solved14 - i.e., the entries of the vector 2*[ 1 r -r -1 -r...

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S 14.1 (P 3.11) _______________ x = [ a b c d e f ].' X = [ A B C D E F ].' For x_1: x_1[n] = (-1)^n * x[n] = v^(3*n) * x[n] By Property 6 (multiplication by Fourier sinusoid in time domain), X is circularly shifted by +3 indices: X_1 = P^(-3)*X = [ D E F A B C ].' For x_2: x_2 = (x + x_1)/2 By linearity of the DFT (Property 1) X_2 = (1/2)*[ A+D B+E C+F D+A E+B F+C ].' For x_3: x_3 = R*x, i.e., circular reversal of x By Property 4 (circular reversal in time domain), X_3 = R*X = [ A F E D C B ].' For x_4: x_4 = (P^3)*x, i.e., circular delay of x by 3 units By Property 5 (circular delay in the time domain), X is multiplied elementwise by w^(3*k) = (-1)^k: X_4 = [ A -B C -D E -F ].' For x_5: x_5 = P*x_3, i. .e, circular delay of x_3 by 1 unit: By Property 5, X_3 is multiplied elementwise by w^k, i.e., the entries of the vector [ 1 r-js -r-js -1 -r+js r+js ].' where r = 1/2, s = sqrt(3)/2. The result is X_5 = [ A (r-js)*F (-r-js)*E -D (-r+js)*C (r+js)*B ].' For x_6: x_6 = (P + P^5)*x By Property 5, X is multiplied elementwise by w^k + w^(5k) = w^n + w^(-k) = 2*cos(pi*k/3),
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Unformatted text preview: i.e., the entries of the vector 2*[ 1 r -r -1 -r r ].' to yield X_6 = 2*[ A r*B -r*C -D -r*E r*F ].' For x_7: x_7[n] = sin(pi*n/3) * x[n] = (v^n - v^(-n))*x[n]/(2j) By Property 6, X_7 = (P*X - P^(-1)*X)/(2j) = (-j/2)*[ F-B A-C B-D C-E D-F E-A ].' For x_8: x_8 = X so by duality (Property 7), X_8 = NR*x = 6*[ a f e d c b ].' S 14.2 (P 3.13) _______________ y[n] = x[n]*cos(pi*n/4) = x[n]*(v^n + v^(-n))/2 where v = exp(j*2*pi*n/8) = exp(j*pi*n/4). In other words, y = (F + F^(-1))*x/2 and thus Y = (P + P^(-1))*X/2 = (1/2)*[0 0 0 0 -1 2 -1 0].' + . .. (1/2)*[0 0 -1 2 -1 0 0 0].' = [0 0 -1/2 1 -1 1 -1/2 0].' S 14.3 (P 3.15) _______________ (i) y is a circular shift of x by 6 positions: y = (P^6)*x Therefore Y = (F^(-6))*x i.e., Y[k] = exp(-j*2*pi*6*k/12)*X[k] Since exp(-j*2*pi*6*k/12) = (-1)^k, we have Y = [x0 -X1 X2 -X3 X4 -X5 X6 -X7 X8 -X9 X10 -X11].' (ii) s = -(F^6)*x, i.e., s[n] = -(-1)^n*x[n] Therefore s = -P^6*x = -[X5 X7 X8 X9 X10 X11 X0 X1 X2 X3 X4 X5].'...
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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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solved14 - i.e., the entries of the vector 2*[ 1 r -r -1 -r...

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