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HW 5 Solution
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(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 quartercycle 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.
 Spring '08
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