# How to calculate to power of a uniformly distributed noise?

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.

• 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 $\frac{q}{12}$. Mar 18 at 20:10
• 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.
– Ole
Mar 18 at 21:32
• @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).
– MBaz
Mar 18 at 21:36
• 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?
– Ole
Mar 18 at 21:42
• 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".
– MBaz
Mar 18 at 21:58

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 $$\tau=0$$.
• So for continuous-time white noise, the autocorrelation function is $$R_N(\tau) = \begin{cases}q/12~ \text{(sic)} & \tau = 0,\\0, &\text{otherwise}\end{cases} ~~??$$ Mar 18 at 20:05