17
\$\begingroup\$

Two sound frequencies differing in only a few hertz can create ‘beats’ and this has been popularized in music designed to produce ‘binaural beats’. Can two lasers be tuned in a similar fashion so that a radio frequency representing the ‘beats’ is created where the light beams overlap/interfere?

\$\endgroup\$
2

5 Answers 5

14
\$\begingroup\$

Yes, however you would need a nonlinear optical medium to get mixing such that sum and difference frequencies are generated.

In a linear medium like air or vacuum, adding and subtracting does not generate sum and difference frequencies.

\$\endgroup\$
5
  • 2
    \$\begingroup\$ Since a photodiode is a "nonlinear" detector (it responds to optical power), then two incident beams will mix, producing sum & difference frequency. However, they should have very similar wavelengths so that the difference frequency is detectable at photodiode pins. \$\endgroup\$
    – glen_geek
    Commented Apr 18 at 12:34
  • \$\begingroup\$ Thank you but i am still stupid. From your reply i conclude that the beat frequency would manifest as a modulation of the light intensity. Now i am at a loss about the use of the words ‘linear’ and ‘nonlinear’. But i do understand the need for a ‘transducer’ the light waves have to be absorbed by something and summed in that ‘something’ which then emits the sum. Pun alert: it is ‘sum-thing’ special. Reminds me of compton scattering somehow (sum-how?) \$\endgroup\$ Commented Apr 18 at 16:56
  • 2
    \$\begingroup\$ Here and here are a couple of references to THz generation from light. I believe the second one (photomixing) is along the lines of @glen's comment. \$\endgroup\$ Commented Apr 18 at 17:27
  • 3
    \$\begingroup\$ "nonlinear" here means that the signal (here it's voltage) that the thing (photodiode) outputs is not proportional to the electric field of the light, but instead to some power of it (squared, for a photodetector). That's why mixing two beams of light on a photodiode causes an RF beat to be detected on that photodiode's signal. We use this all the time in the lab, it's an extremely common technique to extract RF signals from optical frequencies. Search for "Optical heterodyne detection" \$\endgroup\$
    – CharlieB
    Commented Apr 20 at 12:23
  • 1
    \$\begingroup\$ You're using the optical heterodyne detection that @CharlieB describes right now: it's used in the coherent optical links that form Internet backbones. \$\endgroup\$
    – user71659
    Commented Apr 21 at 6:17
10
\$\begingroup\$

The assumption in the question is: If one wave at a high frequency \$f_1\$ and another wave at a similarly high frequency \$ f_2 = f_1 + \epsilon \$ are added together, it's possible to obtain a new signal at a much lower frequency, \$ \epsilon \$, due to constructive and destructive interference alone. Thus, it should be possible to downconvert two optical signals to a radio frequency signal, or downconvert two radio signals to a near-DC baseband signal, just by beaming two waves to the same location.

But it's not the case. Adding or summing two waves is a linear operation, Maxwell's equations are also linear in a vacuum. Thus, it's not possible to downconvert two optical or radio frequency signals to the baseband (at a much lower frequency) just by summing them together. To achieve frequency conversion to baseband, there must be a nonlinear component to create a new low frequency, either at the transmitter, the medium, or at least at the detector.

The source of the confusion is that the term "beat" is ambiguous in audio, because it has several possible related but different meanings. Depending on the frequencies of the signals and how they're detected, new "frequencies" can be created, but the definition of a "frequency" and the mechanism responsible for creating them are not always clearly explained.

This is what confused me originally in the question Why is Frequency Conversion Additive in Acoustics Beating, but Multiplicative in Heterodyning?, and I'm sure it has confused many other people as well. In this answer I'll attempt to clarify their differences.

Case 1: Two Independent Frequencies

Two sine waves at different frequencies are turned on simultaneously.

Using an ideal spectrum analyzer with a sufficiently long window size, no new frequencies can be detected. Similarly, if an (ideal and linear) resonator is used, one can find resonances only at two frequencies.

If two frequencies are close to each other, practically, only one frequency centered at the midpoint of both frequencies may be detected with a broadened spectral line due to the limited resolution bandwidth of a spectrum analyzer (ideally, the line should be infinitely thin). Similarly in resonators, a limited quality factor (Q) causes bandwidth broadening, spreading out from its center frequency.

Case 2: Periodic Amplitude Variations

If the superimposed sine wave is viewed from the time domain, one can see interference. When destructive interference occurs, the amplitude of the superimposed wave becomes a null. When constructive interference occurs, its amplitude reaches a maxima. This is especially noticeable when the frequencies of both waves are close to each other.

If we play the first wave as audio, it's perceived as a pure tone with a stable volume. If then, we play the second close-in-frequency wave as audio, now the superimposed wave is perceived still as a pure tone (the two frequencies are too close to be distinguishable), but it's now unstable with periodic volume changes.

Similarly, when both signals are fed into the antenna of a radio receiver, tuned to a close frequency, the needle of its signal strength meter (S-meter) may start toggling left and right at a low frequency. In electronics, this technique is sometimes used for frequency calibration of oscillators, called Zero Beating. In the old times, it was often used for calibrating local oscillator of a radio receiver by eyes or ears without doing any measurement. It's only rarely used these days.

Because the superimposed tone's volume varies periodically, or "beats", one can say there's a "beat frequency". Further, it can be found from the following trigonometric identity:

$$ \cos (\omega_1 t) + \cos (\omega_2 t) = 2 \cos\left[ \frac{(\omega_1 + \omega_2)}{2} t \right] \cdot \cos\left[ \frac{(\omega_1 - \omega_2)}{2} t \right] $$

One may argue that this process actually produces a new wave with frequency \$ (\omega_1 + \omega_2) / 2 \$ with an amplitude modulation of \$ (\omega_1 - \omega_2) / 2 \$. The first term is the frequency of the superimposed tone (too close-in-frequency to be heard), and the second term is beat frequency responsible for the periodic volume change we hear.

But no downconversion occurs, and no new frequencies are actually generated - at least not before its detection.

One argument is invoking the superimposed signal's waveform in the time domain: The low-frequency signal is the amplitude envelope of the superimposed wave, but the envelope of the wave is not the actual wave, and does not represent a frequency. Unless, of course, you can detect the shape of this envelope to create a new signal. But until that happens, the envelope itself is not a frequency component of the signal. The fact that both humans and S-meters can detect a slow variation of signal's volume, shows that they're indeed such kind of detectors.

Another argument is dogmatic: In the frequency domain, the signals are defined to be linear superposition of sine waves at constant amplitudes and phases, but the right hand side of the equation is a multiplication \$ a \times b\$, so it cannot be understood as two signals at two frequencies in their original forms, but only the result after applying modulation, which inevitably changes the signal's original frequencies to those on the left-hand side. Unless, of course, we have a detector at the receiver that can separate the product back into \$ a + b\$.

It's true that new frequencies can be created from the existing frequencies if you have a modulated an existing signal at \$ (\omega_1 + \omega_2) / 2 \$ using another existing signal \$ (\omega_1 - \omega_2) /2 \$ to create \$ \cos (\omega_1 t)\$ and \$ \cos (\omega_2 t) \$. But it's not correct to say that because it's possible to create \$ \cos (\omega_1 t)\$ and \$ \cos (\omega_2 t) \$ by modulation, it means that \$ (\omega_1 + \omega_2) / 2 \$ and \$ (\omega_1 - \omega_2) / 2 \$ must already exist in the spectrum.

Case 3: Heterodyning, or Frequency Conversion

When the term "beat" is used in an electronics context, it usually means "heterodyning", rather than superposition.

If two waves are combined directly into a new wave, it's just a linear sum of both waves, no new frequencies are generated. To generate new frequencies, you must somehow combine two waves using a method other than a linear sum, which requires a non-linear component called a frequency mixer, such as an analog multiplier or other kinds of modulator, often at the transmitter's side. When two signals are fed into a mixer, two frequencies are generated: a sum frequency \$ \omega_1 + \omega_2 \$, and a difference frequency \$ \omega_1 - \omega_2 \$ (unfortunately, mixer is also an ambiguous term because it also refers to an audio mixer, which only sums the tones linearly...).

$$ \cos(\omega_1 t) \cdot \cos(\omega_2 t) = {\cos[(\omega_1 - \omega_2) t] \over 2} + {\cos[(\omega_1 + \omega_2) t] \over 2} $$

(Note that the two signals right-hand side of the equation are connected with a + sign, so it indeed represents two standalone frequencies, unlike the previous case).

Another possible solution is to sum two waves linearly to create periodic amplitude variation. Then, a suitable "detector" is used to produce a new signal based on this variation, i.e. detecting the envelope of the waveform. Thus, it's also possible to achieve frequency conversion at receiver's side. For example, in radio, a diode is the simplest possible frequency mixer, and a diode-capacitor circuit is the simplest envelope detector. A diode has square-law characteristics at low voltages, making its output proportional to the power of the input signals rather than their amplitudes. This way, new frequencies are created during signal detection.

Sometimes, the non-linear component is the medium itself. For example, in optics, one can beam lights or microwaves into a non-linear material to change their frequencies. For another example, if water is excited by two acoustic signals with extremely high amplitudes (e.g. a shockwave), it becomes non-linear and behaves like a frequency mixer, generating their sums and difference frequencies.

Curiously, quantum mechanics predicts that vacuum itself is non-linear under extremely strong field strengths. When the E&M field strength is greater than the Schwinger limit (electric field greater than \$1.32 \times 10^{18} \text{ V/m} \$, or magnetic field greater than \$4.41 \times 10^{9} \text{ T} \$), Maxwell's equations become invalid and electromagnetic waves are no longer linear. But this is far beyond what's physically achievable by humans in the foreseeable future.

Case 4: Combination tones (or Tartini tones)

If the first experiment at the beginning of the answer is repeated, but now both signals are not close in frequency, a different phenomenon may occur, causing three different tones to be perceived: \$ \omega_1 \$, \$ \omega_2 \$, their difference tone \$ \omega_1 - \omega_2 \$. It's sometimes claimed that a fourth summation tone \$ \omega_1 + \omega_2 \$ can also be heard, but "usually harder to hear than difference tones, and there is doubt that they occur at all." (Handbook for Acoustic Ecology, Barry Truax).

This phenomenon occurs even when the audio sources and mediums are almost completely linear. Thus, this demonstrate that the human ear is a nonlinear detector of acoustic signal. When two audio signals are applied to an ear, the resulted tones is assumed to be nonlinear distortion generated by the inner ear. The frequency mixing process does not exist independent from human physiology. Still, it can be considered as "true" heterodyning instead of an auditory illusion, because it's believed to be caused by the real oscillations by the inner ear.

Unfortunately, because combination tones are easy to produce, they're sometimes used to simulate the effect of heterodyning for educational purposes. This is not a problem in itself, but how non-linearity is created in this case must be clarified. For example, this mistake occurred in the video The Superheterodyne Radio by Technology Connections, which failed to explain source of non-linearity, the human ear acts like the the diode detector in a radio.

Case 5: Binaural Beats

While it's easy to explain the cause of combination tones as the non-linearities in the human inner ear causes frequency mixing. But it can be demonstrated that an effect similar to combination tones still occurs, if two frequencies are applied separately, one to each ear. This is unexpectedly, because in this case, the human inner ear cannot be the source of non-linearities - frequency mixing only occurs if two signals are applied to the same detector.

Thus, it's considered as a separate phenomenon and is given the name Binaural Beats, and is considered as an auditory illusion created by the human brain.

Conclusion

Adding two frequencies together may create four possible effects in addition to the trivial two separate frequencies. Periodic Amplitude Variations does not generate new frequencies by itself, but can be created by a receiver with a suitable non-linear detector, including being perceived as a volume variation if both frequencies are close. Heterodyning is creation of new frequencies (sum and difference) by the use a non-linear frequency mixer at the source, a non-linear medium in the middle, or a non-linear detector at the receiver. When the frequencies are not close to each other, combination tones may be perceived, causing the perception of a new third summation or difference tone, as if heterodyning occurred. This effect is believed to have a physical cause due to non-linearties in the human inner ear. Finally, even when two tones are applied to each ears separately of which nonlinear mixing cannot occur mechanically. Thus, it's considered as a auditory illusion created by the human ear.

Unfortunately, all of the four effects can be described as "beats" or "beating", it's easy to confuse them as if they're same thing. It's probably even possible to create all these four effects during the same experiments...

\$\endgroup\$
6
  • \$\begingroup\$ "Likewise, when it's played as audio, no new tones are perceived. It's only perceived as an unstable tone with constant volume changes." This seems self-contradictory, and also quite easily demonstrably false. If you play a pure 440Hz tone and a pure 442Hz tone at the same time into a human ear, it will hear a (new!) 441 Hz tone modulated by a 1Hz beat. And an electronic frequency analyzer (using an analysis window of < 1 second) will register the same thing, because that's what physically happens. The waveforms are mathematically identical: \$\cos(440t)+\cos(442t)=2\cos(441t)\cos(1t)\$. [1/3] \$\endgroup\$ Commented Apr 20 at 15:28
  • \$\begingroup\$ … What's curious about binaural (that's Latin for "two-eared") beats is that humans will hear the beating 441Hz tone even if the 440Hz tone and the 442Hz tone are played into different ears. Which is surprising, since each ear has its own spiral organ that basically acts as a biomechanical frequency analyzer, and presumably the nerve signals from neither ear report any beating, yet the brain still somehow synthesizes the beat frequency anyway. But you don't mention any of that in your answer, which makes your description of "binaural beats" not only incomplete but simply wrong. [2/3] \$\endgroup\$ Commented Apr 20 at 15:28
  • \$\begingroup\$ … In particular, the video you linked demonstrates normal acoustic beats, not binaural beats: you can still hear the beating even if you listen with just one headphone. (Here's a random video I found that does demonstrate actual binaural beats.) Sorry if I sound harsh, but IMO, your answer spreads more misconceptions than it dispels. If you're going to write a long and somewhat off-topic answer to try to clear up confusion, please try to get your facts right so you don't end up creating more confusion instead. [3/3] \$\endgroup\$ Commented Apr 20 at 15:29
  • \$\begingroup\$ @IlmariKaronen First, I disagree with your definition of "frequency". The conventional definition of a spectrum is the linear superposition of sinusoids. The spectrum only contains two frequencies on the left side of the equation: cos(440t) and cos(442t). The right side of the equation, cos(441t) x cos(1t), describes the input signals (into the modulator) for generating both frequencies, but the product itself is not a frequency that actually exists at a modulator's output (sure, an electronic frequency analyzer may still detect more frequencies if the FFT window is truncated). [1/2] \$\endgroup\$ Commented Apr 20 at 23:08
  • \$\begingroup\$ @IlmariKaronen Next, it appears that I still underestimated the ambiguities of the term "beats" - it has 5 meanings, not 4. The auditory phenomena exist in two forms: two tones to the same ear, and two tones to different ears. New frequencies are perceived by brain in both cases. In the first case, it's conceptually the same as a diode demodulator: detecting the "loudness" of a waveform's envelope is necessarily a nonlinear process, creating new frequencies at the receiver side. I'll update the answer to take that into account. [2/2] \$\endgroup\$ Commented Apr 20 at 23:29
9
\$\begingroup\$

By interference (meaning linear addition) it is not possible to generate an RF (radio frequency) signal because linear addition does not generate new frequency components. However, it is possible by using some sort of "non-linear" modulator to mix two optical frequencies to generate a radio wave.

The following IEEE paper https://ieeexplore.ieee.org/document/9019867 uses a modulated optical signal at 193.4 THz and a constant optical tone at 193.1 THz (difference 300 GHz) to generate a 300 GHz radio wave signal.

To mix the two optical tones, a Photomixer is used:

Two continuous wave lasers with identical polarisation are required, the lasers with frequency ω1 and ω2 are spatially overlapped to generate a terahertz beatnote. The co-linear lasers are then used to illuminate an ultra fast semiconductor material such as GaAs. The photonic absorption and the short charge carrier lifetime results in the modulation of the conductivity at the desired terahertz frequency ωTHz = ω1 - ω2. An applied electric field allows the conductivity variation to be converted into a current which is radiated by a pair of antenna. A typical photoconductive device or 'photomixer' is made from low temperature GaAs with a patterned metalized layer which is used to form an electrode array and radiating antenna.

\$\endgroup\$
6
\$\begingroup\$

In vacuum and linear media (read: most situations), two lasers of different frequencies do linearly superimpose, meaning that the sum of two electromagnetic waves with different frequencies simply contains both original frequencies, and no mixing products, so no lower or higher frequencies than were there before.

Now, do the same in a nonlinear medium (like, glass at high powers, or some specific crystals, different oils…), and you do get the intermodulation products.

You can also do it by having a nonlinear detection method for the laser field amplitude (which you usually do!), and then you get the mixing in the detection. Which is exactly how coherent laser communication works: you mix down your received signal, which is some data-containing signal modulated onto a laser-frequency carrier, down with a reference laser in the receiver by superimposing them and feeding the sum into your detector; you get the frequency difference as IF.

\$\endgroup\$
0
\$\begingroup\$

No.

Just like with sound, the "beats" would be an RF-frequency amplitude modulation of the light, not a new signal. The amplitude of the combined signal varies as the two original signals constructively and destructively add together.

\$\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.