z transform (1)

7/31/2019 z Transform (1)

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The z-Transform

Content

z Introduction

z z-Transform

z Zeros and Poles

z Region of Convergence

z mportant z- rans orm a rs

z Inverse z-Transform

z z-Transform Theorems and Properties

z System Function


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The z-Transform

Introduction

Why z-Transform?

z A generalization of Fourier transform

z Why generalize it?

FT does not converge on all sequence

Notation good for analysis

the discrete-time signals and systems


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The z-Transform

z-Transform

Definition

z Thez-transform of sequencex(n) is defined by

=

=n

nznxzX )()(

Fourier

Trans orme z = e .

( ) ( )j j n

n

X e x n e

=

=


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z-Plane

Re

Im

z = ej

=

=n

nznxzX )()(

j j n

n=

Fourier Transform is to evaluate z-transform

on a unit circle.

z-Plane

X(z)

Re

Im

z = ej

Re

Im


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Periodic Property of FT

X(z)

X(ej)

Re

Im

Can you say why Fourier Transform isa periodic function with period 2?

The z-Transform


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Definitionz Give a sequence, the set of values ofz for which the

z-transform converges, i.e., |X(z)|


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Stable Systems

Im

z A stable system requires that its Fourier transform is

uniformly convergent.

z Fact: Fourier transform is to

evaluatez-transform on a unit

Re

1.

z A stable system requires the

ROC ofz-transform to includethe unit circle.

Example: A right sided Sequence

)()( nuanx n=

x(n)

1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8

n. . .


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Example: A right sided Sequence

)()( nuanx n=

n

n

n znuazX

== )()(

For convergence ofX(z), we

require that

=

Im Im

Which one is stable?

aaRe

1aa

Re

1


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Example: A left sided Sequence

)1()( = nuanx n

1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8

n

x(n)

...


)1()( = nuanx n

n

n

nznuazX

= = )1()(

For convergence ofX(z), we

require that


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ROC forx(n)=anu( n1)

||||,)( azaz

zzX

=

Im

ROC is bounded b the

Rea

pole and is the exterior

of a circle.


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)1()( = nuanx n ||||,)( azaz

zzX 0)

or (ifmz

)1( nu 11 z1||

)1( nuan 111

az|||| az


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Z-Transform Pairs

Sequence z-Transform ROC

)(][cos0 nun 21

0

1

0

]cos2[1

][cos1

+

zz

z1|| >z

)(][sin 0 nun 210

1

0

]cos2[1

][sin

+

zz

z1|| >z

1]cos[1 zr

)(]cos[0 nunr

n 2210 ]cos2[1

+ zrzrrz >

)(]sin[ 0 nunr

n

2210

1

0

]cos2[1

]sin[

+

zrzr

zrrz >||

otherwise0

10 Nnan

11

1

az

zaNN

0|| >z

The z-Transform

-


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The z-Transform

z-Transform Theorems

Linearity

xRzzXnx = ),()]([Z

yRzzYny = ),()]([Z

yx RRzzbYzaXnbynax +=+ ),()()]()([Z

Overlay of

the above two

ROCs


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ShiftxRzzXnx = ),()]([Z

x

n

RzzXznnx =+ )()]([0

0Z

Multiplication by an Exponential Sequence

+


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Differentiation of X(z)

xRzzXnx = ),()]([Z

xRzdz

zdXznnx = )()]([Z

Conjugation

xRzzXnx = ),()]([Z

xRzzXnx = *)(*)](*[Z


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ReversalxRzzXnx = ),()]([Z

xRzzXnx /1)()]([1 = Z

Real and Imaginary Parts

xRzzXnx = ),()]([Z

xRzzXzXnxe += *)](*)([)]([ 21R

xjRzzXzXnx = *)](*)([)]([

21Im


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Initial Value Theorem

0for,0)(


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Convolution of Sequences

=

=k

knykxnynx )()()(*)(

=

=

=n

n

k

zknykxnynx )()()](*)([Z

=

=

=k

n

n

zknykx )()(

=

=

=k

n

n

kznyzkx )()(

)()( zYzX=

The z-Transform

S stem Function


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Shift-Invariant System

h(n)

x(n) y(n)=x(n)*h(n)

X(z) Y(z)=X(z)H(z)H(z)

Shift-Invariant System

X(z) Y(z)

)(

)()(

zX

zYzH =


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Nth

-Order Difference Equation

==

=M

r

r

N

k

k rnxbknya00

)()(

=M

r

r

Nk

k zbzXzazY )()(== rk 00

==

=

N

k

k

k

M

r

r

r zazbzH00

)(

Representation in Factored Form

=

=

M

r

rzcA

1

1)1(

Contributespoles at 0 andzeros at cr

=

N

k

rzd1

1)1(

Contributeszeros at 0 andpoles at dr


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Stable and Causal Systems

=

=

M

r

rzcA1

1)1(

Im

Causal Systems : ROC extends outward from the outermost pole.

=

N

k

rzd

1

1)1(Re

Stable and Causal Systems

=

=

M

r

rzcA

1

1)1(

Im

Stable Systems : ROC includes the unit circle.

1

=

N

k

rzd1

1)1(e


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ExampleConsider the causal system characterized by

)()1()( nxnayny +=

1=

Im

1

11 azRea

)()( nuanh n=

Determination of Frequency Responsefrom pole-zero pattern

z A LTI system is completely characterized by itspole-zero pattern.

1zz

Example:

0j

Im

p1

))(( 21 pzpzz

=

))(()(

21

1

00

0

0

pepe

zeeH

jj

jj

=

Re

z1

p2


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Determination of Frequency Response

from pole-zero pattern


1zz

Example:

0j

Im

p1

|H(ej)|=? H(ej)=?

))(( 21 pzpzz

=

))(()(

21

1

00

0

0

pepe

zeeH

jj

jj

=

Re

z1

p2

Determination of Frequency Responsefrom pole-zero pattern


Example:

0j

Im

p1

|H(ej)|=? H(ej)=?

| | 2

Re

z1

p2

e =| | | | 1 3

H(ej) = 1(2+ 3 )


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Example11

1)(

=

azzH

Im

0 2 4 6 8

-10

0

10

20

dB

Re

a0 2 4 6 8

-2

-1

0

1

2

z transform (1)

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