CE 40763 Digital Signal Processing Fall 1992 Optimal FIR Filter Design

CE 40763Digital Signal Processing

Fall 1992

Optimal FIR Filter Design

Hossein SametiDepartment of Computer Engineering

Sharif University of Technology

Definition of generalized linear-phase (GLP):

Let’s focus on Type I FIR filter:

Optimal FIR filter design

2

)()()( jm eHH

)(),1()(),(12,0 LnNhnhoddLN

)()cos()()(0

GnnaHL

nm

• It can be shown that

0)(2)()()0(

nnLhnaLha

jeGH )()(

(L+1) unknown parameters a(n)

Hossein Sameti, Dept. of Computer Eng., Sharif University of Technology

Problem statement for optimal FIR filter design

3

p s

2

11

11

• Given 21,,, ps determine coefficients of G(ω) (i.e. a(n))

such that L is minimized (minimum length of the filter).

2 0


G(ω) is a continuous function of ω and is as many times differentiable as we want.

How many local extrema (min/max) does G(ω) have in the range ?

In order to answer the above question, we have to write cos(ωn) as a sum of powers of cos(ω).

Observations on G(ω)

4

1)(cos2)2cos( 2

sin)2sin(cos)2cos()2cos()3cos(

cos3cos4)3cos( 3

)cos( n : sum of powers of cos(ω)

)()cos()()(0

GnnaHL

nm

],0[


5

Observations on G(ω)in

iin

0)(cos)cos(

)()cos()()(0

GnnaHL

nm

])(cos)[()(0 0

L

n

in

iinaG

nL

nnG )(cos)()(

0

Find extrema

0)(

ddG

0)(sin)(cos)( 1

0

nL

nnn

0)(cos)()(sin 1

0

nL

nnn


6

Observations on G(ω)0)(cos)()(sin 1

0

nL

nnn

0)(cos)(

or00sin1

0

nL

n

nn

xcos Polynomial of degree L-1

Maximum of L-1 real zeros

Max. total number of real zeros: L+1

Conclusion: The maximum number of real zeros for(derivative of the frequency response of type I FIR filter) is L+1, where (N is the number of taps). 2

1

NL

ddG )(


Problem Statement for optimal FIR filter design

7

p s

2

11

11

• Given 21,,, ps determine coefficients of G(ω) (i.e. a(n))

such that L is minimized (minimum length of the filter).

Problem A Problem B Problem C

2 0 Problem A


Problem B

8

• Given

determine coefficients of G(ω) (i.e. a(n)) such that

is minimized.

21,,, KLps

2

21,,, ps

2

1

KCompute Guess L

Algorithm B

'2),(' na

2?

'2

Increase L by 1

Decrease L by 1

Yes

Stop!

2'2 2

'2

9

Problem C

21 IIF

p s

1I 2I

],0[:1 pI ],[:2 sI

Define F as a union of closed intervals in ],0[

0


10

Problem C)]()()[()( DGWE

where

2

1

1

1)(

I

IKW W is a positive weighting function

L

nnnaG

0)cos()()(

2

1

01

)(II

D Desired frequency response

Find a(n) to minimizeF

EMax

)( 21 IIF

(same assumption as Problem B)

We start by showing that

Problem C= Problem B?

11

2)(

F

EMax

)]()()[()( DGWE

)]()([)()( DG

WE

)()()()(

WEDG



12

p s

F

EMax )(By definition:

)(E

21 IIF

1I 2I



13

p

s

K

By definition:

K

)()(

WE

2

1

1

1)(

I

IKW

F

EMax )(



14

ps

K1

)()()()(

WEDG

2

1

01

)(II

D

K1

)(G



15

ps

K1

K1

p

2

11

11

)(G

2

1

K

2

in Problem C

)(G in Problem B

2s

Conclusion:


16

Find a(n) such that is minimized.2Problem B:

Find a(n) such that is minimized.

Problem C:

Problem B= Problem C

Problem A= Problem C We now try to solve Problem C.


Assumptions: F: union of closed intervals G(x) to be a polynomial of order L:

D = Desired function that is continuous in F. W= positive function

Alternation Theorem

17

L

k

kk xaxG

0)(

)]()()[()( xGxDxWxE

Fx

xEE

)(max

2

1

01

)(II

D

2

1

1

1)(

I

IKW

The necessary and sufficient conditions for G(x) to be unique Lth order polynomial that minimizes

is that E(x) exhibits at least L+2 alternations, i.e., there are at least L+2 values of x such that

18

Alternation Theorem

E

221 ... Lxxx

)(xE

E E E

E E E

for a polynomial of degree 4

𝑬 ( 𝒙𝒊 )=−𝑬 (𝒙 𝒊+𝟏 )=+¿−‖𝑬‖


Number of alternations in the optimal case

19

• Recall G(ω) can have at most L+1 local extrema.

• According to the alternation theorem, G(ω) should have at least L+2 alternations(local extrema) in F.

Contradiction!?

1I2I

21 IIF


20

• can also be alternation frequencies, although they are not local extrema.

ps ,

• G(ω) can have at most L+3 local extrema in F. 21 IIF

Ex: Polynomial of degree 7



21

According to the alternation theorem, we have at least L+2 alternations.

According to our current argument, we have at most L+3 local extrema.

Conclusion: we have either L+2 or L+3 alternations in F for the optimal case.


22

Example: polynomial of degree 7

Extra-ripple case


23

Example: polynomial of degree 7


For Type I low-pass filters, alternations always occur at If not, we potentially lose two alternations.

Optimal Type I Lowpass Filters

24

ps ,


Optimal Type I Lowpass Filters

25

,0Equi-ripple except possibly at


Summary of observations

26

For optimal type I low-pass filters, alternations always occur at

If not, two alternations are lost and the filter is no longer optimal.

ps ,

Filter will be equi-ripple except possibly at ,0


Parks-McClellan Algorithm (solving Problem C)

27

• Given determine coefficients of G(ω) (i.e. a(n))

such that is minimized. 221,,, KLps

2,...,2,1 Li

)()( 1 ii EE 1,...,2,1 Li

)]()()[()( GDWE

21)1()]()()[( i

iii GDW

2,...,2,1 Li)()(

)1()( 2

1

ii

i

i DW

G

At alternation frequencies, we have:

𝑬 (𝝎𝒊 )=+¿−𝜹𝟐


28

Parks-McClellan Algorithm

L

nii nnaGEq

0)cos()()()1.(

)()(

)1()()2.( 21

ii

ii D

WGEq

)()(

)1()cos()( 2

1

0i

i

iL

ni D

Wnna

2,...,2,1 Li

Equating Eq.1 and Eq.2


29


)()(

)1()cos()( 2

1

0i

i

iL

ni D

Wnna

2,...,2,1 Li

).cos()(....)2.cos()2()1.cos()1()0.cos()0(1 1111 LLaaaai

)()(

)1(1

1

211

DW

).cos()(....)2.cos()2()1.cos()1()0.cos()0(2 2222 LLaaaai

)()(

)1(2

2

212

DW

).cos()(....)2.cos()2()1.cos()1()0.cos()0(2 2222 LLaaaaLi LLL

)()(

)1(2

22

12

L

L

LD

W

30


)(

)1()0(

La

aa

)()1(

)()1(

)()1(

22

12

22

121

211

L

L

W

W

W

)(

)()(

2

2

1

LD

DD

L+2 linear equations and L+2 unknowns

BXA


31


unknownsLna 1)(

unknown12

unknownsL 2

equationsLi 2


Remez Exchange Algorithm

32

2),( na i

2

1

1

2

12

)()1(

)(

L

k k

kk

L

kkk

Wb

Db

2

1 coscos1L

kii ik

kb

It can be shown that if 's are known, then can be derived using the following formulae:

i 2



33

)(G

is an Lth-order trigonometric polynomial. We can interpolate a trigonometric polynomial through L+1 of the L+2 known values of or G. Using Lagrange interpolation formulae we can find the frequency response as:


34

Now is available at any desired frequency, without the need to solve the set of equations for the coefficients of .

If for all in the passband and stopband, then the optimum approximation has been found. Otherwise, we must find a new set of extremal frequencies.



Flowchart of P&M

Algorithm

35

Example of type I LP filter before the optimum is found

36

Original alternation frequency

Next alternation frequency


App. estimate of L:

App. Length of Kaiser filter:

Comparison with the Kaiser window

37

324.2

13)log(102 21LM

ps

2.2

81 AN 10log20A

• Example: 6.0,4.0 sp

001.0,01.0 21

• Optimal filter: 2712 LN

• Kaiser filter: 38N


Demonstration

38

6.0,4.0 sp

001.0,01.0 21

26,10 MK

Does it meet the specs?


Demonstration

39

Increase the length of the filter by 1.

6.0,4.0 sp

001.0,01.0 21

27,10 MK

Does it meet the specs?


Documents

CE 40763 Digital Signal Processing Fall 1992 Optimal FIR Filter Design