Problem 2.62 Answers Part (a) The cross correlation r_yx[tau] between x[n] and y[n] at delay tau is the inner product between y[n] and x[n+tau] (for real signals). This is also equivalent to an inner product between y[n-tau] and x[n]. Suppose x[n] has length N, that is x[n] = 0, for n < 0 and n > N-1. Then r_yx[tau] = x[0]y[-tau] + x[1]y[1-tau] + ... + x[N-1]y[N-1-tau] is the inner product between the 1xN vectors X = [x[0] x[1] ... x[N-1]] and Y(tau) = [y[-tau] y[1-tau] ... y[N-1-tau]], that is r_yx[tau] = XY(tau)^T. The Cauchy-Schwartz inequality for length N vectors is abs(XY(tau)^T) < sqrt(XX^T) sqrt(Y(tau)Y(tau)^T) with equality only if X = cY(tau) (and X^T is the transpose of X) for some constant c. The right side of the above inequality can be replaced by just a constant C = sqrt(XX^T) sqrt(YmYm^T) since X is unchanging over tau, and where Ym is the vector Y for tau which has maximum power, that is sqrt(Y(tau)Y(tau)^T) is maximized. So ignoring noise, we expect that the inner product abs(XY^T) will be maximized when Y = aX, which happens when tau = D. When there is noise, it will contribute a term which will hopefully not make it maximum at the wrong value of tau. Brief Summary and Conclusions The method described above is limited by the noise variance. The longer our pulse x[n] is, the greater gain it has against the noise. Hence we see that usually the method fails to find the delay D when sigma = 1 for the short sequences, but when x[n] is the length 127 pulse, it is possible to find D even for relatively large noise. The method can work even when the pulse seems completely buried in the signal y[n].