q, ,qL, which is called the forced response.

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This is in accordance with the solutions of the difference equation as demonstrated before.

WhenXz andHz have one or more poles in common or whenHz and/orXz contain multiple-order poles, thenYz will have multiple-order poles. Consequently, the partial fraction expansion ofYz will contain factors of the form1/1 pz

,

k 1,2,,m where

m is the pole order. The inversion

of these factors will produce the terms of the formnp in the outputyn of the system, as shown inthe previous section.Response of Pole-Zero System with Nonzero Initial Conditions. Suppose that a signalxn is appliedto the pole-zero system atn 0. Thus, the signalxn is assumed to be causal. The effect of all previousinput signals to the system are reflected in the initial conditionsy 1,y 2, ,y N. The one-sidedz-transform can then be used to determine the outputyn forn 0 so that the initial conditions can bedealt with. Thez-transform of the difference equation 3.21 becomes

Yz az Yz ynz bzXzM

(3.46)

Thus, 3.46 can be expressed as

Yz bzM1 az Xz az y nz1 az (3.47a)

Yz HzXzNzAz (3.47b)whereNz az y nz. From 3.47a and 3.47b, it can be observed that thetransform of the response with nonzero initial conditions has two parts:

The output of the system due to the input without initial condition, or the zero-stateresponseYz. The output of the system due to the initial conditions without input, or the zero-inputresponse

Yn.

Hence the total response is the sum of these two output components, which can be expressed in the timedomain by determining the inversez-transforms ofYz andYz separately and adding the results.It can be separately shown that the effect of initial conditions is to slightly alterA in

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Yz A1 pz Q1 qzL

(3.44)

or the natural response of the system.Example 3.32 Determine the unit step response of the system described by the difference equationyn 0.9yn 10.81yn 2xn under the following initial conditions:(a)y 1 y2 0 (b)y 1 y2 1 Answers:

(a)yn 1.099 1.0880.9cos 1.66 un (b)yn 1.099 0.10680.9cos un Transient and Steady-State Response. The natural response of a causal system has the form

yn Apun (3.48)If all the system poles are less than one, then the natural response decays to zero asn approaches infinity.In such a case, the natural response is referred to as the transient response. The rate at which the naturalresponse decays to zero depends on the magnitude of the poles. The farther the poles away from theorigin, the slower the decay of the response.

The forced response of the system has the form

yn QqunL

(3.49)

If all of the input poles are inside the unit circle, the forced response will decay to zero asn approachesinfinity. But when the input poles are on the unit circle (in the case of step input or sinusoidal input) theforced response persists forn 0. In this case, the forced response is called the steady-state response ofthe system.

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Example 3.33 Determine the transient and steady-state responses of the system characterized by the difference equationyn 0.5yn 1 xn when the input signal is

xn 10cos un. The system is initially at rest.

Answer: yn 6.30.5un yn 13.56cosn4 0.5 un

Causality and Stability. As defined previously, a causal linear time-invariant system is one whose unitimpulse responsehn satisfies the conditionhn 0 forn 0. In terms of its system function, alinear, time-invariant system is causal if and only if the ROC of the system function is the exterior of a circleof radiusr , including the pointz .Also, it is stated before that a system is BIBO stable if its impulse response is absolutely summable, whichimplies that its impulse response decays to zero as time approaches infinity. In terms of the systemfunction, a linear time-invariant system is BIBO stable if and only if the ROC of the system includes the unitcircle.