The Gaussian noise in discrete signal models is usually assumed to be independent and identically distributed variables (i.i.d.). Does this mean that the signal must be sampled with the sampling rate of 2W (where W is the bandwidth)? If the sampling rate is faster than 2W, the noise becomes correlated. Now I want to generate correlated noise samples in Matlab, but I am quite confused about how to describe the correlated noise theoretically? I also don't know how to generate correlated noise samples in Matlab. Could anyone explain this to me or give me any tips?

  • $\begingroup$ Could you clarify what you mean by "oversampling"? Your first sentence states a discrete time model, which implies its already sampled. Do you mean upsampled? $\endgroup$
    – user28715
    Commented Nov 7, 2019 at 18:58
  • $\begingroup$ @StanleyPawlukiewicz Sorry for my confusing text, I want to ask about how to describe the noise samples if the sampling rate is faster than the Nyquist rate. $\endgroup$ Commented Nov 8, 2019 at 6:46
  • $\begingroup$ The sample rate kind of is the Nyquist rate, at least we call it that when we graph a digital spectrum. When we sample above the effective bandwidth of the signal, the digital spectrum falls off at the higher frequencies, so the noise isn’t white. The noise is band limited. You simulate it by filtering white digital noise. $\endgroup$
    – user28715
    Commented Nov 9, 2019 at 1:16

1 Answer 1


Assume that the input to a simple up-sampler (say by $L$) is $x[n]$, hence $$y[n] = \begin{cases} { x[n/L] ,\qquad n=kL \\ 0 , \qquad \qquad \ \text{ else } }\end{cases}$$ If you compute the auto-correlation sequence, defined as $$\phi_{xx}(i,j) = \mathbb{E}(x[i]x[i+j])$$ for an iid white Gaussian process, we have that \begin{align} \phi_{xx}(i,j) &= \begin{cases} \sigma^2, \qquad & \text{if } j=0 \\ 0, \qquad & \text{else } \end{cases} \\ \phi_{yy}(i,j) &= \begin{cases} \sigma^2, \qquad & \text{if } j=0 \text{ and }i = kL \\ 0, \qquad & \text{else } \end{cases} \end{align} Unlike i.i.d Gaussian, where $\phi_{xx}(i,j)$ is determined by the lag $j-i$; $\phi_{yy}(i,j)$ is not fully determined by the lag. That said, $y[n]$ is not WSS, whereas $x[n] $ is. With respect to correlation, the quantities $\mathbb{E}(x[n]x[m]) = 0$ for $n \neq m$ is due to the uncorrelation assumption. You can say the same thing on $y[n]$, namely $\mathbb{E}(y[n]y[m]) = 0$ for $n \neq m$.


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