solvedB30

# solvedB30 - X(exp(j*w =(1-exp-j*w(M 1(1-exp-j*w for w...

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Sheet1 Page 1 S 30.1 (P 4.1) ______________ (d[.] = delta[.], the unit impulse function) (i) x[n] = d[n+2] - d[n+1] + 5d[n] - d[n-1] + d[n-2] In other words, x[0] = 5 x[1] = x[-1] = -1 x[2] = x[-2] = 1 x[n] = x[-n] = 0 for n>2 The DTFT of d[n-m] equals exp(-j*m*w), so X(exp(j*w)) = exp(j*2*w) - *exp(j*w) + 5 - *exp(-j*w) + exp(-j*2*W) = 5 - 2*cos(w) + 2*cos(2*w) (ii) This is a generalization of part (i) (we had L=2 there). X(exp(j*w)) = b0 + b1*(exp(j*w)+exp(-j*w)) . .. + b2*((exp(2*j*w)+exp(-2*j*w)) ,,, + bL*(exp(j*L*w)+exp(-j*L*w)) = b0 + 2*b1*cos(W) + 2*b2*cos(2*w) + . .. + 2*bL*cos(L*w) S 30.2 (P 4.2) ______________ (i) The DTFT of x[n] equals X(exp(j*w))) = 1 + exp(-j*w) + exp(-j*2*w) + . .. + exp(-j*M*w) which is a geometric sum with common ratio exp(-j*w). Therefore

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Unformatted text preview: X(exp(j*w)) = (1-exp(-j*w*(M+1))/(1-exp(-j*w)) for w taking values other than multiples of 2*pi. (Otherwise, X(exp(j*W)) = M+1.) The same sum was encountered in Section 3.8.2, as the inner product <v,s> where v = exp(j*w*n) , n = 0:M(i.e., L=M+1) s = exp(j*0*n) = 1, n = 0:M(i.e., w0 = 1) The final expression for that sum was exp(-j*M*w/2) * F(w) Sheet1 Page 2 where F(w) = sin((M+1)*w/2)/sin(w/2), a symmetric function of w. (ii) x = ones(24,1) would be the simplest such definition. fft(x,500) would then compute the DFT of a vector of 24 ones padded with 477 zeros. This is the same as the DTFT of the given (finite-duration) sequence, computed at 500 equally spaced frequencies ranging from 0 to 2*(0.998)*pi....
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solvedB30 - X(exp(j*w =(1-exp-j*w(M 1(1-exp-j*w for w...

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