Laser rangefinder by phase-shift

[DariusP]

I've read that there are laser rangefinders who measure the distance not only by ToF (time-of-flight) method but also by phase-shift.

I think I can understand that. Depending how much the phase of the input pulse shifts from the output pulse we can translate it into how much distance the pulse has traveled.

But a question immediately arose: what happens if the phase shift is over 2pi?

 

[davenn]

Summary: What happens if the phase shift is over 2pi?

I've read that there are laser rangefinders who measure the distance not only by ToF (time-of-flight) method but also by phase-shift.

I work with 2 different models of high precision distance/angle measuring equip
the Trimble SPS730/930 ... accuracy of 1mm 1 sec of arc ( +- 3mm / 3sec) guaranteed over 300m
a little less out to 1km

and the Trimble S8 High Precision, 1mm/1sec ( +- 1mm/1sec) over the same range

The SPS series use a pulse system for distance measuring
The S series use the phase shift method as it is more accurate

from the manual
" the phase measurement needs multiple modulation frequencies to reach a non-ambiguous result
because the phase measurement can only measure the phase of the wave rest piece"

I'm not deeply enough into waves and phases etc to fully understand what "
the phase of the wave rest piece" is ?

There are a number of links on google if you search on that term.
There may also be others here that can clarify that term.

Suffice it to say that the " multiple modulation frequencies " is part of what it used to combat
problems like you suggested. I would suggest that different frequencies will have different phase
shifts and that there is most likely some sort of averaging done to come up with a good resultDave

 

[Ibix]

If you know the wavelength is $\lambda$ and the phase shift is $\phi$ then the distance is $(n+\phi/2\pi)\lambda$. You can't determine $n$.

But if you have two wavelengths and measure the phase shifts over the same distance, then you can assert that $(n_1+\phi_1/2\pi)\lambda_1=(n_2+\phi_2/2\pi)\lambda_2$ and find integers $n_1$ and $n_2$ that satisfy this. There will be multiple solutions. But if you have a rough idea of the distance (from a time-of-flight measure, perhaps) and more than two wavelengths then you can do this for every pair of wavelengths and find the distance which is near your rough estimate and there is a cluster of solutions for the pairs of wavelengths.

At least, that's how I'd approach the problem.

 

[berkeman]

But if you have two wavelengths and measure the phase shifts over the same distance, then you can assert that (n1+ϕ1/2π)λ1=(n2+ϕ2/2π)λ2(n1+ϕ1/2π)λ1=(n2+ϕ2/2π)λ2(n_1+\phi_1/2\pi)\lambda_1=(n_2+\phi_2/2\pi)\lambda_2 and find integers n1n1n_1 and n2n2n_2 that satisfy this. There will be multiple solutions. But if you have a rough idea of the distance (from a time-of-flight measure, perhaps) and more than two wavelengths then you can do this for every pair of wavelengths and find the distance which is near your rough estimate and there is a cluster of solutions for the pairs of wavelengths.

Very interesting. I hadn't heard about using phase shift measurements for absolute distance measurements over longer distances. I'm familiar with using laser interferometry for relative distance measurements to high precision, but this is the first I've seen about using phase in absolute measurements.

Relative distance measurements with laser interferometry:

https://www.keysight.com/en/pc-1000...ter-position-measurement-systems?cc=US&lc=eng

 

[davenn]

At least, that's how I'd approach the problem.

and as I stated ... they use multiple frequencies ... thanks for the maths backup

 

[Charles Link]

I believe the "CW" laser is being modulated at r-f (radio wave) frequencies, and the phase shift is a phase difference of the return signal in the r-f (modulation) signal. $\\$ It is not a phase shift of the laser (visible wavelength), as occurs in a Micheson interferometer. With a Michelson interferometer, you can get accuracies of plus or minus several nanometers, but it does not measure things at long range. $\\$ At least this is my educated guess of how this "phase" type laser rangerfinder most likely operates. I am fairly certain that it is not measuring a phase shift of any visible wavelength.$\\$ With an r-f modulation, you could do essentially the same thing with radio waves (or microwaves). Here the photodiode, even though it detects the visible laser, puts out an r-f signal (because of the r-f modulation) that is amplified and processed. Using an r-f modulated visible laser and a photodiode is somewhat easier than using an r-f radio transmitter and r-f receiver. $\\$ And yes, techniques that @Ibix mentioned in post 3 would work, but this would all be done with e.g. r-f modulation of multiple CW lasers, or even a single CW laser with a slightly complex r-f modulation scheme. $\\$ If simply a single r-f sinusoidal modulation frequency were used, the phase would only be known as $\phi=\phi_o+n (2 \pi)$, (with unknown $n$ ), because the received sinusoidal signal is compared to the emitted sinusoidal signal, and the only thing you get in comparing the two signals is $\phi_o$, with no information about the integer $n$.

 

[davenn]

If simply a single r-f sinusoidal modulation frequency were used, the phase would only be known as ϕ=ϕo+n(2π)ϕ=ϕo+n(2π) \phi=\phi_o+n (2 \pi) , (with unknown nn n ), because the received sinusoidal signal is compared to the emitted sinusoidal signal, and the only thing you get in comparing the two signals is ϕoϕo \phi_o , with no information about the integer nn n .

which is why I have stated twice already, they don't do it that way

 

[Charles Link]

The major point I am trying to make is that they are measuring a phase shift with frequencies around $f=10^9$ Hz (radio wave/microwave), resulting in $\Delta t \approx 10^{-10}$ seconds that are measured, with $\Delta s \approx$ 1 centimeter, rather than phase shifts with frequencies $f \approx 5 \cdot 10^{14}$ Hz (visible light) with $\Delta t \approx 10^{-15}$ seconds, which would result in a $\Delta s \approx 100$ nanometers. $\\$ The $f \approx 5 \cdot 10^{14}$ Hz and $\lambda \approx 500$ nm of the visible light really is not part of the picture here. Frequencies $f \approx 10^9$ Hz can be processed with simple electronics. (@jim hardy Please correct me if this is not the case). There is no electronics than can do anything with $f \approx 10^{14}$ Hz.

 

[DariusP]

The major point I am trying to make is that they are measuring a phase shift with frequencies around $f=10^9$ Hz (radio wave/microwave), resulting in $\Delta t \approx 10^{-10}$ seconds that are measured, with $\Delta s \approx$ 1 centimeter, rather than phase shifts with frequencies $f \approx 5 \cdot 10^{14}$ Hz (visible light) with $\Delta t \approx 10^{-15}$ seconds, which would result in a $\Delta s \approx 100$ nanometers. $\\$ The $f \approx 5 \cdot 10^{14}$ Hz and $\lambda \approx 500$ nm of the visible light really is not part of the picture here. Frequencies $f \approx 10^9$ Hz can be processed with simple electronics. (@jim hardy Please correct me if this is not the case). There is no electronics than can do anything with $f \approx 10^{14}$ Hz.

Is my thinking correct that higher modulation frequencies can give better resolution but lower maximum distance and vice-versa (lower modulation frequencies can give lower resolution but higher maximum distance)?

 

[Charles Link]

Is my thinking correct that higher modulation frequencies can give better resolution but lower maximum distance and vice-versa (lower modulation frequencies can give lower resolution but higher maximum distance)?

I think you have the basic idea that you can get higher resolution with the higher frequency. I don't know that the maximum distance is limited, but it does take a little extra effort to determine just how far away the target is with high precision (higher frequency) equipment.

 

[Charles Link]

One additional comment is as previously stated, these devices are not doing interferometry to measure the distance. For the sake of completeness, this Insights article that I authored explains how interferometry works: https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/
Interferometry is a very interesting subject, and it is used to make high precision distance measurements, but it requires precision optics including a precision (optically flat) beamsplitter. Once you understand some of the details of these interferometric measurements, you should be able to see why employing an interferometer is a much more costly undertaking than these commercially available rangefinders that do not employ these interferometric techniques.