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Unformatted text preview: frequencies in Hz (for the continuoustime signal) are approximately (13/80)*800 = 130 Hz and (20/80)*800 = 200 Hz (Note: Sampling at a rate >2*f1 and >2*f2 guarantees that both discrete sinusoids have frequencies in [0,pi).) S 25.3 (P 3.31) _______________ (i) f1 = 164/640 and f2 = 182/640 cycles/sample. (ii) For N = 200, the Fourier frequencies will be the multiples of 2*pi/200, or 2*pi*0.005. f1 = 164/640 = 0.25625, so nearest Fourier frequency is 2*pi*(0.255), corresponding to k = 51 f2 = 182/640 = 0.284375, so nearest Fourier frequency is 2*pi*(0.285), corresponding to k = 57 (iii) We need 164/640 and 182/640 to both be of the form k/N, where N = M+200 164/640 = 41/160 and 182/640 = 91/320 so the smallest value of N for which both frequencies are expressible as k/N is 320. This means that M = 120. Then f1 = 82/320 (i.e., k=82) and f2 = 91/320 (i.e., k=91)....
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 Spring '08
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