Frequency Response of the Running Average Filter ... The frequency response of an LTI system is the DTFT of the impulse response,. H(ω) = ∑(m = − ∞ to ∞) h(m) ...SkiptomaincontentUCBerkeleyEECSDept.HomePiazzzabSpaceLogisticsOverviewTextbookScheduleTopic notesSidebartopicsExamsResourcesHomeworkReadingLabsEECS20N:SignalsandSystemsHome Up to Topics Previous NextWeek12Freq.ResponseContinuoustimeMoving AverageSignal SpacesTransformsSymmetryInverseExamplesLinearityUsing LinearityConstantsExponentialsSinusoidDiscreteFourier TransformsConvolutionSignalproductsFrequencyResponseoftheRunningAverageFilter ThefrequencyresponseofanLTIsystemistheDTFToftheimpulseresponse,H(ω)=∑(m=−∞to∞)h(m)e−jωm.TheimpulseresponseofanL-samplemovingaverageish(n)=1/L,forn=0,1,...,L−1h(n)=0,otherwiseSincethemovingaveragefilterisFIR,thefrequencyresponsereducestothefinitesumH(ω)=(1/L)∑(m=0toL−1)e−jωm..WecanusetheveryusefulidentitytowritethefrequencyresponseasH(ω)=(1/L)(1−e−jωL)/(1−e−jω).wherewehaveleta=e−jω,N=0,andM=L−1.Wemaybeinterestedinthemagnitudeofthisfunctioninordertodeterminewhichfrequenciesgetthroughthefilterunattenuatedandwhichareattenuated.BelowisaplotofthemagnitudeofthisfunctionforL=4(red),8(green),and16(blue).Thehorizontalaxisrangesfromzerotoπradianspersample.Noticethatinallthreecases,thefrequencyresponsehasalowpasscharacteristic.Aconstantcomponent(zerofrequency)intheinputpassesthroughthefilterunattenuated.Certainhigherfrequencies,suchasπ/2,arecompletelyeliminatedbythefilter.However,iftheintentwastodesignalowpassfilter,thenwehavenotdoneverywell.Someofthehigherfrequenciesareattenuatedonlybyafactorofabout1/10(forthe16pointmovingaverage)or1/3(forthefourpointmovingaverage).Wecandomuchbetterthanthat.TheaboveplotwascreatedbythefollowingMatlabcode:omega=0:pi/400:pi;H4=(1/4)*(1-exp(-i*omega*4))./(1-exp(-i*omega));H8=(1/8)*(1-exp(-i*omega*8))./(1-exp(-i*omega));H16=(1/16)*(1-exp(-i*omega*16))./(1-exp(-i*omega));plot(omega,[abs(H4);abs(H8);abs(H16)])axis([0,pi,0,1]) Copyright©2000--UniversityofCalifor