Discrete-time sampling of filtered white noise

Question

I am trying to understand how I can relate a discrete-time random process to a continuous-time random process sampled at discrete times.

Suppose I have a noise source $N_\tau(t)$ which is derived from unit-amplitude additive white Gaussian noise N(t) that is fed through a single-pole low pass filter $H(s) = \frac{1}{\tau s +1}$.

I understand that the power spectral density of $N_\tau(t)$ is $S_{N_\tau}(\omega) = |H(\omega)|^2 = \frac{1}{(\omega\tau)^2 +1}$. So far so good.

Now I want to sample $N_\tau(t)$ at regular intervals $T$ to obtain a signal $n_\tau[k] = N_\tau(kT)$ -- how would I figure out the standard deviation and spectral density of the discrete samples $n_\tau[k]$?

Answer 1

Hmm. For standard deviation, I see https://dsp.stackexchange.com/a/8632/829 which states for uniform power spectral density $N_0/2$, the standard deviation is

$$\sigma^2 = \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,\mathrm df$$

which in my case would yield

$$\begin{align} \sigma^2 &= \int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,df \\ &= \int_{-\infty}^\infty \frac{N_0}{2}|H(\omega/2\pi)|^2\,d(\omega/2\pi) \\ &= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{N_0}{2}|H(\omega)|^2\,d\omega \\ &= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{N_0}{2}\frac{1}{(\omega\tau)^2+1}\,d\omega \\ &= \frac{1}{2\pi\tau}\int_{-\infty}^\infty \frac{N_0}{2}\frac{1}{(\omega\tau)^2+1}\,d(\omega\tau) \\ &= \frac{1}{2\pi\tau}\int_{-\infty}^\infty \frac{N_0}{2}\frac{1}{u^2+1}\, du \\ &= \frac{N_0}{4\pi\tau}\left[\tan^{-1} u\right]_{-\infty}^{\infty} \\ &= \frac{N_0}{4\pi\tau}\left[\pi/2 - (-\pi/2)\right] \\ &= \frac{N_0}{4\tau}, \end{align} $$

and with unit WGN $N_0/2 = 1$ so $\sigma = \sqrt{1/2\tau}$.

Not sure how to compute the PSD of a sampled random process, however.

Answer 2

If you sample a finite-power continuous time WSS random process $x(t)$, the auto-correlation of the sampled process $y[k]=x(kT)$ equals the sampled auto-correlation of the continuous-time process:

$$y[k]=x(kT)\;\Longleftrightarrow \;R_y[k]=R_x(kT)\tag{1}$$

Since the power spectrum is the Fourier transform of the auto-correlation, the power spectrum of the sampled process is a periodically continued version of the power spectrum of the continuous-time process:

$$S_y(e^{j\omega T})=\frac{1}{T}\sum_{k=-\infty}^{\infty}S_x\left(\omega-\frac{2\pi k}{T}\right)\tag{2}$$

Clearly, we have $R_y[0]=R_x(0)$, so if $x(t)$ is zero mean it follows that $\sigma_y=\sigma_x$.