hwsolB05 - Therefore Ts = k*T/3 = k*(7/30) sec (k=1,2,. ..)...

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Sheet1 Page 1 HW 5 Solution _____________ (i) Clearly, A = 4. The period T equals 0.7 sec, therefore f = 10/7 Hz and W = 20*pi/7 cos(phi) = 2*sqrt(3)/A = sqrt(3)/2, therefore phi = pi/6 or -pi/6. By inspection, t=0 corresponds to a point in the last quarter-cycle of the cosine waveform (assuming the cycle begins at a peak). Thus the correct value of phi is -pi/6. (ii) x[n] = x[n+3] for all n if and only if w = 2*pi*(Ts/T) satisfies 3*w = k*(2*pi) for integer k
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Unformatted text preview: Therefore Ts = k*T/3 = k*(7/30) sec (k=1,2,. ..) (iii) Setting w = pi/7 = 2*pi*(Ts/T), we obtain Ts = T/14 = 0.05 sec and more generally, Ts = 0.05 + k*(0.7) sec (k=0,1,2,. ..) (iv) cos((16*pi/21)*n - phi) = cos((-16*pi/21)*n + phi). Using the negative value for w = 2*pi*(Ts/T), we obtain Ts = -(8/21)*T = -8/30 sec Then the positive values of Ts that produce the given x[n] are Ts = (0.7 - 8/30) + k*(0.7) = 13/30 + k*(0.7) sec...
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This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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