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ece431notesCh2_2007_09_13 - 2.1 The Discrete Fourier...

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CHAPTER2 TheDFT,Convolution,andWindowing InChapter1,wesawthatsignalprocessingforcontinuous-time,bandlimitedwaveforms andsystemscanbeaccomplishedbydiscrete-timesignalprocessing. However,computers canonlyevaluate finite sums. Recallthatforaninfinitesequence x n ,itsDTFTis X ( f ) : = m = - x m e - j 2 π fm M m = - M x m e - j 2 π fm forlarge M . Thesumontherightcontains2 M + 1terms. Lateritwillbemoreefficientto usesumswithanevennumberofterms.Forthisreason,weusetheapproximation X ( f ) M - 1 m = - M x m e - j 2 π fm . In order to write this as a sum starting from zero, we can make the change of variable n = m + M toget X ( f ) 2 M - 1 n = 0 x n - M e - j 2 π f ( n - M ) . Althoughtheforegoingapproximationinvolvesonlyafinitesum,acomputercannotevalu- ateitforallvaluesof f inoneperiod,say | f |≤ 1 / 2.Instead,thecomputercanonlyevaluate thesumforfinitelymanyvaluesof f .Let N : = 2 M ,andlet f = k / N toget 1 X ( k / N ) N - 1 n = 0 x n - M e - j 2 π k ( n - M ) / N = e j 2 π kM / N N - 1 n = 0 x n - M e - j 2 π kn / N . (2.1) Regarding this last sum as a function of k , observe that it has period N ; i.e., replacing k with k + N doesnotchangethevalueofthesum. Soweonlyneedtoevaluatethesumfor k = 0 ,..., N - 1. 2.1. TheDiscreteFourierTransform(DFT) Givenafinitesequence y 0 ,..., y N - 1 ,its discreteFouriertransform (DFT)is Y k : = N - 1 n = 0 y n e - j 2 π kn / N . (2.2) 1 Since N = 2 M , e j 2 π kM / N = e j π k = ( - 1 ) k . 12 2.1 TheDiscreteFourierTransform(DFT) 13 It is easy to show that Y k is a periodic function of k with period N . We show below in Section2.1.5thatthesequence y n canberecoveredfromtheDFTsequence Y 0 ,..., Y N - 1 by the inverseDFT (IDFT) y n = 1 N N - 1 k = 0 Y k e j 2 π kn / N . (2.3) Ofcoursetheright-handsideisaperiodicfunctionof n withperiod N . Hence, although y n is only defined for n = 0 ,..., N - 1, we often think of it as being an infinite-duration periodicsignalwithperiod N . 2.1.1. SummingaPeriodicSequenceoveraPeriod Let z n haveperiod N .Thenforany m , m +( N - 1 ) n = m z n = N - 1 n = 0 z n . Thisismosteasilyseenpictorially. Forexample,if z n hasperiod N = 5,weseefromthe diagram n : 0 1 2 3 4 5 6 7 8 9 z n : a 0 a 1 a 2 a 3 a 4 a 0 a 1 a 2 a 3 a 4 that z 0 + ··· + z 4 and z 3 + ··· + z 7 arebothequalto a 0 + ··· + a 4 . AsimpleapplicationoftheforegoingistotheIDFTformula(2.3)when N = 2 M + 1. Then y n = 1 N M k = - M Y k e j 2 π kn / N , wherewehaveusedthefactthatsince Y k and e j 2 π kn / N haveperiod N ,sodoestheirproduct. 2.1.2. ComputationoftheDFTinM ATLAB If y =[ y 0 ,..., y N - 1 ] ,thentheDFTof y , Y =[ Y 0 ,..., Y N - 1 ] ,canbecomputedinM ATLAB withthecommand Y = fft(y) .Here FFT standsfor fastFouriertransform .TheFFT isaspecialalgorithmthatcomputestheDFTveryquickly. SincetheDFTisperiodic, Y - 1 = Y - 1 + N = Y N - 1 Y - 2 = Y - 2 + N = Y N - 2 . . . Y - N / 2 = Y - N / 2 + N = Y N / 2 . So,toplot Y k for k = - N / 2to k = N / 2 - 1,weneedtotake Y = [ Y 0 ,..., Y N / 2 - 1 , Y N / 2 ,..., Y N - 1 ] andconvertitto [ Y N / 2 ,..., Y N - 1 , Y 0 , Y 1 ,..., Y N / 2 - 1 ] . ThisisdonewiththeM ATLAB command fftshift . Thecorrespondingvectorof k val- uescanbegivenby k=[0:N-1]-N/2 , andwecouldthenusethecommand plot(k, fftshift(Y)) . If we are approximating X ( k / N ) in (2.1), then we would use k/N to havethehorizontalaxisrunfrom - 1 / 2to1 / 2.
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