Chapter 9 Analysis and Design of Digital Filterymzhang/EE422/EE422-9.doc · Web viewChapter 9...

EE 422G Notes: Chapter 9 Instructor: Zhang

Chapter 9 Analysis and Design of Digital Filter

9-1 Introduction

What designs have we done in this course? What do we mean by filters here? What do we mean by filters design?

Given specifications (requirements) => H(z)

Let’s see how we can implement a digital filter (processor) if its H(z) is given?

9-2 Structures of Digital Processors

1. Direct-Form Realization

The function is realized! What’s the issue here?

Count how many memory elements we need!

Page 9-1


Can we reduce this number?If we can, what is the concern?

Denote Implement H2(z) and then H1(z) ?

Why H2 is implemented?(1)

(2)

H2 is realized!

Can you tell why H1 is realized?

Page 9-2


What can we see from this realization? Signals at and : always the same Direct Form II Realization

Example

Solution:

Page 9-3


Important: H(z)=B(z)/A(z) (1) A: 1+…. (2) Coefficients in A: in the feedback channel

2. Cascade Realization

Factorize

General Form

Apply Direct II for each!

3. Parallel Realization (Simple Poles)

Page 9-4


Example 9-1

cascade and parallel realization!Solution:

(1)Cascade:

(2)Parallel

In order to make deg(num)<deg(den)

Page 9-5

Partial-Fraction Expansion for s


z = anything other than 0, ½, 1/8, 1 B = -112 For example, z=2

Example 9-2: System having a complex conjugate pole pair at Transfer function

Page 9-6


How do we calculate the amplitude response and ?

How the distance between the pole and the unit circle influence |H| and H ?

How the distance between the pole and the unit circle influence ?

How the pole angle influence and ?

See Fig. 9-7

Page 9-7


Page 9-8


HW: 9-2(a), 9-11, 9-13, 9-14

-9


9-3 Discrete-Time Integration

A method of Discrete-time system Design: Approximate continuous-time system Integrator a simple system system input Output

Discrete-time approximation of this system: discrete-time Integrator

1. Rectangular Integration change of y from t0 to t :

t = nT, t0 = nT- T

T small enough =>

A discrete-time integrator: rectangular integrator

Page 9-10

Constant

Constant y(nT) : System outputx(nT) : System input , to be integrated


2. Trapezoidal Integration

or

3. Frequency Characteristics

3.1 Rectangular Integrator

Frequency Response

Or

Page 9-11

Constants


Amplitude Response Phase Response

3.2 Trapezoidal Integrator

Frequency Response

Amplitude: Phase:

3.3 Versus Ideal IntegratorIdeal (continuous-time ) Integrator

when T=1 second (Different plots and relationships will result if T is different.)

Page 9-12


Low Frequency Range (Frequency of the input is much lower than the sampling frequency: It should be!)

High Frequency: Large error (should be)

Example 9-4 Differential equation (system)

Page 9-13


Determine a digital equivalent.

Solution (1) Block Diagram of the original system (2) An equivalent (3) Transfer Function Derivation

Design: 9-4 Find Equivalence of a given analog filter (IIR): Including methods in Time Domain and Frequency Domain.

9-5 No analog prototype, from the desired frequency response: FIR 9-6 Computer-Aided Design

Page 9-14


9-4 Infinite Impulse Response (IIR) Filter Design (Given H(s) Hd(z) )

9-4A Synthesis in the Time-Domain: Invariant Design1. Impulse – Invariant Design(1) Design Principle

(2) Illustration of Design Mechanism (Not General Case)Assume:

(1) Given analog filter (Transfer Function)

(a special case)

(2) Sampling Period T (sample ha(t) to generate ha(nT))

Derivation:(1) Impulse Response of analog filter

(2) ha(nT): sampled impulse response of analog filter

(3) z-transform of ha(nT)

Page 9-15


Sampled impulse response of analog filter

(4) Impulse-Invariant Design Principle

Digital filter is so designed that its impulse response h(nT)equals the sampled impulse response of the analog filter ha(nT)

Hence, digital filter must be designed such that

(5) (scaling)

=>

(3) Characteristics(1) when T 0 frequency response of digital filter

(2)

(3) Design: Optimized for T = 0 Not for T 0 (practical case) (due to the design principle)

(4) Realization: Parallel

Page 9-16

z-transfer function of the digital

filter. Of course,the z-transfer function of its impulse response.


(5) Design Example Solution:

2. General Time – Invariant Synthesis(1) Design Principle

(2)

Derivation

Page 9-17


Given: Ha(s) transfer function of analog filterXa(s) Lapalce transform of input signal of analog Filter

T sampling periodFind H(z) z-transfer function of digital filter

(1) Response of analog filter xa(t)

(2) ya(nT) sampled signal of analog filter output

(3) z-transform of ya(nT)

(4) Time – invariant Design Principle

Digital filter is so designed that its output equals the sampledoutput of the analog filter

Incorporate the scaling :

z-transfer function of digital filter

(5) Design Equation

special case X(z)=1, Xa(s) = 1 (impulse) => (6) Design procedure

A: Find (output of analog filter) B: Find C: Find D:

Example 9-5

Find digital filter H(z) by impulse - invariance. Solution of design:

(1) Find

Page 9-18

T


(2) Find

(3) Find

(4) Find z-transfer function of the digital filter

use G = T

(5) Implementation

Characteristics

(1) Frequency Response equations: analog and digital

Analog :

Digital :

(2) dc response comparison ( )

Page 9-19


Analog:

Digital:

Varying with T (should be)

,

for example , good enough

(3) versus : Using normalized frequency

Page 9-20


(4) versus

(5) Gain adjustment when => frequency response inequality

adjust G => at a special for example

If G = T = 0.3142 => If selecting G = T/1.0745 =>

3. Step – invariance synthesis

Example 9-6 . Find its step-invariant equivalent.

Solution of Design

Page 9-21


Comparison with impulse-invariant equivalent.

9-4B Design in the Frequency Domain --- The Bilinear z-transform

1. Motivation (problem in Time Domain Design)

Introduced by sampling, undesired!

x(t) bandlimited ( )0)2/( sffX

Page 9-22


for

for

Consider digital equivalent of an analogy filter Ha(f): )

Ha(f): bandlimited => can find a Hd(z) Ha(f): not bandlimited => can not find a Hd(z) Such that 2. Proposal: from axis to axis

( : given sampling frequency) (s plane to s1 plane)

Page 9-23


Observations: (1) Good accuracy in low frequencies (2) Poor accuracy in high frequencies (3) 100% Accuracy at a given specific number such that 0.01

Is it okay to have poor accuracy in high frequencies? Yes! Input is bandlimited! What do we mean by good, poor and 100% accuracy? Assume (1) (originally given analogy filter) (2) The transform is

Then, is a function of . Denote .

Good accuracy:

Poor Accuracy

100% Accuracy (Equal)

Is )(~

1H bandlimited? That is, can we find a 1 such that )(~1H =0?

Yes, 2/1 . We have no problem to find a digital equivalent

Page 9-24


without aliasing! Let’s use as a number (for example 0.2) representing any low frequency, Then, because is a good approximation of , should be a good approximation. A digital filter can thus be designed for an analogy filter which is not bandlimted!

Two Step Design Procedure: Given: analogy filter (1) Find an bandlimited analogy approximation ( ) for (2) Design a digital equivalent for the bandlimited filter . Because of the relationship between ( ) for , is also digital equivalent of . The overlapping (aliasing) problem is avoided! The designed digital filter can approximate (for and take the same value) at low frequency. 3. axis to axis (s plane to s1 plane) transformation

Requirement : ( is given sampling frequency.)

Proposed transformation :

Effect of C:We want the transformation map

(for example, ) to

=>

i.e. when the sampling period T is given, C is the only parameter which determines what will be mapped into axis with the same value.

Example:

not bandlimited

If we want to map to

Page 9-25

Variable in domain

Variable in domain

Constant


Hence, for any given T or

is bandlimited as a function of by

when at low frequencies.

Further Exactly holds!

How to select or sampling frequency at which ?

(1) should be small?

why? ,

(The accuracy should be good at low frequencies) (2) When is given or determined by application, should be large

enough such that to ensure the accuracy in the frequency range including

When

since for small x.

4. Design of Digital Filter using bilinear z-transform A procedure: (1)

(not bandlimited, (bandlimited,original analog) analog) or (2)

Page 9-26


(Transfer replace by z function of digital filter)

* Question: Can we directly obtain Hd(z) from Ha(s) ? Yes! (But how?)

Bilinear z-transform

Preparation : (1)

(2)

Hence,

Replace by s , by s1 ( )

Replace for digital filter

direct transformation from s to z (bypass s1)

Example

Digital Filter

2122212

212

21

1

21

212

2

)1()1(2)1()1(

112

)1()1(

)(

zzzCz

zz

zzC

zH

cc

c

cc

cd

C: only undetermined parameter in the digital filter.To determine C: (1)

(2) (related to the frequency range of interest)

Page 9-27

Bilinear z – transformation


Example 9-7

: break frequency Take Consider

C determined => Hd(z) determined

2

21

171573.01292893.0585786.0292893.0)(

z

zzzH d

, , To compare the frequency response with the original analog filter Ha : )8()2()2()(

4

ca

ff

saaa rfjHrfjHfjHjHcs

( replace s by ) ,

Too low fs => poor accuracy in fc.

Page 9-28


9-4C Bilinear z-Transform Bandpass Filter

1. Construction Mechanism(1)From an analog low-pass filter Ha(s)

to analog bandpass filter

i.e., replace s by to form a bandpass filter

For example low-pass

band-pass

Why? Original low-pass

Low => High Gain High => Low Gain

After Replacement

high => high => low gain

low => high => low gain

(2)From analog to digital

Replace s in by

for example

Page 9-29

In the low pass filter


2. Bilinear z-transform equationAnalog Low-pass Bandpass (analog)

s b

c

ss 22

)1()1()1(

11

11

2

212212

1

1

22

1

1

zCzzC

zzC

zzC

b

c

b

c

2

2122

22

22

2

212212

1

21

)1()1()1(

z

zzCC

CC

zCzzC

c

c

b

c

b

c

2

21

11

z

zBzA

with

3. How to select ( ) for bandpass filter (design)

Page 9-30

digital bandpass filter

Direct Transformation s (in low-pass)

s (in low-pass)


Important parameters of bandpass(1) center frequency (2) upper critical frequency(3) low critical frequency

Selection of for bandpass:

Design of ( )

We want , Also want

one parameter C => impossible

solution 2

tan2

tan22 TTC lu

c

bandwidth 2

tan2

tanT

CT

C lub

Hence, A and B can be determined to perform the transform.

4. Convenient design equation

why no C?

2tan

2tan1

2tan

2tan1

2

2tan

2tan

2tan

2tan

222

22

TT

TT

TTCC

TTCC

B

lu

lu

lu

lu

Page 9-31


2)cos(

2)cos(

2)cos()cos(

tantan1tantan1

T

T

lu

luyxyx

yxyx

Example : )1()1(1

11

11221

2

2

21

zzBzA

z

zzBzA

s

5. In the normalized frequencyReference frequency: sampling frequency

s

u

s

uu f

fr

s

l

s

ll f

fr

=> ,

=>

)(cos)(cos

2

)(cot

lu

lu

lu

rrrr

B

rrA

s =>

Example 9-9 Lowpass Transfer function of bandpass digital filter

A and B? Determined by design requirements. sampling frequency fs = 5000Hz

, ,

Page 9-32

In low-pass

In low-pass


Page 9-33


9-5 Design of Finite-Duration Impulse Response (FIR) Digital Filter

Direct Design of Digital Filters with no analog prototype. Can we also do this for IIR? Yes!

Using computer program in next section.9-5A A few questions1. How are the specifications given?

By given and )(

2. What is the form of FIR digital filter?Difference Equation

(What is T ? sampling period)

Transfer function 3. How to select T ?4. After T is fixed, can we define the normalized frequency r and

and )(r ? Yes!

Can we then find the desired frequency response

)()()( rjerArH ? Yes!5. Why must H( r ) be a periodic function for

digital filter? H( r ) = H ( n + r ) ? Why? What is its

period?

Page 9-34

period sampling he tnot period, )'(( 1 srHTr

r is not time


6. Can H( r ) be expressed in Fourier Series ? Yes! How?

See general formula :

In our case for H(r):

2/1

2/1

2

2

)(

)(

drerHX

eXrH

rjnn

n

rjnn

What does this mean? Every desired frequency response H(r) of digital filter can be expressed into Fourier Series ! Further, the coefficients of the Fourier series can be calculated using H(r)!

9-6B Design principle

n

rjnneXrH 2)(

Denote

Consider a filter with transfer function

What’s its frequency response ? )()())(( 2 2 rHenThenTh

n

rnjd

n

nrjd

given specification of digital filter’s frequency response!9-6 C Design Procedure

(1) Given H(r)

(2) Find H(r)’s Fourier series

where (3) Designed filter’s transfer function

What’s hd(nT) ? Impulse response!

Page 9-35

0 : sampling frequency? No! 0T : period of )(tx


Example 9-10: ) 2cos1(21)( rrH

Solution :(1) Given : done(2) Find ’s Fourier series

n

rjnd enThrrH 2)()2cos1(

21)(

where n = 0

212cos

21

21)2cos1(

21)0(

2/1

2/1

2/1

2/1

2/1

2/1

rdrdrdrrhd

2/1

2/1

)1(22/1

2/1

)1(2

2/1

2/1

2 2 22/1

2/1

2

41

41

221

21)(

dredre

dreeedrenTh

rnjrnj

nrjrjrj

rjnd

1,00)( 41

41)(

41

41)(

2/1

2/1

2/1

2/1

nnTh

drTh

drTh

d

d

d

2||,0)(,41

)(,21

)0(

41

21

41

)()( 2 2 2

nnThThh

eeenThrH

ddd

rjrj

n

rnjd

(3) Digital Filter111

41

21

41)()0()()(

zzzThhzThznTh dddn

nd

9-6 D Practical Issues : Infinite number of terms and non-causal(1)

Truncation => ))()(()()(

n

ndr

M

Mn

ndnc znThnwznThzH

Rectangular window function

MnMn

nwr ||0||1

)(

Truncation window

Effect of Truncation (windowing):Time Domain: Multiplication ( h and w )Frequency Domain: Convolution

Page 9-36

2M+1 terms


rrMeew

M

Mn

rnjrjr sin

)12(sin)( 2 2

(After Truncation: The desired frequency Hr

rr wH frequency response of truncated filter )The effect will be seen in examples!

(2) Causal Filters:

M

Mn

nMd

M

Mn

nd

Mc znThznThzzH )()()()(

k = n+M

Define )( MtkThL dk

Relationship: Frequency Response

MrrrrArA

eeHeHncc

nccMrjrjnc

rjc

2)()()()(

)()( 222

Design Examples

Hamming window:

Mn

MnMn

nwh

||0

||cos46.054.0)(

Example 9-11 Design a digital differentiatorStep1 : Assign

should be the frequency response of the analog differentiator H(s) = s

=> Desired rfjjrH srff

fs

2|)( 2

Step2 : Calculate hd(nT)

Page 9-37


njnj

fn

nf

eenj

fee

nf

enjn

fe

njnf

enj

renf

rderenf

rrdenf

drnrejnf

drerfjdrerH

drerfjdrerHnTh

ss

njnjsnjnjs

njsnjs

r

r

nrjnrjs

nrjr

r

nrjs

nrjs

nrjs

nrjs

nrj

nrjs

nrjd

sin22

cos22

22

21

21

21

21

21

)(

2

) 2()(

) 2()()(

2

2

2/1

2/1

2 2

2/1

2/1

22/1

2/1

2

2/1

2/1

2

2/1

2/1

2

2/1

2/1

22/1

2/1

2

2/1

2/1

22/1

2/1

2

i.e.,

Step 3: Construct nc filter with hamming window (M=7)

Page 9-38

nrj

b

a

b

a

b

a

evru

vduuvudv 2 ,

0


)()(

)()1()()1()(

7

7

1

1

7

zHzzH

znwnf

znwnf

zH

c

n

n

nh

nsn

n

nh

ns

Example 9- 12: Desired low- pass FIR digital filter

characteristic

5.0||15.00

15.0||1)(

rr

rH

nn

eenj

drenTh jjrnjd 3.0sin

1)(

21)( 3.03.015.0

15.0

2

3.01

3.0cos3.0lim) (/) (

) (/) 3.0(sinlim)0(0 0

nndnd

ndndThnnd

Page 9-39


NC filter with 17 weight’s window: ,

Page 9-40


Example 9-13 (90o phase shifter)

nnn

nee

n

en

en

drejdrjenTh

jnjn

jnjn

rnjrnjd

1

cos

1)(

21

)1( 2

1)1( 2

1

)()(2/1

0

20

2/1

2

n = 0 =>

=>

Filter: 7)()()(7

7

MznwnhzHn

nhdNC

Page 9-41


Fig. 9-32 Amplitude response of digital 90 degree phase shifter

9-6 Computer-Aided Design of Digital Filters

9-6A Command yulewalk for IIR

Example 9-14

9-6B Command remez for FIR

Example 9-15

Chapter 9 Homework

2, 4, 6, 11, 13, 14, 25, 26, 27, 29, 30, 31, 44, 48, 53, 61, 62, 64, 66, 68, 69

Page 9-42

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