Is it possible to compute the thermal noise of an electronic system given only it's transfer function (phase + magnitude)? In my case I have a black box circuit of passive elements. I can't measure the thermal noise directly as any amplifier used to take the measurement would include current, voltage and thermal noise.

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    $\begingroup$ In case you have a known input and you can measure the output, you can extract the noise by subtracting the expected output from the measured one, right? $\endgroup$ May 27, 2022 at 20:27

1 Answer 1


General, such measurements can be made by applying multiple inputs.

For example, you say the box is full of passives, so it's a linear device: scale the input a factor $a$, you get the same output, but also scaled by $a$, whilst additive noise (which thermal noise is) adds power independent of the input amplitude.

$$ U_{out} \propto U_{in}$$

For a signal of a single frequency, your system will have a fixed gain/attenuation $A$. So,

$$ U_{out} = A\cdot U_{in}.$$

So, apply a sine at any given frequency with one amplitude, measure the output power. Apply the same frequency, twice the amplitude (i.e., four times the power), measure the output power.

\begin{align} P_{out,1} &= A^2 P_{in,1} + P_{noise}\\ P_{out,2} &= A^2 P_{in,2} + P_{noise}\\ &= A^2 4P_{in,1} + P_{noise}\\ P_{out,2}-P_{out,1} &= A^2 4P_{in,1} + P_{noise}- A^2 P_{in,1} - P_{noise}\\ &=A^2(4-1)P_{in,1}\\ &=3A^2P_{in,1} \end{align}

Great! Since you set $P_{in}$, and measured $P_{out,2}-P_{out,1}$, you can now calculate $A$. The difference between $A\cdot P_{in,k}$ and $P_{out,k}$ is $P_{noise}$. (Of course, the measured noise power is a stochastic thing – you need to measure / average long enough to get a good estimate!) Note that you can also get $A$ as the magnitude of your transfer function, but a) where's the fun in that, and b) if you actually knew that, you'd probably just be throwing $k_BTR$ at the problem, as you could, based on knowledge of the characteristic impedance of your measurement system, calculate both an equivalent noise bandwidth and an effective resistance.

Sadly, that noise power is the sum of both your device's noise and your measurement setup's noise. But that's not too bad! You can take a resistor that properly terminates your measurement device, measure its temperature, and thus calculate the Johnson-Nyquist Noise power that it has. Measure only the resistive terminator, and you get the noise your measurement setup adds. (This also works with passive attenuators between your device under test and your measurement setup etc.; there's many ways to do such measurements, and what I draft is but a very naive one.)

The powers you'll have to measure are pretty low. Thermally isolating (i.e., foam, non-electrostatic blankets, making sure there's not an air draft) the devices under test is a good idea to get repeatable measurements.


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