the_beber

You need to lock in

https://feddit.org/post/26623643

Mar 4 at 9:28amWeb
Show threadthe_beberMar 4
Oh no… I just noticed an additional “ whoever finds it, can keep it (or make their own perfect version).
Show threadThorryMar 4
Show threadeinfach_orangensaftMar 4
quality meme
Show threadjaydevMar 5

I feel like I should know this meme, but don’t. I have 2 main questions:

  • Why does the multiplication of the 2 signals result in an extra fluctuation in the product?
  • Why would the low pass filter produce a flat line? Is it because it’s letting such low frequencies pass that none of the frequencies in the product signal are not included?
    • Show threadthe_beberMar 5

    The device, I‘m referencung here is called a lock-in amplifier. When you try to measure an extremly noisy signal without all the noise, you can use one of these. If you‘re dealing with a DC-signal, you can chop it at the reference frequency.

    Here‘s a great write up on the priciples of this technique: zhinst.com/…/zi_whitepaper_principles_of_lock-in_…

    But TLDR: After the reference signal is adjusted to have same frequency (and therefore constant phase difference), you get a signal that oscillates with ω_\text{in} - ω_\text{ref} and ω_\text{in} + ω_\text{ref}. Crucially, in the case, where ω_\text{in} = ω_\text{ref} the term becomes constant U(t) = U_0 |e^{i \theta}| while the other terms from other frequency components (Fourier-series) still oscillate. This is where the averaging comes in. An oscillating signal will average (roughly) 0 over a long enough duration. The output is then the amplitude of the desired signal without all the noise.

    Show threadBigDanishGuyMar 5

    TBH it’s been more than a decade since I graduated electronics engineering, I really didn’t like filter theory, so I went with embedded electronics for my specialization, and went to work in an entirely different field so…

    What’s the purpose of this filter, if all you’re getting out is DC? I mean wouldn’t you want to set the cutoff frequency high enough to actually get a signal through? If all you’re looking for is the DC component, then wouldn’t you be better off (as in a more simple, thus cheaper, solution) just doing a rolling average filter?

    Again, I really didn’t follow along in signal processing that well, so I may just be exposing my ignorance here.

    Show threadthe_beberMar 6
    The (effective) low-pass is only applied after the signals have been multiplied. This filters out the signal with the same frequency (and phase) as the reference. I have a more detailed comment on how this works in response to another person here in this comment section. This is useful for applications, where the signal can be drowned out by the noise, but you know what to look for, and where the phase information is important. (e. g. a DC signal that is chopped at a selected frequency or an AM-signal).