<p>Absolutely this can work as can be seen in  <a href="https://www.mwatelescope.org/" rel="nofollow noreferrer">The Murchison Widefield Array</a> or MWA in Western Australia. This is essentially exactly how it works. You definitely do loose information (the first stage of combining antennas significantly reduces the field of view, and the number of interferometric baselines you get is reduced).</p>
<p>From <a href="https://www.mwatelescope.org/telescope" rel="nofollow noreferrer">this page</a>:</p>
<pre><code>Frequency Range            70-300 MHz
Number of Receptors        2048 dual-polarization dipoles
Number of antenna tiles    128
Number of baselines        8128
</code></pre>
<p>There are 32 receptors per tile (2 polarizations for <span class="math-container">$4x4=16$</span> elements), and <span class="math-container">$N = 128$</span> tiles. The number of baselines is <span class="math-container">$N(N-1)/2 = 8128$</span>.</p>
<p>The advantages is a reduction in the data rates, so reduced digital processing. Note the first stage would be treated as beam forming - you cannot do interferometry between the first stage and somehow do interferometry between the results. Ie the first stage is summing the voltages (with appropriate phase corrections) and the second stage would be interferometry.</p>
<hr />
<p>One of the tiles making up the 32T, a prototype instrument for <a href="https://en.wikipedia.org/wiki/Murchison_Widefield_Array#System_overview" rel="nofollow noreferrer">the Murchison Widefield Array</a> (<a href="https://en.wikipedia.org/wiki/File:MWA_32T_Tile.jpg" rel="nofollow noreferrer">source</a>)</p>
<blockquote>
<p>An MWA antenna consists of a four by four regular grid of dual-polarisation dipole elements arranged on a 4m x 4m steel mesh ground plane. Each antenna (with its 16 dipoles) is known as a &quot;tile&quot;. Signals from each dipole pass through a low noise amplifier (LNA) and are combined in an analogue beamformer to produce tile beams on the sky. Beamformers sit next to the tiles in the field. The radio frequency (RF) signals from the tile-beams are transmitted to a receiver, each receiver being able to process the signals from a group of eight tiles. Receivers therefore sit in the field, close to groups of eight tiles; cables between receivers and beamformers carry data, power and control signals. Power for the receivers is provided from a central generator.</p>
</blockquote>
<p><a href="https://i.sstatic.net/xppTT.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xppTT.jpg" alt="One of the tiles making up the 32T, a prototype instrument for the Murchison Widefield Array." /></a></p>

CC BY-SA 4.0
2021-05-27T11:02:06.027Z
Welcome to astronomy SE! Nice answer. Would you mind editing in a link to MWA, please?
B--rian
CC BY-SA 4.0
2021-05-27T17:11:38.913Z
Welcome to Stack Exchange! I've added some more/better links to MWA to better support your answer. Please feel free to edit further. *Thanks!*
uhoh
CC BY-SA 4.0
2021-05-27T10:21:39.577Z
2021-05-27T17:10:44.113Z
2021-05-27T17:10:44.113Z
Chris
<p><strong>update:</strong> See <a href="https://astronomy.stackexchange.com/a/44029/7982">@Chris' cool answer about MWA</a> for something thad does work!</p>
<hr />
<blockquote>
<p>In principle, would this work?</p>
</blockquote>
<p>I can't say &quot;no in principle it could never work&quot; but the combination in the layers loses information than it would almost certainly decrease image quality somewhat.</p>
<blockquote>
<p>Is it more efficient and/or easier than traditional methods?</p>
</blockquote>
<p>It depends what &quot;more efficient&quot; means. If the performance of your array is worse by a factor X but you saved a fraction of the money Y, is this a &quot;more efficient&quot; way to do science?</p>
<hr />
<p>In an imaging optical telescope (or any imaging system including eyes) every pixel is illuminated simultaneously and directly by all areas of the aperture. From a given point in the distance a telescope will (try to) preserve the phase of all paths reaching the pixel so that the resulting <em>intensity</em> corresponds to the incoming power. This allows the system to obtain the best resolution.</p>
<p>Once you perform the interference and measure the resulting intensity, you lose the phase information forever (in conventional systems).</p>
<p>In the same way, in a radio telescope array usually all signals from all elements are combined together in a device called a &quot;correlator&quot; which in modern radio astronomy is a digital computer. Each pixel in the final image is calculated from correlations of every possible pair of elements in the array.</p>
<p>For example, from Alma Observatory's <a href="https://www.almaobservatory.org/en/about-alma-at-first-glance/how-alma-works/technologies/the-alma-main-array-correlator/" rel="nofollow noreferrer">Correlator</a> page:</p>
<blockquote>
<p><strong>The ALMA Main Array Correlator</strong></p>
<p>To make images from millimetric wavelengths joined by multiple antennas, we need an absolutely colossal amount of computer power. <strong>Signals from each antenna pair – there are 1225 possible pairs alone in the main array of antennas (50)- should be mathematically compared billions of times per second.</strong> You would need millions of laptop computers to perform the number of operations that ALMA carries out every second! This need resulted in the construction of one of the fastest supercomputers in the world, the ALMA Correlator.</p>
<p>The Correlator, installed in the AOS Technical Building at an altitude of 5,000 meters above sea level, is the last component in the cosmic wavelength collection process. It is a very large data processing system, composed of four quadrants, each of which can process data from up to 504 antenna pairs. The complete Correlator has 2,912 printed circuits, 5,200 interface wires and over 20 million welding points. The Correlator is made up of Tunable Filter Bank (TFB) cards. The distribution requires four TFBs for data that arrives from a single antenna. These cards have been developed and optimized by the Bordeaux University in France.</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Combination#Number_of_k-combinations" rel="nofollow noreferrer">Note that</a>:</p>
<p><span class="math-container">$$1225 = \frac{50 \times 49}{2}.$$</span></p>
<p>Again, once you perform correlation, you lose the phase information.</p>
<p>If you did that in each branch of your diagram, you would never be able to do the subsequent correlations properly because you'd lose phase information along the way, and could therefore no longer perform correlation of every possible pair.</p>
<p>There may still be some lossy algorithms that allow you to do some reduced amount of imaging the way that you propose, but the point of building such a large and expensive array would be to get the maximum amount of information.</p>
<p>So in reality the signal from each element is heterodyned with a local oscillator (<a href="https://astronomy.stackexchange.com/q/28666/7982">How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?</a>) to a baseband of a few GHz, digitized (<a href="https://astronomy.stackexchange.com/q/18052/7982">Why are the ALMA receivers' ADCs only 3-bits?</a>) and then sent along a digital fiber optic cable to the main correlator computer building, with the original phase information from each dish still intact (albeit in digital form).</p>
<hr />
<p><strong>Important Caveat:</strong> However, in your diagram you <em>could imagine</em> that each of the green elements in the top layer is a &quot;patch&quot; of reflector on one dish, and the combination boxes in red (middle layer) is the collecting feed horn of one dish. So phase information from within the aperture of a single dish is indeed lost forever.</p>
<p>In that sense yes, it does work, and the resolution of an array is limited by the distances that separate the centers of each of the dishes, and not by the size of the dishes.</p>
<p>To help think more about that, see <a href="https://astronomy.stackexchange.com/q/21121/7982">How do ASKAP's focal plane phased array feeds interact with the entire array phasing?</a></p>

CC BY-SA 4.0
2020-01-01T22:24:11.727Z
Of the three currently linked question, only one has any answers posted, and for that one no answer has yet to be accepted.
uhoh
CC BY-SA 4.0
2020-01-01T23:40:06.247Z
Thanks for such a detailed answer! Just a clarification - my measure of "efficiency" here would be - would it be computationally easier for each individual correlator? I would assume that each of the red and green boxes would have an easier time interfering or combining 3 signals than 9 - am I correct in assuming this?
sforsingh
CC BY-SA 4.0
2020-01-02T00:05:33.220Z
@Calc-You-Later that sentence is not incorrect, but doing that would not let you generate a final radio image of the same quality (less=easier). It would be more efficient in terms of money alone if you build a cheaper system, but you will lose performance. If you *need* or *want* that lost performance, then it would not be more efficient because zero results divided by lots of money spent is zero efficiency. It's possible that you might be better off hurting your results by using fewer dishes than hurting your results by using localized pre-processing, at least that's my understanding.
uhoh
CC BY-SA 4.0
2020-01-02T00:07:55.860Z
@Calc-You-Later However, if your preprocessing boxes are local correlators and you pass along not only the raw data and its phase information, but also some of the local correlation results by *increasing* the bandwidth of the interconnects, that might work, but it's not any cheaper and may be more expensive. It would be basically moving some processors from the central correlators out towards the edges; you don't save anything and you introduce more complexity and more long-distance high-bandwidth connections.
uhoh
CC BY-SA 4.0
2020-01-02T00:09:28.950Z
This is all very helpful and makes sense - thank you for the responses! I'll take those factors into account
sforsingh
CC BY-SA 4.0
2020-01-02T00:10:46.430Z
@Calc-You-Later thank you for the accept but it may be too early for that; others may add additional answers. I'm no expert on the topic, let's see how other people vote and comment before you accept this answer. Thanks!
uhoh
CC BY-SA 4.0
2020-01-02T00:13:26.400Z
Ah, understood - I'll wait for other answers.
sforsingh
CC BY-SA 4.0
2020-01-01T22:13:14.850Z
2021-05-27T17:13:12.017Z
2021-05-27T17:13:12.017Z
uhoh
<p>tl;dr - Is splitting up the process of interferometry as shown in the diagram possible, and if so, is it more efficient and/or easier than traditional methods?</p>

<p>I have been doing some research into radio interferometry, and I had a question about it - I know that we can combine the signals from multiple telescopes into one "image" using interferometry, but what if I was able to do that multiple times?</p>

<p>Let me describe this a bit better. Say I have nine dishes. I arrange three of them in an equilateral triangle, and then I do the same with the other 6, and then, I arrange those three groups of three into a larger equilateral triangle. Then, I combine the signals from each three of the telescopes, and then I combine the three resulting signals. Here's a diagram that might help - the blue dishes are the "telescopes", and the red and green boxes are the "processors" or where the signals would be interfered.</p>

<p><a href="https://i.sstatic.net/HY5oX.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HY5oX.png" alt="enter image description here"></a></p>

<p>In principle, would this work? And, in the context of radio astronomy, would it be easier to combine or interfere only three signals rather than nine?</p>

<radio-astronomy><interferometry>
Would "layered" radio interferometry work?
sforsingh
# Would "layered" radio interferometry work?

(Asked by: username_0 on 2020-01-01)

tl;dr - Is splitting up the process of interferometry as shown in the diagram possible, and if so, is it more efficient and/or easier than traditional methods?

I have been doing some research into radio interferometry, and I had a question about it - I know that we can combine the signals from multiple telescopes into one "image" using interferometry, but what if I was able to do that multiple times?

Let me describe this a bit better. Say I have nine dishes. I arrange three of them in an equilateral triangle, and then I do the same with the other 6, and then, I arrange those three groups of three into a larger equilateral triangle. Then, I combine the signals from each three of the telescopes, and then I combine the three resulting signals. Here's a diagram that might help - the blue dishes are the "telescopes", and the red and green boxes are the "processors" or where the signals would be interfered.

[![enter image description here](https://i.sstatic.net/HY5oX.png)](https://i.sstatic.net/HY5oX.png)

In principle, would this work? And, in the context of radio astronomy, would it be easier to combine or interfere only three signals rather than nine?

**Tags:** radio-astronomy, interferometry


---

## 2 answers

### Answer by username_1 on 2021-05-27. Score: 2 | Accepted answer

Absolutely this can work as can be seen in  The Murchison Widefield Array or MWA in Western Australia. This is essentially exactly how it works. You definitely do loose information (the first stage of combining antennas significantly reduces the field of view, and the number of interferometric baselines you get is reduced).
From this page:
Frequency Range            70-300 MHz
Number of Receptors        2048 dual-polarization dipoles
Number of antenna tiles    128
Number of baselines        8128

There are 32 receptors per tile (2 polarizations for $4x4=16$ elements), and $N = 128$ tiles. The number of baselines is $N(N-1)/2 = 8128$.
The advantages is a reduction in the data rates, so reduced digital processing. Note the first stage would be treated as beam forming - you cannot do interferometry between the first stage and somehow do interferometry between the results. Ie the first stage is summing the voltages (with appropriate phase corrections) and the second stage would be interferometry.

One of the tiles making up the 32T, a prototype instrument for the Murchison Widefield Array (source)

An MWA antenna consists of a four by four regular grid of dual-polarisation dipole elements arranged on a 4m x 4m steel mesh ground plane. Each antenna (with its 16 dipoles) is known as a "tile". Signals from each dipole pass through a low noise amplifier (LNA) and are combined in an analogue beamformer to produce tile beams on the sky. Beamformers sit next to the tiles in the field. The radio frequency (RF) signals from the tile-beams are transmitted to a receiver, each receiver being able to process the signals from a group of eight tiles. Receivers therefore sit in the field, close to groups of eight tiles; cables between receivers and beamformers carry data, power and control signals. Power for the receivers is provided from a central generator.

> **Comments:**
> **username_2**: Welcome to astronomy SE! Nice answer. Would you mind editing in a link to MWA, please?  
> **username_3**: Welcome to Stack Exchange! I've added some more/better links to MWA to better support your answer. Please feel free to edit further. *Thanks!*  

---

### Answer by username_3 on 2020-01-01. Score: 1

**update:** See [@username_1' cool answer about MWA](https://astronomy.stackexchange.com/a/44029/7982) for something thad does work!

---

> In principle, would this work?

I can't say "no in principle it could never work" but the combination in the layers loses information than it would almost certainly decrease image quality somewhat.

> Is it more efficient and/or easier than traditional methods?

It depends what "more efficient" means. If the performance of your array is worse by a factor X but you saved a fraction of the money Y, is this a "more efficient" way to do science?

---

In an imaging optical telescope (or any imaging system including eyes) every pixel is illuminated simultaneously and directly by all areas of the aperture. From a given point in the distance a telescope will (try to) preserve the phase of all paths reaching the pixel so that the resulting *intensity* corresponds to the incoming power. This allows the system to obtain the best resolution.

Once you perform the interference and measure the resulting intensity, you lose the phase information forever (in conventional systems).

In the same way, in a radio telescope array usually all signals from all elements are combined together in a device called a "correlator" which in modern radio astronomy is a digital computer. Each pixel in the final image is calculated from correlations of every possible pair of elements in the array.

For example, from Alma Observatory's [Correlator](https://www.almaobservatory.org/en/about-alma-at-first-glance/how-alma-works/technologies/the-alma-main-array-correlator/) page:

> **The ALMA Main Array Correlator**
>
> To make images from millimetric wavelengths joined by multiple antennas, we need an absolutely colossal amount of computer power. **Signals from each antenna pair – there are 1225 possible pairs alone in the main array of antennas (50)- should be mathematically compared billions of times per second.** You would need millions of laptop computers to perform the number of operations that ALMA carries out every second! This need resulted in the construction of one of the fastest supercomputers in the world, the ALMA Correlator.
>
> The Correlator, installed in the AOS Technical Building at an altitude of 5,000 meters above sea level, is the last component in the cosmic wavelength collection process. It is a very large data processing system, composed of four quadrants, each of which can process data from up to 504 antenna pairs. The complete Correlator has 2,912 printed circuits, 5,200 interface wires and over 20 million welding points. The Correlator is made up of Tunable Filter Bank (TFB) cards. The distribution requires four TFBs for data that arrives from a single antenna. These cards have been developed and optimized by the Bordeaux University in France.

[Note that](https://en.wikipedia.org/wiki/Combination#Number_of_k-combinations):

$$1225 = \frac{50 \times 49}{2}.$$

Again, once you perform correlation, you lose the phase information.

If you did that in each branch of your diagram, you would never be able to do the subsequent correlations properly because you'd lose phase information along the way, and could therefore no longer perform correlation of every possible pair.

There may still be some lossy algorithms that allow you to do some reduced amount of imaging the way that you propose, but the point of building such a large and expensive array would be to get the maximum amount of information.

So in reality the signal from each element is heterodyned with a local oscillator ([How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?](https://astronomy.stackexchange.com/q/28666/7982)) to a baseband of a few GHz, digitized ([Why are the ALMA receivers' ADCs only 3-bits?](https://astronomy.stackexchange.com/q/18052/7982)) and then sent along a digital fiber optic cable to the main correlator computer building, with the original phase information from each dish still intact (albeit in digital form).

---

**Important Caveat:** However, in your diagram you *could imagine* that each of the green elements in the top layer is a "patch" of reflector on one dish, and the combination boxes in red (middle layer) is the collecting feed horn of one dish. So phase information from within the aperture of a single dish is indeed lost forever.

In that sense yes, it does work, and the resolution of an array is limited by the distances that separate the centers of each of the dishes, and not by the size of the dishes.

To help think more about that, see [How do ASKAP's focal plane phased array feeds interact with the entire array phasing?](https://astronomy.stackexchange.com/q/21121/7982)

> **Comments:**
> **username_3**: Of the three currently linked question, only one has any answers posted, and for that one no answer has yet to be accepted.  
> **username_0**: Thanks for such a detailed answer! Just a clarification - my measure of "efficiency" here would be - would it be computationally easier for each individual correlator? I would assume that each of the red and green boxes would have an easier time interfering or combining 3 signals than 9 - am I correct in assuming this?  
> **username_3**: @Calc-You-Later that sentence is not incorrect, but doing that would not let you generate a final radio image of the same quality (less=easier). It would be more efficient in terms of money alone if you build a cheaper system, but you will lose performance. If you *need* or *want* that lost performance, then it would not be more efficient because zero results divided by lots of money spent is zero efficiency. It's possible that you might be better off hurting your results by using fewer dishes than hurting your results by using localized pre-processing, at least that's my understanding.  
> **username_3**: @Calc-You-Later However, if your preprocessing boxes are local correlators and you pass along not only the raw data and its phase information, but also some of the local correlation results by *increasing* the bandwidth of the interconnects, that might work, but it's not any cheaper and may be more expensive. It would be basically moving some processors from the central correlators out towards the edges; you don't save anything and you introduce more complexity and more long-distance high-bandwidth connections.  
> **username_0**: This is all very helpful and makes sense - thank you for the responses! I'll take those factors into account  
> **username_3**: @Calc-You-Later thank you for the accept but it may be too early for that; others may add additional answers. I'm no expert on the topic, let's see how other people vote and comment before you accept this answer. Thanks!  
> **username_0**: Ah, understood - I'll wait for other answers.  

---

