# Time derivative of signal - effect on noise distribution

I have an angular velocity measurement that has a certain amount of ripple that yields an histogram shaped like a gaussian curve. I quantify that ripple using a standard deviation. I am interested in knowing more about the angle ripple.

What would be the relationship between the two ripples?

My guess would be that the derivative (from angle to angular velocity) push noises at higher frequency and that the gaussian curve will get a little wider (bigger standard deviation). How can I validate that and what mathematics are involved here?

Let's call $$a[n]$$ the samples of the angular measurement. Then, an estimate of the angular velocity (in angle/sec) is given by discrete differences, i.e., $$v[n] = (a[n]-a[n-1])/t_0$$, where $$t_0$$ is the sampling time.
Let's assume a signal pluse noise process so that $$a[n] = a_0[n] + w[n]$$. Inserting, we obtain $$v[n] = v_0[n] + \bar{w}[n]$$, where $$\bar{w}[n] = (w[n]-w[n-1])/t_0$$. Certainly, if $$w[n]$$ is zero mean, so is $$\bar{w}[n]$$ You can now compute the variance of your effective noise samples $$\bar{w}[n]$$, namely $$\mathbb{E}\{\bar{w}[n]^2\} = \frac{1}{t_0^2}\left(\mathbb{E}\{\bar{w}[n]^2\} + \mathbb{E}\{\bar{w}[n-1]^2\} -2 \mathbb{E}\{\bar{w}[n]\bar{w}[n-1]\}\right).$$ If your noise samples $$w[n]$$ are i.i.d. and have constant variance $$\sigma_w^2$$, this leads to $$\mathbb{E}\{\bar{w}[n]^2\} = \frac{2\sigma_w^2}{t_0^2}$$. So there is a doubling of variance, but also some kind of rescaling. This depends of course on the unit of your velocity (whether it's angle/second or angle/sample, in the last case your variance is just $$2\sigma_w^2$$). NB: If they are correlated, this changes to $$2\sigma_w^2(1-\rho)$$, where $$\rho$$ is the correlation coefficient.