solved04 - i.e., f = -20 + 100*k The only possible value...

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S 4.1 (P 1.21) ______________ We have x[n] = cos(w*n), where w = 2*pi*f/800 (and f is in Hz) For x[n] to be the same as cos(0.4*pi*n) = cos(-0.4*pi*n) for all n, it must be that f/800 = 0.2 + k (k integer) or f/800 = -0.2 + k (k integer) Thus f = 160 + k*800 or f = -160 + k*800 The aliases in the range 0 to 3,000 Hz are therefore f = 160, 640, 960, 1440, 1760, 2240 and 2560 Hz S 4.2 _____ x(t) = cos(2*pi*f*t + 1.8) where f is between 700 and 800 Hz. The sampling period is T_s = .01 sec, so x[n] = x(n*T_s) = cos(2*pi*f*0.01*n + 1.8) = (i) If x[n] also equals cos(2*pi*0.35*n + 1.8), it follows that f*0.01 = 0.35 + k where k is an integer. Thus f = 35 + 100*k and the only possible value for f in the given interval is f = 735 Hz. (ii) If x[n] equals cos(2*pi*0.2*n - 1.8), it follows that f*0.01 = -0.2 + k
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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 discrete-time 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 is-0.6*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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solved04 - i.e., f = -20 + 100*k The only possible value...

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