Yogananda Isukapalli

Discrete - Time Signals and Systems

Z-Transforms 1

A/DConverter

C/DConverter

A/DConverter

D/CConverter

Signal Sensors

Processing Actuators

There are many mathematical descriptions for the general discrete-time system shown above

Discrete-Time SystemFig.10.1

Fig.10.2

0 0

).

[ ] [ ]N M

k kk k

a Difference equations

a y n k b x n k= =

- = -å å

).

[ ] [ ] [ ] [ ] [ ]k k

b Convolution sums for FIR systems

y n x k h n k h k x n k= - = -å å).

( ) ( ) [ ]

( )n

n

c System functions

Y zH z h n z

X z

¥-

=-¥

= = å

Z Transform: Definition

Given a discrete-time sequence{ }

2 1 1..... .....

[ ] ....., [ 2], [ 1], [0], [1], [2],.....( ), [ ] ;

( ) [ ]

[ 2] [ 1] [0] [1]

n

n

x n x x x x x

X z Z transform of x n is defined as

X z x n z

x z x z x x z

¥-

=-¥

-

= - -

-

=

= + - + - + + +

å

X(z) is a complex valued function evaluated in acomplex plane

Whenever an infinite series converges, the z-transform X(z) has a finite value in some region Aof the complex plane. This region is termed as the Region of Convergence (ROC)

A

Complex z -Plane

Re(z)

Im(z)

Fig.10.3

For z=rejw & r=1 thus making |z|=1, this contour in the z-plane is a circle of unity radius and is termed as the unit circle

Unit Circle

|z|=1

Re(z)w

Im(z)

[ ] ( )x n X z«

Fig.10.4

Z-Transform can be applied to FIR Filter

Why Z-transform?The Z-transform of a finite discrete-time signal results in a polynomialThe mathematics of polynomials is a well developed concept. The analysis part is simplified in Z-domain

[ ] ( )

( ) ( ) ( )

h n H zConvolution will become simple multiplicationY z X z H z

«

=

2 1 0 1 2

0

( ) [ ]

..... [ 2] [ 1] [0] [1] [2] .....

[ ] .... 0 0.... 1 0 0.......

1

n

n

R C

n

n

O all Z

X z x n z

x z x z x z x z x z

n z zd

¥-

=-¥

- -

¥-

=-¥

=

= + - + - + + + +

= = + + + + +

=

å

å

Example 11 0

[ ] [ ]0 0n

x n nn

d=ì

= = í ¹î 1

n0Fig.10.5• • •••• 1 2 3-2 -1-3

Example 2

1 1

1 , 0

[ 1]

[ 1]

( ) [ ]

[ 1] ....0 0.... 1 0....

[ ]

.

ROC

n

n

n

n

all Z except Zn

n

X z x n z

n z z z

z

x n

d

d

d

¥-

=-¥

¥- - -

=-¥

- =«-

-

=

= - +

=

= + + =

å

å

1

n1•0•• -1-2 • • •2 3 4Fig.10.6

Example 3

1

1

,

[ 1

[ 1]

]

( ) [ ]

[ 1] ....0 0.... 1 0....

[ ]

n

n

n

OC

n

R all Z except Z

n

n

X z x n z

n

z

x

z

z

n

z

d

d

d

¥-

=-¥

¥-

=-¥

= ¥

+

+ «

=

=

= + = + + +

=

å

å

1

n-1•-2•• -3-4 • • •0 1 2Fig.10.7

Example 4

2

1

3

-1

-1 0 1 2

n

x[n]

2101

2101

21012

312 .......03120...

.....]2[]1[]0[]1[]2[.....

][)(

--

--

--

¥

-¥=

-

+-+=

+++-+++=

++++-+-+=

= å

zzzz

zzzz

zxzxzxzxzx

znxzXn

n

Fig.10.8

Example 5: Infinite Unit Sequence[ ] [ ] 1 , x n u n for all n n 0 = = ³

1

0 1 2 3 4 5 … … n

2 1 0 1 2

1 2 3

0

1 1

( ) [ ]

..... [ 2] [ 1] [0] [1] [2] .....

[ ] 1 ...

1 ( ) 1 1

n

n

n n

n n

ROC z

X z x n z

x z x z x z x z x z

u n z z z z z

zusing sum of infinite seriesz z

¥-

=-¥

- -

¥ ¥- - - - -

=-¥ =

- >

=

= + - + - + + + +

= = = + + + +

= =- -

å

å å

Fig.10.9

2 3

0

1 1 ...

1 1;

n

n = only ifa a a aa

a¥

=

= + + + + +-

<å

Example 6: Finite Unit Sequence [ ] [ ] 0 11 x or nn u n f N= £ £ -=

1

0 1 2 3 4 … … … N-1 n

1 1

0 01 2 ( 1)

1 , 0

( ) [ ]

[ ]

1 ...1 ( ) 1

n

n

N Nn n

n n

N

N

ROC all z except z

X z x n z

u n z z

z z z

z using sum of serfinite iesz

¥-

=-¥

- -- -

= =

- - - -

-

- =

=

= =

= + + + +

-=

-

å

å å

Fig.10.10

12 1

0

1 ...

11

NN

n

n

N=

a a a a

aa

--

=

= + + + +

-

-

å

Example 7: Infinite Right handed sequence

( )

1 0 1 2

0

1 1 2 2 3 3

0

1

[ ] [ ]

( ) [ ]

..... [ 1] [0] [1] [2] .....

[ ]

1 ...

1 (

1

n

n

n

n n n n

n n

n

n

x n a u n

X z x n z

x z x z x z x z

a u n z a z

az az a z a z

using sum of infiniteaz

¥-

=-¥

- -

¥ ¥- -

=-¥ =

¥- - - -

=

-

=

=

= + - + + + +

= =

= = + + + +

=-

å

å å

å

) z

seriesz a

=-

Example 7 continued...

( ) 2 3

0

1

1 ...

1

1

1;

1

1

,

n

n

= only if

higher

Consider the sThe issue of co

order ter

um of infinite geomet

ms will become zero when

az or z a

ricnvergen

for th

sere

ies

r

c

e p

a a a a

aa

a

¥

=

-

= + + + + +

-<

<

< >

å

( )

z a ROCis known as Region of Convergence

oblem at hand

>

Example 7 continued...

[ ] [ ], (

[ ]

)

n

n ROC

zx n a u n X z

z az

a u nz a

z a

= =-

«-

>

‘A’ ROC

, z a pole=

{ }m zÁ

{ }e z´

z a<

z a>

0, z zero=!

Fig.10.11

Example 8: Finite Right handed sequence

( )

1 1

0 0

11 1 2 2 ( 1) ( 1)

0

-1

1

( ) [ ]

[ ]

1 .....

1 ( ( )

1

[ ] [ ]

)

0 1

n

n

N Nn n n n

n n

NN

n

n

Nn

N

ROC all

X z x n z

a u n z a z

az az a z a z

azusing sum of series

a

x n a u

z

n

finite

for n N¥

-

=-¥

- -- -

= =

-- - - - - -

=

-

=

= =

= = + + + +

-=

-

= £ £ -

å

å å

å

Z

Example 9: Left handed sequence

( )1

1

[ ] [ 1]

( ) [ ]

[ 1]

0, 1

' ' , -

1

n

n

n

n n

n

n n

n

n n

n

x n a u n

X z x n z

a u n z

a z n

The sign of n can be changed n n

a z

n

¥-

=-¥

¥-

=-¥

--

=-¥

¥-

=

= - - -

=

= - - -

= - ³ -¥ < £ -

=

= -

- -

å

å

å

å

!

( )

1

0

0

1

1

0 0

1

1

1

1

1 1 ( )

1

1 1

1

1

1

n n

n

n n

n

n

n n

n

n n

n

n

a z

= a z

using sum of infinite seriesa z

a z

a z a z

a z

¥-

=

¥-

=

¥

=

-

¥-

=

¥- -

=

-

= -

= -

- -

-

= --

+ -

æ ö+ç ÷

è øæ ö

+ =ç ÷è ø

=

å

å

å

å

å

Example 9 continued...

1

1

1

[ 1]

[ ]

1 11

,

1

,

n

n

ROC

ROC

za u n

z az

a z za

a u na

z a

z

z

In the infinite sum

a z or z a

Notice the inte

Region of conver

resting result

gence

z a

z a

Same Z transform f

-

-

-

- - - «-

«

- -

- -

->

=

-

=

< <

<

'

twoor different signals withdifferent ROC s

Example 9 continued...

[ ] [ 1], ( )

[ 1]

n

n ROC

zx n a u n X z

z az

a u nz a

z a-

= - - - =

-

-

- - « <

, z a pole=´

‘A’ ROC

z a<

!0, z zero=

z a>{ }m zÁ

{ }e zÂ

Fig.10.12

Example 10: Mixed Sequences

( )

( )

( )

[ ] [ ] [ 1]

( ) [ ]

[ ]

[ ] [ 1]

[ ] [ 1]

[ 1]

n n

n

n

n

n

n

n n

n n

n n

n n n n

n n

x n a u n b u n

X z x n z

z

a u n z

a u n b u n

a u n z b u n z

b u n z

¥-

=-¥

¥-

=-¥

¥

=-¥

¥ ¥-

=-¥ =-¥

- -

-

= - - -

=

=

=

=

- - -

- - -

+ - - -

å

å

å

å å

( )

! !

.

.

[ ]

( )=

[

]

1

ROC ROC

n n

n n

n n

Right H Sequence Left Handed Sequ

z

e

a z b

nce

a u

z zz

n z

a z b

From the previous two examples

X z

ROC has to be the overlapping

b u n

area of the

t

z

wo

¥ ¥-

=-¥ =-¥

-

> <

-

=

-

+

+

- - -å å"#$#% "###$###%

{ } , zregi n bs ao z> <&

Example 10 continued...

Example 10 continued ...

, ( ) '

If there is no overlapping region hence

X z doesn t

b

t

a

exis

<

If b a>

ROC is bounded by the

largest and smallest poles

{ }m zÁ

{ }e zÂ

‘A’ ROC: |a|<|z|<|b|

two zeros

ab

´ ´polez b=polez a=

!

Fig.10.13

Example 11: Mixed Sequence

1

.

0

1 0

.

( ) [ ]

[ ]

[ ] ,

;

[ ] [ 1] [ ]

( )

' '

n

Left H sequence R Hande

n

n

n n

n n n n

n n

n

n

d

n n n

n

X z x n z

x n can be split into two sequences

x n b u

x n b

n b u n

X z b z b z

b

for

b z

z

all n¥

-

=-¥

-

- ¥- - -

=-¥ =¥ ¥

-

= =

=

= - +

=

= +

=

-

+

å

å å

å å

!"#"$ !#$

! !

0

2

1

1

0

1

( ) 1 ( )

1 1 11 1

1

(1 )

1

(1 )( )

n n

n n

ROC z bROC zb

bz bz

bz bzbz zb

ROC b z

z z b

z bbz z b

b

¥ ¥-

=

-

<

=

>

= - +

= - +- -

= +

<

- -

-=

- -

<

å å

Example 11 continued...

, '1 If ROC db oesn t exist>

< 1 0If b <

Example 11 continued...

{ }m zÁ

{ }e zÂ

‘A’ ROC:

two zeros

b1 b

´ ´ 1pole

zb

=polez b=!

1 b zb

< <

Fig.10.14

Example 12: Two right handed signals

[ ]

( ) [ ]

,

1[ ] 1

1 1[ ] 7 [ ] 6 [ ]

3

33 1 3

1[ ] 1 2

2

2

1 2

n

n

n

n

ROC

n

RO

n

C

n

ROC

za u n

z a

z

z

z

z

X z x n z

Since both are rig

x n

ht handed signals

z a

u n z

u n z

u n u n

¥-

=-¥

«-

«-

«-

=

>

=

>

-

>

æ öç

æ ö æ öç ÷ ç ÷è ø è ø

÷è ø

æ öç ÷è ø

å

{ }

6

2

, 2

1

1

7( )1 3

1 3 1 2

z

z

ROC z

zX zz

z z

-

>> > Þ

= --

!

Example 12 continued...

{ }m zÁ

{ }e zÂ

‘A’ ROC:

two zeros

1 3

1 2

´ ´1 3,z pole=

!

1 2z >

1 2,z pole=

The shape is drawnarbitrarily

Fig.10.15

Example 13

( )

( )

[ ]

[ ] [ ]

[ ] [

( ) [ ]

[

]

]

( ) 1

n

j

j

n

n

ROC

n j n jR C

RO

n

O

C

j

j

n

nj

za u n

z a

z

z aez

z ae

x n Aa u n

x n A u n

X z x n z

z a

Aa e u n z ae

z aX z e

e

ae

q

q

q q

q

q

q

¥-

=-¥

«-

«-

-

=

>

>

>

=

=

\ = =

å

!

2

0 0

1 1

1 (2 2 cos )

2 2 cos 1

cos[ ]2

( ) [ ]2

1

22 2

[ ] cos[ ]

[

]

j n j n

j n j nn

n

ROC z RO

j n n j n n

n n

j j

C z

z z

z z

e en

e eX

x

z z u n

e

n n u n

z e z

z zz e z e

a a

a a

a a

a a

a

a

a

a-

-¥-

=-¥

¥ ¥- - -

= =

>

-

>

-=

- +

+=

æ ö+= ç ÷ç ÷

è øæ ö

= +ç ÷è ø

= +- -

æè

=

å

å å

!"#"$ !"#"$

1ROC z >öç ÷

ø

Example 14

Reference

James H. McClellan, Ronald W. Schafer and Mark A. Yoder, “7.1-7.2,--Signal Processing First”, Prentice Hall, 2003