# Discrete-time sampling of filtered white noise

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 $$n_\tau[k]$$?

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.

• Why is $N_0/2$ equal to $1$? Does this have something to do with the claim in your question that it is "unit-amplitude" white noise that you are considering whatever that adjective means to you? And if so, please explain "unit-amplitude" means, it is not something that I have encountered before in the context of white noise. – Dilip Sarwate Jun 16 '20 at 11:51
• Normalized might be the better term – Dan Boschen Jun 16 '20 at 15:17
• Normalized -- yes, I can always scale it to the real amplitude. I just want to understand the relationship between the abstract AWGN source in continuous time and its effect on a discrete-time system. – Jason S Jun 16 '20 at 19:29

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

• "the power spectrum of the sampled process is an aliased version of the power spectrum of the continuous-time process" OK that's what I was missing – Jason S Jun 16 '20 at 19:30
• @JasonS: Actually, that formulation isn't really great. I should have said "a periodically continued version" because aliasing doesn't need to occur if the noise is band-limited and if the sampling rate is sufficiently high. – Matt L. Jun 16 '20 at 19:38