Suppose we receive $R(t)=X(t)+W(t)$, where $X(t)$ is band-limited to $[-B/2, B/2]$ and $W(t)$ is white Gaussian noise with autocorrelation $R_W(\tau)=\frac{N_0}2\delta(\tau)$. If we filter $R(t)$ with an ideal LPF, i.e. $h(t)=B\cdot \mathrm{sinc}(Bt)$, and then sample it at Nyquist rate, i.e. $Y_n = \left. Y(t)\right|_{T=n/B}=X_n + Z_n$, where $Y(t)=R(t)\ast h(t), Z(t)=W(t)\ast h(t)$ and $Z_n=Z(nT)$, then it is known that $\{Z_n\}$ is also white, i.e. $\mathbb E[Z_n Z_m^*]=0$ if $m\ne n$.

Question: However, if we sample $Y(t)$ at a higher rate, i.e. $T<1/B$, is it true that the sampled noise sequence $\{Z_n\}$ become colored (correlated)?

I searched for a while, but didn't get concrete confirmation. Only some hints were found, e.g. How to describe correlated noise after the signal is oversampled?. So I want to confirm it. Here's what I've tried:

First, recall that $h(t)=B\, \mathrm{sinc}(Bt)$ and $H(f)=\text{rect}(\frac{f}B)=\left\{\begin{array}{lr} 1, & -\frac{B}2\le f \le \frac{B}2\\ 0, & \text{o.w.} \end{array} \right.$ Hence

$$Z_n = Z(nT)=\int_{-\infty}^{\infty} W(\tau)B\, \text{sinc}\left(B(nT-\tau)\right)d\tau=\int_{-\infty}^{\infty} W(\tau)\phi_n(\tau)d\tau$$

where ${\phi}_n(\tau)\triangleq B\,\mbox{sinc}(B(nT-\tau))=B\,\mbox{sinc}(B(\tau-nT))$. When $T=1/B$, it's widely known that $\{\phi_n(\tau)\}$ is a set of orthogonal functions and hence $\{Z_n\}$ are uncorrelated/white.

However, when $T<1/B$, this doesn't seem to be true any more: With Parseval's theorem,

$$\int_{-\infty}^{\infty}\phi_n(\tau)\phi_m^*(\tau)d\tau=\int_{-\infty}^{\infty}\text{rect}\left(\frac{f}B\right)e^{-j2\pi fnT}\text{rect}\left(\frac{f}B\right)e^{j2\pi fmT} df=\frac{\sin(\pi (m-n)BT)}{\pi(m-n)T}$$

This is not zero in general, is it? For example, with 2X oversampling, i.e. $T=\frac1{2B}$, we have $\int_{-\infty}^{\infty}\phi_0(\tau)\phi_1^*(\tau)d\tau= \frac{2B}\pi$, and hence $\mathbb E[Z_0 Z_1^*]=\frac{N_0 B}{\pi}\ne 0$, i.e. correlated.

  • $\begingroup$ Is the original noise also bandlimited to $B$ or does it go to $2B$ and beyond. What anti aliasing filter are you using when you sample at $2B$ ? $\endgroup$
    – Hilmar
    Commented Mar 21 at 11:27
  • $\begingroup$ @Hilmar $W(t)$ is AWGN with $R_W(\tau)=N_0/2\, \delta(\tau)$, so its bandwidth is infinite. As for 2X oversampling, it's with respective to the bandwidth of $X(t)$, so the anti-aliasing filter is still the same, i.e. $h(t)=B\, \text{sinc}(Bt)$. $\endgroup$
    – syeh_106
    Commented Mar 21 at 13:17

1 Answer 1


I think you proved that it is true.

Another way to prove is to approach from the frequency domain.


  1. A white noise has uniform power spectral density.
  2. The spectrum of a signal $X(f)$ sampled at rate $1/T$, $\sum_{k\in \mathbb Z} X(f - k/T)$

Then, a signal with spectrum $\textrm{rect}(f/B)$ will be white if, and only if, $T=n/B$ for some $n \in \mathbb N$.

Proof: The only way to keep the spectrum uniform is when for each rising edge of the terms $\sum_{k\in \mathbb{Z}} X(f - k/T)$ there is a falling edge of another term. i.e. for every integer $k_1$ exists another integer $k_2$ such that $k_1/T - k_2/T = B$, that can be rewritten as $(k_1 - k_2) = B\cdot T$, the right hand side is constant and the left hand side is integer.

This is analogous to fitting bricks without gaps.

  • $\begingroup$ I appreciate the answer. This perspective indeed affords a simple proof, considering the fact that the autocorrelation of the samples of a WSS random process are the samples of the autocorrelation of the random process. Equivalently, the power spectrum of the sampled random process (i.e. a DT random sequence) is the sum of shifted power spectrum of the original random process. $\endgroup$
    – syeh_106
    Commented Mar 25 at 6:56
  • $\begingroup$ BTW, is it a typo? I assume you meant "a signal with spectrum $\text{rect}(f/B)$ will be white if, and only if, $T=n/B$ for some $n\in \mathbb N$." $\endgroup$
    – syeh_106
    Commented Mar 25 at 7:01
  • $\begingroup$ You are right, thank you $\endgroup$
    – Bob
    Commented Mar 25 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.