solved17A

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 &gt;2*f1 and &gt;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)....
View Full Document

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online