Basics on Digital Signal Processing

z - transform - Digital Filters

Vassilis Anastassopoulos

Electronics Laboratory, Physics Department,

University of Patras

2/31

Outline of the Lecture

1. The z-transform

2. Properties

3. Examples

4. Digital filters

5. Design of IIR and FIR filters

6. Linear phase

3/31

z-Transform

The z-transform is more general than the DFT

Time to frequency

1

0

2

)()(N

n

nkNj

enxkX

1

0

)()(N

n

nznxzX

Time to z-domain (complex)

Transformation tool is the

complex wave

With amplitude |ejω|=1

Nkjj ee /2

Transformation tool is

With amplitude ρ

changing(?) with time

jj eez

4/31

R

I z=R+jI -z

|z|=1

The DFT is an expression of the z-transform on the unit circle.

z-Transform For z=ejω i.e. ρ=1 we work on the

unit circle

And the z-transform degenerates

into the Fourier transform.

1

0

)()(N

n

nznxzX

The quantity X(z) must exist with finite value on the unit circle i.e. must

posses spectrum with which we can describe a signal or a system.

5/31

z-Transform convergence

We are interested in those values of z for which X(z) converges.

This region should contain the unit circle.

R

I

|a|

ROC

Why is it so?

The values of z for which X(z) diverges are

called poles of X(z).

1

0

1

0

)(1/ )( )()(N

n

nN

n

n znxznxzX

At z=0, X(z) diverges

6/31

R

I

|a|

ROC

a

a2

a3

a4

x(n)

n

1

z-Transform example

Which is the z-transform and the ROC of a discrete time sequence

x(n)=an for n0 and a<1 ?

0

1

0

1

0

)( )()(n

n

n

nnN

n

n azzaznxzX

which for |az-1|<1 or |z|>|a| converges to az

z

azzX

11

1)(

The pole z=a, is never included in the ROC

7/31

Poles and zeros of X(z)

R

I

|a|

ROC

Poles: X(z)= Zeros: X(z)=0

For finite sequences, X(z) converges

everywhere except at z=0

For infinite sequences, X(z) converges

everywhere outside the circle with radius the

pole with maximum value.

1

3

5

3

1

n

x(n)

X z x n z z z z z zn

n

( ) ( )

1 2 3 4 53 5 3

For stable, causal digital systems the region of convergence includes the

Unit Circle so that the system possesses spectrum

8/31

M

N

N

N

zbzbzbb

zazazaazX

2

2

1

10

2

2

1

10)(

z-Transform general form

Those values of z (zi) that make the nominator zero are called zeros.

While the poles are the values of z (pi) that make the denominator zero

and thus X(z) diverges.

)())((

)())(()(

21

21

N

N

pzpzpz

zzzzzzkzX

For stable, causal digital systems the region of convergence includes the

Unit Circle so that the system possesses spectrum

H e H z h n z h n ej T

z e

n

n z e

j T

n

j T

j T

( ) ( ) ( ) ( )

9/31

A simple way to evaluate the signal from the X(z) is to perform the

division

X zz z

z z( )

.

1 2

1 0356

1 2

1 2

X z z z z( ) . . 1 3 36439 257561 2 3

The signal is x(0)=1, x(1)=3, x(2)=3.6439, x(3)=2.5756

Inverse z-Transform

10/31

z-Transform properties

Delay or Shift )()(

)()(

zXzmnx

zXnx

m

x n X z

x n X z

ax n bx n aX z bX z

1 1

2 2

1 2 1 2

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

Linearity

y n h k x n k Y z H z X zk

( ) ( ) ( ) ( ) ( ) ( )

Convolution

11/31

y(n-Μ)

x(n-Ν)

y(n-Μ+1)

x(n-Ν+1) x(n-2)

y(n-2) y(n-1)

x(n-1)

y(n)

x(n) Ts

Ts Ts Ts

b1

a1 a0

b2

a2 aΝ-1

bΜ-1 bΜ

aΝ

+

Ts Ts

z-Transform and digital systems

y n a x n k b y n kk

k

N

k

k

M

( ) ( ) ( )

0 1

N

k

k

k

N

k

k

k

zb

za

zH

1

0

1

)(

12/31

y n a x n k b y n kk

k

N

k

k

M

( ) ( ) ( )

0 1

Y z a z X z b z Y zkk

k

N

kk

k

M

( ) ( ) ( )

0 1

M

k

k

n

nN

k

k

n

n

n

n knybzknxaznyz10000

)()()(

0100

)()()(n

nM

k

k

n

nN

k

k knyzbknxzazY

M

k

k

N

k

k kzYbkzXazY10

)()()(

z-transform both sides

Replace transforms

Interchange

summations

Shift property

Digital system z-Transform derivation

Common factors

M

k

k

k

N

k

k

k zbzYzazXzY10

)()()(

13/31

Digital system z-Transform derivation

)(

1)(

)(

1

0 zH

zb

za

zX

zYN

k

k

k

N

k

k

k

N

k

k

k

M

k

k

k zazXzbzY01

)()1)((

Common factors

N

k

k

k

M

k

k

k zazXzbzYzY01

)()()(

y n a x n k b y n kk

k

N

k

k

M

( ) ( ) ( )

0 1

y(n-Μ)

x(n-Ν)

y(n-Μ+1)

x(n-Ν+1) x(n-2)

y(n-2) y(n-1)

x(n-1)

y(n)

x(n) Ts

Ts Ts Ts

b1

a1 a0

b2

a2 aΝ-1

bΜ-1 bΜ

aΝ

+

Ts Ts

Time

Frequency

14/31

Application

Find the impulse response h(n) and the input-output relationship of the filter

described by H z

z

z( )

.

1

1 05

1

1

Examine the stability of the filter and find its frequency response.

H zY z

X z

z

z

Y z Y z z X z X z z

y n y n x n x n

y n x n x n y n

Z

( )( )

( ) .

( ) . ( ) ( ) ( )

( ) . ( ) ( ) ( )

( ) ( ) ( ) . ( )

1

1 0 5

0 5

0 5 1 1

1 0 5 1

1

1

1 11

Solution

y(n)

x(n)

z-1

z-1

-1

-0.5

( ) / ( . ) . . .1 1 05 1 15 0 75 03751 1 1 2 3 z z z z z

h(n) is obtained from the terms of the polynomial which results after the division

15/31

Application

Im

Re

|z|=1

1

H zz

z( )

.

1

1 05

1

1Stable

0 0.1 0.2 0.3 0.40

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Frequency

Mag

nitud

e

0 0.1 0.2 0.3 0.40

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

FrequencyPh

ase

Frequency response

using MATLAB

16/31

Im

Re

|z|=1

0

π/2 or Fs/4

π or Fs/2

z=ejωt

H zz

z( )

.

1

1 05

1

1

The region from 0 to π corresponds to the region 0-

Fs/2 of the Frequency response

Frequency on the Unit Circle

17/31

Digital Filters

•They are characterized by their

Impulse Response h(n), their

Transfer Function H(z) and their

Frequency Response H(ω).

•They can have memory, high

accuracy and no drift with time and

temperature.

•They can possess linear phase.

•They can be implemented by digital

computers.

18/31

Digital Filters - Categories

IIR y n a k x n k b y n kk

N

k

k

M

( ) ( ) ( ) ( )

0 1

H z

a z

b z

kk

k

N

kk

k

M( )

0

1

1

y(n-Μ)

x(n-Ν)

y(n-Μ+1)

x(n-Ν+1) x(n-2)

y(n-2) y(n-1)

x(n-1)

y(n)

x(n) Ts

Ts Ts Ts

b1

a1 a0

b2

a2 aN-1

bΜ-1 bΜ

aN

+

Ts Ts

19/31

FIR

Digital Filters - Categories

y n h k x n k a k x n kk

N

k

N

( ) ( ) ( ) ( ) ( )

0 0

H z a z h k zkk

k

Nk

k

N

( ) ( )

0 0

x(n-Ν+1) x(n-Ν+2) x(n-2) x(n-1)

y(n)

x(n) Ts

aΝ-1 ή

h(Ν-1)

aΝ-2 ή

h(Ν-2) a2 ή

h(2)

a1 ή

h(1)

a0 ή

h(0)

+

Ts Ts

•Stable

•Linear phase

20/31

Digital Filters - Examples

0 0.1 0.2 0.3 0.40

0.5

1

1.5

Frequency

Ma

gnitud

e II

R0 0.1 0.2 0.3 0.4

-3

-2

-1

0

1

Frequency

Pha

se

IIR

0 0.1 0.2 0.3 0.40

0.5

1

1.5

Frequency

Ma

gnitud

e F

IR

0 0.1 0.2 0.3 0.4-4

-2

0

2

4

FrequencyP

ha

se

FIR

H za a z a z

b z b z( )

0 11

22

11

221

H z h k z k

k

( ) ( )

0

11

a0=0.498, a1=0.927,

a2=0.498, b1=-0.674

b2=-0.363.

h(1)=h(10)=-0.04506

h(2)=h(9)=0.06916

h(3)=h(8)=-0.0553

h(4)=h(7)=-.06342

h(5)=h(4)=0.5789

21/31

IIR filter design

Design using the Bilinear z-transform, BZT

Design using the position of poles and zeros

on the unit circle

Im

3Fs/4

Fs/4

-Fs/2

Fs/2 0 or Fs |z|=1

1

•The frequency response is zero at the

points of zeros

•The frequency response takes a peak at

the position of poles.

•In order to have real coefficients of the

filter, the poles must appear in pairs. The

same happens for the zeros as well.

22/31

IIR Filter Design - position poles and zeros on the unit circle

Design a band-pass filter with the following specifications

Sampling frequency 1000Hz, full rejection at dc and 500 Hz, Narrow pass-band at

250 Hz, 20Hz 3dBs bandwidth.

0 0.1 0.2 0.3 0.4 0

2

4

6

8

10

12

14

16

18

Frequency

Magnitude

Zeros at 0 and Fs/2 or 180ο (dc 0, 500/1000 Fs/2).

Poles at 250/1000 Fs/4 or 90ο and complex conjugate at -90ο

The radius of poles should be smaller than 1 (stability)

r bw Fs 1 1 20 1000 0937( / ) ( / ) .

Im

Re

|z|=1

1

23/31

IIR Filter Design - position poles and zeros on the unit circle

y(n)

x(n)

0.87

z-2

z-2

Im

Re

|z|=1

1

H zz z

z re z re

z

z

z

zj j( )

( )( )

( )( ) . ./ /

1 1 1

0877969

1

1 08779692 2

2

2

2

2

937.0)1000/20(1 )/(1 sFbwr

0 0.1 0.2 0.3 0.40

2

4

6

8

10

12

14

16

18

Frequency

Magn

itude

0 0.1 0.2 0.3 0.4-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Frequency

Phas

e

y n x n x n y n( ) ( ) ( ) . ( ) 2 0877969 2

24/31

FIR filters

y n h k x n k a k x n kk

N

k

N

( ) ( ) ( ) ( ) ( )

0 0

H z a z h k zkk

k

Nk

k

N

( ) ( )

0 0

x(n-Ν+1) x(n-Ν+2) x(n-2) x(n-1)

y(n)

x(n) Ts

aΝ-1 ή

h(Ν-1)

aΝ-2 ή

h(Ν-2) a2 ή

h(2)

a1 ή

h(1)

a0 ή

h(0)

+

Ts Ts

•Stable

•Linear phase

Design Methods

• Optimal filters

• Windows method

• Sampling frequency

25/31

Main categories of ideal filters

0

0

1

|Η(ω)|

2π ωc π 0

1

|Η(ω)|

2π ωc1 ωc2 π 0

1

|Η(ω)|

2π ωc π

0

j

Η(ω)

2π π

-j

j Η(ω)

2π π

Low-pass Band-pass High-pass

Differentiator Hilbert Transformer

26/31

FIR filter design – basic concept

The filter coefficients result from the Inverse Fourier

Transform of the Desired Frequency Response.

h n H e d

e d

f n

n

D Dj n

j n

c c

c

c

c

( ) ( )

sin( )

1

2

1

2

2

h nn

n( )

sin( / )

( / )

1

2

2

2

c / 2

27/31

-π/2 π/2

1

-π π 0

|HD(ω)|

1

-π π 0

|H(ω)|

1

-π π 0

|Ht(ω)|

1

-π π 0

|W(ω)|

0.5

0.5

ht(n)

w(n) 1

0

0.5

-8

-8

-8

h(n)

8

8

8

Ts

... ...

hD(n)

FIR filter design – Windows method

0

)2

cos()1()( N

naa

nwR

Gibbs phenomenon

Hamming Window

28/31

FIR filter design – Optimal filters

Parks and McClellan method.

Distributes Approximation error from the discontinuity

all over the frequency band.

29/31

0 100 200 300 400 500

-80

-60

-40

-20

0

Frequency

Ma

gnitud

e

0 100 200 300 400 500-5

0

5

Frequency

Pha

se

0 5 10 15 20 25 30 35 40 45-0.2

0

0.2

0.4

Time

IR

0 100 200 300 400 500-90

-80

-70

-60

-50

-40

-30

-20

-10

0

Frequency

Ma

gni

tud

e

Windows: 61 Coefficients

Parks and McClellan method: 46

and equiripple

FIR filter design – Comparisons

30/31

FIR filters – Linear phase

Linear phase is strictly related with the symmetry of the Impulse Response

H h k e

h h e h e h e h e h e h e

e h e h e h e h h e h e h

j kT

k

j T j T j T j T j T j T

j T j T j T j T j T j T

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

0

6

2 3 4 5 6

3 3 2 2

0 1 2 3 4 5 6

0 1 2 3 4 5

( )

( )( ) ( )( ) ( )( ) ( )

( )cos( ) ( )cos( ) ( )cos( ) ( )

6

0 1 2 3

2 0 3 2 1 2 2 2 3

3

3 3 3 2 2

3

e

e h e e h e e h e e h

e h T h T h T h

j T

j T j T j T j T j T j T j T

j T

( ) a ( ) b a

h n h N n( ) ( ) 1Condition

The phase is introduced by the term e-j3ωT and equals

θ(ω)=3ωΤ=ωΤ (7-1)/2

31/31

Linear phase - Same time delay

180o

If you shift in time one signal, then you have to shift the other signals the

same amount of time in order the final wave remains unchanged.

For faster signals the same time interval means larger phase difference.

Proportional to the frequency of the signals. -αω

5*180o

3*180o