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| Computers Forum Index » Computer - DSP » Fourier coefficients of N WGN samples... |
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| sauravrt... |
 Posted: Mon Oct 26, 2009 4:02 am |
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I have N WGN samples y[0],y[1]....y[N-1]
W(e^(jw)) is the DTFT of these N samples.
With my analysis, I have come up with following results
E(W(e^(jw)) = 0
E(W(e^(jw1)).W*(e^(jw2))) =
e^(j(w1-w2)(N-1)/2).sin((w1-w2)N/2)/sin((w1-w2)/2)
For the Fourier coefficients at w1 and w2 to be uncorrelated, w1-w2 =
k(pi/N)
Can anyone please confirm this result?
*E(X) -> Expectation of X
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| Dilip Warrier... |
| Posted: Wed Nov 04, 2009 3:27 pm |
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Guest
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On Oct 25, 7:02 pm, "sauravrt" <sauravtulad... at (no spam) hotmail.com> wrote:
Quote: I have N WGN samples y[0],y[1]....y[N-1]
W(e^(jw)) is the DTFT of these N samples.
With my analysis, I have come up with following results
E(W(e^(jw)) = 0
E(W(e^(jw1)).W*(e^(jw2))) > e^(j(w1-w2)(N-1)/2).sin((w1-w2)N/2)/sin((w1-w2)/2)
For the Fourier coefficients at w1 and w2 to be uncorrelated, w1-w2 > k(pi/N)
Can anyone please confirm this result?
*E(X) -> Expectation of X
S
Use the vector representation:
W(\omega) = \sum y[k] \exp(-j*\omega*k) = \Omega^* \textbf{y}
where
\textbf{y} = ( y[0], y[1], \ldots, y[N-1] )^T
and
\Omega = ( \exp(j*\omega*0), \exp(j*\omega*1), \ldots, \exp(j*\omega*
(N-1)) )
Use the linearity of expectations and the facts that
E[\textbf{y}] = \textbf{0}
and
E[\textbf{y} \textbf{y^*}] = 2 \sigma^2 \textbf{I}
The results you need should come out of that quite easily. Your
correlation formula is correct but the condition for correlation=0 is
off by a factor of 2. |
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