Copyright © 2005. Shi Ping CUC Chapter 8 FIR Filter Design Content Introduction Properties of...

Chapter 8FIR Filter Design

Content

Introduction

Properties of Linear-Phase FIR Filters

Window Design Techniques

Introduction

The advantages of the FIR digital filter

The phase response can be exactly linear;

They are relatively easy to design since there are no stability problems;

They are efficient to implement;

The DFT can be used in their implementation.

Introduction

The advantages of a linear-phase response

Design problem contains only real arithmetic and not complex arithmetic;

Linear-phase filter provide no delay distortion and only a fixed amount of delay;

For the filter of length N (or order N-1) the number of operations are of the order of N/2.

The basic technique of FIR filter design

Window design techniques;

Frequency sampling design techniques;

Optimal equiripple design techniques.

Introduction

return

The system function and frequency response

The system function of FIR filters

Let h(n), n=0,1,…,N-1 be the impulse response of length N. Then the system function is

nznhzH

It has (N-1) poles at the origin and N-1 zeros located anywhere in the z-plane.

The frequency response of FIR filters

)( jeH Magnitude response function

Amplitude response function)(H

)( jeH )(H The difference between and

)( jeH is always positive and the associated phase response is a discontinuous function.

may be both positive and negative and the associated phase response is a continuous function.

Consider the following example:

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

)cos21(1

3/20 )(

0 ,cos21)( jeH

cos21)(H

Linear-phase conditions

)()( )(jj eHeH

d )( A constant group delay

nznhzHFor

)( For

The phase response is through the origin.

The phase response is not through the origin.

)1()( nNhnh

)( For

Frequency response of linear-phase FIR filters

)()]([

)]1([)()(

zmhzzmh

znNhznhzH

)()( 1)1( zHzzH N

)]()([2

zHzzHzH

Symmetric impulse response )1()( nNhnh

1cos[()(

1cos[()()(

Antisymmetric impulse response )1()( nNhnh

1sin[()(

1sin[()()(

Type 1: symmetric impulse response, N is odd

The properties of amplitude function )(H

1cos[()()(

)cos()(

)cos()2

1cos[()(2)

)cos()()(2/)1(

1,,2,1 ),

The middle sample

Type 2: symmetric impulse response, N is even

1cos[()()(

1(cos[)()(

2,,2,1 ),

Type 3: antisymmetric impulse response, N is odd

1sin[()()(

)sin()()(2/)1(

1,,2,1 ),

Type 4: antisymmetric impulse response, N is even

1sin[()()(

1(sin[)()(

2,,2,1 ),

Zero locations of linear-phase FIR filters

)()( 1)1( zHzzH N

If has a zero at)(zH jrezz 1

Then for linear phase there must be a zero at

For a real-valued filter, there must be zeros at

jrezz 1je

return

Basic window design idea

Choose a proper ideal frequency-selective filter (which always has a noncausal, infinite-length impulse response);

Then truncate (window) its impulse response to obtain a linear-phase and causal FIR filter.

The emphasis is on

Selecting an appropriate ideal filter;

Selecting an appropriate windowing function.

Denote an ideal frequency-selective filter by )( jd eH

|| ,1)(

)](sin[21

)()( 1

deeeHFnhc

ccnjjjdd

)()( nRnw N

otherwise ,0

10 ),()()()(

Nnnhnwnhnh d

otherwise ,0

)( NnN

Windowing

The effect of Window function

deWeHeH jjd

j )()(21

sin()(

njjR eW

njj enweW

sin()(

d eHeH

|| ,1)(

)()()(21

)()(21

eHdWHe

deWeHeH

dWHH Rd )()(

The conclusion

The periodic convolution produces a smeared version of the ideal response )( j

Since the window has a finite length equal to N, its response has a peaky main lobe whose width is proportional to 1/N, and has side lobes of smaller heights.

The main lobe produces a transition band in whose width is responsible for the transition width. This width is then proportional to 1/N. The wider the main lobe, the wider will be the transition width.

)( jeH

The side lobes produces ripples that have similar shapes in both the passband and stopband.

Windowing functions Rectangular window

sin()(

)()( ),()()

eWeWnRnw

This is the simplest window function but provides the worst performance from the viewpoint of stopband attenuation. The width of main lobe is N/4

Gibbs phenomenon

The truncation of the infinite length will introduce ripples in frequency response . The oscillatory behavior near the band edge of the filter is called the Gibbs phenomenon.

)(H)(nhd

When the N is increased:The transition band of the filter will decreaseBut the relative amplitude of the peaky values will remain constant.

Bartlett window

)1 ,1( ,

Since the Gibbs phenomenon results from the fact that the rectangular window has a sudden transition from 0 to 1 (or 1 to 0), Bartlett suggested a more gradual transition in the form of a triangular window. The width of main lobe is N/8

Hanning window

This is a raised cosine window function given by:

(25.0)(5.0)(

NWWW RRR

The width of main lobe is:N

1)( nR

π nnw N

Hamming window

2cos46.054.0)( nR

π nnw N

This is a modified version of the raised cosine window.

(23.0)(54.0)(

NWWW RRR

Blackman window

4cos08.0

2cos5.042.0)( nR

π nnw N

This is a 2-order raised cosine window.

(25.0)(42.0)(

Kaiser window

This is one of the most useful and optimum windows.

Where is the modified zero-order Bessel function, and is a parameter that can be chosen to yield various transition widths and stopband attenuation. This window can provide different transition widths for the same N.

window Blackman 5.8

window Hamming 44.5

windowr rectangula 0

The design equations for Kaiser window

Given spsp AR , , ,

ps The norm transition width:

95.7sANThe filter order N:

dB21 0

dB5021dB )21(07886.0)21(5842.0

dB50 )7.8(1102.04.0

Summary of window function characteristics

Window name

Window function Filter

Peak value of side lobe

The width of main lobe

Transition width

Min. stopband attenuation

Rectangular -13 dB -21 dB

Bartlett -25 dB -25 dB

Hanning -31 dB -44 dB

Hamming -41 dB -53 dB

Blackman -57 dB -74 dB

Design procedure

Given the ideal frequency response )( jd eH

Compute the impulse response of ideal filter)(nhd

Determine the window shape and N from the minimum stopband attenuation and the transition width

Compute the impulse response of the designed filter)()()( nwnhnh d

Compute the frequency response of the designed filter and verify the performance

)( jeH

Examples of FIR linear-phase filter design

Digital FIR lowpass filter

Design a digital FIR lowpass filter:

sec)/rad(1032

sec)/rad(105.12

sample

Example

Solution:

Compute the digital frequencies

2.0 ,3.02

sample

Derive the frequency response of ideal FIR lowpass filter

||3.0 ,0

3.0|| ,

|| ,)(

Compute the impulse response of the ideal filter

Determine the window shape and N

)](3.0sin[

)](sin[

ndeenh

ccnjjd

dB 50sA

1 ,33 ,2.0

Hamming

Compute the impulse response of the designed filter

cos46.054.0

2cos46.054.0)(

33 nRπ n

π nnw N

)]16(3.0sin[)(

cos46.054.0)16(

)]16(3.0sin[

)()()(

33 nRπ n

nwnhnh d

Compute the frequency response of the designed filter

)]([DTFT)( nheH j

Verify the performance of the designed filter

dB 073.0

It is not satisfied by this design

Let N = 34 and redesign

dB 048.0

Digital FIR highpass filter

An ideal FIR highpass filter can be obtained from two ideal FIR lowpass filters, provided they have the same phase response.

The frequency response of an ideal FIR highpass filters

|| ,)(

The impulse response of an ideal FIR highpass filters

)](sin[

)](sin[2

deeHnh

Example

Design a digital FIR highpass filter :dB 5.0 ,6.0

dB 60 ,4.0

Solution:

2.0 ,5.02

6.0 ,4.0

Derive the impulse response of ideal FIR highpass filter

)]27(5.0sin[

)]27(sin[

)](sin[

)](sin[)(

155 ,55 ,2.0

dB 60sA Blackman

Note: the N must be odd for FIR highpass filters

)()()( nwnhnh d

4cos08.0

2cos5.042.0

4cos08.0

2cos5.042.0)(

55 nRπ nπ n

π nnw N

)]([DTFT)( nheH j

dB 0039.0

It is satisfied by this design

Digital FIR bandpass filter

An ideal FIR bandpass filter can be obtained from two ideal FIR lowpass filters, provided they have the same phase response.

0 1c 2c

The frequency response of an ideal FIR bandpass filters

otherwise ,0

|| ,)( 21 cc

)](sin[

)](sin[2

deeHnh

Example

Design a digital FIR bandpass filter :

dB 1 ,35.0

dB 60 ,2.0

Solution:

15.0)](),min[(

725.02

,275.02

dB 60 ,8.0

dB 1 ,65.0

Derive the impulse response of ideal FIR bandpass filter

)5.36(

)]5.36(275.0sin[

)5.36(

)]5.36(725.0sin[

)](sin[

)](sin[)( 12

nnh cc

174 ,74 ,15.0

dB 60sA Blackman

)()()( nwnhnh d

4cos08.0

2cos5.042.0

4cos08.0

2cos5.042.0)(

74 nRπ nπ n

π nnw N

)]([DTFT)( nheH j

dB 73 dB, 73

dB 003.0 dB, 003.0

Digital FIR bandstop filter

An ideal FIR bandstop filter can be obtained from three ideal FIR lowpass filters, provided they have the same phase response.

0 1c 2c

The frequency response of an ideal FIR bandpass filters

otherwise ,0

|| ,|| 0 ,)( 21

)](sin[

)](sin[2

)( )( )(

dedede

deeHnh

njnjnj

Example

Design a digital FIR bandstop filter :

dB 40 ,4.0

dB 5.0 ,3.0

Solution:

1.0)](),min[(

,35.02

dB 5.0 ,7.0

dB 40 ,6.0

Derive the impulse response of ideal FIR bandstop filter

163 63 ,62 ,1.0

dB 40sA Hanning

)](35.0sin[

)](65.0sin[

)](sin[

)](sin[)( 12

nnh cc

Note: the N must be odd for FIR bandstop filters

)()()( nwnhnh d

63 nRπ n

π nnw N

)]([DTFT)( nheH j

dB 44 dB, 44

dB 0884.0 dB, 0884.0

return

0 2/3 1

Magnitude Response

Frequency in pi units

0 2/3 1

Piecewise Phase Response

0 2/3 1

Amplitude Response

0 2/3 1

Phase Response

return

0 0.5 1 1.5 2-10

0phase Response

Frequency in pi units0 1 2 3 4 5 6 7 8 9 10

h(n), N is odd

0 1 2 3 4 5 6 7 8 90

h(n), N is even

return

)1()( nNhnh

0 0.5 1 1.5 2

Phase Response

Frequency in pi units0 1 2 3 4 5 6 7 8 9 10

h(n), N is odd

0 1 2 3 4 5 6 7 8 9

h(n), N is even

return

)1()( nNhnh

0 5 10

Type-1 Impulse Response

0 0.5 1 1.5 2-20

frequency in pi units

Type-1 Amplitude Response

-1 0 1

Real Part

Type-1 Pole-Zero Plot

return

2 , ,0 偶对称

0 5 10

0 0.5 1 1.5 2-40

-1 0 1

Real Part

return

奇对称

0|)( H

不能用来设计高通、带阻滤波器

0 5 10

0 0.5 1 1.5 2-40

-1 0 1

Real Part

return

2 , ,0 奇对称0|)( 2,,0 H

不能用来设计低通、高通、带阻滤波器

0 5 10

0 0.5 1 1.5 2-10

-1 0 1

Real Part

return

2 , ,0 偶对称0|)( 0H

不能用来设计低通滤波器

jIm[z]

Re[z]0

return

Quadruplet

-10 -5 0 5 10 15 20 25 30-0.05

return

)()( nRnw N

-10 -5 0 5 10 15 20 25 30-0.05

)()( nRnw N

return

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-5

sin()(

return

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-5

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

return

20N)(dH 5.0c

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-5

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

5.0 ,20 cN

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

Amplitude response

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

Magnitude response

)( jeH

return

0 5 10 15 20 250

Rectangular window: N=25

n-1 -0.5 0 0.5 10

Magnitude response

-1 -0.5 0 0.5 1

Amplitude response

frequency in pi units-1 -0.5 0 0.5 1

Amplitude response of filter: wc=0.5pi

0 5 10 15 20 250

Rectangular window: N=25

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 1

Amplitude response

Magnitude response of filter: wc=0.5pi

return

-1 -0.5 0 0.5 1

N = 21

-1 -0.5 0 0.5 1

N = 51

N = 101

(a) (b)

0 10 20 300

Triangular window: N=35

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 1

20Amplitude response

return

0 10 20 300

Hanning window: N=35

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 10

Amplitude response

return

0 10 20 300

Hamming window: N=35

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 10

return

0 10 20 300

Blackman window: N=35

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 1

Amplitude response

return

-10 0 100

Kaiser window: N=35,beta=7.865

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 1

-10 0 100

Kaiser window: N=35,beta=9.5

n-1 -0.5 0 0.5 1

Magnitude response

-1 -0.5 0 0.5 1

return

0 5 10 15 20 25 30

Ideal Impulse Response

n 0 5 10 15 20 25 300

Hamming Window

0 5 10 15 20 25 30

Actual Impulse Response

n 0 0.2 0.4 0.6 0.8 1-100

0Magnitude Response in dB

return

0 5 10 15 20 25 30

n 0 5 10 15 20 25 300

Hamming Window

0 5 10 15 20 25 30

n 0 0.2 0.4 0.6 0.8 1-100

return

0 10 20 30 40 50-0.4

0.6Ideal Impulse Response

n 0 10 20 30 40 500

Blackman Window

0 10 20 30 40 50-0.4

0.6Actual Impulse Response

n 0 0.2 0.4 0.6 0.8 1-120

return

1s 1p 2p 2s

0 2.0)(

35.0 65.0 8.0

0 20 40 60

n 0 20 40 600

Blackman Window

0 10 20 30 40 50 60 70

n 0 0.2 0.4 0.6 0.8 1

return

1s1p 2p2s

0 3.0)(

4.0 6.0 7.0

dB .50

0 10 20 30 40 50 60-0.2

0.8Ideal Impulse Response

n 0 10 20 30 40 50 600

Hanning Window

0 10 20 30 40 50 60-0.2

0.8Actual Impulse Response

n 0 0.2 0.4 0.6 0.8 1-70

return

Copyright © 2005. Shi Ping CUC Chapter 8 FIR Filter Design Content Introduction Properties of...

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