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].