# Optical chopper

1. Oct 7, 2011

### cepheid

Staff Emeritus
What is the purpose of using a chopper to modulate an optical source? The Wikipedia article is not that informative.

2. Oct 7, 2011

### MrNerd

From the wikipedia page, I believe it's something like pulsing the light. Counting the pulses in a given amount of time can give you speed for something like a wheel.

3. Oct 7, 2011

You can get a symmetrically pulsed light source without the constraints of your electrical circuit (for example rise/fall time on an LED).

4. Oct 7, 2011

### G01

An optical chopper is used in spectroscopy experiments all the time. You can use the chopper to "chop" or pulse your signal at a given frequency. Then, you measure the signal with a lock-in amplifier set to that frequency.

It is a very common way of measuring a signal while getting rid of a lot of the noise, since the lock-in will only amplify signals modulated at that specific frequency.

CAVEAT: The lock-in may also amplify harmonics of your chopper frequency as well, so it's always best to avoid any multiple of 60hz, etc.. In our lab, we set the lock-in reference frequency to a prime number to avoid landing on any high harmonic of some noise source.

5. Oct 7, 2011

### Andy Resnick

We use a chopper for the same reason as G01 mentioned- a sync or reference signal that is used with a lock-in amplifier.

6. Oct 7, 2011

Cool trick!

7. Oct 7, 2011

### G01

Yes! Prime Numbers = Math = Awesome!

8. Oct 7, 2011

### cepheid

Staff Emeritus
Thanks for this G01! With this explanation, choppers begin to make sense in the contexts in which I hear them talked about. Now I want to go read up on lock-in amplifiers.

9. Oct 7, 2011

### Andy Resnick

Yep- that's also why submarine propellers use a prime number of blades- no subharmonic noise that can be detected by sonar.

10. Oct 7, 2011

### G01

Anytime!

Now THAT is cool!

11. Oct 8, 2011

### Redbelly98

Staff Emeritus
Wait, frequencies have units (cycles/s or if you prefer s-1), so it seems nonsensical that it could be set to a prime number, or for that matter any (unitless) number.

Can you clarify what you mean?

12. Oct 8, 2011

### G01

Why does a prime number have to be unitless? You can have three apples and you have a prime number, 3, with a unit, "apples." It has a unit, but it doesn't change that fact that 3 is prime. You can still only divide the apples two ways without breaking individual apples into pieces. i.e. into 3 groups of 1 or 1 group of 3

Consider this example:

Imagine you are chopping your signal at 120 Hz. So, you set the lock-in to isolate and amplify signals at 120Hz. That's fine, but the point of the lock-in is to eliminate noise, and with this configuration, the second harmonics of 60Hz noise sources will still get through.

So, instead you chop your signal at 389Hz. You can now amplify your signal and be sure your not landing on some harmonic of 60Hz or some other large noise source, as 389 is a prime number.

Of course, theoretically, you can have a noise source with frequency 194.5Hz whose second harmonic will get through. So, the method isn't a full proof way to eliminate all noise from a measurement. Still though, this trick is a good easy way to make sure your not inadvertently amplifying some harmonic of a large, obvious noise source like the 60 Hz AC in the nearby wall outlet.

13. Oct 8, 2011

### G01

The basic principle behind a lock-in is actually pretty simple. The lock-in will generate a reference frequency:

$$V_r=A_{r} sin(\omega_r t+\phi_r)$$

We mutiply that reference by each fourier component of the incoming signal, which each have the form:

$$V_s=A_{s} sin(\omega_s t+\phi_s)$$

leaving us with:

$$V_rV_s=A_rA_{s} sin(\omega_r t+\phi_r) sin(\omega_s t+\phi_s)=\frac{A_rA_s}{2}(cos((\omega_r+\omega_s)t+ \phi_r+\phi_s)+cos((\omega_r-\omega_s)t+ \phi_r-\phi_s)$$

This resulting signal is run through a low pas filter. In most cases, both sum and difference terms will be attenuated and we will be left with nothing.

However, if there is any frequency component in the signal such that $\omega_r\approx \omega_s$ we are left with a low frequency, ideally DC, signal that gets through the filter and goes on to be amplified:

$$V_{measured}=\frac{A_rA_s}{2}cos(\Delta \phi)$$

Lock-ins can also match the phase of signal and reference, so we can now easily get out the signal amplitude, $A_s$.

As for the details behind how the lock-ins do all of this electronically, I'd need to do some review myself. But hopefully, this illuminates the conceptual idea behind how the lock-in can hone in on and amplify a chopped signal.

Last edited: Oct 8, 2011
14. Oct 9, 2011

### Redbelly98

Staff Emeritus
To my thinking, when I see "number", it implies a unitless value. But now I understand what you meant, so thank you.

15. Oct 9, 2011

### johng23

Sorry if I'm being slow, but I don't understand what you're saying at all.

It sounds to me like you're arguing that some extraneous device in the lab is more likely to put out noise at an integer frequency than a random decimal frequency. Maybe I can buy that, but where do the harmonics come into this? If you chop at 120 Hz, wouldn't you be more worried about some noise at 120 Hz rather than a harmonic of 60? The lock-in is designed to filter out harmonics as well, so if I understand you, you're saying the 60 Hz outlet puts out some noise at 120 Hz, before ever getting to the lock in. The outlet seems like an extreme example - I would think most devices don't put out such well defined noise spikes. In that case whatever fundamental noise source was present at your chopping frequency would dominate over harmonics, and the prime number argument shouldn't matter at all.

16. Oct 9, 2011

### cepheid

Staff Emeritus
I grasp the conceptual idea, if not necessarily all of the mathematical details, which I'd like to sit down and work through myself. Question -- do you set the reference frequency to match the chopping frequency? Do you want the low pass to have a very narrow bandwidth in order to isolate frequency components near ωr and get rid of the rest?

Also, on other subject of discussion in this thread, maybe a possible rephrasing of johng23's question (and I am wondering about this as well), is, why is it assumed that the output from a source at frequency ω will also preferentially contain higher order harmonics of that frequency? Is it just because it is periodic at that fundamental frequency, but not necessarily perfectly sinusoidal?

17. Oct 9, 2011

### G01

The amplitudes of 60Hz "hum" harmonics is not small by any means:

http://en.wikipedia.org/wiki/Mains_hum

http://en.wikipedia.org/wiki/File:Mains_hum_spectrum.png

60Hz will have a much larger contribution to the overall noise than any intrinsic noise source. (See below.)

Yes, I am. Perhaps the outlets are a bit extreme, but remember that any electronic gives off 60 Hz hum. My lab, as an example, has 3 computer towers, two monitors, two laser control boxes and a refrigeration/pumping unit to pipe cooling water into our pump laser. All of these devices are always on and flooding the room with a 60Hz background. Also remember that this is a laser lab and we are talking about the experiments that involve measuring the light in a laser beam. So, all the ambient lighting in the room, powered by (you guessed it) the 60 AC coming from the wall, is going to be picked up along with the measurement signal.

When I say my lab is flooded with 60Hz, I mean that almost literally.

See the graph from wiki above. 60Hz noise has very well defined frequency spikes.

You are severely underestimating the amount of 60Hz and it's harmonics present in an average lab.

Let's think about what types of "fundamental" noise we are dealing with. Well there is: 1/f noise (which some people call "pink noise"), Johnson noise, and shot noise to name a few.

Shot noise is only a problem in really small signals. It appears when the finite numbers of particles that make up your signal (electrons or photons) is so small that the random nature of the detection process becomes important. Essentially, you start seeing uncertainties resulting from the Poisson distribution that describes your random measurement "events." As you'd expect, this a ridiculously tiny effect. It has a much smaller power contribution than the 60Hz background.

1/f noise is intrinsic to all electronic devices. It's power spectrum falls off with the inverse of the frequency. Once the 60 Hz noise is gone, you may have to worry about it, but again, 60Hz and it's harmonics are dominant.

Johnson noise results from thermal agitation of charge carriers and for frequencies lower than a few THz it's power spectrum is linearly proportional to frequency, the resistance of your measuring device, Temperature, and Boltzman's constant. The presence of Boltzman's constant essentially ensures this contribution to the noise power is no where even close to 60Hz.

The point is that 60Hz isn't really noise in the sense that the other sources are. It is an actual "signal" human beings are generating everywhere (we just don't care about it). Thus, it's contribution to the measurement can be relatively huge. Once we get rid of it, we can start worrying about the relatively small fundamental noise sources described above. They aren't even on the same scale as the 60Hz background.

Yes, that is the idea.

Ideally, I would think so, but I'd have to check to see if this is actually the case in a real Lock-in amplifier.

I think this question is best answered by actually looking at the actual spectrum of 60Hz noise from wiki that I posted above. You can see there are lot's of higher order contributions, and the 60Hz background hum is not as simple as it's name suggests.

Last edited: Oct 9, 2011
18. Oct 10, 2011

### johng23

I'm not trying to be pedantic, just trying to understand you. So it's not really the case that the chopping frequency needs to be prime, it just needs to be some frequency that isn't heavily influenced by the 60 Hz hum. I suppose choosing a prime number is a decent way to meet this criterion, but your original post made it seem like there was something intrinsically special about prime frequencies, which still doesn't seem to be the case. And looking at the graph, there are clearly prime frequencies within the broad band around 60 Hz.

So what is the best frequency for a 1 kHz laser? Anything other than a clean subharmonic, and you end up with strange effects due to spatially chopping the beam. Seems like that could be worse than dealing with some 60 Hz hum.

19. Oct 10, 2011

### Redbelly98

Staff Emeritus
Well, you don't actually set the reference frequency by "dialing it in" on the lock-in. There would be a problem if the chopper were to slow down or speed up, even slightly, and drift from the set reference frequency.

Instead, lock-ins have an input that takes a square wave, and it uses the frequency of the square wave as the reference frequency. You can use part of the beam that has passed through the chopper to generate the square wave, insuring that the reference frequency exactly matches the frequency of the chopped beam.

20. Oct 10, 2011

### AlephZero

I'm happy with G01's description of the problem in terms of sidebands, but I don't "get" this comment at all.

Certainly there are good reasons why water propellors usually have a small number of blades, reasons to do with rotordynamics would exclude 2, and to as lesser degree higher even numbers. So in practice 3 or 5 blades are common choices, though 7 blade props have also been used for subs.

Do you have a reference for the "no subharmonic noise" assertion? AFAIK the main source of water prop noise is unwanted cavitation, not subharmonics of anything.

21. Oct 10, 2011

### G01

Yes. Ideally you'd want to choose a reference, prime or otherwise, far away from 60Hz to avoid that broadband noise. Choosing a prime is then just a good way to ensure that your not landing on one of the higher frequency narrowband harmonics in the 60Hz hum spectrum.

There's nothing "intrinsically special" going on here. As berkman said, it's just a "cool trick." Nothing more.

Depends on a number of factors. The higher your chopping frequency is, the less 1/f noise you will pick up, etc. So, the higher the better for that reason. If you have other amplifiers in your system, that will affect the ideal chopping frequency. For instance, the current preamplifier in my experiment has a large roll off in gain with frequency, so after a certain point, the loss in signal due to this roll off is no longer compensated by the roll off of the 1/f noise, and my signal to noise ratio goes down with increased chopping frequency.

Also, your upper limit for chopping/reference frequency will be limited by the maximum mechanical speed of your optical chopper (unless you have some way to chop electronically).

I know the lock-in I use can take reference waves on the order of 1KHz, so if you have a pulsed laser with a rep rate in the KHz, you could potentially use the pulse of the laser as your "chop." There may be other reasons why this is a bad idea, though. I'm not sure, as I haven't really thought about it.

My laser has a rep rate in the MHz, which is way too high for a lock-in. The lock-in essentially sees a continuous beam. I need to chop externally.

Could you elaborate on what you mean?

Last edited: Oct 10, 2011
22. Oct 10, 2011

### Andy Resnick

You are correct, cavitation is a major noise source. There are many design elements that go into submarine propellers (Table 1):

The authors present a 8-blade here.

I've also noticed that many hubcaps on cars have a prime number of struts- 3, 5, and 7 are very common, although there are many 6- and 8-spoke designs.

23. Oct 11, 2011

### johng23

Yeah that was unclear. I'm doing pump probe experiments, so you want to chop only one of your beams to minimize background (either scatter from the pump, or the signal generated by the probe alone). The pulses are <100 fs, so the only way to modulate the beam at something other than a subharmonic is to clip it spatially every few pulses. This affects the overlap between pump and probe, so it's not a very clean thing to do.

Anyway, I'm not sure how significant the mains hum will be when you are using the chopper to eliminate a huge laser background.

24. Jul 12, 2012

### tufnatufna

I have a question related to optical chopper, too. Why is the chopper used in dispersed infrared spectrometer?

25. Jul 13, 2012

### Claude Bile

Choppers are a form of modulation for optical signals.

Modulation is awesome if you are trying to detect a signal in the presence of noise, because you can synchronise your detector to the chopper frequency.

This is a layer of insurance insofar as you can be sure that if you modulate something at frequency X, the signal you detect at frequency X can be reasonably attributed to your source.

Long story short: Chopping your signal allows you to differentiate it from noise.