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I hope someone can help me understand the meaning of variance in this case. I am reading an old research paper by Ephraim and Malah (for my DSP project) and it says:

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Knowing that $D_k$ is the FFT of white noise, what I am trying to find out is $\lambda_d(k)$. Can anyone explain to me the meaning of $E\lbrace|D_k|^2\rbrace$? I think it is the variance of a particular $D_k$ with respect to the mean, but since this is white noise, shouldn't it be zero?

Please help, I am stuck!!!!

Thanks.

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$E\{|D_k|^2\}$ is the expected value of the squared magnitude of the noise in the $k^{th}$ frequency bin. Since they assume that the noise is zero mean, this equals the noise variance. Remember that the variance of a random variable $X$ is defined as $\sigma_X^2=E\{|X-\mu_X|^2\}$, where $\mu_X=E\{X\}$ is the mean of $X$. So if $\mu_X=0$ (as assumed here for the noise) then $\sigma^2_X=E\{|X|^2\}$. Note that $E\{|D_k|^2\}$ is also the noise power in the $k^{th}$ bin, so it can't be zero unless there is no noise at all in that bin.

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  • $\begingroup$ Thank you very much... That makes sense now. ... Many thanks again $\endgroup$
    – Mona
    Commented Nov 22, 2014 at 19:15
  • $\begingroup$ Indeed that was very helpful : $\endgroup$
    – Mona
    Commented Nov 24, 2014 at 17:18
  • $\begingroup$ I am reviewing my MMSE code, and I just to make sure that I am using the variance function correctly. From the above comments, I assumed that E{|Dk|2} = |Dk|2 and in matlab, it is just (abs(Dk))^2 . Is this correct? Thanks in advance. $\endgroup$
    – Mona
    Commented Nov 30, 2014 at 19:47
  • $\begingroup$ @Mona: No, $E\{|D_k|^2\}$ is the average noise power, which needs to be estimated. To keep things simple, you could assume that your input data contain a segment with only noise from which you can obtain an estimate of $E\{|D_k|^2\}$ by averaging $|D_k|^2$ over all noise frames. Obviously, this assumes that the noise is stationary, i.e. its characteristics don't change over time. In a more realistic setting you would use a noise tracking algorithm such as this one. $\endgroup$
    – Matt L.
    Commented Nov 30, 2014 at 20:13
  • $\begingroup$ Thanks for clarification, I just read the abstract of the paper and browsed over its algorithm. It seems clear, but I must admit I am new to this. I am sure if I implement any additional thing on this project, my professor will give me bonus. However, let me go through Malah’s paper first. Yes, I agree, we are keeping things simple; i.e. taking white noise and adding it to the original speech and then doing reconstruction according to Malah’s algorithm. So, E{|Dk|2} is the difference between the kth element and the average noise of a particular frame? $\endgroup$
    – Mona
    Commented Nov 30, 2014 at 23:00

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