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In a Pound-Drever-Hall lock, an electronic signal is produced of the form:

enter image description here

We are interested in electronically isolating the term: enter image description here

My question is: how can we filter to get exactly this term (listed above) without any time-dependent terms?

I know that I can use a bandpass filter around the frequency $\Omega$ will isolate this term:

enter image description here

We don't want the time-dependent terms in the signal and instead we want the imaginary term without frequency:

enter image description here

In the literature, it says that we can combine this signal (using a mixer) with a frequency \$cos(\Omega t)\$ - and a low-pass filter.. but how exactly does this work?

Mixing the signals will produce a term \$cos(\Omega)^2\$. I understand that the average of the signal will have a DC-offset. But if my electronics has a high bandwidth, then this isn't really helping eliminate the time-dependence in these terms, right?

So, in summary, can I use a mixer and a low-pass filter to extract out that particular term? And, if so, how exactly does it work?

Additional information:

(This isn't that important for answering the question, but if you are interested) the function F(omega) is: enter image description here And we are interested in isolating this particular term because it produces a good looking "error signal" that looks like this picture: enter image description here

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  • \$\begingroup\$ I only skimmed what you wrote in about 10 seconds, or so. But one thing to look for is a way to convert sine and cosine terms into sine-squared plus cosine-squared terms (because the sum of the two is 1 and eliminates time, if you can achieve it.) Complex conjugate is a way (squaring magnitude.) But I'd need to spend more time than I have right now to read through the details. Just wanted to toss that out to see if it sticks for you. You've spent more time on this, so my apologies for spending so little of my own so far. I may return later. \$\endgroup\$ – jonk Mar 13 at 2:31
  • \$\begingroup\$ for those interested en.wikipedia.org/wiki/… \$\endgroup\$ – analogsystemsrf Mar 13 at 2:46
  • \$\begingroup\$ Is \$\Omega\$ a fixed value you know in advance, or something that can vary? \$\endgroup\$ – The Photon Mar 13 at 22:59
  • \$\begingroup\$ @The Photon: This is basically my doubt (answered by OP below) posed in a different manner. \$\endgroup\$ – Dirceu Rodrigues Jr Mar 14 at 0:47
  • \$\begingroup\$ @DirceuRodriguesJr, No, I really mean is it a fixed value, not just one that's available in some signal in the system. Because it looks to me like you could extract the desired signal by just bandpass filtering around frequency \$\Omega\$ and then using a peak detector or power detector, with no need for the downmixing proposed in Edgar's answer. \$\endgroup\$ – The Photon Mar 14 at 1:02
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To answer the specific question in your title: no, a low pass filter does not “average higher order frequencies” it attenuates them.

To answer your actual question: yes, the basic procedure of demodulation and filtering will do what you want.

The math is rather straightforward. Simply multiply by \$ \sin(\Omega t) \$ and integrate over one period and you will see the term you want. Given your equation, using \$ \sin(\Omega t) + \cos(\Omega t)\$ instead as the multiplier should produce a cleaner result.

The issue is in how to implement the integration in a way that does not compromise your feedback loop stability. The simplest way is to use a low pass filter with a cutoff frequency of at most \$ \Omega / 2\$ to ensure the removal of feedthrough and double-frequency components.

Do note that a real-life continuous integrator is always just a low-pass filter. An actual integrator has infinite gain at DC, real components with real finite gains and leakage will limit this gain to perhaps \$ 10^5 \$ if not less, which directly implies a low-pass response.

Of course, much lower cutoff frequencies are commonly used, so as to have low-order practical filters. How much attenuation is necessary will depend on your application, but in general you would use filter that has a bandwidth not much higher than that of your signal of interest (in this case your feedback loop response speed).

You can also design a filter with specific zeros at the problematic frequency components (\$ \Omega \& 2\Omega \$). But in most cases, a very simple low pass is much more than enough.

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  • \$\begingroup\$ Could you go more into detail about this 'low-pass = integration' part? First you say: "a low pass filter does not “average higher order frequencies” it attenuates them." - but then you say that a low pass filter can obtain the term of interest through "integration." You say low-pass filter can implement this "integration", but it doesn't have to do with some of the frequency modes reducing to their time averages? \$\endgroup\$ – Steven Sagona Mar 13 at 8:32
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    \$\begingroup\$ First note that formally you require integration over one (or more) full periods, this is not what you get unless you explicitly implement it. Second, a continuous integrator is just a first-order low-pass whose cutoff frequency has been set to zero. Third, the “average” of any frequency component (except DC) is zero not a “DC offset”. Fourth, mere examination of the modulated equation shows that ideally all you need is the removal of the double-frequency component, an ideal low-pass filter does that, a real filter just attenuates it. \$\endgroup\$ – Edgar Brown Mar 13 at 12:46
  • \$\begingroup\$ @Edgar Brown: I do not know the details of this laser-focused experience in this way. So, just one question: Is the signal \$sin (\Omega t)\$ continuously available in the experiment (in order to be multiplied with \$P_{inc}\$) or would it need to be generated and synchronized? \$\endgroup\$ – Dirceu Rodrigues Jr Mar 13 at 14:54
  • \$\begingroup\$ @DirceuRodriguesJr you are asking the wrong person. Given the way the question was formulated I simply assumed it was available. If not, it’s possible to derive it from the signal itself with some compromises. \$\endgroup\$ – Edgar Brown Mar 13 at 16:35
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    \$\begingroup\$ @EdgarBrown, Yes. I understand that. I could not find an appropriate LPF in my lab, so we decided to make one on our own with a resistor and capacitor of appropriate values. Thank you! \$\endgroup\$ – Sonali Gera Mar 14 at 23:37

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