EE 313 Linear Signals & Systems (Fall...
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EE 313 Linear Signals & Systems (Fall 2018) Solution Set for Homework #7 on Infinite Impulse Response (IIR) Filters CORRECTED By: Mr. Houshang Salimian and Prof. Brian L. Evans Prolog for the Solution Set. Both discrete-time finite impulse response (FIR) filters and discrete-time infinite impulse response (IIR) filters are widely used in practice. They are both used in speech and audio processing. Speech compression/decompression algorithms on smart phones use IIR filters. Splitting the audible range into sub-woolfer (20-200 Hz), wolfer (200-2000 Hz) and tweeter (2000- 20000 Hz) bands in audio speakers is implemented using IIR filters. FIR filters are more common than IIR filters in image and video processing; however, IIR filters are used in image display and printing. Analog-to-digital converters use IIR filters, and digital-to-analog converters use both FIR and IIR filters. Both FIR and IIR filters are used in communication systems. As the passband becomes more and more narrow, the number of FIR coefficients increases at a much faster pace vs. the number of IIR coefficients. For example, for a bandpass filter that passes wolfer frequencies of 200 Hz to 2000 Hz in a discrete-time audio signal sampled at 48000 Hz would require an FIR filter with 4352 coefficients or an IIR filter with 33 coefficients (obtained from 16 poles, 16 zeros, and 1 gain). Each coefficient represents a multiplication when computing an output sample; hence, the wolfer FIR filter requires 100 times more multiplications than the wolfer IIR filter. When we had analyzed the frequency response of the averaging filter on homework #6, tuneup #6, and tuneup #7, we had seen that phase response is linear vs. frequency. Lecture slide 9-9 derives the magnitude and phase response for the two-point averaging filter. A linear phase response corresponds to a fixed delay (called the group delay) through the filter. See lecture slide 9-8. Although we didn’t have the chance to go over lecture slides 9-8 and 9-9, we did see a linear phase response for a 3-point lowpass filter on lecture slide 9-6. Also, the bandpass FIR filter based on the Hamming window in mini-project #2 had linear phase (constant group delay). FIR filters can be designed to have linear phase over all frequencies. Although this is not possible for IIR filters, IIR filters can be designed to have approximate linear phase over the passband. Audio, imaging, and certain communication systems are very sensitive to phase distortion. IIR filters have one significant drawback— when implemented, they can become bounded-input bounded-output (BIBO) unstable even though their design is BIBO stable. FIR filters are always BIBO stable. Please see lecture slides 11-12 and 11-13 and Handout H on BIBO Stability. Problem 1 Prolog: This problem asks you to analyze seven linear time-invariant (LTI) filters in the frequency and z domains, and match their magnitude response to one of six example responses. This problem is intended to help you make connections between the time, frequency and z domains. In particular, the problem is intended to build experience on the impact of poles and zeros in the transfer function in the z-domain on the magnitude response of a filter. Each LTI filter is observed for n ≥ 0. All of the initial conditions are zero as a necessary condition for LTI to hold. MATLAB code is given at the end of the problem solution.
Transcript of EE 313 Linear Signals & Systems (Fall...
EE313LinearSignals&Systems(Fall2018)
SolutionSetforHomework#7onInfiniteImpulseResponse(IIR)FiltersCORRECTED
By:Mr.HoushangSalimianandProf.BrianL.Evans
PrologfortheSolutionSet.Bothdiscrete-timefiniteimpulseresponse(FIR)filtersanddiscrete-timeinfiniteimpulseresponse(IIR)filtersarewidelyusedinpractice.Theyarebothusedinspeechandaudio processing. Speech compression/decompression algorithms on smart phones use IIR filters.Splitting theaudible range intosub-woolfer (20-200Hz),wolfer (200-2000Hz)and tweeter (2000-20000Hz)bandsinaudiospeakersisimplementedusingIIRfilters.FIRfiltersaremorecommonthanIIRfiltersinimageandvideoprocessing;however,IIRfiltersareusedinimagedisplayandprinting.Analog-to-digital converters use IIR filters, and digital-to-analog converters use both FIR and IIRfilters.BothFIRandIIRfiltersareusedincommunicationsystems.
As the passband becomesmore andmore narrow, the number of FIR coefficients increases at amuchfasterpacevs.thenumberofIIRcoefficients. Forexample, forabandpassfilterthatpasseswolferfrequenciesof200Hzto2000Hzinadiscrete-timeaudiosignalsampledat48000Hzwouldrequire an FIR filter with 4352 coefficients or an IIR filter with 33 coefficients (obtained from 16poles,16zeros,and1gain).Eachcoefficientrepresentsamultiplicationwhencomputinganoutputsample;hence,thewolferFIRfilterrequires100timesmoremultiplicationsthanthewolferIIRfilter.
Whenwehadanalyzedthefrequencyresponseoftheaveragingfilteronhomework#6,tuneup#6,andtuneup#7,wehadseenthatphaseresponseislinearvs.frequency.Lectureslide9-9derivesthemagnitude and phase response for the two-point averaging filter. A linear phase responsecorresponds to a fixed delay (called the group delay) through the filter. See lecture slide 9-8.Althoughwedidn’thavethechancetogooverlectureslides9-8and9-9,wedidseealinearphaseresponsefora3-pointlowpassfilteronlectureslide9-6.Also,thebandpassFIRfilterbasedontheHammingwindowinmini-project#2hadlinearphase(constantgroupdelay).
FIRfilterscanbedesignedtohavelinearphaseoverallfrequencies.AlthoughthisisnotpossibleforIIR filters, IIR filters can be designed to have approximate linear phase over the passband.Audio,imaging,andcertaincommunicationsystemsareverysensitivetophasedistortion.
IIR filters have one significant drawback— when implemented, they can become bounded-inputbounded-output(BIBO)unstableeventhoughtheirdesignisBIBOstable.FIRfiltersarealwaysBIBOstable.Pleaseseelectureslides11-12and11-13andHandoutHonBIBOStability.
Problem1
Prolog: Thisproblemasks you toanalyze seven linear time-invariant (LTI) filters in the frequencyandzdomains,andmatchtheirmagnituderesponsetooneofsixexampleresponses.Thisproblemisintendedtohelpyoumakeconnectionsbetweenthetime,frequencyandzdomains.Inparticular,theproblemisintendedtobuildexperienceontheimpactofpolesandzerosinthetransferfunctioninthez-domainonthemagnituderesponseofafilter.EachLTIfilterisobservedforn≥0.AlloftheinitialconditionsarezeroasanecessaryconditionforLTItohold.
MATLABcodeisgivenattheendoftheproblemsolution.
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
Concerningtheconnectionsbetweenthetime, frequencyandzdomains fordiscrete-timefilters, Irecorded a YouTube video in spring 2014 for the EE 445S Real-TimeDigital Signal Processing Labcourse.Pleasewatchfromthe1:29marktothe22:25markandfrom43:01totheend(50:51)at
https://www.youtube.com/watch?v=WWEKNvvJBvs&list=PLaJppqXMef2ZHIKM4vpwHIAWyRmw3TtSf
Polesorzerosattheorigininthez-domainhavenoimpactonthemagnituderesponsebetweenthedistancefromanypointontheunitcircle𝑧 = 𝑒!!totheoriginis1forallvaluesofthediscrete-timefrequency𝜔.Whenagroupofpolesisseparateinanglefromagroupofzeros,theanglesofpolesneartheunitcircleindicateapassbandofthefilterandtheanglesofzerosonorneartheunitcircleindicateastopbandofthefilter.ThisisthecaseforallofthefiltersbelowexceptfilterS2.FilterS2hasonepoleandonezeroatthesameangle.Theirradiiareinverselyrelated,andthisgivesanall-passfilter.PleaseseeHandoutIonAll-PassFiltersat
http://users.ece.utexas.edu/~bevans/courses/signals/handouts/Appendix%20I%20All%20Pass%20Filters.pdf
Solutions
FilterS1isgivenasadifferenceequation.ThisproblemwasthesubjectofTune-UpTuesday#9.
1 1 1 1[ ] 0.9 [ 1] [ ] [ 1] [ ] 0.9 [ 1] [ ] [ 1]2 2 2 2
y n y n x n x n y n y n x n x n= − + + − ⇒ − − = + −
( )1 1 1 11 1 1 1( ) 0.9 ( ) ( ) ( ) ( ) 1 0.9 ( )2 2 2 2
Y z z Y z X z z X z Y z z X z z− − − −⎛ ⎞− = + → − = +⎜ ⎟⎝ ⎠
H (z) = Y (z)X (z)
=
121+ z−1( )
1−0.9z1=
12z +1( )
z −0.9=12
z +1z −0.9⎛
⎝⎜
⎞
⎠⎟
Tocalculatezerosandpoles,wecalculaterootsofnumeratoranddenominator,respectively.
Zeros: 1 0 1z z+ = ⇒ = − .Zeroontheunitcircleatangleπ indicatesthatfrequenciesat–πandπ rad/samplearezeroedoutandareinthestopband.
Poles: 0.9 0 0.9z z− = ⇒ = Apoleneartheunitcircleindicatesthatapeakinthemagnituderesponseoccursatthepoleangle,whichis0rad/sample.Thepassbandiscenteredat0rad/sample.
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
FrequencyresponsematchesfigureA.Frequencyresponseshowsthatthefrequenciesaroundzeroarepassedandthemagnitudedecreasesforhigherfrequencies,hence,S1isalow-passfilter.
FilterS2isalsogivenasadifferenceequation.
[ ] 0.9 [ 1] 9 [ ] 10 [ 1] [ ] 0.9 [ 1] 9 [ ] 10 [ 1]y n y n x n x n y n y n x n x n= − − + + − ⇒ + − = + −
( ) ( )1 1 1 1( ) 0.9 ( ) 9 ( ) 10 ( ) ( ) 1 0.9 ( ) 9 10Y z z Y z X z z X z Y z z X z z− − − −+ = + → + = +
1
1
( ) 9 10 9 10( )( ) 1 0.9 0.9
Y z z zH zX z z z
−+ += = =
+ +
Zero:9 10 0 10 / 9 1.111z z+ = ⇒ = − = − .Radiusistheinverseoftheradiousofthepole.
Pole: 0.9 0 0.9z z+ = ⇒ = − .Poleandzeroareatthesameanglewillinteract.Seetheprolog.
ThefrequencyresponsematchesfigureD.Thisfilterpassesallfrequencieswiththesamemagnitude,i.e.magnitude=10,soS2isanall-passfilter.
FilterS3isgivenbyitstransferfunctioninthezdomain:
( ) ( )1
1
1 11 12 2( )1 0.9 0.9
z zH z
z z
−
−
− −= =
+ +
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
Zero: 1 0 1z z− = ⇒ = .Zeroontheunitcircleatangle0 indicatesthatfrequencyat0rad/sampleiszeroedoutandisinthestopband.Infact,thestopbandiscenteredat0rad/sample.
Pole: 0.9 0 0.9z z+ = ⇒ = − .Apoleneartheunitcircleindicatesthatapeakinthemagnitude
responseoccursatthepoleangle,whichisπrad/sample.Thepassbandiscenteredatπrad/sample
ThisplotmatchesfigureB,whichisahighpassfilter.
FilterS4isgivenasadifferenceequation:
1 3 1[ ] [ ] [ 1] [ 2] [ 3] [ 4]4 2 4
y n x n x n x n x n x n= + − + − + − + −
1 2 3 41 3 1( ) ( ) ( ) ( ) ( ) ( )4 2 4
Y z X z z X z z X z z X z z X z− − − −= + + + +
4 3 2
1 2 3 44
1 3 1( ) 1 3 1 4 2 4( )( ) 4 2 4
z z z zY zH z z z z zX z z
− − − −+ + + +
= = + + + + =
Wecancalculatetherootsofthenumeratorbyusingroots()inMATLAB.
p=[1/413/211/4];roots(p)
MATLABfindfouruniquerootscloseto-1:
Zeros: 4 3 21 2,3 4
1 3 1 0 1.0002, 1 0.0002, 0.99984 2 4z z z z z z j z+ + + + = ⇒ = − = − ± = −
Thecorrectanswerisarepeatedrootatz=-1,whichiscorrectlyindicatedinthepole-zeroplot.
Havingfourrepeatedzerosatangleofπmeansthatthereductioninmagnitudeatfrequenciesnearπ rad/sampleareraisedtothefourthpower.Thesmallmagnitudevaluesbecomemuchsmaller.
Poles: 4 0 0z z= ⇒ = Fourpolesatz=0.Polesattheorigininthezdomainhavenoeffectonthemagnituderesponse.
FrequencyresponseissimilartoFigureF,whichisalowpassfilter.
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
FilterS5isgivenasatransferfunctioninthezdomain:
4 3 2 11 2 3 4
4
1( ) 1 z z z zH z z z z zz
− − − − − + − += − + − + =
Zeros: 4 3 2 11,2 3,41 0 0.3090 0.9511, 0.8090 0.5878z z z z z j z j− + − + = ⇒ = − ± = ±
Poles:Fourpolesatz=0.
ThisplotisshowninFigureE.Accordingtothefrequencyresponse,wehaveahighpassfilter,becauseithashighgainathigherfrequenciesandlowgainatlowerfrequencies.ForS3,wehadanotherhighpassfilter,therethemagnitudeinstopbandwasverylowwhileforS5wecanseethatevenat0,considerablemagnitudecanbedetected.
FilterS6isgivenasadifferenceequation:
3 3 3
0 0 0[ ] [ ] ( ) ( ) ( )k k
k k ky n x n k Y z X z z X z z− −
= = =
= − ⇒ = =∑ ∑ ∑
3 231 2 3
30
( ) 1( ) 1( )
k
k
Y z z z zH z z z z zX z z
− − − −
=
+ + += = = + + + =∑
Zeros: 3 21 2,31 0 1,z z z z z j+ + + = ⇒ = − = ±
Poles: 3 0 0z z= ⇒ = Threepolesatz=0.
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
ThefrequencyresponseislikefigureC,anditrepresentsalowpassfilter.
FilterS7isgivenasadifferenceequation:
[ ] [ ] [ 1] [ 2] [ 3] [ 4] [ 5]y n x n x n x n x n x n x n= + − + − + − + − + −
1 2 3 4 5( ) ( ) ( ) ( ) ( ) ( ) ( )Y z X z z X z z X z z X z z X z z X z− − − − −= + + + + +
5 4 3 21 2 3 4 5
5
( ) 1( ) 1( )
Y z z z z z zH z z z z z zX z z
− − − − − + + + + += = + + + + + =
Zeros: 5 4 3 21 2,3 4,51 0 1, 0.5 0.866, 0.5 0.866z z z z z z z j z j+ + + + + = ⇒ = − = ± = − ±
Poles: 5 0 0z z= ⇒ = Fivepolesatzero
Thefrequencyresponseshowsalowpassfilter,butitdoesnotmatchanyofthesixcandidatemagnituderesponses.
MATLABcode:
%S1 %feedforwardCoeffs = [ 1/2 1/2 ]; %feedbackCoeffs = [ 1 -0.9 ]; %------------------------------------------ %S2 %feedforwardCoeffs = [ 9 10 ]; %feedbackCoeffs = [ 1 0.9 ]; %------------------------------------------- %S3 %feedforwardCoeffs = [ 1/2 -1/2 ]; %feedbackCoeffs = [ 1 0.9 ]; %------------------------------------------- %S4 %feedforwardCoeffs = [ 1/4 1 3/2 1 1/4 ]; %feedbackCoeffs = 1; %------------------------------------------- %S5 %feedforwardCoeffs = [ 1 -1 1 -1 1 ]; %feedbackCoeffs = 1; %-------------------------------------------
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
%S6 %feedforwardCoeffs = [ 1 1 1 1 ]; %feedbackCoeffs = 1; %------------------------------------------- %S7 feedforwardCoeffs = [ 1 1 1 1 1 1 ]; feedbackCoeffs = 1; %------------------------------------------- figure; zplane(feedforwardCoeffs, feedbackCoeffs); W = -pi : 0.001 : pi; [H, W] = freqz( feedforwardCoeffs, feedbackCoeffs, W ); figure; plot(W, abs(H)); xlim([-pi pi]) xlabel('Radian Frequency') ylabel('Magnitude')
Problem2:a)ThissystemisformedbycascadingtwoLTIsystems.Inthez-domain:
12 1
2
( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )W z H z X z
Y z H z H z X z H z X zY z H z W z
=⎧⇒ = =⎨
=⎩
2 1( ) ( ) ( )H z H z H z=
b) 15[ ] [ ] [ 1]6
y n x n x n= + −
11
5( ) ( ) ( )6
Y z X z z X z−= +
111
( ) 5( ) 1( ) 6
Y zH z zX z
−= = +
( )1 2 1 1 2 32 1
5 7 2 5( ) ( ) ( ) 1 2 1 16 6 3 6
H z H z H z z z z z z z− − − − − −⎛ ⎞= = − + + = − − +⎜ ⎟⎝ ⎠
c) ( ) ( ) ( )Y z H z X z=
7 2 5[ ] [ ] [ 1] [ 2] [ 3]6 3 6
y n x n x n x n x n= − − − − + −
d)MATLABcode
h1=[15/6];h2=[1-21];h=conv(h2,h1);zplane(h)
Fall2018EE313Homework7solution|TheUniversityofTexasatAustin
e) [ ] [ ]y n x n= so ( ) ( ) ( ) 1Y z X z H z= ⇒ =
f)Accordingtoparte,H(z)=1toguaranteeinputandoutputareequal.Also,frompart(b)
2 1( ) ( ) ( )H z H z H z=
12
51 ( ) 16
H z z−⎛ ⎞= +⎜ ⎟⎝ ⎠
21
1( ) 516
H zz−
=+
g)H2willbestableifitspolesareinsidetheunitcircle.H2(z)isinverseofH1(z),sozerosofH1(z)arepolesofH2(z).Therefore,zerosofH1(z)shouldbeinsidetheunitcircle.
Epilog:Parts(f)and(g)concernthedesignoffilterH2compensateforthefrequencydistortionintroducedbyfilterH1.Equalizingfrequencydistortioninanunknownsystemhasmanyapplications,includingdisplay/printingofimages,calibratingbiomedicalinstrumentation,compensatingphasedistortioninananalog-to-digitalconverter,andcommunicationreceivers.Inacommunicationreceiver,thechannelequalizercompensatesforfrequencydistoritoninthecommunicationchannelaswellasintheanalog/RFcircuitsinthetransmitterandreceiver.
