0
\$\begingroup\$

I was told to try out a modulated optical sensor as a better and more "robust" option than the usual (unmodulated) optical sensors, saying the modulated ones are less subject to interference. Conceptually, I don't know what that means or why that is.

My understanding of modulation is to shift a signal of lower frequency spectrum to a higher frequency band, such as in the case of FM radio where a 20-20kHz audio signal is modulated by ~100MHz carrier which allows the signal to be transmitted wirelessly, and to be demodulated and reconstructed with minimal loss at the receiver side.

The visible light spectrum already has very high frequency, from approximately 480THz to 750THz. If we modulate these lights with an even higher frequency wave, the carrier wave will practically be in the ultra-violet or even in the micro-wave range - but that can't be true, since the light coming out of a modulated optical sensor is still visible and has the same color. So the frequency of the LED light has not changed.

If the modulation means PWM, where the light is emitted in pulses of lower frequency, now not only the light itself will be interfered with by other THz waves, the pulses of lower frequency will also be subject to interference by signals of lower frequency, such as radio. So the PWM should increase the chance of interference rather making it more immune to interference. If there isn't for communication purposes where information can be encoded in the pulses, we should avoid PWM at all.

Perhaps there is a different meaning to modulation, or my understanding is incorrect. Why is a modulated optical sensor more immune to interference?

\$\endgroup\$
1
  • \$\begingroup\$ This whole idea is already used widely in emergency vehicle access to street light control, so that they can change the lights as they approach an intersection. Just get with it. These systems use crystal controlled narrow-band filters and it works just fine. The answers you have below already are good enough on the topic. \$\endgroup\$
    – jonk
    Commented Aug 25, 2021 at 6:51

4 Answers 4

2
\$\begingroup\$

Your basic problem is that you have an odd image of modulation.

Take your description of a frequency modulation scheme

... the case of FM radio where a 20-20kHz audio signal is modulated by ~100MHz carrier which allows the signal to be transmitted wirelessly ...

That is backwards. You modulate the 100 MHz carrier with the low frequency signal. You seem to assume that the modulating signal must be at a higher frequency than the base signal.

I recommend you read a bit more about modulation, perhaps starting with something simple like amplitude modulation.

The datasheet for the modulated optical sensor you linked to describes the modulation it uses:

How it works: Instead of being on continuously like our other optical sensors, the LED light source in the ROSM is modulated on and off at a high frequency. That high frequency light signal is reflected back at the sensor for processing while all other sources of light are filtered out and/or ignored.

This is a form of amplitude modulation. The transmitter turns the light on and off many thousand times per second. The receiver sees the light and records the intensity. The intensity of the received light pulses many thousand times per second.

This is often used in infrared remote controls. The transmitter turns the infrared LED on and off at 35kHz (35000 times per second.) The receiver detects the intensity of the infrared light. When the infrared detector "sees" the transmitter, a stream of pulses at 35kHz comes out of it. The receiver itself is only interested in the 35kHz pulses. If 35kHz is there then the transmitter is on, else it is off. The transmitter pulses the 35kHz signal to convey the keypress codes. The signal is doubly modulated, if you will. Keypress information onto 35kHz, 35kHz onto infrared.

The modulated sensors are similar.

The transmitter modulates (pulses) the LED with a 35kHz signal. The detector "sees" the pulses, and has an output that says "35kHz is present" or "35kHz is not present."

If you reflect the transmitter signal back to the receiver, then the output will signal "35kHz present."

The interesting bit is that you now aim the transmitter at a rotating object that has a small mirror or white spot on it. When the white spot comes around, it reflects the transmitted signal back to the receiver. As it rotates, it sometimes reflects the signal back and sometimes doesn't.

The output of the sensor changes state as the object rotates. It changes constantly from "35kHz present" to "35kHz not present" as the white spot passes by.

Again, it is doubly modulated. The light is pulsed at 35kHz, then what the receiver "sees" is pulsed by the rotating object.

There are many sources of light that the detector can see. It can't tell light from the LED from light from a light bulb from light from the sun. They all contain red light.

Pulsing the light from the LED makes it distinct. You are no longer looking for just the presence of red light. You are looking for pulses of red light at a specific frequency. You can filter the output of the detector so that only that one particular frequency gets through.

Natural sources of light don't turn on and off thousands of times a second. Light bulbs flicker some, but at a relatively low frequency that is quite distinct from the higher frequency used to pulse the sensor LED.

Pulsing the transmitter signal allows the receiver to ignore all other sources of light and "see" only its own transmitter.


Your sensor may or may not really be using 35kHz. It may be using a higher pulse rate.

Then again, it might just be using 35kHz. That's a commonly used frequency for remote controls and such, so that detector ICs would be readily available. 35kHz translates to 583 pulses per revolution at 250000 RPM, so it ought to be quite detectable.

\$\endgroup\$
1
  • \$\begingroup\$ ah that clears up some of the wrong concept I had. If the receiver focuses on the 35kHz pulses and filters out rest of the frequency just to detect whether the 35kHz signal is present or not, then it really doesn't matter what color of laser the optical sensor emits. Be it red, green or violet, only the modulated 35Hz matters \$\endgroup\$
    – KMC
    Commented Aug 25, 2021 at 8:20
2
\$\begingroup\$

You're modulating with a frequency that's very much lower than several THz — probably it's in the kHz. So you don't have to worry about the sidebands being in the ultraviolet!

The principle is simple: if you use an unmodulated LED to illuminate the target, what the photosensor sees is the sum of the reflected LED light and any ambient light of the same color. If the ambient light is bright enough, then variations in the ambient light might swamp the signal that you're trying to measure.

But now imagine turning the LED on and off at 10kHz. What the photosensor sees is still the sum of the ambient light and the reflected LED light, but the reflected LED light is now a 10kHz square wave! It can measure the amplitude of that 10kHz square wave, and ignore the ambient component entirely. Unless the ambient light also varies at 10kHz (which is unlikely), it will contribute almost no noise to the system. Only the part of the return signal that correlates with the 10kHz modulation matters.

\$\endgroup\$
2
  • \$\begingroup\$ Are you implying that the receiver only cares for 10kHz signal (ON if there's a 10kHz pulses and OFF otherwise) and just ignores the intensity or the frequency of the light source? \$\endgroup\$
    – KMC
    Commented Aug 25, 2021 at 7:10
  • \$\begingroup\$ @KMC Sort of. The frequency still matters (there's an optical filter, which helps improve SNR), but the modulation bandwidth is very small compared to the filter bandwidth. And the intensity still matters, in that it has to be enough to be detected (but it can be weaker, relatively speaking, than it could have without the modulation). \$\endgroup\$
    – hobbs
    Commented Aug 25, 2021 at 13:49
1
\$\begingroup\$

Whilst light is a very high frequency, there are plenty of other sources visible to the sensor. Modulating the light source of interest means the sensor can distinguish between the correct light source and others (like sunlight). The modulation would only be a few kHz or so. The sensor will have a bandpass filter for the modulation frequency. This is a common technique - tv infrared remote controls modulate the signal at around 38kHz.

\$\endgroup\$
4
  • \$\begingroup\$ Modulate a THz light with a kHz? That would shift the THz to the kHz band and light would not be visible, but I can still see the light coming out from the sensor. If the light is visible, that has to be in the THz visible spectrum. \$\endgroup\$
    – KMC
    Commented Aug 25, 2021 at 6:40
  • 1
    \$\begingroup\$ @KMC no it wouldn't. If you modulate 500THz with 10kHz, your sidebands are at 499.99999999 THz and 500.00000001 THz. \$\endgroup\$
    – hobbs
    Commented Aug 25, 2021 at 6:44
  • \$\begingroup\$ @hobbs, shouldn't the upper band be 10kHz + 500THz and lower band 10kHz - 500THz (well into the negative frequency) instead if I modulate 500THz with 10kHz? Only if I modulate a 10kHz with 500THz do I get the narrower 499.99999999 THz to 500.00000001 THz band. \$\endgroup\$
    – KMC
    Commented Aug 25, 2021 at 6:53
  • \$\begingroup\$ @KMC no A) you've got the sense of what is modulated with what consistently backwards in your question and all of your comments, B) negative frequency is meaningless in this context, so even if you do it the other way around, you take the absolute value and get the same result anyhow. \$\endgroup\$
    – hobbs
    Commented Aug 25, 2021 at 6:56
1
\$\begingroup\$

If you simply turn a light source ON or OFF to transmit information, or block and unblock a constant light source to sense movement, then your receiver has to distinguish between the ambient light level (OFF) and the ambient light plus your low power light emitter.

Both of these conditions form your baseband signal, and leave you with the problem of distinguishing two slightly different DC levels; and a typical photodiode amplifier's noise increases at DC, making things worse.

That is difficult and prone to interference, even if you can cut out most of the ambient light.

If you modulate the emitter with a high frequency, (38kHz is common) then you merely need to detect the presence or absence of a 38kHz component in the signal from the photodiode, which is a much more robust means of detection.

Modulation is simply OOK (On/Off Keying) or 100% amplitude modulation at 38 kHz; nothing more than switching the light source on or off at that rate.

As the detector is a photodiode, it isn't a synchronous detector; its actual frequency is irrelevant. The only form of frequency selection possible (without exotics) is a coloured filter (or infra red filter) which is pretty broad, though you could use a prism as seen on a Pink Floyd album cover and move the detector to select a specific colour to watch.

Likewise (without exotics) FM is not possible, the sidebands being 38kHz from a THz signal will be indistinguishable from it, so OOK is it. (It also allows you to increase the peak power applied to the lamp, as long as the mean power including OFF period, is within its ratings.)

You can in principle modulate the intensity of the ON pulses (or their width, to modulate the transmitted power) and use that to carry a low quality analog signal, but this is rarely done. You would need AGC in the receiver to keep the mean output level constant, then variations in that level would form the analog signal.

\$\endgroup\$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.