hwsolB04

# hwsolB04 - number of periods of each of the two components...

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Sheet1 Page 1 HW 4 Solution _____________ (i) For x[.], we have w=3*pi/4 = (3/8)*(2*pi). Thus the fundamental period of x[.] equals N = 8. N = 8 n = 0 : 4*N-1 x = cos(3*pi*n/4 + 2*pi/3) stem(n,x) (ii) The values of w such that cos(w*n) is periodic with fundamental period N=8 are of the form w = (k/8)*(2*pi) where k has no common factors with 8. In other words, k = 1, 3, 5, 7 and the corresponding values of w are pi/4, 3*pi/4, 5*pi/4 and 7*pi/4 (iii) v[n] = y[2*n] = cos(2*pi*n/5 + 2*pi/3) Thus v[n] has frequency 2*pi/5 and period N = 5. (iv) If x[n] has fundamental period N1 and y[n] has fundamental period N2, then the pair (x[n], y[n]) will have fundamental period N = least common multiple of N1 and N2 (This is the smallest number of samples which contains a whole

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Unformatted text preview: number of periods of each of the two components.) In this case, lcm(8, 10) = 40 If (x[n], y[n]) = (x[n+40], y[n+40]) then clearly also x[n]+y[n] = x[n+40]+y[n+40] The fundamental period of x[n]+y[n] equals 40 (not a submultiple thereof). This is always the case with sums of sinusoids of different frequencies, and in this problem it can be verified directly by examining a vector of 40 consecutive samples of x[n]+y[n]. Sheet1 Page 2 (v) z[n] = x[n]^2 ==> z[n] = cos(3*pi*n/4+2*pi/3)^2 = 0.5 * cos(2 * (3*pi*n/4+2*pi/3) ) + 0.5 z[n] = 0.5 * cos(3*pi*n/2+4*pi/3) + 0.5 z[n] is periodic with period 4. Note that the period of z[n] is half the period of x[n]....
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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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hwsolB04 - number of periods of each of the two components...

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