Noise variance of DT samples from CT track-and-hold signal

Question

(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.

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.

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?

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.

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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 $B$.