What causes the peak in the thermal noise spectrum of an RLC circuit?

[BillKet]

Hello! Below is a figure of the thermal noise spectrum of an RLC circuit from a certain experiment. I am not sure I understand the shape. The spectrum is peaked around the resonant frequency. However, in thermal equilibrium, I would expect that the electrons to have random motions, so for a given energy (and hence frequency) all the phases would be equally possible, and that would mean that the spectrum would be flat. For example, the Fourier transfrom of 2 signals out of phase is flat (because the 2 signals cancel each other so the signal itself is flat). So, even if the motion at the resonant frequency is amplified, the motions at the same frequency but opposite phase would cancel. So what exactly gets amplified on resonance? Thank you!

https://www.physicsforums.com/attachments/281641

 

[Dale]

The circuit is acting as a filter. You are starting with a flat spectrum and then running it through a band pass filter.

 

[BillKet]

The circuit is acting as a filter. You are starting with a flat spectrum and then running it through a band pass filter.

But that would select the frequencies, no? Wouldn't equal frequencies (which would both be passed through the filter), but out of phase by $\pi$ cancel each other, so the output of the two would be zero?

 

[Dale]

If it canceled like that then it wouldn’t be noise, it would be 0.

 

[BillKet]

If it canceled like that then it wouldn’t be noise, it would be 0.

Fair point, but then I guess my question is, why aren't all the phases equally likely?

 

[Dale]

All phases are equally likely. But that is not what you were describing. You are describing equal and opposite, not equally likely.

 

[BillKet]

All phases are equally likely. But that is not what you were describing. You are describing equal and opposite, not equally likely.

Isn't it the same thing? Aren't a phase $\theta$ and a phase $\theta + \pi$ are equally likely for any $\theta$?

 

[DaveE]

If you decompose the noise spectrum each "source" will have a frequency and a phase (and maybe a direction if that's not constrained). All have to match (within the limits of your detector) to interfere completely. That can happen, I'm sure, but it's not the majority of the signals. It also is more of a characterization of the noise source(s), not the filter response.

 

[Dale]

Isn't it the same thing?

No, it isn’t the same thing.

If I have a dice, then 1 through 6 are equally likely. But that doesn’t mean that every time a 1 comes up that a 6 also comes up. Equally likely does not mean that both are always rolled together. I can roll a 1 without also rolling a 6 at the same time.

 

[BillKet]

No, it isn’t the same thing.

If I have a dice, then 1 through 6 are equally likely. But that doesn’t mean that every time a 1 comes up that a 6 also comes up.

True, but if you throw that dice many times, the difference between the times 1 or 6 appears go to zero. What you are saying is that it is the very small difference in occurrences, given that we don't have an infinite number of electrons that leads to that signal?

 

[Twigg]

Sorry if I'm beating a dead horse, but I thought a little math might be helpful here.

As you say, all the phases are equally likely, and so the amplitude will on average be zero. In other words, if you took a Fourier transform of a thermal noise voltage-vs-time signal $V(t)$, that Fourier transform $\tilde{V}(\omega)$ will be zero on average for all frequencies $\omega$. In other words, $\langle \tilde{V}(\omega) \rangle = 0$. This is because there is total destructive interference between all the phases present in the thermal noise at frequency $\omega$.
I can't see the figure you're referring to, but I'd bet a kidney the y-axis is a power spectrum $P(\omega)$, which is NOT the Fourier transform of $V(t)$, but the Fourier transform of $V(t)^2$. $P(\omega)$ isn't zero, because the powers associated with each random phase will add (just like how optical intensities add for incoherent light sources, or how variances add).

Edit: tl;dr when I said $\langle \tilde{V}(\omega) \rangle$ you can just read that as $\tilde{V}(\omega)$. They're the same thing.
I realized that my bad math habits as an experimentalist are showing. Why am I talking about time averaging Fourier transforms? Aren't those both integrals over time from $-\infty$ to $\infty$? Yeah, they are. It's just that in my pea brain I tend to think about these things in terms of finite data sets like you'd find in the lab, where you have to take multiple measurements of the same signal and average them out. In pure mathematical formalism, there's no need to take a time average over a Fourier transform because it already samples all time. Sorry about that. Old habits and all.

 

[BillKet]

Sorry if I'm beating a dead horse, but I thought a little math might be helpful here.

As you say, all the phases are equally likely, and so the amplitude will on average be zero. In other words, if you took a Fourier transform of a thermal noise voltage-vs-time signal $V(t)$, that Fourier transform $\tilde{V}(\omega)$ will be zero on average for all frequencies $\omega$. In other words, $\langle \tilde{V}(\omega) \rangle = 0$. This is because there is total destructive interference between all the phases present in the thermal noise at frequency $\omega$.
I can't see the figure you're referring to, but I'd bet a kidney the y-axis is a power spectrum $P(\omega)$, which is NOT the Fourier transform of $V(t)$, but the Fourier transform of $V(t)^2$. $P(\omega)$ isn't zero, because the powers associated with each random phase will add (just like how optical intensities add for incoherent light sources, or how variances add).

Edit: tl;dr when I said $\langle \tilde{V}(\omega) \rangle$ you can just read that as $\tilde{V}(\omega)$. They're the same thing.
I realized that my bad math habits as an experimentalist are showing. Why am I talking about time averaging Fourier transforms? Aren't those both integrals over time from $-\infty$ to $\infty$? Yeah, they are. It's just that in my pea brain I tend to think about these things in terms of finite data sets like you'd find in the lab, where you have to take multiple measurements of the same signal and average them out. In pure mathematical formalism, there's no need to take a time average over a Fourier transform because it already samples all time. Sorry about that. Old habits and all.

Thanks a lot for this! So if it is the Fourier transform of $V(t)^2$ then it makes perfect sense. But now I am a bit confused about what you measure in practice. Going to the light example, if I have 2 electric fields out of phase, the sum of their squares is obviously not zero, but what I measure in an interferometry experiment as the power, is the square of the sum, which in this case is zero. Am I using this analogy the wrong way?

 

[Twigg]

The key is what you'd call coherence in the optics context. There isn't a word for it in electrical engineering that I know of, because almost every source in typical electronics is coherent. As a general procedure, think about the source of the signal (be it optical or electrical). In many cases, you can think about the signal source as a collection of a lot of microscopic sources. Ask yourself, are the microscopic sources synchronized (coherent) or random (incoherent)? If they're synchronized (coherent), the amplitudes add. If they're random (incoherent), the powers (squares of amplitude) add.

Example #1: You have electrons flowing from a radio antenna being driven by a pure sinusoidal E-field at frequency $f$. The "microscopic sources" are the electrons in the antenna. All the electrons in the antenna are being driven to oscillate by the same electric field, so they have the same phase relative to each other. The current you get out of the antenna is a well-defined sinusoid. If you wired two antennas to the same circuit, their amplitudes would add, not their intensities.

Example #2: A resistor has some thermal noise due to its temperature. The "microscopic sources" are the electrons in the resistor. The motion they get from thermal noise is governed by Maxwell-Boltzmann statistics, and is completely random. To get more technical, the velocities of any two electrons in the resistor are totally uncorrelated. Their is no phase relationship between any of the electrons. So if you wired two resistors together, their powers would add, not their amplitudes.

Example #3: Laser light used in a two-slit interferometer. The photons that leave the laser are all stimulated-emitted from the gain medium, so they inherit the phase of the seed light. So all the photons have the same phase and are correlated. As a result, the amplitudes of the light going through each slit add, and you see fringes.

Example #4: Light from an incandescent bulb used in a two-slit interferometer (without spatial filtering). The photons from the light bulb originate from thermal motions of the electrons in the filament. These thermal motions are random and uncorrelated, so there is no phase relationship between the photons. As a result, the intensities (squared E-field) of the light going through each slit add, and you don't see fringes.

Does that help?

Also, there is a big difference between two signals being "out of phase" and two signals being uncorrelated. The former implies the signals are identical after a $\pi/2$ phase shift, and the later implies that knowing one signal doesn't tell you anything about the other.

 

[Dale]

True, but if you throw that dice many times, the difference between the times 1 or 6 appears go to zero.

If you are recording a signal then you are not throwing that dice many times (once for each electron in the circuit). You are throwing it once for each sample you record. If you record a signal many times then you are throwing the dice many times and if you average those signals then it goes to zero.

It is wrong to think of each electron as providing a separate noise signal which is averaged to obtain the circuit noise. The noise comes from the whole circuit and the noise power does not depend on the size of the circuit. For a purely resistive ideal circuit the noise power is $k_B T \Delta f$. There is ideally no dependence on mass or volume.

 

[hutchphd]

The former implies the signals are identical after a π/2 phase shift

I think you mean $\pi$. That's a complicated way to say (-1)!

 

[Twigg]

@Dale, you're absolutely right thermal noise not scaling with circuit size. Just to clarify for other's benefit, the only point to thinking about the electrons in thermal noise is to convince yourself that the noise power spectrum is the quantity of interest instead of the noise amplitude spectrum. Otherwise, as Dale says, it can be a misleading mental picture. Thanks Dale!

@hutchphd, you're right! Whoops! What's a factor of two among friends?