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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.

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  • $\begingroup$ 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
    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
    Mar 21 at 13:17

1 Answer 1

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I think you proved that it is true.

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

If

  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.

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  • $\begingroup$ 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
    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
    Mar 25 at 7:01
  • $\begingroup$ You are right, thank you $\endgroup$
    – Bob
    Mar 25 at 14:50

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