(I will be referring to info from these slides in this post, for your reference.)

Consider first for comparison a typical CT low-pass filter, without any sampling, for example implemented by the RC circuit below.

rc filter

Here the CT input source $v_n(t)$ with transform $V_n(s)$ is the noise generated by the resistor. The noise is attenuated by the filter at the output according to the transfer function

$$ V_{no}(s) = \frac{1}{1+RCs}V_n(s) = H(s) V_n(s) $$

The resistor's thermal noise has a one-sided PSD $S_n(f)=4kTR$. We also have a theorem (whose name I don't know) that $S_{no}(f)=|H(f)|^2S_n(f)$. By Parseval's theorem, $\overline{v_{no}(t)^2}=\int_0^\infty S_{no}(f) \, df$. Substituting in, we arrive at the typical result that $\overline{v_{no}(t)^2}=kT/C$.

Now, consider the same circuit except with a switch that implements a track-and-hold response, as illustrated below. (Located at slides 25 and 26 in the PDF). $\phi$ indicates the clock signal controlling the switch: when $\phi$ is high, the switch is on and the output tracks the input. When $\phi$ is low, the switch is off and the output is disconnected from the input and held constant.

track and hold


Now I would like to consider the discrete-time sequence of samples $v_{out}[n]$ as illustrated in the figure, and I would like to find the quantity $\overline{v_{out}[n]^2}$. This is the average value of the squared time-domain output samples, where again this output is due to the noise alone. The lecture slides perform the following derivation (slide 28):


In other words, it's exactly the same as the completely continuous-time case, both in the steps taken and the final result. Why is this the case? Is there a rigorous justification for why this correct or is there an implicit approximation here? How is the discrete-time nature of the samples being taken into account? If this assumes instantaneous settling of the output once the tracking phase starts, how would one go about a more exact derivation?

  • $\begingroup$ There's a lotta inconsistency in notational convention depicted here. My recommendation is that lower-case be used for functions of time and upper-case be used for functions of frequency. Also a slightly different notation $x[n]$ be used for discrete-time (or discrete-frequency) signals than for continuous-time signals $x(t)$. $\endgroup$ Commented Aug 17, 2023 at 19:17
  • $\begingroup$ @robertbristow-johnson Thanks for the note, I've updated the post to try to correct this. I can't edit the figures taken from the PDF slides directly, but I can try to clarify if there's any confusion there. $\endgroup$
    – Halleff
    Commented Aug 17, 2023 at 19:27

1 Answer 1


Okay, this is coming in installments. The fancy name for this RC low-pass filtered (or "leaky integrated") white noise process is the Ornstein-Uhlenbeck process.

White noise, $w(t)$, has a constant measure of power per unit frequency.

$$ S_w(f) = \frac{\eta}{2} \qquad \forall f \in \mathbb{R} $$

That means that theoretical white noise which has infinite bandwidth also has infinite power. This, of course doesn't exist anywhere physically.

In fact the thermal noise of resistors does have a power spectrum that rolls off somewhere in the terahertz, but we model it as a constant amplitude in $f$ for all practical frequencies.

$$\begin{align} S_n(f) &= \frac{4 \pi \hbar R |f| }{e^{2 \pi \hbar |f|/(k_\mathrm{B} T)}-1} \\ \\ & \approx 2 k_\mathrm{B} T R \qquad \qquad |f| \ll \frac{k_\mathrm{B} T}{\hbar} \\ \end{align}$$

At room temperature $\frac{k_\mathrm{B} T}{\hbar} \approx$ 39 THz, so the constant evaluation is justified for thermal noise, $\eta = 4 k_\mathrm{B} T R$, and in SI units, this would be in volts${}^2$ per Hz.

Now the RC low-pass filter is described by this first-order differential equation:

$$ \frac{\mathrm{d}y(t)}{\mathrm{d}t} = \frac{1}{RC}\big(-y(t) + x(t)\big) $$

has Laplace Transform

$$ sY(s) = \frac{1}{RC}\big(-Y(s) + X(s)\big) $$

and transfer function

$$ H(s) = \frac{Y(s)}{X(s)} = \frac{1}{1+RCs} $$

and frequency response

$$ H(j2\pi f) = \frac{1}{1+j2\pi RC f} $$

With magnitude-squared

$$ \Big|H(j2\pi f)\Big|^2 = \frac{1}{1+(2\pi RC f)^2} $$

So if $x(t)$ and $y(t)$ are finite energy signals (not finite power signals) and we define the energy density spectra as $S_x(f) = |X(j2\pi f)|^2$ and $S_y(f) = |Y(j2\pi f)|^2$, then it's pretty clear that output and input spectra are related as

$$ S_y(f) = \Big|H(j2\pi f)\Big|^2 S_x(f) $$

Now, if the input is a power signal (infinite energy because it's going forever), then also the output is a power signal and the power spectra are related just the same. In relating mean power to energy, both input and output have the same $\frac1T$ scaling factor as the limit goes to $\infty$.

Now, if the input $x(t)$ is white noise, then $x(t)=w(t)$ and $S_x(f)=S_w(f)=\frac{\eta}{2}$. The output power spectrum is:

$$\begin{align} S_y(f) &= \Big|H(j2\pi f)\Big|^2 S_x(f) \\ \\ &= \frac{1}{1+(2\pi RC f)^2} 2 k_\mathrm{B} T R \\ \end{align}$$

Now that is the power spectrum of the output noise, and unlike theoretical white noise, this spectrum has a finite area under the curve and is finite power.

$$\begin{align} \int\limits_{-\infty}^{\infty} S_y(f) \ \mathrm{d}f &= \int\limits_{-\infty}^{\infty} \frac{2 k_\mathrm{B} T R}{1+(2\pi RC f)^2} \ \mathrm{d}f \\ \\ &= 2 k_\mathrm{B} T R \int\limits_{-\infty}^{\infty} \frac{1}{1+u^2} \ \frac{\mathrm{d}u}{2\pi RC} \\ \\ &= \frac{k_\mathrm{B} T}{\pi C} \int\limits_{-\infty}^{\infty} \frac{1}{1+u^2} \ \mathrm{d}u \\ \\ &= \frac{k_\mathrm{B} T}{C} \\ \end{align}$$

If I did everything correctly and if you use SI units, the result should be volts${}^2$. That would be the variance of any random sample of that voltage.

Now, the track-and-hold (we used to call it a "sample-and-hold") has significance in that it holds the value constant until the next sample (as opposed to returning to zero or something like that). That leaves the variance unchanged.

If you were to divide that that power by the peak value of $S_y(f)$, you would have the equivalent noise bandwidth of the system.

$$\begin{align} 2B &= \frac{\frac{k_\mathrm{B} T}{C}}{2 k_\mathrm{B} T R} \\ \\ &= \frac{1}{2RC} \end{align}$$

I am calling it "$2B$" to be consistent with the convention that the bandwidth $B$ is one-sided (only the positive frequency) and $2B$ is the two-sided bandwidth. That would be the same power or variance you would get with the same white noise strength $\eta$ (which is not variance, despite what Royi or someone else might call it) and an open channel of spectrum with a one-sided bandwidth of $B$.

  • $\begingroup$ This is essentially finished, but there might be a coda I tack on regarding autocorrelation of this RC low-pass filtered white noise. $\endgroup$ Commented Aug 19, 2023 at 0:27
  • $\begingroup$ Thanks a lot for the answer- I'm wondering about this part: "Now, the track-and-hold ... holds the value constant until the next sample ... That leaves the variance unchanged." Could you elaborate on why this is? Are there specific conditions that lead to this being the case? For example, suppose the product $RC$ is large compared to half the clock period, meaning the low-pass nature of the filter becomes very apparent (see the linked PDF pages 32-34 for what I mean, where $T_s$ is the clock period, and so $T_s/2$ is the duration of the track/hold phases). Would this change the variance? $\endgroup$
    – Halleff
    Commented Aug 19, 2023 at 2:58
  • 1
    $\begingroup$ Even though the sampling that happens at the instance we go from track to hold occurs at regular times, we are getting for each sample a random number that has the same statistics as if we sampled that process at a random time. There is no synchronization of the sampling instances with any parameters in the random process (which are constant). We get a random number that has an expected value (the statistical mean) of zero and a variance (also the mean square) of $\frac{k_\mathrm{B} T}{C}$. $\endgroup$ Commented Aug 20, 2023 at 4:48

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.