eee507_2DFIRDesign2.pdf

7/24/2019 eee507_2DFIRDesign2.pdf

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EEE 507 - Lecture 13Copyright 2004 by Prof. Lina Karam

FIR Filter Design Using Transformations

Objectives! Reduces M-D FIR design to 1-D FIR design

! Efficient realization of designed filters

Motivation

! M-D designs based on L! (Chebyshev) norm are difficult! 1-D designs very well studied and understood

Strategy (M = 2; similar for any M-D design)! Given ideal 2-D specifications I(!1, !2)

" Transform 2-D specs into appropriate 1-D specs by choosing appropriate 1-D filter I(!) 1-D filter calledprototype filter.

" Design 1-D FIR filter H(!) approximating I(!) using known 1-D techniques

" Transform the designed 1-D prototype FIR filter H(!) into a 2-D FIR filterH(!1, !2) by using an appropriate 2-D mapping function

" Want to choose 1-D prototype I(!) and 2-D mapping function F(!1, !2) suchthat H(!1, !2) ~ I(!1, !2).

H(!) H(!1, !2)

F(!1, !2)

mapping1-D FIR prototype Resulting designed2-D FIR filter


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Two main issues

! How can we transform a 1-D FIR filter into an 2-D FIR filter ?

! How do we choose the 1-D filter (prototype filter) and mapping function

F(!1

, !2

) ?


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How to transform 1-D FIR filter into a 2-D FIR filter?

! Only symmetric 1-D filters can be transformed by the proposed

McClellan Transformation method.

! Background:

" Mcclellan transformation method maps 1-D positive-symmetric FIR filters into

2-D (positive-symmetric) FIR filters.

" Consider a 1-D zero-phase symmetric filter of length 2N+1

=

=N

Nn

njenhH )()(

=

=N

n

nna

0

cos)(

=

++=N

n

njnj eenhh1

])[()0(

=

=

++=N

n

nj

Nn

njenhenhh

1

1

)()()0(

==

Nnnh

nhna

1),(2

0),0()(

where


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How to transform 1-D FIR filter into a 2-D FIR filter? (continued)! Background: Chebyshev Polynomials

" It is well-known that can be expressed as a polynomial ofdegree n in the variable

where

" Thus:

)(coscos nTn =

11),()(2)(

)(1)(

polynomialChebyshev)(

11

1

0

=

=

=

=

+ xxTxxTxT

xxTxT

nxT

nnn

th

n

=

=N

n

nTnaH0

)(cos)()(

ncoscos

nnaH

N

n== 0 cos)()(


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How to transform 1-D FIR filter into a 2-D FIR filter? (continued)

! Background: Derivation of the Chebyshev Recursion

" Let , then

or

)cos(

2

1)cos(

2

1coscos BABABA ++=

)1(and == nBA

)2cos(cos)1cos(cos2 += nnn

)2cos()1cos(cos2cos += nnn)(cos)(coscos2)(cos 21 = nnn TTT

)()(2)( 21 xTxxTxT nnn =


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How to transform 1-D FIR filter into a 2-D FIR filter? (continued)

! The Transformation:

"McClellan suggested that we can obtain a 2-D zero-phase FIR filter

if we make the substitution

Then

=

=N

n

n FTnaH0

2121 )],([)(),(

),(cos 21 Freplace by

=

=N

n

nTnaH0

)(cos)()(

nnaH

N

n== 0 cos)()(


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How to transform 1-D FIR filter into a 2-D FIR filter? (continued)! The Transformation:

" The mapping function F(!1,!2) should be real (zero-phase) and

since F(!1,!2) has to replace cos(!) and has to take the samevalues of cos(!) since argument of Chebyshev polynomial=> H(!1,!2) takes same exact values of designed 1-D FIRprototype filter H(!).

Consequence:

- values of H(!1,!2) are determined by values of H(!).

- H(!1,!2) only takes values of H(!).


1),(1or1),( 2121 FF

=

=N

n

n FTnaH0

2121 )],([)(),(

=

=N

n

nTnaH0

)(cos)()(


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" The mapping function F(!1,!2) should be chosen to be the frequencyresponse of a zero-phase FIR filter

" If F(!1,!2) is real and a(n) real H(!1,!2) is real zero-phase

" If F(!1,!2) is (2Q+1) x (2Q+1) H(!1,!2) is (2NQ+1) x (2NQ+1) becauseH(!1,!2) is a polynomial of degree Nin F(!1,!2) same size as FN

" Size of support of h(n1,n2) depends on Q (order of transformation subfilter F)and on N (order of 1-D prototype filter)

)12)(12(

)14)(14(*

)12()12(

2

++

++

++

NQNQFfff

QQFff

QQFf

N

TimesN

!"!#$ %

&

!!! "!!! #$"#$gaFa

n nQnQFTg

nn ofsupportofsize...

)12()12()(

0

++

++)12()12(),(

0...

21 ++

++

NQNQHh

bFb nn

!"!#$


FT


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" The mapping function F(!1,!2) determines the symmetries of thedesigned 2D FIR filter H(!1,!2) => H(!1,!2) have the samesymmetries of F(!1,!2).

" Example: For zero-phase, even-symmetric H(!1,!2), F(!1,!2)should be zero-phase and even-symmetric of the form:

==N

n

n FTnaH0

2121 )],([)(),(

=

=N

n

nTnaH0

)(cos)()(

)sin()sin(

)cos()cos(),(

2

0 0

1

2

0 0

121

jis

rqtF

I

i

J

j

ij

Q

q

R

r

qr

= =

= =

+

=

Note: Order of F(!1,!2) = max(Q,R,I,J).


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" The mapping function F(!1,!2) determines the shape of thecontours of the designed 2D FIR filter H(!1,!2) => Contours ofH(!1,!2) have the same shape as the contours of F(!1,!2).

" Let

"Consequence: Shape of contours of H(!1,!2) determined by F(!1,!2).

Value C determined by {a(n)}, n = 1 to N, which depends on thedesigned 1-D prototype H(!) => prototype H(!) determines thevalues of H(!1,!2) along contours.

==N

n

n FTnaH0

2121 )],([)(),(

=

=N

n

nTnaH0

)(cos)()(

1,),( 21 = CCF

CCTnaHN

n

n ==

=021 ][)(),(

1

constCF ==2

constCH == '21 ),(


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Flexible and Modular:

- Change F(!1,!2) to change contour shape

- Change (prototype) to change values on contours

Procedure:

1) Design transformation subfilterF(!1,!2) :

a. to get desired symmetries

b. to produce or approximate desired contours

c. OR, if prototype filter already specified, to map values of prototype todesired 2-D locations or contours

2) Design prototype to get desired values along contours


{ })(na


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Example:

Consider

Let A=-B=-C=-1/2, D=E=1/4 (proposed by McClellan)


212121 coscos2

1)cos(

2

1)cos(

2

1

2

1),( +++=F

symmetrysamehas),(designed

symmetry)(quadrant),(),(),(),(

21

21212121

H

FFFF

=>

===

cos()cos()cos()cos(),( 1212121 +++++= EDCBAF

),( 21 nnf1n

8

1

4

1

8

14

1

2

1

4

18

1

4

1

8

1

2n


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EEE 507 - Lecture 13

Copyright 2004 by Prof. Lina Karam


Example: (continued)

Contours of F(!1,!2):

Note:1)

=>Prototype exactly mapped along axis

2)

=>Prototype exactly mapped along axis

)()(cos)()0,( 10

11 HTnaHN

n

n == =

01 =

22 cos),0( =F )()(cos)(),0( 20

22 HTnaHN

n

n == =

02 =

11 cos)0,( =F

2

Tend to be rectangular

Circular around origin

),( 21 F

1

212121 coscos2

1)cos(

2

1)cos(

2

1

2

1),( +++=F


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To get lowpass 2-D filter, use

To get bandpass 2-D filter, use

c

)(H

)(H

L H


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Designed 2D Filter H(!1, !2)

Prototype Filter H(!)Mapping F(!

1,!

2)


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Mapping F(!

1,!

2) Prototype Filter H(!

)


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Prototype Filter H(!)

Desired I(!1, !2) Mapping F(!1, !2)



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Design Example: We want to design a Fan filter using transformation

Requirement: 1) Use a 1st order transformation F(!1,!2) function => Q=R=1

2) Use a lowpass 1-D prototype


2

1

),( 21 I

2/2/

I(!)


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Example: Design of a Fan Filter (continued)

Choice of Transformation Function

- quadrant symmetric => needs to be quadrant symmetric

so, of the form:

Note: since quadrant symmetric => need to consider onlyone quadrant


),( 21 I ),( 21 F

),( 21 F

211120111000

2

0 0

121

coscoscoscos

)cos()cos(),(

tttt

rqtFQ

q

R

r

qr

+++=

= = =

2

1

),( 21 I

0

2

1


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! Determination of parameters:Typically done by:

1. Exploiting other symmetries

2. Mapping values of prototype appropriately

21112011100021 coscoscoscos),( ttttF +++=


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1. Exploiting Other Symmetries:

symmetric about line

=> Choose to be symmetric about same line

3 unknowns => need 3 equations to solve for them

),( 21 F

),( 21 I += 12

),(),( 1221 =FF

121110121000

211120111000

coscoscoscos

coscoscoscos

tttt

tttt

+=

+++

=

=

0110

1001

tt

tt

211121100021 coscos)cos(cos),( tttF ++=

21112011100021 coscoscoscos),( ttttF +++=

2

1

),( 21 I


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2. Map values of prototype:

Since used 1-D prototype is lowpass, appropriatemapping is as follows:

1-D point ! maps into 2-D point (!1,!2)

Note: To map value at 1-D point !to a desired 2-D location (!1,!2),simply set F(!1,!

2)=cos(!

).

211121100021 coscos)cos(cos),( tttF ++=

2

1

),( 21 I

)2

,2

(2

)0,(

),0(0

=

=

=


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a)

b)

c)

211121100021 coscos)cos(cos),( tttF ++=),0(),(0 21 ==

1)0cos(),0( ==F

1cos0cos)cos0(cos 111000 =++ ttt

)0,(),( 21 ==

1)cos()0,( == F

10coscos)11( 111000 =++ ttt2/124)2()1( 1010 == tt

)2/,2/(),(2/ 21 ==02/cos)2/,2/( == F

000

=t

1)11( 111000 =++ ttt12 111000 =+ ttt

12 111000 = ttt

(1)

(2)

0122/1

01011

10

00 ==

=

=tt

t

t

2121 cos2/1cos2/1),( = F( using (1) )


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2121 cos2/1cos2/1),( =F

Nnnhnaha

FTna

FTnhhH

N

n

n

N

n

n

==

=

+=

=

=

1),(2)(),0()0(where

)),(()(

)),(()(2)0(),(

0

21

1

2121

)cos(cos2/1),( 2121 =F2

1

0

coscos 12

12

>=>

F

)(H

2

2

)(oflength)12( nhN =+

F < 0

cos(!) >0 cos(!) < 0


25/30




Review of design procedure

: from 1-D prototype design

: frequency response of low-order FIR zero-phase filter

! If h(n) is 2N+1 points long and f(n1,n2) is (2Q+1)x(2Q+1), then

h(n1,n2) is (2NQ+1)x(2NQ+1).

! Contour shape and symmetries depends only on F.

! Amplitude associated with a particular contour depends upon the

1-D prototype

{ })(na

),( 21 F

=

=N

n n

FTnaH0 2121

)],([)(),(

=

=N

n

nTnaH0

)(cos)()(


26/30



Implementation of Transformed FIR Designs

Since the filter is FIR, a direct convolution can be used to implementthe filter

A DFT can also be used

There also exists a special implementation that exploits the design! Based on the recursion of Chebyshev polynomials

! Thus, if we have and , we can get ,because

F(!1,!2) is an FIR filter.

)],([)],([),(2)],([ 2122112121 FTFTFFT nnn =

1)],([ 210 =FT

),()],([ 21211 FFT =][FTn][1 FTn ][2 FTn

+ +

-

][1 FTn][FTn

][2

FTn

),(2 21 F


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From the Chebyshev polynomials recursion, we obtain:

In addition, we can get and from the polynomial thatcome before and so on until we get back to and .

][2 FTn][1 FTn

++

-

][1 FTn ][FTn

][2 FTn

),(2 21 F

][0 FT][1 FT

F 2F 2F

-1

-1

+ +

][0 FT ][1 FT ][2 FT ][3 FT


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Finally, the frequency response H(!1, !2) of the 2D designed FIRfilter is just a linear combination of the :

where

Note: This structure will work for filters of any dimensionality.

To realize an M-dimensional filter ,we simply substitute an M-dimensional mapping function

=

=N

n

n FTnaH0

2121 )],([)(),(

][FTi

.1),(2)(and)0()0( Nnnhnaha ==

F 2F 2F

-1

-1

+ +

+ + +out

IN

h(0) 2h(1)2h(2) 2h(3)

),...,,( 21 MF

),...,,( 21 MH


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Consider the first-order transformation:

Direct implementation of F requires 5 multiplications per output and8 adds per output

! Note: Multiplications reduced by combining terms

)cos()cos(coscos),( 21212121 +++++= EDCBAF

==

==

==

==

==

=

121

121

21

21

11

21

,1,2

,1,2

1,0,2

0,1,2

0,

),(

nnnE

nnnD

nnC

nnB

nnA

nnf

222

22

222

DCE

BA

B

ECD

),( 21 nnf of size 3x3


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Computational Complexity

! To implement each filter F of size (2Q+1)x(2Q+1) requires in general:

Multiplications/Output

Additions/ OutputRows of storage

! For the whole filter H(!1, !2), we have Nfilters F:

Mults/OutputAdds/Output

2)12( +Q

1)12(

2

+Q)12( +Q

)1()12(

2

+++ NQN)1()12( 2 ++ NQN

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eee507_2DFIRDesign2.pdf