My question concerns the dimensions of the autocorrelation function of the white noise.

First is a short summary of what I understand about the random signals. If we have a stationary random signal $x(t)$ whose amplitude is a random variable distributed according to some probability law : $p(x)$, then :

  • its autocorrelation function is an ensemble average of $x(t)\,x(t+\tau)$ : $$ R_{xx}(\tau) = \int_{-\infty}^{\infty}\ x(t)\,x(t+\tau)\,p(x)dx $$ -if the signal is ergodic, we can replace the ensemble average by the time average to obtain another expression for the signal autocorrelation : $$ R_{xx}(\tau) = \lim_{T \rightarrow \infty}\frac{1}{T}\int_{-T/2}^{T/2}\ x(t)\,x(t+\tau)\,dt $$ From both of these expressions, the dimensions of $R_{xx}$ are $[x^2]$ whatever are units of $x$. Suppose also that the signal is zero-mean, then $R_{xx}(0) = \sigma^2$, the variance of $x$. If $x(t)$ is a white-noise, then any source I read informs me that : $$ R_{xx}(\tau) = \sigma^2\,\delta(\tau) $$ But according to this last expression, the units of $R_{xx}$ are now : $[\frac{x^2}{s}]$.

Here is the question : how come that we change the dimensions of $R_{xx}$ by passing from the general expression to this particular expression for white-noise?

I would greatly appreciate your help. Let me know if I was not clear enough somewhere.

Hello everyone,

Thank you for your answers and corrections. I will put here the corrected version of $R_{xx}(\tau)$ in terms of an ensemble average :

$$ R_{xx}(\tau) = \iint_{-\infty}^{\infty}\ x_1\,x_2\,p(x_1,x_2,t_1,t_2)dx_1\,dx_2 $$

where $t_1 = t$, $t_2 = t + \tau$, $x_1 = x(t)$, $x_2 = x(t+\tau)$. I am assuming here that the $R_{xx}$ depends only on the separation between two different time moments and not their particular values.

This correct expression does not change the dimensions of $R_{xx}$ however, it is still $[x^2]$ or $V^2$ for example.

Unfortunately I am not quite sure I understand your answers. So, I will try ask one more question, which is related to my first one.

Suppose we have an ergodic bandwidth limited Gaussian noise $x(t)$ of zero-mean and finite variance $\sigma^2$ :

$$ x \sim \mathcal N (0, \sigma^2) $$

In addition, the values of the signal at any two moments of time are not correlated. From the time average form of $R_{xx}(\tau)$ we get :

$$ R_{xx}(\tau) = \left \{ \begin{aligned} \sigma^2\ , \tau = 0\\ 0\ , \text{otherwise} \end{aligned} \right . $$

The power spectral density $PSD$ is constant for any frequency :

$$ PSD(f) = \frac{\sigma^2}{F_s}\quad ,\text{where}\ -F_s/2 < f < F_s/2 $$

where $F_s$ is the sampling frequency. The area under the PSD gives the power of the signal :

$$ \text{Power} = \frac{\sigma^2}{F_s} F_s = \sigma^2 $$

So, for the signal considered the variance is finite, is not

$$ R_{xx}(\tau) = \sigma^2\,\delta(\tau) $$

and $R_{xx}(\tau)$ has dimensions of $V^2$.

Now imagine we let $F_s$ to increase infinitely to obtain a white Gaussian noise which has infinite frequency content. For me, $R_{xx}$ will remain the same. No matter how big is $F_s$. The only thing which will change, is $PSD$, which will become smaller and smaller, but in the way that the area under $PSD$ remains equal to $\sigma^2$.

Is there anything wrong with this reasoning? I do understand that to get $R_{xx}(\tau) = \sigma^2\,\delta(\tau)$ one needs to fix the PSD : $PSD = \sigma^2$. But this does not make any sens to me, as the area under the horizontal line does not have the units of the power any more.

Thanks in advance for your help.

  • $\begingroup$ Your first displayed integral is incorrect; $x(t)$ and $x(t+\tau)$ are two different random variables and so you need to multiply by the joint density of both random variables and do a two-dimensional integral in computing the expected value of $x(t)x(t+\tau)$. Also, white noise has infinite variance $\endgroup$ Commented Dec 17, 2021 at 3:48

2 Answers 2


For the sake of better understanding and without loss of generality, assume $x(t)$ has the unit Volts. This is en par with the electrical engineering nomenclature, which is common to this topic.

To your question: This is a convention. For any non-periodic power signal of infinite length, the first two definitions will yield $R_{xx}(\tau)=\infty\delta(\tau)$ because all of the signal's energy (which is infinite) will be concentrated in $R_{xx}(0)$. As usual when dealing with power signals (as opposed to energy signals), rather than considering energy, we consider power as the defining measure. White noise is defined by $\sigma^2$, its power. It might help to imagine $R_{xx}$ as "autocorrelation per second" for stationary noise signals, rather than thinking the whole thing has the unit $\text{V}^2/s$.


You are missing the time unit from $dt$. The units of the first one are $[x]^2 \cdot s$ . If you think of [x]^2 as power than $[x]^2 \cdot s$ is energy and indeed $R_{xx}(0)$ is the total energy of the signal.

That doesn't work for white noise, since it has infinite energy. Hence we switch to a power definition (in your second form) and there $R_{xx}(0)$ is indeed the power with units of $[x]^2$.

  • $\begingroup$ The $dt$ is eliminated by the $1/T$ before the integration. The units he stated are correct imo. $\endgroup$
    – Max
    Commented Dec 17, 2021 at 7:26

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