My math may be a little rusty, so I would like confirmation or correction or my calculations here.

Given white noise samples, $x_i$, which are IID and zero-mean, and variance $\sigma^2_x$. I want to show that the outputs of any pair of DFT filters is also IID.

Knowing covariance of the outputs should be enough to draw a conclusion. So I examine the following:

Given the outputs $X_k$ and $X_{k'}$ defined as

$$X_k=\Sigma_{n=0}^{N-1}x_ne^{-2\pi ikn/N}$$

$$X_{k'}=\Sigma_{m=0}^{N-1}x_me^{-2\pi i{k'}m/N}$$

We can calculate the expected value of $X_k X^*_{k'}$

$$\begin{eqnarray*} E(X_k X^*_{k'}) &=&E\left(\sum_{n=0}^{N-1}x_ne^{-2\pi ikn/N}\sum_{m=0}^{N-1}x_me^{2\pi i{k'}m/N}\right)\\ &=&E\left(\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x_n x^*_m e^{-2\pi i(kn-{k'}m)/N}\right)\\ &=&\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}E\left(x_n x^*_m \right) e^{-2\pi i(kn-{k'}m)/N}\\ \end{eqnarray*}$$

We know that $E(x_k x^*_{k'})=\sigma^2_x$ when $k={k'}$ and $E(x_k x^*_{k'})=0$ when $k\neq {k'}$. So we can simplify:

$$\begin{eqnarray*} E(X_k X^*_{k'}) &=&\sigma^2_x\sum_{n=0}^{N-1}e^{-2\pi i(k-{k'})n/N}\\ &=&N\sigma^2_x|_{k={k'}} \end{eqnarray*}$$

But if I'm right so far, I still am fuzzy on the case where $k \neq {k'}$. I expect it to equate to zero. Can someone help me out with that part?


1 Answer 1


Let $m=k-k'$. Then the sum in your final equation becomes

$$\sum_{n=0}^{N-1}e^{-2\pi imn/N}=\frac{1-e^{-2\pi im}}{1-e^{-2\pi im/N}},\quad m\neq 0\tag{1}$$

where I've used the formula for the geometric series. Noting that $e^{-2\pi im}=1$ for any integer $m$ shows that the sum in $(1)$ equals zero.

[Note that your sums should only have $N$ elements, i.e., their indices should go from $0$ to $N-1$ (instead of $N$).]

  • $\begingroup$ Thank you, Matt. That's very straightforward. And thanks for noting the error. I attribute that to Mathjax blindness. I'll fix that. $\endgroup$
    – Jim
    Commented Oct 9, 2016 at 15:34
  • $\begingroup$ Your proof shows only that the two terms are uncorrelated which is quite different from showing that they are independent, let alone identically distributed as the OP wants. Unless of course you assume that the white noise process is Gaussian. $\endgroup$ Commented Oct 10, 2016 at 0:00
  • $\begingroup$ @DilipSarwate: I actually didn't intend to proof anything; I just helped the OP with a technical detail (as requested in the last sentence of the question), namely to show that the last expression is zero for $k\neq k'$. I left the implications up to the OP, but it's good that you point out the requirement that the noise be Gaussian, in case the OP wasn't aware of this. $\endgroup$
    – Matt L.
    Commented Oct 10, 2016 at 7:11

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