1
\$\begingroup\$

My understanding of lock-in used to be only that of an analog lock-in amplifier.

Now, I am thinking about digital lock-in technique.

Reading this has been helpful.

However, I am having trouble identifying the digital lock-in's equivalent for the integration time found in the analog lock-in's multiplier unit.

Is this not the right way to think about the more modern lock-in, which actually doesnt have a multiplier unit but achieves high SNR measurement with the use of a mixer and LPF.

I would like to apply 'the integration of the product of the reference signal and the input signal' picture to the modern lock-in if there is a way. (thinking about the LPF in the time domain may be the way to do exactly that but I can't exactly put my finger on it)

\$\endgroup\$
2
  • \$\begingroup\$ Am unsure of your conception of a modern digital lock-in. Many old analog lock-ins used a switching mixer that was sensitive to odd-harmonic input signals - not ideal. Most high-performance modern digital lock-ins use a mixer (in the digital domain) that is not sensitive to odd-harmonic input signals - they do this by simulating a sinusoidal (or cosine) reference signal. This case fits the lock-in block diagram in your reference document very closely. \$\endgroup\$
    – glen_geek
    Oct 1, 2022 at 13:24
  • \$\begingroup\$ @glen_geek I recently learned that LPF is represented as a transfer function and \$Q_{out}(\omega) = H(\omega)Q_{in}(\omega)\$ in the frequency domain. It is also a convolution of the transfer function and the input signal. From this perspective, I am at peace with the LPF being averaging. My feeling uneasy comes from my inability to tie the duration of measurement to any of the values or variables in the time domain picture \$\endgroup\$
    – Blackwidow
    Oct 1, 2022 at 13:43

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.