# dimensions of the autocorrelation function of white noise

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.

• 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 Dec 17, 2021 at 3:48

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 $$[x]^2$$.
• The $dt$ is eliminated by the $1/T$ before the integration. The units he stated are correct imo.