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254 chapter 10A 1.5mp1.0lit 0.5ud 0.0eWindowed-sinc filter kernel1.0Am 0.8pl 0.6it 0.4ud 0.2e0.0Windowed-sinc frequency responsex10 3 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.41.0Am 0.8pl 0.6it 0.4ud 0.2e0.0Sample numberIdeal frequency response2x10A 1.5mp1.0lit 0.5ud0.0e0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45FrequencyTruncated-sinc filter kernelx10 3 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.40.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50FrequencySample number2x10Figure 10.7 Lower left: Frequency response for an ideal lowpass filter. Lower right:Truncated-sinc filter kernel (time domain). Upper left: Windowed-sinc filter kernel. Upper right:windowed-sinc filter frequency response.Here rect(ω) is the rectangular function (Figure 10.8). Although maybe not obvious,a rectangular pulse in the frequency domain has a Fourier transform that isproportional to the sinc function in the time domain [Smi 91, Wiki]∫ +∞( tdω e −iωt defrect(ω)=sinc =2)sin(πt/2) , (10.68)πt/2−∞where the π’s are sometimes omitted. Consequently, we can filter out the highfrequencycomponents of a signal by convoluting it with sin(ω c t)/(ω c t), a techniquealso known as the Nyquist–Shannon interpolation formula. In terms of discretetransforms, the time-domain representation of the sinc filter ish[i]= sin(ω ci). (10.69)iπAll frequencies below the cutoff frequency ω c are passed with unit amplitude, whileall higher frequencies are blocked.In practice, there are a number of problems in using this function as the filter.First, as formulated, the filter is noncausal; that is, there are coefficients at negativetimes, which is nonphysical because we do not start measuring the signal untilt =0. Second, in order to produce a perfect rectangular response, we would haveto sample the signal at an infinite number of times. In practice, we sample at (M +1)−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 254