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        riddles >> hard >> Data Corruption
        
(Message started by: william wu on Oct 24^th, 2004, 11:21am)

Title: Data Corruption
Post by william wu on Oct 24^th, 2004, 11:21am

Never was any interest in this old problem, but I'm just writing this solution so I don't have to rederive it a third time after I forget the answer again ;)

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a) By examining the above graphs, estimate n₀, the index of the sequence value that has been corrupted. Explain.

One sequence value is corrupted. Thus y[n] can be written as

y[n] = x[n] + c[cdot][delta][n - n₀].

where c is some constant. The DTFT of this signal yields

Y([omega]) = X([omega]) + c[cdot]e[sup](-jn₀[omega]).

Applying Euler's formula this becomes

Y([omega]) = X([omega]) + c[cdot]{ cos(n₀[omega]) - j sin( n₀[omega] }.

Note that X([omega]) is said to be bandlimited to [pi]/3. Thus the signal we see in the ([pi]/3 to [pi) range must be due to the shifted delta in time. Observe that the real and imaginary parts of the DTFT in this range follow the quadrature of a cosine and sine with the same frequency and phase shift. Analyzing either sinusoid we see that the normalized period is 1/4. Returning to our expression for Y([omega]), we thus have 1/4 = 2[pi]/n₀ ---> n₀ = 8.

b) If y[n₀] = x[n₀] + c, estimate c. Explain your reasoning.

The sinusoids have magnitude 5, so c should have magnitude 5. But now the issue remains of whether it is +5 or -5. Observe that at [omega]=[pi], the cosine in the real part of the DTFT is +5. This suggests c = -5, since if c = +5 we would have 5cos([pi]) = -5. We can further verify the guess by checking it against the sine wave in the imaginary part of the DTFT. We should be observing

Im{ (-5){ cos([omega]n₀) - j sin([omega]n₀) } = 5 sin([omega]n₀)

Normally for a quadrature cosine-sine pair, the sine wave will peak [pi]/2 units after the cosine. But now that the cosine has been inverted by the (-5), the sine peaks [pi]/2 units before the cosine.

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