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FIR Filter DesignWindow method

Spring 2012

© Ammar Abu-Hudrouss -Islamic University Gaza

Slide ٢Digital Signal Processing

Introduction

c

cH,0,1

)(

0sin

0,

nnn

nnh

c

c

An ideal low-pass filter has a frequency response characteristics

The impulse response of this filter

As we can see h(n) is infinite and hence we need to truncate the higher coefficients.

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Slide ٣Digital Signal Processing

Paely-Weiner Theorem

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For practical implementation, the filter function should be causal.Paely – Weiner theorem summarize the conditions that should besatisfied to guarantee causality which are :

1. The frequency response H() cannot be zero except at afinite set of points at frequency.

2. The magnitude H() cannot be constant in any finite range offrequencies

3. The Transition between the passband and stopband cannot beinfinitely sharp .

4. Real part and Imaginary part of H() are interdependent byHilbert Transform.

Slide ٤Digital Signal Processing

Practical Filter

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Slide ٥Digital Signal Processing

FIR Filter

1

0

M

kknxkbny

122

110

MM zbzbzbbzH

A FIR system can be described by the following difference equation

The transfer function is given by

The impulse response is

Mbbbbnh ,,,, 210

Then H(z) can be expressed as

1

0

)(M

k

kzkhzH

Slide ٦Digital Signal Processing

In this type, the coefficients of impulse response are symmetric

And N = M-1 is even

Type I Symmetric FIR filter

nMhnh 1

Example: consider

But as h(0)=h(6), h(1)=h(5), h(2)=h(4)

65

4321

6543210

zhzhzhzhzhzhhzH

342516 32110 zhzzhzzhzhzH

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Slide ٧Digital Signal Processing

Or

Type I Symmetric FIR filter

3210 1122333 hzzhzzhzzhzzH

Converting to DTFT

3cos22cos13cos023 hhhheH j

R

j HeH 3

HR() is a real-valued function but it can be positive or negative, then the phase is given by

0303

R

R

HH

Slide ٨Digital Signal Processing

The result can be generalized

Type I Symmetric FIR filter

2/22/cos212/cos12/cos022/

NhNhNhNh

eH Nj

R

Nj HeH 2/

the phase is given by

02/02/

R

R

HNHN

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Slide ٩Digital Signal Processing

In this type, the coefficients of impulse response are symmetric

And N =M -1 is odd

Type II Symmetric FIR filter

nMhnh 1

After mathematic manipulation, we get

2/1

1

2/ 5.0cos212

N

n

Nj nnNheH

The phase is given by

02/02/

R

R

HNHN

Slide ١٠Digital Signal Processing

In this type, the coefficients of impulse response are anti-symmetric

And M =N -1 is even

Type III Symmetric FIR filter

nMhnh 1

After mathematic manipulation, we get

2/

1

2/ sin2

2N

n

Nj nnNheH

The phase is given by

02/02/

R

R

HNHN

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Slide ١١Digital Signal Processing

In this type, the coefficients of impulse response are anti-symmetric

And M =N -1 is even

Type IV Symmetric FIR filter

nMhnh 1

After mathematic manipulation, we get

2/1

1

2/ 5.0sin212

N

n

Nj nnNheH

The phase is given by

02/02/

R

R

HNHN

Slide ١٢Digital Signal Processing

Properties of Linear Phase FIR Filters

All the four type have constant group delay (linear phase)

2/Ndd

1. Type I FIR filters have either an even number of zeros or no zeros at z = 1 and z = −1.

2. Type II FIR filters have an even number of zeros or no zeros at z = 1 and an odd number of zeros at z = −1.

3. Type III FIR filters have an odd number of zeros at z = 1 and z = −1.

4. Type IV FIR filters have an odd number of zeros at z = 1 and either an even or odd number of zeros at z = −1.

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Slide ١٣Digital Signal Processing

Properties of Linear Phase FIR Filters

•Type I and II are suitable only for lowpass filters•Type III is suitable for designing bandpass filters•Type IV is used mainly for highpass and bandpass

Slide ١٤Digital Signal Processing

Window method

The magnitude responses of four ideal classical types of digital filters are shown in Figure.

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Slide ١٥Digital Signal Processing

An ideal low-pass filter has a frequency response characteristics:

To get FIR filter:

hLP (n) is truncated to be defined only between n = -N and n = N

This is equivalent to multiplying h(n) by w (n). Where

Windows method

0sin

0,

nnn

nnh

c

c

LP

otherwise,0

,1)(

NnNw

Slide ١٦Digital Signal Processing

Rectangular window

The resultant functions is given by

The new Hd (), will not be the same as the ideal H (). Gibbs phenomena is raised by this truncation of h(n). The overshot can be reduced by increasing M but the oscillation increases.

nhnwnh LPd

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Slide ١٧Digital Signal Processing

other windows

To reduce gibbs phenomena, researchers have use different type of windows such as

1) Bartlett windows

2) Hann Windows

3) Hamming windows

10;1212

1

MnM

Mnnw

101

2cos121

Mn

Mnnw

101

2cos46.054.0

Mn

Mnnw

Slide ١٨Digital Signal Processing

4) Blackman window

5) Kaiser window

6) Lanczose window

101

4cos08.01

2cos5.042.0

Mn

Mn

Mnnw

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Slide ١٩Digital Signal Processing

4) Tukey window

Slide ٢٠Digital Signal Processing

Impulse response for HP, BP and BS Filers

The impulse response of the highpass, bandpass and bandstopare derived by the same method

0sin

0,

nnn

nnh

c

c

0sinsin1

0,

12

12

nnnn

nnh

cc

cc

0sinsin1

0,

21

12

nnnn

nnh

cc

cc

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Slide ٢١Digital Signal Processing

Examples

Example 1: Design a bandpass filter using hamming window of length 11. given that c1 = 0.2 and c1 = 0.6

04.0sin6.0sin10,4.0

nnnn

nnh

90102cos46.054.0

nnnw

Solution: M = 11 = 2N + 1

N =5.

substituting in h (n) for bandpass filter and w (n) for hamming window and shifting the result by 5

0,0289.0,1633.0,2449.0

,1156.0,4.0,1156.0,2449.0,1633.0,0289.0,0nh

Slide ٢٢Digital Signal Processing

Examples

Example 1:

90112cos46.054.0

nnnw

substituting in w(n) for hamming window

08.0,1679.0,0379.0,6821.0

,9121.0,0.1,9121.0,6821.0,0379.0,1679.0,08.0nw

The result is achieved by hw(n) = w(n)h(n)

0,0.0049,0.0650-

,0.1671-,0.1055,4.0,0.1055,0.1671-,0.0650-,0.0049,0nhw

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Slide ٢٣Digital Signal Processing

Hamming Window Examples

Example 2 a) Calculate the filter coefficients for a 3-tap FIR lowpass filter

with a cutoff frequency of 800 Hz and a sampling rate of 8,000 Hz using Hamming window

b) Determine the transfer function and difference equation of the designed FIR system.

c) plot the magnitude frequency response

1871.0)1()1(,2.00 hhh

The coefficients of the lowpass filter has been calculated before

2.08000/80022 scc Tf

Slide ٢٤Digital Signal Processing

Hamming Window Examples

nnw cos46.054.0)(

01497.0)2()1(,2.00 www hhh

For Hamming window with M = 3

and for n = 0 ,1, 2

Then the windowed impulse response coefficients are

08.0)2()1(,10 hamhamham www

convert to z-domain 21 1497.02.001497.0)( zzzH

Then the difference equation is )2(01497.0)1(2.0)(01497.0)( nxnxnxny

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Slide ٢٥Digital Signal Processing

Hamming Window Examples

201497.02.001497.0)( jj eeH

)01497.02.001497.0()( jjj eeeH

)cos02994.02.0()( jeH

cos02994.02.0)( H

0cos02994.02.0if0cos02994.02.0if

)(

H

Slide ٢٦Digital Signal Processing

Hamming Window Examples

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Slide ٢٧Digital Signal Processing

Hamming Window Examples

Example 3: a) Design a 5-tap FIR band reject filter with a lower cutoff frequency of

2,000 Hz, an upper cutoff frequency of 2,400 Hz, and a sampling rateof 8,000 Hz using the Hamming window method.

b) Determine the transfer function.

0

sinsin

0,

nnn

nn

nnh

LH

LH

2N + 1 = 5 which leads that N = 2, h (n) is given by

09355.0)4()0(01558.0)3()1(

,9.02

hhhh

h

Then the five samples of h (n ) shifted by 2 is equal to

Slide ٢٨Digital Signal Processing

Hamming Window Examples

2

cos46.054.0)( nnw

00748.0)4()0(00841.0)3()1(

,9.02

ww

ww

w

hhhh

h

For Hamming window with M = 5

and for n = 0 ,1 ,2, 3, 4

Then the windowed impulse response coefficients are

,08.0)4()0(,54.0)3()1(

,12

hamham

hamham

ham

wwww

w

Convert to z-domain 4321 00748.000841.09.000841.000748.0)( zzzzzH