Comment on You need to lock in
jaydev@lemmy.world 4 days ago
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?
the_beber@feddit.org 4 days ago
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.