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I heard that Lock-in amplifiers (LIAs) especially play to their strengths when the signals are weak compared to the noise level. But then I talked with someone about it, who understands the principles of lock-in amplification, and she said - which makes sense to me now - that of course the signal amplitude still needs to be larger than the level of noise. Otherwise we couldn't represent the signal V_s like this: $$ V_{s} = R\cdot cos(\omega_{s} t + \phi) $$ Is that correct? I find the formulation "buried in noise" a bit confusing then...

PS: I often get criticized for not explaining enough about the basics of the topic that I ask my question about. Since I don't want my question to get closed again, I would like to refer you to this page, which I used to learn about it: https://www.zhinst.com/others/en/resources/principles-of-lock-in-detection Also, to forestall criticism that I just stipulate that "buried in noise" is an existent phrase in this context, I would refer you to this page, where you can see some examples of this phrase: https://preview.tinyurl.com/y64re9ln (secure URL: only preview of website, that would otherwise redirect to a Google domain)

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What you're missing is the bandwidth, both of signal and noise.

If you look at, let's say, a 1 V rms sinewave signal, together with 10 V rms noise on an oscilloscope, you'll see only noise.

However, if the noise occupies a 1 MHz bandwidth, and is flat with frequency, and you pass the signal + noise through a 1 kHz bandwidth filter centred on the signal, then you will eliminate 99.9% of the noise power, dropping its amplitude to 0.3 V rms. The signal will then be clearly visible.

A lock-in-amplifier is a neat way to make a very narrow filter centred on the frequency you feed in as the reference.

You can use the same principle even without sine waves. Spread spectrum systems like CDMA and GPS use a pseudo-random square wave signal as the reference, and call the 'multiply and average' process convolution or correlation. As long as the reference is is the same as the underlying signal, and as long as the averaging process produces an effective bandwidth small enough to drop the noise power, the signal can be 'dug out of the noise'. A lock-in-amplifier is a special case of the more general 'correlation with a reference' that's used for CDMA.

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    \$\begingroup\$ Nice answer. The OP may be interested to know that noise levels are specified in "volts per square-root Hertz" for exactly this reason. Those units make it clear that if you reduce your Hertz, you reduce your volts. It also makes it clear that there's a square-law response in there, so you need 100x less bandwidth for 10x less noise. Or as per your example, a 1000x reduction in bandwidth gives a 31.6x reduction in noise. (Not 99.9% though - I think you could improve your answer by removing that, because it makes things less clear.) \$\endgroup\$
    – Graham
    Commented Nov 5, 2020 at 9:41
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    \$\begingroup\$ @Graham Tahnks for pointing that out, the word 'power' was in my mind as I wrote 99.9%, I just forgot to include it. The 0.3 Vrms is of course 'engineering rounded'. I prefer to express noise as both power and amplitude, as then if anyone starts to do sums with them, the square will be memorable if it's the first time they've thought about it. \$\endgroup\$
    – Neil_UK
    Commented Nov 5, 2020 at 11:32
  • \$\begingroup\$ CDMA works quite differently than lock-in amplifiers. For one, the CDMA signal can be below the noise floor in a per Hertz fashion. So filtering wouldn’t help you there. CDMA utilizes the fact that the channel capacity is linear in the bandwidth and logarithmic in the signal to noise ratio, as given by the Shannon-Hartley theorem. Hence one can decrease the power of the received signal exponentially while increasing the bandwidth linearly maintaining the same bit error rate. \$\endgroup\$
    – user110971
    Commented Nov 5, 2020 at 17:37
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    \$\begingroup\$ @user110971 an LIA is a special case of the CDMA 'correlate with a reference then low-pass filter' method, or 'projection onto orthogonal bases'. Filtering and LIA use sinusoidal bases. CDMA uses generally pseudorandom bases, often termed 'sequency' to compare and contrast with 'frequency', and I've seen the word 'liftering' used to mean sequency filtering. The important thing is the underlying correlation that's common to them both. Frequency and sequency are orthogonal, so to say signal is below a perHz noise floor only matters if we're filtering, it's still extractable if liftering. \$\endgroup\$
    – Neil_UK
    Commented Nov 5, 2020 at 19:30
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NASA will acquire distant, or weak, satellite signals, buried in the noise and having some frequency uncertainty, by sweeping the receiver over the range of expected frequencies.

Once acquired, such systems can tighten the Phase_Locked_Loop bandwidth even more, as long as the Transmitted signal has low phase noise.

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Part of the challenge of such circuits/systems, given the need to implement an almost PURE mathematically exact CORRELATION, is the DISTORTION of the mixer or however the internal signal_model and the real signal_plus_noise are processed to generate the "We have a correlation event".

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  • \$\begingroup\$ It's also important to point out that, especially in the Nasa case, you can use arbitrary amounts of Forward Error Correction (like Reed-Solomon Error Coding) to trade speed for resiliency. Every time you double the amount of error correction you add, you get a 3 dB 'coding gain'. \$\endgroup\$
    – antiduh
    Commented Nov 5, 2020 at 19:53
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the signal amplitude still needs to be larger than the level of noise

For an LIA to be effective, the signal amplitude in its bandwidth of interest needs to be somewhat bigger than the prevailing noise in that same bandwidth.

When viewed on a scope the signal may still appear to be "buried in noise" but not if you applied a tight band limiting filter. Then the signal would be much more clearly represented on your scope image. That is something along the lines of an analogy to a LIA.

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    \$\begingroup\$ @ Andy Not true. A modern LIA can detect and measure signals that are 120 dB below the background noise in the same bandwidth (ref: zhinst.com/americas/en/resources/…). The LIA effectively reduces the signal bandwidth down to DC so that the noise can be filtered out with a LP filter. \$\endgroup\$
    – Barry
    Commented Nov 4, 2020 at 17:14
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    \$\begingroup\$ @Barry if you're going to make claims like that you need to be very careful about how you define terms like "bandwidth", and you need to keep in mind that if a LIA vendor is explaining how it works, there will be hype, even if it's a subdued, engineerish sort of hype. "Effectively reducing the signal bandwidth down to DC" can only be done if the actual effective signal bandwidth is effectively zero. Because there is no magic, and doing anything else would require magic. \$\endgroup\$
    – TimWescott
    Commented Nov 4, 2020 at 17:19
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    \$\begingroup\$ @Barry this quote from their site disagrees with what you said: The best instruments on the market today have a dynamic reserve of 120 dB [5], which means they are capable of accurately measuring a signal in the presence of noise up to a million times higher in amplitude than the signal of interest. - you said "in the same bandwidth" and that is not substantiated by the website quote further down the page you linked. \$\endgroup\$
    – Andy aka
    Commented Nov 4, 2020 at 18:00
  • \$\begingroup\$ Sorry, but I cannot follow you all: If the signal is, let's say, 1000 times smaller in amplitude than the noise, then how do we get a function $V_{s}$ as given above? Say, we have an optical chopper and a solar cell, then how can the chopper lead to a signal input that has a cosine (or square wave) shape ? (The noise would still be 1000 times larger, no matter if the buried signal is 0 or 100%...) \$\endgroup\$
    – AskingBecauseIHaveQuestions
    Commented Nov 4, 2020 at 23:40
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Say, we have an optical chopper and a solar cell, then how can the chopper lead to a signal input that has a cosine (or square wave) shape ? (The noise would still be 1000 times larger, no matter if the buried signal is 0 or 100%...)

A more concrete example might help here. Suppose you have a signal that is 1-10 microvolt and constant. You try to measure it, but find that you get noise of 100 microvolts in your measurement. The signal is buried in the noise of your first measurement, but you can do better.

Take 100 measurements and average them. Your noise is random and will tend to average out. Your signal is constant and will not. After 100 measurements, your average has its noise reduced to 10 microvolts. Now do 10,000 measurements and average. Your noise is now 1 microvolt. Do 1,000,000 measurements. Your noise is now 0.1 microvolts and you can easily measure your signal.

In this case, by averaging 1,000,000 times, you have made your measure 1,000,000 times longer, and thus reduced its bandwidth by the same factor. Since your signal is constant (zero bandwidth) and your noise is not, you can get the SNR as high as you want by measuring long enough (reducing bandwidth).

A lock-in amplifier is a clever device for reducing the bandwidth of a measurement. In the real world it would be hard to average 1 million measurements because other things than the noise would start to be a problem (DC drift, correlated noise in your measurement device, etc). The lock-in, by locking in to the modulated signal of your chopper, can side step a lot of these problems and perform measurements with a very, very low bandwidth.

But then I talked with someone about it, who understands the principles of lock-in amplification, and she said - which makes sense to me now - that of course the signal amplitude still needs to be larger than the level of noise.

In the above example, you could average and get the signal back out because the signal was constant and the noise was not. Viewed in terms of signal per unit of bandwidth, clearly the signal was much more bigger than the noise. If you had a time varying signal so that you could only average 100 measurements, then the signal would be truly buried in the noise and you would not be able to recover it.

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Buried in noise would mean the signal level is less than the noise level.

To get around that problem you repeat the signal many times and then add them all up so that the noise levels effectively stay the same and the signal gets larger than the noise.

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  • \$\begingroup\$ Not really. You may be describing oversampling, where the "noise" is ADC sampling error. The point of oversampling isn't that sampling error stays the same, it's that the averaged result can give you a more accurate reading. Alternatively you may be describing a type of filter called a "moving average", which if used as a low-pass filter can use the same principle to reduce the bandwidth as described by Neil_UK above. Both only apply in digital processing. Either way, you're describing one possible solution without getting the principles of it, or at least not explaining them to the OP. \$\endgroup\$
    – Graham
    Commented Nov 5, 2020 at 10:00

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