04 01 z transforms slides presentation

Math Transform MethodsTopic 04 : Z Transforms

Lectures 2733 : Z-Transforms

Dr Kieran Murphy

This module and thesenotes were developedby Dr Pardaig Kirwanwith only minor modi-fications on my part.

Credits:

Department of Computing and Mathematics,Waterford Institute of Technology.

([email protected])

Autumn Semester, 2014

OutlineDiscrete vs continuous functions.

Definition and properties of the Z-transform.

Convert a discrete function in time domain (t) to a function in z-domain, and back.

Solving constant linear, constant coefficient difference equations using Z-transforms.

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Sequences

Sequences Lecture 27

The ztransform is the major mathematical tool for analysis in such topics as digitalcontrol and digital signal processing.

SequencesA sequence is a set of numbers formed according to some definite rule. Forexample the sequence

{1, 4, 9, 16, 25, . . .}is formed by the squares of the positive integers.If we write

y1 = 1, y2 = 4, y3 = 9, . . .

then the general or nth term of the sequence is yn = n2. The notations y(n) and y[n]are also used sometimes to denote the general term. The notation {yn} is used as anabbreviation for a whole sequence.

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Sequences


An alternative way of considering a sequence is to view it as being obtained bysampling a continuous function. In the above example the sequence of squares canbe regarded as being obtained from the function

y(t) = t2

by sampling the function at t = 1, 2, 3, . . . as shown in Figure 1.

1 2 31

4

9

The notation y(n), as opposed to yn, for the general term of a sequence emphasizesthis sampling aspect. Here the sequence {2n} are the sample values of thecontinuous function y(t) = 2t at t = 1, 2, 3, . . .

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Sequences


Defining a sequenceAn alternative way of defining a sequence is as follows:

(i) give the first term y1 of the sequence(ii) give the rule for obtaining the (n+ 1)th term from the nth.

A simple example isyn+1 = yn + d y1 = a

where a and d are constants.It is straightforward to obtain an expression for yn in terms of n as follows:This sequence characterised by a constant difference between successive terms

yn+1 yn = d n = 1, 2, 3, . . .

is called an arithmetic sequence.

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Sequences


Example 1Calculate the nth term of the arithmetic sequence defined by

yn+1 yn = 2, y1 = 9

Write out the first 4 terms of this sequence explicitly.Suggest why an arithmetic sequence is also known as a linear sequence.

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Sequences


NomenclatureThe equation

yn+1 yn = dis called a difference equation or recurrence equation or more specifically a firstorder, constant coefficient, linear, difference equation.The sequence whose nth term is

yn = a+ (n 1)d

is the solution of for the initial condition y1 = a.The coefficients are the numbers preceding the terms yn+1 and yn so are 1 and +1respectively.The classification first order for the difference equation follows because thedifference between the highest and lowest subscripts is (n+ 1) (n) = 1.

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Sequences


NomenclatureNow consider again the sequence

{yn} = {2n}

Clearlyyn+1 yn = 2n+1 2n = 2n

so the difference here is dependent on n i.e. is not constant. Hence the sequence{2n} = {2, 4, 8, . . .} is not an arithmetic sequence.

Example 2For the sequence yn = 2n calculate yn+1 2yn. Hence write down a differenceequation and initial condition for which {2n} is the solution.

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Sequences


NoteMore generally it follows that

yn+1 Ayn = 0, y1 = A

has solution sequence {yn} with general term

yn = An

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Sequences


A second order difference equationSecond order difference equations are characterised, as you would expect, by adifference of 2 between the highest and lowest subscripts.A famous example of a constant coefficient second order difference equation is

yn+2 = yn+1 + yn

The solution {yn} is a sequence where every term is the sum of the two precedingones.

Example 3What additional information is needed if the above difference equation is to besolved?

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Sequences


Example 4Find the first 6 terms of the solution sequence of

yn+2 = yn+1 + yn

for each of the following sets of initial conditions(a) y1 = 1, y2 = 3(b) y1 = 1, y2 = 1

Fibonacci SequenceThe sequence {1, 1, 2, 3, 5, 8, 13, . . .} is a very famous one; it is known as theFibonacci Sequence. It follows that the solution sequence of the difference equationyn+2 = yn+1 + yn with initial conditions y1 = y2 = 1 is the Fibonacci sequence.What is not so obvious is what is the general term yn of this sequence.

One way of obtaining yn in this case, and for many other linear constant coefficientdifference equations, is via a technique involving Ztransforms which we shallintroduce shortly.

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Basics of Ztransform theory

Basics of Ztransform theory Lecture 28

In this Section, which is absolutely fundamental, we define what is meant bythe Ztransform of a sequence.

We then obtain the Z transform of some important sequences and discussuseful properties of the transform.

Most of the results obtained are tabulated at the end of the Section.

The Ztransform is the major mathematical tool for analysis in such areas asdigital control and digital signal processing.

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The Ztransform plays a similar role for discrete systems, i.e. ones wheresequences are involved, to that played by the Laplace transform for systems wherethe basic variable t is continuous.

The Laplace transformRecall that

1 the Laplace transform definitioninvolves an integral

2 applying the Laplace transform tocertain ordinary differentialequations turns them into simpler(algebraic) equations

3 use of the Laplace transform givesrise to the basic concept of thetransfer function of a continuous(or analog) system.

The ZtransformSpecifically:

1 the Ztransform definitioninvolves a summation

2 the Ztransform converts certaindifference equations to algebraicequations

3 use of the Ztransform gives riseto the concept of the transferfunction of discrete (or digital)systems.

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Definition 5The Ztransform of a sequence {yn} is given by the infinite series

Z(yn) = y0 +y1z+

y2z2+

y3z3+ . . . =

n=0

ynzn

We use the notation Y(z) = Z(yn) to mean thatY(z) is the Ztransform of thesequence {yn}.We shall also call {yn} the inverse Ztransform of Y(z) and write symbolically

{yn} = Z1(Y(z))

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Note1 The Ztransform only involves the terms of the sequence yn where

n = 0, 1, 2, . . . .2 Terms y1, y2, . . . whether zero or non-zero, are not involved.3 The infinite series must converge for Y(z) to be defined as a precise function of

z. We shall discuss this point later.4 The precise significance of the quantity (strictly the variable) z need not

concern us except to note that it is complex and, unlike n, is continuous.

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Unit impulse sequence (delta sequence)This is a simple but important sequence denoted by n and defined as

n =

{1 n = 00 n = 1,2,3, . . .

n

n

-3 -2 -1 1 2 3

1

0The significance of the term unit impulse is obvious from this definition.

Example 6Determine the Ztransform of the following sequences.

1 n2 n33 nm

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Unit step sequenceThe unit step sequence is

un ={

1 n = 0, 1, 2, . . .0 n = 1,2,3, . . .

n

un

-3 -2 -1 0 1 2 3

1

Example 7Determine the Ztransform of the following sequences.

1 un2 un33 unm

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Example 8Determine the Ztransform of the geometric sequences.

fn ={

an n = 0, 1, 2, . . .0 n = 1,2,3, . . .

Example 9Determine the Ztransform of the following geometric sequences.(a) 2n

(b) (1)n, the unit alternating sequence(c) en

(d) en where is a constant.

The basic Ztransforms obtained have all been straightforwardly found from thedefinition. To obtain further useful results we need a knowledge of some of theproperties of Ztransforms.

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Properties of Ztransforms

Properties of Ztransforms Lecture 29

Linearity PropertyThe Ztransform of a sum of sequences is the sum of their respectiveZtransforms, i.e.

Z(vn + wn) = Z(vn) + Z(wn)

The Ztransform of a multiple of a sequence is the multiple of theZtransform of the sequence, i.e.

Z(avn) = aZ(vn)

These can be expressed as a single rule as follows:

Z(avn + bwn) = aZ(vn) + bZ(wn)

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We can now use the linearity property and the exponential sequence {en} toobtain the Ztransforms of hyperbolic and of trigonometric sequences relativelyeasily.

Example 10Determine the Ztransform of the following sequences.(a) {sinh(n)}(b) {sinh(n)}(c) {cosh(n)}(d) {cos(n)}

(e) {sin(n)}(f) {cos(npi)}(g) {sin ( npi2 )}(h) {cos ( npi2 )}

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Multiplication of a sequence by an

Suppose fn is an arbitrary sequence with Ztransform F(z). Then

Z(anfn) = F(za

)That is, multiplying a sequence {fn} by the sequence {an} does not change the formof the Ztransform F(z); we merely replace z by

za

in that transform.

Example 11Write down the Ztransform of the sequence {vn} where

1 vn = e2n sin(3n)2 vn = en cos(n)

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Ramp sequenceThe ramp sequence {rn} is given by

rn ={

n n = 0, 1, 2, . . .0 n = 1,2,3, . . .

n

rn

-3 -2 -1 0 1 2 3

1

2

3

Example 12Determine the Ztransform of the ramp sequence {rn}.

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Recall now that the unit step sequence has Ztransform

Z{un} = zz 1 = U(z)

Example 13

Obtain the derivative of U(z) =z

z 1 with respect to z.

Note:Therefore

Z{rn} = Z{nun} = zdU(z)dzi.e.

Z{nun} = zdZ{un}dz

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This property is still valid when we replace the unit step sequence by an arbitrarysequence {fn}.

Multiplication by nShow that

Z{nfn} = zdZ{fn}dz

Example 14By differentiating the Ztransform R(z) of the unit ramp sequence obtain theZtransform of the causal sequence {n2}.

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Table of Z-Transforms of main sequences

fn F(z) = Z{fn}n 1

Unz

z 1n

z(z 1)2

n2z(z+ 1)(z 1)3

anz

z aean

zz ea

cos(n)z2 z cos()

z2 2z cos() + 1sin(n)

z sin()z2 2z cos() + 1

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Shift Theorems

Shift Theorems Lecture 30

We now consider perhaps the most important properties of the Ztransform.

These properties relate the Ztransform Y(z) of a sequence {yn} to theZ-transforms of

(i) left shifted or advanced sequences {yn+1}, {yn+2}, etc.(ii) right shifted or delayed sequences {yn1}, {yn2}, etc.

The results obtained, formally called shift theorems, are vital in enabling us tosolve certain types of difference equation and are also invaluable in theanalysis of digital systems of various types.

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Shift Theorems


Example 15Recall that the sequences {yn+1}, {yn+2}, . . . denote the sequences obtained byshifting the sequence {yn} by 1, 2, . . . units to the left respectively. LetY(z) = Z{yn}.(a) Determine the Ztransform of {yn+1} in terms of Y(z).(b) Determine the Ztransform of {yn+2} in terms of Y(z).

Example 16Recall that the sequences {yn1}, {yn2}, . . . denote the sequences obtained byshifting the sequence {yn} by 1, 2, . . . units to the right respectively. LetY(z) = Z{yn}.(a) Determine the Ztransform of {yn1} in terms of Y(z).(b) Determine the Ztransform of {yn2} in terms of Y(z).

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Shift Theorems


The shift theoremsThe left shift theorems or advance theorems are:

Z{yn+1} = zY(z) zy0Z{yn+2} = z2Y(z) z2y0 zy1

......

...Z{yn+m} = zmY(z) zmy0 zm1y1 zym1

The right shift theorems or delay theorems are:

Z{yn1} = y1 + 1z Y(z)

Z{yn2} = y2 + y1z +1z2Y(z)

......

...

Z{ynm} = ym + ym1z + +y1zm1

+1zm

Y(z)

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Ztransforms and Difference Equations

Ztransforms and Difference Equations Lecture 31

In this we apply Ztransforms to the solution of certain types of differenceequation.

We shall see that this is done by turning the difference equation into anordinary algebraic equation.

We investigate both first and second order difference equations.

A key aspect in this process in the inversion of the Ztransform.

As well as demonstrating the use of partial fractions for this purpose we showan alternative, often easier, method using what are known as residues.

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Example 17Consider

yn+1 3yn = 4, n = 0, 1, 2, . . .

with initial condition y0 = 1. Multiply both sides by1zn

and sum each side over all

positive integer values of n and zero.

MethodTo solve a linear constant coefficient difference equation, three steps are involved:

1 Replace each term in the difference equation by its Ztransform and insert theinitial condition(s).

2 Solve the resulting algebraic equation for Y(z).3 Find the inverse Ztransform of Y(z).

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Example 18Solve

yn+2 7yn+1 + 10yn = 0 with y0 = 5, y1 = 16

Example 19By solving

yn+2 = yn+1 + yn, n = 0, 1, . . .

subject to the initial conditions y0 = y1 = 1 obtain the general term yn of theFibonacci sequence.

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Example 20Obtain the Ztransform of

{fn} = {nan}Hence, solve

yn+2 6yn+1 + 9yn = 0, n = 0, 1, 2, . . .subject to y0 = 0, y1 = 1.

Class ExerciseSolve the difference equation

yn+2 + yn = 0

with y0, y1 arbitrary.

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Residues

Residues Lecture 32

The method of inversion of Ztransforms using residues has its basis in abranch of mathematics called complex integration.

You may recall that the Z quantity of Ztransforms is a complex quantity,more specifically a complex variable.

However, it is not necessary to delve deeply into the theory of complexvariables in order to obtain simple inverse Ztransforms using what are calledresidues.

In many cases inversion using residues is easier than using partial fractions.Hence reading on is strongly advised.

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Residues

Residues Lecture 32

Pole of a function of a complex variableIf G(z) is a function of the complex variable Z and if

G(z) =G1(z)

(z z0)k

where G1(z0) 6= 0 and finite then G(z) is said to have a pole of order k at z = z0.

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Residues

Residues Lecture 32

For example if

G(z) =6(z 2)

z(z 3)(z 4)2then G(z) has the following 3 poles.

(i) pole of order 1 at z = 0.(ii) pole of order 1 at z = 3

(iii) pole of order 2 at z = 4.(Poles of order 1 are sometimes known as simple poles.)Note that when z = 2,G(z) = 0. Hence z = 2 is said to be a zero of G(z).

Example 21Write down the poles and zeros of

G(z) =3(z+ 4)

z2(2z+ 1)(3z 9)State the order of each pole.

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Residues

Residues Lecture 32

Definition 22The residue of a complex function G(z) at a first order pole z0 is

Res(G(z), z0) =[G(z)(z z0)

]z0

The residue at a second order pole z0 is

Res(G(z), z0) =

[d(G(z)(z z0)2

)dz

]z0

You need not worry about how these results are obtained or their full mathematicalsignificance. (Any textbook on Complex Variable Theory will provide the details.)

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Residues

Residues Lecture 32

Example 23Determine the residues of the complex function

G(z) =3(z+ 4)

z2(2z+ 1)(3z 9)

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Inverse Ztransform formula using residues

Inverse Ztransform formula using residues Lecture 33

Inverse Ztransform formula using residuesRecall that, by definition, the Ztransform of a sequence {fn} is

F(z) = f0 +f1z+

f2z2+ + fn

zn+

If we multiply both sides by zn1 where n is a positive integer then we obtain

F(z)zn1 = f0zn1 + f1zn2 + f2zn3 + + fnz +fn+1z2

Using again a result from complex integration it can be shown from this expressionthat the general term fn is given by

fn = sum of the residues of F(z)zn1 at its poles

The poles of F(z)zn1 will be those of F(z) with possibly additional poles at theorigin.

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Inverse Ztransform formula using residues

Inverse Ztransform formula using residues Lecture 33

To illustrate the residue method of inversion we shall re-do some of the earlierexamples that were done using partial fractions.

Example 24Determine the inverse Z-transform of the following complex functions.

1 Y(z) =z2

(z a)(z b)2 F(z) =

6z2 160zz2 7z+ 10

3 F(z) =6z2 9z(z 3)2

4 F(z) =z2

z2 + 1

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04 01 z transforms slides presentation

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sequence yn

arithmetic sequence

linear sequence

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notation yn

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