数字信号处理答案9.docx
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数字信号处理答案9
Chapter9Solutions
9.1Substitutingh[n]fory[n]andδ[n]forx[n]:
h[n]=0.1δ[n]+0.5δ[n–1]+0.9δ[n–2]+0.5δ[n–3]+0.1δ[n–4]
n
0
1
2
3
4
5
6
7
8
9
h[n]
0.1
0.5
0.9
0.5
0.1
0.0
0.0
0.0
0.0
0.0
Theimpulseresponsehasafinitenumberofnon-zerotermsandisthereforeFIR.Thelengthoftheimpulseresponseis5,whichisonegreaterthanthemaximumdelayinthefilter.Thisisageneralrelationshipforallimpulseresponsesthatstartatn=0:
thelengthoftheimpulseresponseisonesamplelargerthanthemaximumdelayinthefilter.
9.2
（a）
（b）
（c）
（d）
9.3（a）Fromquestion2,
Ω
0
π/8
π/4
3π/8
π/2
5π/8
3π/4
7π/8
π
20log|H（Ω）|
0.0
–2.9
–16.9
–14.5
–16.9
–20.5
–16.9
–31.4
–16.9
θ（Ω）（）
0
–68
–135
–12
–90
23
–45
68
0
|H（Ω）|
Ω
θ（Ω）
Ω
（b）Inthepassband,whichlieswithinthefirst“bump”ofthemagnitudespectrum,thephasespectrumfollowsastraightline.Thismeansthephaseislinearinthisregion,whichinturnmeansthatnodistortionwilloccur.
9.4（a）Thedifferenceequationforathree-termmovingaveragefilteris
Thefilterfindstheaverageofeachgroupofthreeinputsamples.Thefirstsixteenoutputsare:
n
0
1
2
3
4
5
6
7
y[n]
3.33
3.67
4.00
1.00
1.00
1.00
1.00
1.00
n
8
9
10
11
12
13
14
15
y[n]
1.00
1.00
1.00
4.00
4.00
4.00
1.00
1.00
y[n]
n
Thefilterhastheeffectofsmoothingoutthelargespikesintheinput.Boundaryeffectsareevidentinthefirstfewsamplesoftheoutput.
（b）Thefrequencyresponseforthethree-termmovingaveragefilteris
Ω
0
π/4
π/2
3π/4
π
|H（Ω）|
1.0000
0.8047
0.3333
0.1381
0.3333
Themagnituderesponseforthefiltershowsacut-offfrequencyatapproximately1radian.Thispointdefinesthepassbandedgeforthissimplefilter.
|H（Ω）|
Ω
（c）Toexaminethephaseresponseinthepassband,phasesbetween0and1radianarecalculatedinsmallsteps.
Ω
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
θ（Ω）
0.0
–0.1
–0.2
–0.3
–0.4
–0.5
–0.6
–0.7
–0.8
–0.9
–1.0
Thevaluesinthetablecertainlyappeartodescribeastraightline.Thetruephaseresponse,plottedbelow,confirmsthatphaseislinearinthepassband.
θ（Ω）
Ω
9.5Forthispassbandedgefrequencytheimpulseresponseforanideallowpassfilteris
Thesamplevaluesaregiveninthetable.Notethatatn=0,itiseasiesttoevaluateh1[n]ifitisre-expressedas
Atn=0,thesincfunctionhasthevalue1,soh1[0]=0.5/π.
n
h1[n]
–8
–0.0301
–7
–0.0160
–6
0.0075
–5
0.0381
–4
0.0724
–3
0.1058
–2
0.1339
–1
0.1526
0
0.1592
1
0.1526
2
0.1339
3
0.1058
4
0.0724
5
0.0381
6
0.0075
7
–0.0160
8
–0.0301
9.6（a）Theimpulseresponsefortheideallowpassfilterisgivenby
Thesamplesoftheimpulseresponsebetweenn=–3andn=3are
n
–3
–2
–1
0
1
2
3
h1[n]
0.0750
0.1592
0.2251
0.2500
0.2251
0.1592
0.0750
Aftertruncation,theimpulseresponseforthenon-deallowpassfilterconsistsofthesevennon-zerosampleslistedinthetableandzerosamplesforothervaluesofn.Thefigurebelowshowsthetruncatedimpulseresponse.
h1[n]
n
（b）Theimpulseresponsemustbeshiftedthreestepstotherighttomakeitcausal.Thecausalimpulseresponseisshowninthefigure.Itsequationis
h[n]=0.0750δ[n]+0.1592δ[n–1]+0.2251δ[n–2]+0.2500δ[n–3]
+0.2251δ[n–4]+0.1592δ[n–5]+0.0750δ[n–6]
h[n]
n
（c）Thefrequencyresponsefortheimpulseresponseis:
Ω
0
π/4
π/2
3π/4
π
|H（Ω）|
1.1685
0.4622
0.0683
0.0378
0.0319
|H（Ω）|
Ω
Anideallowpassfilterwithacut-offfrequencyofπ/4radiansandapassbandgainidenticaltothenon-idealfilterissuperimposedontheplot.
9.7（a）Thepassbandrippleisthemaximumdeviationfromunitygaininthepassband.Theminimumgainthepassbandis0.962;themaximumis1.078.Thus,thepassbandrippleisδp=0.078.
（b）Thegainatthetopofthefirstsbumpinthestopbandis0.1005.Thisisthestopbandrippleδs.
（c）Sincethepassbandrippleis0.078,thegainattheedgeofthepassbandis1–0.078=0.922,or–0.705dB.
（d）Thegainattheedgeofthestopbandis0.1005,or–19.96dB.
（e）Thestopbandattenuationisthedifferencebetweenthegainattheedgeofthepassbandandthegainattheedgeofthestopband,indB.Thestopbandattenuationis–0.705–（–19.96）=19.26dB.
（f）Thepassbandrippledefinestheedgeofthepassband.Itoccursatthefrequency0.672rads.Theedgeofthestopbandisthefirstfrequencyatwhichthefiltergaindropsbelowthestopbandripple.Thisfrequencyis0.892radians.Thetransitionwidthisthedifferencebetweenthestopbandedgeandthepassbandedge,or0.22radians.
（Themagnituderesponseshownusesa27-termrectangularwindowtocreatealowpassfilterwithitspassbandedgeatπ/4radians.）
9.8（a）Thepassbandrippleispracticallyzero.Theexactvalueofthegainattheedgeofthepassbandis20log（1–δp）=–0.019dB,forapassbandrippleofδp=0.002.
（b）Thestopbandrippleisdeterminedbythegainattheedgeofthestopband,setbythehighestsidelobe.Fromthegraphitappearstobeabout–52dB.Theexactvalueis
–51.5dB.Therefore20logδs=–51.5,whichgivesδs=0.0027.
（c）Thebandwidthissetbythe–3dBfrequencies.Theirexactvaluesare1.608and2.478radians,thoughtheycannotbeidentifiedwiththismuchprecisionfromthegraph.UsingfS=22000,theseconverttoanalogfrequencies5630and8677Hz,forabandwidthof3047Hz.
（d）Thestopbandattenuationisthenegativeofthegainattheedgeofthestopband,or–51.5dB.
（e）Thecenterfrequencyis2.04radians,or7143Hz.
（f）Theupperedgeofthepassband,wherethegainis20log（1–δp）=–0.019dB,occursat2.365radians.Thefrequencyattheedgeofthestopbandoccursat2.661radians.Thus,theupperpassbandandstopbandedgeslieat8281and9317Hz,foratransitionwidthof1036Hz.
（Themagnituderesponseshownusesa71-termhammingwindowtocreateabandpassfilterwithpassbandedgesat0.5πand0.8πradians.）
9.9（a）
|H（f）|
f
（b）Gainattheedgeofpassband=–0.72dBmeans1–δp=0.92,orδp=0.08,whichisalsotherequiredpassbandripple.Stopbandattenuationof25dBisthesameasstopbandgainof–25dB,whichequals0.0562
|H（f）|
f
9.10
w[n]
（a）
n
（b）
w[n]
n
（c）
w[n]
n
（d）
w[n]
n
9.11（a）
Ω
0
π/4
π/2
3π/4
π
|W（Ω）|
9.0
1.0
1.0
1.0
1.0
（b）
Ω
0
π/4
π/2
3π/4
π
|W（Ω）|
4.0
2.0
0.0
0.0
0.0
|W（Ω）|
Ω
（c）
Ω
0
π/4
π/2
3π/4
π
|W（Ω）|
4.4
1.76
0.08
0.08
0.08
|W（Ω）|
Ω
（d）
Ω
0
π/4
π/2
3π/4
π
|W（Ω）|
3.36
2.0
0.32
0.0
0.0
|W（Ω）|
Ω
9.12（a）Arectangularwindowgivestherequiredstopbandattenuation.SinceN=0.91（12000）/1000=10.92,N=11isthebestchoice.
（b）AHammingwindowisthebestchoice.SinceN=3.44（5000）/2000=8.6,N=9isthebestchoice.
（c）Asin（b）,aHammingwindowisthebestchoice.SinceN=3.44（5000）/500=34.4,N=35termsareneeded.
（d）Thestopbandattenuationisthedifferencebetweenthepassbandandstopbandgains,or40dB.AHanningwindowgivestherequiredattenuation.Thetransitionwidthisthedistanceinfrequencybetweenthestopbandandpassbandedges,or1.5kHz.Fromthetable,N=3.32（22000）/1500=48.7.Sincethenumberoftermsmustbeodd,N=49isthebestchoice.
9.13（a）（i）Somepassbandrippleisaddedtothesketchtoshowitexists.
|H（f）|
f
（ii）AHanningwindowcanbeusedtomeetthespecifications.Thetransitionwidthis2.5–1=1.5kHz.N=3.32（12000）/1500=26.6,or27termsmustbeincluded.
（iii）Thepassbandedgefrequencyusedinthedesignshouldbe（desiredpassbandedge）+（transitionwidth/2）=1000+（1500/2）=1750Hz.
（b）（i）Somepassbandrippleisaddedtothesketchtoshowitexists.
|H（f）|
f
（ii）Arectangularwindowcanmeetthespecifications.N=0.91（2000）/330=5.52.Tobesurethespecificationsaremet,chooseN=7.
（iii）Toobtainapassbandedgeat500Hz,thedesignmustproceedwiththepassbandedgefrequency500+330/2=665Hz.
9.14（a）50zeros,50poles,51coefficients
（b）100zeros,100poles,101coefficients
9.15Thetransitionwidthis500Hz.Togetthepassbandedgeintherightplace,thepassbandedgefrequencyforthedesignshouldbepassbandedge+transitionwidth/2=3000+500/2=3250Hz.ThematchingdigitalfrequencyisΩ1=2πf/fS=2π（3250）/12000=0.5417πrads.Thus,theequationfortheideallowpassfilterwiththecorrectpassbandedgeish1[n]=sin（nΩ1）/（nπ）=sin（0.5417πn）/（nπ）.Thisfunctionisstraightforwardtoevaluate,exceptatn=0,whereh1[0]=（Ω1/π）sinc（nΩ1）=0.5417.
Therequiredstopbandattenuationof20log（0.01）=–40dBpointstotheHanningwindow.Thiswindow’sstopbandattenuationleadstoastopbandrippleof10（–43/20）=0.0071whichsatisfiesthestopbandripplerequirements.Thepassbandrippleforthiswindowis–0.06dBor0.007,whichalsosatisfiesthespecifications.Thetransitionwidthforthefilteris500Hz.Thenumberoftermsrequiredis3.32（12000）/500=79.7.Using79termswillproduceafilterslightlypoorerthantheonespecified.Using81termswillproduceafilterthatexceedsthespecifications.Forthiscase,select79terms.Thewindowfunctionisw[n]=0.5+0.5cos（2πn/N–1）=0.5+0.5cos（2πn/78）,definedforn=–39ton=39.
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