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$P = 4kT$ (where $k$ = Boltzmann’s constant, and $T$ = temperature of the instrument ($K$))

And the mean voltage is thus, of course, $V^2/R = 4kT$

And the voltage is distributed as a Gaussian around that mean, with RMS given by $\sqrt{4kRT\Delta f}$.

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Emphasis mine:

I want to plot the expected voltage vs time trace

The expectation (that's a term from probability theory and statistics) of a white process is zero. So, your expected voltage trace is a constant 0.

You might mean that you want to plot an exemplary realization of a random noise process - in that case:

  • For a discrete-time noise process, draw a random number from a Gaussian distribution for any time instant. Make sure your random generator isn't having any correlation between current and any previous samples.
  • For a continuous-time noise process, you can't really draw that, because you'd need to draw uncountable infinitely many random values for any interval, no matter how small that interval.

Notice that in either case, the drawing you made has probability 0 of actually being the realization of the white noise you want to be observing; you can't deterministically draw a random thing.

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Before getting to the answer, several corrections are necessary. First, the units of the Boltzmann constant are J/K, so kT is energy, not power. Thus P is not equal to 4kT. Second, the expected value of the thermal noise voltage, $\mu$, is zero, as required by the second law of thermodynamics. So $V^2/R$ is not equal to 4kT.

When I do computer simulations involving Gaussian white noise, of which thermal noise is a type, I assume that I am sampling a pseudo-random Gaussian noise generator, i.e., the software I am using provides a way to get successive negligibly-correlated samples. I assume a constant sampling rate of $f_s$ samples per second. The constant point spacing, $\Delta t$, is $1/f_s$ seconds. Then the Nyquist frequency, $f_{nyq}$, is $f_s /2$ Hz, which is equivalent to $1/2\Delta t$ Hz.

The Gaussian white noise has zero population mean, $\mu$, and $\sigma^2$ population variance, so the bilateral noise power spectral density, $\eta$, is $\sigma^2 /f_s$. This simply means that the total noise power, $\sigma^2$, is uniformly distributed, because the noise is white, from minus $f_{nyq}$ to plus $f_{nyq}$. This frequency interval is simply $f_s$. Hence

$$ \eta = \sigma^2 /f_s = \sigma^2 /2f_{nyq} = \sigma^2 Δt \tag 1 $$

The unilateral noise power spectral density, $2 \eta$, is $2 \sigma^2 Δt$. Therefore

$$ 2 \eta = 2\sigma^2 /f_s = \sigma^2 /f_{nyq} = 2\sigma^2 Δt \tag 2 $$

For thermal noise below roughly 10 THz (and see here for a bit more on that), the unilateral noise power spectral density is well approximated as $4RkT$. Hence

$$ 4RkT = 2 \eta = 2\sigma^2 Δt = 2\sigma^2 /f_s = \sigma^2 /f_{nyq} \tag 3 $$

Thus

$$ \sigma^2 = 2RkT / Δt = 2RkTf_s = 4RkTf_{nyq} \tag 4 $$

So select R, T and either $f_s$ or $\Delta t$. Then equation (4) gives the population variance for the pseudo-random Gaussian noise generator.

Example: Assume $R = 10 k \Omega$, $T = 300 K$, $k = 1.380649 \times 10^{-23} J/K$ and $\Delta t = 1 \mu s$, so $f_s = 1 MHz$. Then $\sigma^2 = 8.283894 \times 10^{-11}$ $V^2$ and $\sigma = 9.101590 \times 10^{-6} \space V$. In the software I use, the syntax for generating a sample from the pseudo-random Gaussian noise generator is simply $Gaussian(\mu ,\sigma)$. Since $\mu = 0$, and with $\sigma$ as above, this becomes $Gaussian(0,9.101590 \times 10^{-6})$. A specific temporal trace, from 0 to 1.023 ms in $1\mu s$ increments, is

A Gaussian temporal trace

From equation (3), $4RkT = 1.6567788 \times 10^{-16}$ $V^2/Hz$. This is the theoretical unilateral noise power spectral density (PSD). The plot below shows the average of $10^4$ unilateral, mean-subtracted 1024 point PSDs. It is seen that the PSD is white and in excellent agreement with the theoretical value.

PSD

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  • $\begingroup$ Note: the question is cross-posted here, but I obviously cannot cross-post my answer there. $\endgroup$ – Ed V Aug 31 at 14:59
  • $\begingroup$ I am not the OP, but I found your answer very helpful for stuff I am doing. I understand everything except for: Why do you need to use the Nyquist frequency? I.e. why is there a factor of 2 in $\sigma^2 = 2 R k T/\Delta f$ and not a factor of 4? $\endgroup$ – Christian Sep 28 at 9:53
  • $\begingroup$ @Christian Glad it helps! The factor of 2 in the first equality in the expressions in (4) simply follows by rearrangement of the expressions in (3). The Nyquist frequency always arises when sampling takes place, but it is simply half the sampling frequency, so it need not explicitly appear in an equation. Hope this helps clarify things. $\endgroup$ – Ed V Sep 28 at 13:13
  • $\begingroup$ Thanks, do you know if there there is a formal derivation why $\Delta f = f_{Nyq} $ and not just $\Delta f = f_s$? $\endgroup$ – Christian Sep 28 at 13:17
  • $\begingroup$ @Christian Where is $\Delta f$ coming from? If you want to use unilateral PSDs, i.e., PSDs only applicable to positive frequencies, then the full frequency range of the sampled noise is 0 to Nyquist. For bilateral PSDs, the range is minus to plus Nyquist, so it is twice the magnitude, and the PSD is half as large. For either of these, the total variance is simply the integral of the PSD over its associated frequency range. Since the PSDs are constant, the integrations reduce to products and they yield the same result. $\endgroup$ – Ed V Sep 28 at 13:26

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