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Chapter8

DesignofDigitalFilters

Contents

8.1 8.2c asystemmeanstodesignadigitallter.

Goal:givendesired

magnituderesponsejHd(!)j phaseresponse\Hd(!) tolerancespecications(howfarfromideal?),

H(z)=P

M k=0bkz1 PN k=0akz1=gQ i(zzi)Q j(zpj); multiplies,anddelays.

Overview

N=0,FIRorall-zero.linearpassbandphase.

8.1

GeneralConsiderations

Causal(fornow)

Polesinsideunitcircleforstability

8.1.1Causality

AnLTIsystemiscausaliff

impulseresponse:h[n]=0forn<0 frequencyresponse:WhatcanwesayaboutH(!)?

Fact:ifh[n]iscausal,then

interdependent. c

Noperfectlyatregions

whichthefrequencyresponseisaconstant. iscausal. following. pStopband RipplePassband

Stopband1

Transition BandPassband Ripple

PSfragreplacements

1+1 11 p!s 2 !jH(!)j

Oftenonemustiterate.

howwigglytheresponseisinthestopband.

Exampleapplication:CDdigitalcrossover

8.4c 8.2

DesignofFIRFilters

y[n]=M1X k=0b kx[nk]: b k=0toachievethatgoal.

Impulseresponse

ClearlyforanFIRlter

h[n]=bn;n=0;:::;M1

0;otherwise:

(Anexceptionwouldbenotchlters.)

8.2.1SymmetricandantisymmetricFIRlters

Ifocusonthesymmetriccase.

Systemfunction:

H(z)=M1X

n=0h[n]zn:

Howdowemakealterhavelinearphase?

Example.ForM=5:h[n]=fb0;b1;b2;b1;b0g:

Example

.ForM=3andh[n]=n

1=2;1;1=2o

H(!)=1

2+e|!+12e|2!=e|!12e|!+1+12e|!

=e|!(1+cos!); k=0toPM1 c

Proof.SupposeMiseven:

H(z)=M1X

n=0h[n]zn M=21X n=0h[n]zn+M1X n=M=2h[n]zn(splitsum) M=21X n=0h[n]zn+M=21X n

0=0h[M1n0]z(M1n0)(n0=M1n)

M=21X n=0h[n]zn+M=21X n=0h[n]z(M1n)(symmetryofh[n]) M=21X n=0h[n]h z n+z(M1n)i (combine) =z(M1)=2M=21X n=0h[n]h z (M1)=2n+z((M1)=2n)i (splitphase).

Thusthefrequencyresponseis

H(!)=H(z)

z=e|! =e|!(M1)=2M=21X n=0h[n]h e |!((M1)=2n)+e|!((M1)=2n)i =2e|!(M1)=2M=21X n=0h[n]cos !M1 2n e|!(M1)=2Hr(!)

Hr(!)=2M=21X

n=0h[n]cos !M12n :(Realsinceh[n]isreal.)

Phaseresponse:

\H(!)=!M1

2;Hr(!)>0

!M1

2+;Hr(!)<0:

8.6c cannotusetheshiftproperty.

Fromabove,H(z)=z(M1)=2PM=21

n=0h[n]z(M1)=2n+z((M1)=2n)so

Hz1=z(M1)=2PM=21

n=0h[n]z(M1)=2n+z((M1)=2n)=zM1H(z):Thus

H(z)=z(M1)Hz1:

thensoisq.

Example

Re(z)Im(z)2

r1 r

Re(z)Im(z)4

Delays

x a(t)L$esXa(s):

Wecanmakeanapproximation,e.g.,

e s1s+1 2s22 untilthenexttimepoint. c 8.2.2

Designoflinear-phaseFIRltersusingwindows

h d[n]=1 2Z H d(!)e|!nd!:

Example

.Asshownpreviously,ifHd(!)=1;j!j!c;

0;otherwise;=rect

!2!c thenhd[n]=!csinc!cn h[n]=w[n]hd[n]=hd[n]w[n];n=0;:::;M1

0;otherwise;

Whatistheeffectonthefrequencyresponse?

W(!)=M1X

n=0w[n]e|!n

H(!)=W(!)

2Hd(!)=1

2Z H d()W(!)d;(8-1) where

2denotes2-periodicconvolution.

k=12(!2k);aDirac

Soinpracticewemustmaketradeoffs.

Example

.rectangularwindow.w[n]=1;n=0;:::;2

2(n1)andh[n]=n

sinc12;1;sinc12o wheresinc12=2=0:64. cos(!):Picture 8.8c

Example.rectangularwindow.

w[n]=1;n=0;:::;M1

0;otherwise;

W(!)=M1X

n=0e |!n==e|!(M1)=2Wr(!);whereWr(!)=8 :sin(!M=2) sin(!=2);!6=0

M;!=0:

2:

Thefunctionsin(!M=2)

Example

.Hereisthecase!c==4forvariousvaluesofM.

010200

0.2 0.4 0.6 0.8 1 n w[n]

Rectangular Window

-p 0 p -0.2 0 0.2 0.4 0.6 0.8 1 w W r (w) / M

Window Response

-p 0 p0 0.2 0.4 0.6 0.8 1 w |H(w)|

Magnitude Response

010200

0.2 0.4 0.6 0.8 1 n w[n]

Rectangular Window

-p 0 p -0.2 0 0.2 0.4 0.6 0.8 1 w W r (w) / M

Window Response

-p 0 p0 0.2 0.4 0.6 0.8 1 w |H(w)|

Magnitude Response

c

Time-delayindesiredresponse

alsophase-shiftedtocompensate. H d(!)=e|!(M1)=2;j!j!c;

0;otherwise;=)jHd(!)j=1;j!j!c;

0;otherwise;

sothat h d[n]=!c sinc!c nM12

H(!)=1

2Z H d()W(!)d=12Z !c =e|!(M1)=21 2Z !c !csin((!)M=2)sin((!)=2)d:

2T=22=44kHz

iswellwithinthetolerablerangeforaudio. -p 0 p0 0.2 0.4 0.6 0.8 1

Ideal Frequency Response

H d (w) -20-1001020-0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 n h d [n]

Ideal Impulse Response

-20-1001020-0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 n h[n]

Causal FIR Filter 1

-p 0 p0 0.2 0.4 0.6 0.8 1

Magnitude Response 1

|H(w)| -20-1001020-0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 n h[n]

Causal FIR Filter 2

Shifted

by (M-1)/2 -p 0 p0 0.2 0.4 0.6 0.8 1 1.2 |H(w)|

Magnitude Response 2

8.10c

Sidelobes

Therectangularwindow

hashighsidelobesinW(!),and

Solution:usesomeotherwindowfunction.

Example

.Hanningwindow: w[n]=1 2

1cos2nM1

051015200

0.2 0.4 0.6 0.8 1 w[n]

Rectangular Window

M=9 -p 0 p-80 -32 -13 0 |W(w)| / |W(0)| (dB)

Norm. Window Response

051015200

0.2 0.4 0.6 0.8 1 w[n]

Hanning Window

M=9 -p 0 p-80 -32 -13 0 |W(w)| / |W(0)| (dB)

Norm. Window Response

051015200

0.2 0.4 0.6 0.8 1 nquotesdbs_dbs14.pdfusesText_20