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Unformatted text preview: i.e., f = 20 + 100*k The only possible value for f is 780 Hz. S 4.3 (P 1.22) ______________ The cyclic frequency of x(t) is 150 Hz. The sampling rate is f_s = 1/T_s = 500 samples/sec. The resulting discretetime sinusoid is x[n] = cos(0.6*pi*n), i.e, it has w = 0.6*pi. Note that there is no phase shift. (i) We want the aliases of 150 Hz with respect to the sampling rate of 500 samples/sec: f = 150 + k*500 Hz and f = 150 + k*500 Hz In the range 0 to 2000 Hz, we have: 650, 1150, 1650 and 350, 850, 1350, 1850 Hz (ii) An increase in T_s results in an increase in w. The next value of w equivalent to 0.6*pi is0.6*pi + 2*pi = 1.4*pi and is obtained for T_s = (1.4*pi)/(300*pi) = 4.67 ms. Note: This is the same as 6.67 ms (period of x(t))  2.0 ms....
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 Spring '08
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 Frequency, sampling rate, Hz

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