2
$\begingroup$

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 $n_\tau[k]$?

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Commented Jun 16, 2020 at 11:51
  • $\begingroup$ Normalized might be the better term $\endgroup$ Commented Jun 16, 2020 at 15:17
  • $\begingroup$ 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. $\endgroup$
    – Jason S
    Commented Jun 16, 2020 at 19:29
1
$\begingroup$

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 $\sigma_y=\sigma_x$.

$\endgroup$
2
  • $\begingroup$ "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 $\endgroup$
    – Jason S
    Commented Jun 16, 2020 at 19:30
  • $\begingroup$ @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. $\endgroup$
    – Matt L.
    Commented Jun 16, 2020 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.