lock-in amplifier - what's the purpose of the low-pass-filter

Question

I have a question about how a lock-in amplifier works. This is a picture with german tags; I'll translate very quick:

Bandpaßfilter: bandpass filter

Phasenverschieber: phase "shifter"

Mischer: mixer

Tiefpaßfilter: low pass filter

I dont really understand what the purpose of the lowpass filter is.

The mixer creates the cross-correlation between the both signals, where $U_{sig}$ is the incoming (noisy) signal and $U_{ref}$ is the reference signal which phase can be changed with the 'phase-shifter'.

My confusion:

I read that the cross correlation is ZERO for non-equal frequencies. When $U_{sig}$ is a signal with a more or less wide spectrum of frequencies and $U_{ref}$ a (more or less) monofrequent signal, will the mixer already filter the adjusted frequency $U_{ref}$ from the signal-frequency? If so:

What frequencies are cut off by the lowpass-filter?

I really would appreciate some explanations about the lock-in-amplifier. Thank you in advance.

Answer 1

The low pass filter serves two purposes.

First, mixing two signals of the same frequency will not create a signal that is purely DC. You might know the trigonometric relation $$\cos(a)\cos(b) = \frac{1}{2}\cos\left(a-b\right)+ \frac{1}{2}\cos\left(a+b\right).$$ So mixing two frequencies will result in a new signal that has also 2 frequency components, namely the sum and the difference of the original frequencies. If the frequencies are the same, the mixed signal has a DC component (zero frequency) and one at double the original frequency. Since we are only interested in the DC component, the other one is removed by the filter.

Secondly, the low pass filter serves to narrow the bandwidth of the amplifier. Any frequencies in the original signal 'close' to the carrier (or reference) will be mixed to a frequency 'close' to DC. The filter removes them, making the amplifier only sensitive to frequencies very close to the reference.