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I was wondering how I would calculate the power of a noise, if I have a uniformly distributed white noise between $-\frac{q}{2}$ and $\frac{q}{2}$? My approach was to use the formula

$$\int\limits_{-\frac{q}{2}}^{\frac{q}{2}} x^2 \ \frac{1}{q} \ \mathrm{d}x = \frac{q^2}{12}$$

But I'm not sure if that is right. Or do I need to use the autocorrelation function, which I think should be $R_{XX}(\tau) = \frac{1}{q^2} \delta(\tau)$, to calculate the power spectral density function $S_X(f) = \frac{1}{q^2}$? I'm just confused about the exact definition of power here.

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  • $\begingroup$ Are you talking about continuous-time white noise or discrete-time white noise? Also, your calculation of the variance is incorrect: the density has value $\frac 1q$ for $x \in [-\frac q2, \frac q2]$ and the integral for variance evaluates to $\frac{q^2}{12}$, not $\frac{q}{12}$. $\endgroup$
    – Dilip Sarwate
    Commented Mar 18, 2023 at 20:10
  • $\begingroup$ Thank you for your correction, I just copied the fraction from the integral boundary and wasn't paying attention. And yes, I am talking about continuous-time white noise. $\endgroup$
    – Ole
    Commented Mar 18, 2023 at 21:32
  • $\begingroup$ @Ole I assumed the noise was discrete (in fact I assumed the context of your question was uniform quantization). If the noise is white and continuous, then its variance is not defined and its power is infinite (and my answer is not valid). $\endgroup$
    – MBaz
    Commented Mar 18, 2023 at 21:36
  • $\begingroup$ Thank you for your answer, is there another characterization of power in this case? Would it be possible to calculate the (constant?) amplitude of the Fourier transform? $\endgroup$
    – Ole
    Commented Mar 18, 2023 at 21:42
  • $\begingroup$ Only if you restrict the noise bandwidth. Strictly speaking it wouldn't be white anymore, but it is usual to call a PSD that is constant in a certain band "white". $\endgroup$
    – MBaz
    Commented Mar 18, 2023 at 21:58

1 Answer 1

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(This answer is only valid under certain assumptions; see comments below original question)

The definition of power for a random signal is the same regardless of the distribution. For a [update: discrete-time] wide-sense stationary random signal with zero mean, such as yours, the easiest way to calculate the power is to find the variance, which in your case is easy to determine as $q^2/12$.

You can also use the autocorrelation, in which case the power is the autocorrelation evaluated at $\tau=0$.

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    $\begingroup$ So for continuous-time white noise, the autocorrelation function is $$R_N(\tau) = \begin{cases}q/12~ \text{(sic)} & \tau = 0,\\0, &\text{otherwise}\end{cases} ~~??$$ $\endgroup$
    – Dilip Sarwate
    Commented Mar 18, 2023 at 20:05
  • $\begingroup$ @DilipSarwate, truly white noise, in the continuous-time domain, has infinite power. The OP has to maybe say that his uniform-distributed noise is white within a given finite bandwidth. $\endgroup$
    – robert bristow-johnson
    Commented Mar 18, 2023 at 20:21
  • $\begingroup$ There are multiple issues that the OP is missing and that are not being addressed. $\endgroup$
    – robert bristow-johnson
    Commented Mar 18, 2023 at 21:28

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