Using phase difference to find the angle of a signal

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I want to locate an infrared signal using Angle on arrival (AoA), I have elected to use Phase Interferometry to achieve this, I am however struggling to understand how the phase difference (∆ϑ) is found. Can someone explain how I could find this?
 

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  • #2
BvU
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Hello @Euan12345 , :welcome: !

Is this homework ? PF guidelines require your attempt at solution before you can get assistance.

If it is or if it isn't: a few words of context are definitely needed here: there is a source at the top left and there are two receivers, distance ##s## apart that report a phase.

In the drawing, ##\lambda\Delta \theta\over 2\pi## can only be a phase difference if the two ##\theta## are exactly equal.

In other words, it works only if ##\Delta\theta << \theta##.
 
  • #3
berkeman
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an infrared signal using Angle on arrival (AoA)
Also, is this a coherent light source (laser)? What kind of polarization characteristics does it have?
 
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This isn't homework. added some definitions, the sources is a point source so no it isn't coherent. (also, when I said IR signal that was wrong. will probable be using UHF or VHF waves, with wavelength 0.1 m to 1m in length)
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  • #5
berkeman
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the sources is a point source so no it isn't coherent. (also, when I said IR signal that was wrong. will probable be using UHF or VHF waves, with wavelength 0.1 m to 1m in length)
An RF antenna radiates a "coherent" waveform (all of the same phase). What is the spacing between the receive antennas compared to the wavelength? If the spacing is small compared to the wavelength (not too small), you just use a mixer to determine the phase between the two received waveforms.
 
  • #6
Baluncore
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This is UHF radio direction finding 101.
The way to find the phase difference is to employ two identical receiver channels that share the first local oscillator. That way the phase of the IF, (or the baseband, if direct down conversion), will maintain the phase of the RF signals.
There are a few other possibilities, such as doppler DF.
1. Who operates the transmitter?
2. How is the signal modulated?
3. What is the expected signal strength?
4. How accurately do you need the angular result?
5. Is there any interference?
6. Why do you want to do it?
 
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An RF antenna radiates a "coherent" waveform (all of the same phase). What is the spacing between the receive antennas compared to the wavelength? If the spacing is small compared to the wavelength (not too small), you just use a mixer to determine the phase between the two received waveforms.
I don't know the spacing (S), will be less than half the wavelength however. so less than 50cm. What do you mean by mixer. Sorry my knowledge is lacking in this area.
 
  • #8
berkeman
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What do you mean by mixer. Sorry my knowledge is lacking in this area.
A mixer is a multipler for the two signals. If they are the same frequency, you get a DC component out of that multiplicaiton. The magnitude of that DC component depends on the phase -- if they are in phase you get "100%" and if they are 180 degrees out of phase you get "-100%".

Do you really want to build this (fun project!), or is this a paper exercise?

https://www.minicircuits.com/WebStore/Mixers.html

https://www.electronics-notes.com/articles/radio/rf-mixer/theory-mathematics-maths.php
 
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This is VHF radio direction finding 101.
The way to find the phase difference is to employ two identical receiver channels that share the first local oscillator. That way the phase of the IF, (or the baseband, if direct down conversion), will maintain the phase of the RF signals.
There are a few other possibilities, such as doppler DF.
1. What is the expected signal strength?
2. How accurately do you need the angular result?
3. Is there any interference?
The signal strength will be weak don't know exactly. The accuracy will need to be +/-0.5 degrees. I was going to repeat it multiple times then workout an average. There will be some interference but it isn't constant.
 
  • #10
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A mixer is a multipler for the two signals. If they are the same frequency, you get a DC component out of that multiplicaiton. The magnitude of that DC component depends on the phase -- if they are in phase you get "100%" and if they are out of phase you get "-100%".

Do you really want to build this (fun project!), or is this a paper exercise?

https://www.minicircuits.com/WebStore/Mixers.html

https://www.electronics-notes.com/articles/radio/rf-mixer/theory-mathematics-maths.php
thanks for your help, I will build it as part of a larger project
 
  • #11
hutchphd
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If precision becomes an issue adding more elements to the receiving array may be the most cost effective solution for angular acuity.
This is indeed a very useful exercise.
 
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  • #12
tech99
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The only problem you seem to have is how to measure a phase difference between two UHF signals.
One way is to display the two signals on a dual beam oscilloscope and measure the phase difference. This requires an oscilloscope able to work at UHF. Alternatively you can rotate the entire array for max signal, then the phase difference is zero.

No doubt a Vector Network Analyser (now cheaply available) would also give the phase difference and is the modern approach. You need to be very careful about reflected signals from the ground or objects in the room. One way to lessen reflections is to work outdoors and have the array on the ground looking straight up. Then you can position a UHF transmitter, like a walkie talkie, two metres or so above it. Somehow keep the operator well away.
 
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  • #13
tech99
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It just occured to me that the simplest experiment would be to use sound waves. If you had two microphones as the detectors and a small loud speaker as the source, that would work very easily with a simple oscilloscope. I would suggest a frequency of about 1 kHz.
 
  • #14
boneh3ad
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You should take a look at the cross-spectral density between two time series. Alternatively, you can use cross-correlation, though that is mostly appropriate if you have a single frequency of interest. Ultimately, both approaches would be mathematically equivalent, as they are Fourier transform pairs per the Wiener–Khinchin theorem, but their ease of use means they are each best suited for specific types of applications.

And yes, you'll definitely want to keep an eye on sensor spacing. If you get larger than ##\lambda/2## you end up with what is essentially spatial aliasing. You can somewhat overcome this in post-processing, but only if you already know some basic wave properties a priori.
 
  • #15
Baluncore
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I am however struggling to understand how the phase difference (∆ϑ) is found. Can someone explain how I could find this?
This is not a simple problem. You are embarking on a lifelong interest in amateur radio, or a career in RDF and signals intelligence.

1. Two identical antennas, spaced less than a few wavelengths apart, will mutually couple. That coupling will vary across the band, and with the direction of signal arrival.

2. Without some special antenna system, in a controlled antenna environment, the best you will do is about ±2°. That will be because other signal paths and local reflections will sum energy with unknown phase into the signal paths. To significantly improve on the ±2° accuracy, a large and expensive interferometer array, with digital signal processing is needed. The DSP would resolve the different incoming signals into a 2D picture like a side-scan radar image, or radio astronomy VLBI.

3. With small signals, you must down-convert the two UHF signals to a lower frequency using identical mixers, with a single local oscillator. I would probably synchronously digitise the two low-frequency signals, take the FFT of each, then compare the phase of the two signals in each channel of the spectrum.

4. You will need to calibrate the system. A test signal from a very small loop antenna between the antennas, is required to create a phase correction table. Any narrow band selectivity will cause wandering phase errors. Avoid selective narrow-band analogue IF channels if possible. If you need narrow band filters use identical DSP filters.

5. Software defined radios with two identical channels are available that will make it easier.
 
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  • #16
boneh3ad
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This is not a simple problem. You are embarking on a lifelong interest in amateur radio, or a career in RDF and signals intelligence.
No issues with anything else in here, but I'd like to point out that signals processing tasks like this have much broader application than just amateur radio and SIGINT. It's a deep and fascinating rabbit hole, for sure.

My go-to bible on signals processing are the two books by Bendat and Piersol:
[1] Bendat, J. S., & Piersol, A. G. (2010). Random Data (4th ed.). John Wiley & Sons, Inc. https://doi.org/10.1002/9781118032428
[2] Bendat, J. S., & Piersol, A. G. (1993). Engineering Applications of Correlation and Data Analysis (2nd ed.). Wiley Interscience.
 
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  • #17
sophiecentaur
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Measuring the phase between two signals:
Whatever the frequency of the signals, of they are 'continuous tones', i.e. the same frequency, there is an easy method to display their phases (AKA, by inference, distance), using Lissajous Figures on a basic oscilloscope. This wiki article tells you all you need to know and there is a great diagram half way down that shows how the Lissajous display tells you the relative phases (and amplitudes). You don't need a high frequency scope - just a single, stable source with which you mix the two signals (beat them down) to produce a pair of signals with (equal) manageable frequencies. We're talking in terms of a technology that was available before WW2, even. There's actually very little difference, in principle, between that and a high end network vector analyser (millimetre accuracy in that case).

At the other end of things:
It would have helped if you had initially given us an idea of the actual distances involved and the required accuracy. GPS is fantastically good at giving relative distances with a precision of less than a metre. GPS can be used to plot the progress of a walker in wilderness areas or around the garden. Absolute accuracy may not be good but differential accuracy is excellent with off the shelf handheld Garmin (others exist) devices (around 100 quid will get you a good one). Smart phones can do the same job and they have apps with all sorts of built in calculations.
 
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