hwsolB22 - (Note that a negative real number has angle = pi...

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Sheet1 Page 1 HW 22 SOLUTION ______________ (i) Since s is real-valued, S has circular conjugate symmetry, i.e., S = [6 -5-3*j 3+2*j -7 3-2*j -5+3*j ].' (ii) s[0] + . .. + s[6] = S[0] = 6 (iii) We compute the modulus and angle of each entry of S to obtain the amplutude and phase spectra. Again, angle(z=x+j*y) = arctan(y/x) + (0 or pi), according to whether x>=0 or x<0. abs(S) = [ 6 5.8310 3.6056 7 3.6056 5.8310 ].' (Amplitude spectrum is circularly symmetric.) angle(S) = [ 0 -2.6012 0.5880 pi -0.5880 2.6012 ].'
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Unformatted text preview: (Note that a negative real number has angle = pi = -pi. Phase spectrum is circularly antisymmetric.) (iv) s[n] = 1 - (7/6)*(-1).^n . .. + (2*5.8310/6)*cos(pi*n/3 - 2.6012) . .. + (2*3.6056/6)*cos(2*pi*n/3 + 0.5880) (v) s is generated by setting n = (0:5).' and using the equation above (with s replacing s[n]). fft(s) returns the given S (with some roundoff errors in the fifth significant digit): fft(s) ans = 6-5.0001 - 2.9999i 3.0000 + 2.0000i-7 3.0000 - 2.0000i-5.0001 + 2.9999i...
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