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Topic: Data Corruption (Read 1515 times) |
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william wu
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Data Corruption
« on: Oct 24th, 2004, 11:21am » |
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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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| « Last Edit: Oct 24th, 2004, 11:29am by william wu » |
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