eee507_2DFIRDesign1

7/24/2019 eee507_2DFIRDesign1

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Design of 2D Finite-Impulse-Response (FIR) Filters

2-D FIR Filter Design Problem

Let

and: Impulse response of ideal filter

, 21

),( 21 nni

consists of an infinite summation and cannot be realized in

2211

1 2

),(),( 2121njnj

n n

eenniI

=

),( 21 I

Approximate by a finite-degree polynomial function),( 21 I

2211

12 22njnj

N N

111 212

computedbetotscoefficien

2121 ,, Nn Nn= = 43421

FIR Filter

Copyright by Prof. Lina Karam

EEE 507 - Lecture 12


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2-D FIR Filter design using windows

Truncates infinite sum

en ca o - proce ure

Spatial or time-domain method

Review of 1-D Procedure

.

2. Then

.

= dWIH )()(1)(

3. The window w(n) should be chosen such that

- it is N-point long

-

- w(n) should be conjugate symmetric if H() is to be linear-phase

10);1()( * = NnnNwnw




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The 2-D Procedure

. e

2. Then

,.,, 212121 nnwnnnn =

=

2122112121 ),(),(1

),( ddWIH

3. The window w(n1,n2) should be chosen such that (Good

Window criteria)

- 1 2

- its DTFT

- w(n1,n2) should be conjugate symmetric if H(1,2) is to be

-

),(),( 2121 W

10

10);1,1(),(

22

112211

*

21

=

Nn

NnnNnNwnnw




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-

Choosing The 2-D Window

1-D windows often used for generating 2-D windows

- Method 1:

a 2D window formed as the outer roduct of Good 1-D windows

)().(),( 221121 nwnwnnw =

b) has a rectangular region of support

c) Main lobe shape and side lobe heights can be calculated using

1-D results =

),( 21 nnw

- Method 2:

., 221121

)(),(2

2

2

121 nnwnnw +=

a) w(n) is a Good 1-D continous window function

b) w(n1,n2) is a sampled and approximately a circularly rotatedversion of the 1-D window.




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Desi n of 2D Finite-Im ulse-Res onse FIR Filters


Example 1: 11x11 lowpass FIR filter design based on an outer product

window and a 1-D Kaiser window

Perspective Plot of Magnitude Response




6/18

Desi n of 2D Finite-Im ulse-Res onse FIR Filters 2-D FIR Filter design using windows

Example 1: 11x11 lowpass FIR filter design based on an outer

-

Contour Plot of Magnitude Response




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Example 2: 11x11 lowpass FIR filter design based on a rotated window

-

Perspective Plot of Magnitude Response




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Example 2: 11x11 lowpass FIR filter design based on a rotated

w n ow an a - a ser w n ow

Contour Plot of Magnitude Response




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Optimal FIR Filter Design Error measure needed to assess how much the designed filter

deviates from the desired ideal filter ),( 21 H),( 21 I

),(),(),( 212121 IHE =2211

1 2

),(),( 2121njnj

n n

eennhI

+=(n1,n2) in

Choose the filter coefficients to minimize some function of this error L Norm:

finite-extent ROS

Lp Norm:

2/1

21

2

2122}),(

4{ =

ddEE

L

Norm:

p

p ddEE 21212 }),(4{ =

),(max EE =



),( 21 B


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Optimal Least-Squares Designs

=

22 1

ddEE

[ ] =1 2

;),(),( 22121n n

nninnh

4

If R is the region of support of h(n1,n2):

[ ] [ ] +=2

21

2

2121

2

2 ),(),(),( n Rnn Rn nninninnhE

To minimize E22, set

= Rnnnni

nnh),(,),(

),(2121

21

where w(n1,n2) is a rectangular window with support R.

e se,),(),( 2121 nnnni =




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Optimal Design with Constraints

),(]exp[),(),( 21),(

22112121

21

=Rnn

InjnjnnhE

We have been treating the filter coefficients as if they were independent.These may be constrained.

Exam le: Desi n of a zero- hase filter with real coefficients

(i.e., H(1, 2) Symmetric) ),(),( 2121 nnhnnh =

,,,, 21),(

22112121

21

=Rnn




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Example: Design of a zero-phase filter with real coefficients (continued)

=> H(1, 2) & h(n1,n2) even symmetric

),(]cos[),(2)0,0(),( 2122112121 ++= InnnnhhE

where (assuming filter is (2N+1)x(2N+1))

)(),(),()( 211

21 = = IiaF

i

i

),( 21 Rnn

2N+1

F= # of degrees of freedom = # samples in region above1)12( 22 ++N

a(i) = ith free parameter

= th

2==

+++=

==

1)12(,),(2

1,)0,0(

1221 nnNinnh

ih

Note: For any linear constraints, the error can be written in the form (*); but F,

a i will differ for different choices of constraints.

,21i

1)12(),cos( 122211 +++=+= nnNinn




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p ma es gn w ons ra n s

where

),(),()(),( 211

2121 IiaEF

i

i=

=

F= # of degrees of freedom = # of free parameters

a(i) = ith free parameter

=ith basis function),( 21 i

Example: Suppose we want to design a 5x5 circularly symmetric

lowpass FIR filter. The ideal frequency response is given by:

+ R,1 2222

1

otherwise,0, 21

2

2

2

1

2

2

2

121 offunctionJINC"")(),( nnnnfnni +=+=

So, we would like to have:

,,,22

nnhnnhnnnnh ==+=




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Example: Design of a 5x5 circularly symmetric lowpass filter (continued)

),(),()(),( 12212

2

2

121 nnhnnhnnfnnh ==+=

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

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

hhhhh

hhhhhonly 6 out of the 25parameters are free.

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

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

hhhhh

hhhhh

= ),( 21 H




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, 21




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F

==

=i

iiaH

1

2121 ),)),

where F=

=

=

)2(

)1(

a

a

=

=

=

),(

),(

212

211

=)(Fa

M

=),(

,

21

213

F

M




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-

Takin the derivative of E 2 with res ect to the each of the a i and settin the

=

=

21

2

21

1

212

2

2 ),(),()(4

1ddIiaE

F

i

i

result to zero gives the following set of Flinear equations:

=

==F

i

Kik FKIia1

,...,2,1;)(

=

2121212 ),(),(4

ddKiiK

=

2121212 ),(),(1 ddII KK Note: If orthogonal basis , then

KiiK = ;0ii

iIia

= )(

)},({ 21 i

Issues:

Integral may be difficult to solve

No possibility of frequency weighting.




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Optimal Designs with Constraints

Discrete Solutions

Previous difficulties can be partially alleviated by replacing E2by

)],(),([ 21212 mmm mmm IHWE

)],(),()([ 2121 mmmmi

F

m IiaW =

- Weights

- Error can be controlled with the values of the weights and thedensity/locations of the samples.

0MW

1m i=

- The coefficients that minimize E2are given by

=

==F

i

Kik FKIia1

,...,2,1;)(

w ere),(),( 21

*

21 mmKmmi

m

miK W =

,,*

IWI =



m

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