Square wave excitation for lock in amplifier?

[aagum_bae]

TL;DR Summary

Square wave and square wave demodulates to DC why?

Why does square wave for both excitation and demodulation lead to demodulation at DC
While using one of the signals sine the signal gets demodulated to odd harmonics?

I'm unable to do the maths of Fourier expansion and multiplying and passing through low pass filter

 

[DaveE]

After the mixer, either case will give you DC (which could be 0) and all of the other AC mixer products. Normal use, i.e. what makes this a lock-in amplifier and not just a mixer, is the narrow LPF to extract only the DC component at the output.

You don't really need to know about fourier transforms for a basic understanding of this, but you do need to know about how mixers work.

Read through this explanation and ask about the parts you don't understand:
https://www.zhinst.com/americas/en/resources/principles-of-lock-in-detection#:~:text=A lock-in amplifier performs,pass filter to the result.

 

[Baluncore]

Welcome to PF.
You could say, that the biggest odd harmonic, is the fundamental.

 

[aagum_bae]

After the mixer, either case will give you DC (which could be 0) and all of the other AC mixer products. Normal use, i.e. what makes this a lock-in amplifier and not just a mixer, is the narrow LPF to extract only the DC component at the output.

You don't really need to know about fourier transforms for a basic understanding of this, but you do need to know about how mixers work.

Read through this explanation and ask about the parts you don't understand:
https://www.zhinst.com/americas/en/resources/principles-of-lock-in-detection#:~:text=A lock-in amplifier performs,pass filter to the result.

https://www.analog.com/en/analog-dialogue/articles/synchronous-detectors-facilitate-precision.html

Can you take a look at this please? This article suggests that using square wave for both lock in and excitation leads to demodulation at DC, while using sinewave for one of the wave leads to demodulation at odd harmonics

 

[Baluncore]

If a square wave is used as the detection reference, then the harmonics of the square wave will also down-convert signals and noise present on the signal input to the mixer. That energy will be detected, and so be present on the output of the LPF. By using a clean sinewave as the reference, those unwanted spectral sources are rejected by the synchronous detector.

The modulation of the source can be done with a square wave, but the detection should be done with a clean sinewave.

 

[DaveE]

https://www.analog.com/en/analog-dialogue/articles/synchronous-detectors-facilitate-precision.html

Can you take a look at this please? This article suggests that using square wave for both lock in and excitation leads to demodulation at DC, while using sinewave for one of the wave leads to demodulation at odd harmonics

OK, I skimmed it. Analog Devices is a great source of information for this sort of thing. You can assume they are correct.

Unfortunately, I don't fully understand your question. It sounds like you are confused about the basics.

If one input to the mixer is a pure sine wave, then only that corresponding frequency in the two inputs will contribute to the DC component. Any frequency component that appears at both mixer inputs will be converted to DC with the amplitude also being a function of their phase shift. I think this is explained well in both of the previous links.

A common low frequency precision implementation is to use a square wave as the mixer reference because that allows analog switches to create a mixer with a very accurate DC output. This is important for low signal levels. The downside is that odd harmonics are also converted down to DC. Systems like this will often use a synthesized sine wave excitation (the other mixer input) to reduce this problem. Other applications don't really care about the frequency selectivity of the system so that's not as necessary.

 

[aagum_bae]

OK, I skimmed it. Analog Devices is a great source of information for this sort of thing. You can assume they are correct.

Unfortunately, I don't fully understand your question. It sounds like you are confused about the basics.

If one input to the mixer is a pure sine wave, then only that corresponding frequency in the two inputs will contribute to the DC component. Any frequency component that appears at both mixer inputs will be converted to DC with the amplitude also being a function of their phase shift. I think this is explained well in both of the previous links.

A common low frequency precision implementation is to use a square wave as the mixer reference because that allows analog switches to create a mixer with a very accurate DC output. This is important for low signal levels. The downside is that odd harmonics are also converted down to DC. Systems like this will often use a synthesized sine wave excitation (the other mixer input) to reduce this problem. Other applications don't really care about the frequency selectivity of the system so that's not as necessary.

Thanks a lot for your time, if you look at the figures in that link, one of them with both inputs as square wave and the other one with one of the input being a sine wave. Now when you look at the figures where one of the input is a sine wave you see the spectrum at odd harmonics, but when you look at the figure with both inputs as square wave you see the spectrum only at DC, why is that so?

 

[DaveE]

Thanks a lot for your time, if you look at the figures in that link, one of them with both inputs as square wave and the other one with one of the input being a sine wave. Now when you look at the figures where one of the input is a sine wave you see the spectrum at odd harmonics, but when you look at the figure with both inputs as square wave you see the spectrum only at DC, why is that so?

Because for the square waves each of the mixer inputs has the same (multiple) harmonic frequencies, so each harmonic is mixed with itself (modified by the system under test) and thus contributes to some of the DC output.

If one of the inputs is a sine wave, then only that specific frequency in the other input can be mixed down to DC.

Maybe reading a bit about mixers would help you too.

 

[aagum_bae]

present on the signal input to the mixer. That energy will be detected, and so be present on the output of the LPF. By using a clean sinewave as the reference, those unwanted spectral sources are rejected by the synchronous detector.

Because for the square waves each of the mixer inputs has the same (multiple) harmonic frequencies, so each harmonic is mixed with itself (modified by the system under test) and thus contributes to some of the DC output.

If one of the inputs is a sine wave, then only that specific frequency in the other input can be mixed down to DC.

Maybe reading a bit about mixers would help you too.

Here is my problem:

this is the equation at the LPF of the LIA, when multiplying a sine wave and a square wave, setting

the equation reduces to

after integration we get

which means

which means for an impure sinewave with harmonics or some noise at harmonics will get demodulated, but not to the fundamental frequency, this results in the combed filter effect as shown by

My question is when mutliplying two sqaure wave why does the analog devices website show that everything gets demodulated to fundamental DC? When the surae wave itself is composed of infinte sine waves I would expect multiplication of two sqaure waves to also give demodulated output at non zero frequencies.

 

[DaveE]

There is zero chance I'm going to review the math. You'd have to pay me for that. But...

I would expect multiplication of two sqaure waves to also give demodulated output at non zero frequencies.

Yes, exactly. The mixer does that too.

But a lock-in amplifier has a low pass filter that (ideally) rejects everything that isn't DC. If you put two frequencies $f_1$ and $f_2$ into a balanced mixer, the output is at $f_1 + f_2$ and at $f_1 - f_2$. Then you reject anything that isn't DC, so we only care about the signals where $f_1 = f_2$.

If there are multiple frequencies in the inputs then each one mixes with the others. So if you mix frequencies $f_1 + 3f_1$ with $f_2 + 3f_2$, you get mixer outputs at all of the combinations $f_1 + f_2$, $f_1 - f_2$, $f_1 + 3f_2$, $f_1 - 3f_2$, $3f_1 + f_2$, $3f_1 - f_2$, $3f_1 + 3f_2$, $3f_1 - 3f_2$, etc. The ones that subtract to 0 (DC), get through the LPF.

 

[aagum_bae]

Thanks a lot, I finally understand T_T