solved17A - frequencies in Hz (for the continuous-time...

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S 25.1 (P 3.29) _______________ (i) The discrete-time (sampled) signal consists of two Fourier sinusoids at frequencies 7*(2*pi/32) and 11*(2*pi/32) The frequencies of the continuous-time components are found by multiplying the above by fs=500 samples/sec: 7*(2*pi/32)*500 and 11*(2*pi/32)*500 rad/sec or (7/32)*500 = 109.4 and (11/32)*500 = 171.9 Hz (ii) The phase spectrum is not given in this problem, so we cannot write an equation for the time-domain signal. If the phase spectrum were given, and the phases at the two frequencies (k=7 and k = 11) were equal to q1 and q2, respectively, then the continuous time signal would be given by the equation x(t) = (252/32)*cos(7000*pi*t/32 + q1) + . .. (168/32)*cos(11000*pi*t/32 + q2) S 25.2 (P 3.30) _______________ The two peaks at k = 13 (roughly) and k = 20 correspond to angular frequencies (13/80)*(2*pi) = 0.16*(2*pi) and (20/80)*(2*pi) = 0.25*(2*pi) for the sampled signal. Since the sampling rate equals 800 samples/sec, the corresponding
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Unformatted text preview: frequencies in Hz (for the continuous-time 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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solved17A - frequencies in Hz (for the continuous-time...

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