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# solved13 - (ii The synthesis equation for a real-valued...

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S 13.1 (P 3.7) ______________ X = [4 1+j 3-j z1 z2].' (i) From the analysis equation (k=0), we have that x[1] + x[2] + x[3] + x[4] + x[5] = X[0] = 4 (ii) Since x is real-valued, X has conjugate symmetry about index k=5/2. Thus (with ' denoting complex conjugate, as in MATLAB) X[3] = X'[2] , or z1 = 3+j X[4] = X'[1] , or z2 = 1-j (iii) Amplitude spectrum is abs(X): [4 sqrt(2) sqrt(10) sqrt(10) sqrt(2)].' = [4.0000 1.4142 3.1623 3.1623 1.4142].' Phase spectrum is angle(X): [0 atan(1) atan(-1/3) atan(1/3) atan(-1) ].' = [0 0.7854 -0.3218 0.3218 -0.7854].' Therefore: x = (1/5)*4 + (2/5)*1.4142*cos(2*pi*n/5 + 0.7854) + (2/5)*3.1623*cos(4*pi*n/5 - 0.3218) S 13.2 (P 3.8) ______________ x[n] = 3*(-1)^n + 7*cos(pi*n/4 + 1.2) + 2*cos(pi*n/2 - 0.8) (i) Of the eight Fourier frequencies k*(2*pi/8), (k=0,1,. ..,7) only five are present in x[n]: k = 1, 2, 4, 6, 7

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Unformatted text preview: (ii) The synthesis equation for a real-valued N-point vector can be also written as x[n] = X[0]/N + (X[N/2]/N)*(-1)^n + sum (2*|X[k]|/N)*cos(2*pi*k*n/N + angle(X[k])) where the last sum is over 0<k<N/2. Comparing this to the given equation for x[n], we have: |X[k]| = [0 28 8 0 24 0 8 28].' angle(X[k]) = [ 0 1.2 -0.8 0 0 0 0.8 -1.2].' (The angle of X[4] equals 0 because X[4]>0; it would have been equal to pi otherwise.) S 13.1 (P 3.10) _______________ x = [ 2 1 -1 -2 -3 3 ].' We have: x1 = P*x = [3 2 1 -1 -2 -3].' x2 = (P^5)*x = [1 -1 -2 -3 3 2].' x3 = P*x + (P^5)*x = [4 1 -1 -4 1 -1].' R*x = [2 3 -3 -2 -1 1].' x4 = x + R*x = [4 4 -4 -4 -4 4 ].' x5 = x - R*x = [0 -2 2 0 -2 2].' F^3 = diag([1 -1 1 -1 1 -1]), so x6 = (F^3)*x = [2 -1 -1 2 -3 -3].' x7 = x - (F^3)*x = [0 2 0 -4 0 6].'...
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## This note was uploaded on 10/18/2011 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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solved13 - (ii The synthesis equation for a real-valued...

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