solved15 - S 15.1(P 3.14 x =[4 3 2 1 0 1 2 3 X =[A B C D E...

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S 15.1 (P 3.14) _______________ x = [4 3 2 1 0 1 2 3].' X = [A B C D E F G H].' (i) x is real and circularly symmetric, thus X is also real and circularly symmetric. It follows that B=H, C=G, and D=F (ii) y = [0 1 2 3 4 3 2 1].' = (P^4)*x so Y[k] = exp(-j*(2*pi/8)*4*k)*X[k] = (-1)^k * X[k] = [A -B C -D E -F G -H].' = [A -B C -D E -D C -B].' S 15.2 (P 3.16) _______________ (i) X can be also written as X = [1 ; ones(10,1); zeros(43,1); ones(10,1)] and thus R*X = X i.e., X is circularly symmetric. Since X is also real, we conclude that the time domain signal has the same properties, i.e., it is real-valued and circularly symmetric. Thus x = ifft(X) has no imaginary part; a small imaginary part resulting from the IDFT computation is due to round-off. After confirming this fact, x is defined as x = real(ifft(X)) ensuring that it is treated as a real-valued quantity from that point onwards. (ii) Since N = 64, it follows that 3*pi/4 = 24*(2*pi/N), i.e., it is the k=24th Fourier frequency for this vector size.
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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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solved15 - S 15.1(P 3.14 x =[4 3 2 1 0 1 2 3 X =[A B C D E...

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