<p>You'd be completely right that a single Fourier transform would be needed <em>if</em> the interferometer were able to sample the entire <span class="math-container">$(u,v)$</span> plane. Unfortunately, that's not the case; we only have a fairly small, finite number of dishes and a finite amount of time. Coverage of the plane will increase as time passes and the Earth rotates, but it's going to be imperfect. Therefore, we don't observe the true visibility function, <span class="math-container">$V(u,v)$</span>, but rather the visibility function multiplied by the sampling function <span class="math-container">$S(u,v)$</span>:
<span class="math-container">$$V_{\text{measured}}(u,v)=S(u,v)V(u,v)$$</span>
where <span class="math-container">$S(u,v)$</span> is 1 if the point <span class="math-container">$(u,v)$</span> is sampled and 0 otherwise. Therefore, if we Fourier transform this quantity into image space, we won't get the true image <span class="math-container">$I(x,y)$</span> but rather another quantity <span class="math-container">$I^{D}(x,y)$</span>, which is what we call the dirty image. We can also write this as a convolution:
<span class="math-container">$$I^D(x,y)=B(x,y)*I(x,y)$$</span>
where <span class="math-container">$B(x,y)$</span> is the dirty beam, the inverse Fourier transform of <span class="math-container">$S(u,v)$</span>, and <span class="math-container">$I(x,y)$</span> is the true sky brightness.<span class="math-container">$^{\dagger}$</span></p>
<p>In short: if we were able to sample the <span class="math-container">$(u,v)$</span> plane perfectly - meaning that <span class="math-container">$S(u,v)=1$</span> everywhere in the <span class="math-container">$(u,v)$</span> plane - then this wouldn't be an issue. But our coverage will always be incomplete, so we have to take into account the sampling function and the dirty beam, and then use an algorithm like <a href="https://archive.lib.msu.edu/crcmath/math/math/c/c368.htm" rel="nofollow noreferrer">CLEAN</a> to perform a deconvolution on <span class="math-container">$I^D(x,y)$</span> and try to recover <span class="math-container">$I(x,y)$</span> as best as we can.</p>
<p><a href="http://www.alma.inaf.it/images/Imaging_feb16.pdf" rel="nofollow noreferrer">This ALMA talk</a> discusses everything I've said above and goes into some examples, as well as image weighting methods and CLEAN.</p>
<hr />
<p><span class="math-container">$^{\dagger}$</span>Minor note: In the impossible (but ideal) case where we sample the entire <span class="math-container">$(u,v)$</span> plane, the dirty beam should be a two-dimensional delta function, since <span class="math-container">$S(u,v)=1$</span> and the Fourier transform of a delta function yields 1. Then, since the convolution of any function with a delta function yields the original function, we have that
<span class="math-container">$$I^D(x,y)=\delta(x,y)*I(x,y)=I(x,y)$$</span>
as we would expect; perfect sampling means the dirty image is the same as the true sky image.</p>

CC BY-SA 4.0
2021-07-15T15:44:14.073Z
I just wanted to comment that this explanation has been tremendously helpful. There were some points that were subtle to me that I was unable to reconcile, so thanks again!
Astroturf
CC BY-SA 4.0
2021-07-15T16:26:06.937Z
@Astroturf I'm glad to hear that!
HDE 226868
CC BY-SA 4.0
2021-07-14T21:00:29.150Z
2021-07-15T00:09:53.557Z
2021-07-15T00:09:53.557Z
HDE 226868
<p>In radio astronomy, the dirty image is equal to the inverse Fourier transform of the uv-plane coverage times the sampled visibilities. The dirty image is also equal to the convolution of the &quot;dirty beam&quot; or PSF with the true image, so one needs to perform the dirty beam deconvolution to reveal the true image of the sky. One of the most famous equations in radio astronomy connects the visibility function to the sky brightness distribution directly. If this is the case, why are the intermediate (dirty beam and dirty image) steps needed to find recover the true image? Is the true image not the sky brightness distribution or am I missing something?</p>
<p><a href="https://science.nrao.edu/facilities/alma/naasc-workshops/alma_dr/Braatz_Imaging2.pdf" rel="noreferrer">Referenced equations on 3rd page of radio astronomy lecture from NRAO</a></p>

<observational-astronomy><telescope><radio-astronomy><radio-telescope>
If the visibility function is the Fourier transform of the sky brightness distribution, why do you need the dirty beam and dirty image to find it?
Astroturf
# If the visibility function is the Fourier transform of the sky brightness distribution, why do you need the dirty beam and dirty image to find it?

(Asked by: username_0 on 2021-07-14)

In radio astronomy, the dirty image is equal to the inverse Fourier transform of the uv-plane coverage times the sampled visibilities. The dirty image is also equal to the convolution of the "dirty beam" or PSF with the true image, so one needs to perform the dirty beam deconvolution to reveal the true image of the sky. One of the most famous equations in radio astronomy connects the visibility function to the sky brightness distribution directly. If this is the case, why are the intermediate (dirty beam and dirty image) steps needed to find recover the true image? Is the true image not the sky brightness distribution or am I missing something?

[Referenced equations on 3rd page of radio astronomy lecture from NRAO](https://science.nrao.edu/facilities/alma/naasc-workshops/alma_dr/Braatz_Imaging2.pdf)

**Tags:** observational-astronomy, telescope, radio-astronomy, radio-telescope


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## 1 answers

### Answer by username_1 on 2021-07-14. Score: 3 | Accepted answer

You'd be completely right that a single Fourier transform would be needed *if* the interferometer were able to sample the entire $(u,v)$ plane. Unfortunately, that's not the case; we only have a fairly small, finite number of dishes and a finite amount of time. Coverage of the plane will increase as time passes and the Earth rotates, but it's going to be imperfect. Therefore, we don't observe the true visibility function, $V(u,v)$, but rather the visibility function multiplied by the sampling function $S(u,v)$:
$$V\_{\text{measured}}(u,v)=S(u,v)V(u,v)$$
where $S(u,v)$ is 1 if the point $(u,v)$ is sampled and 0 otherwise. Therefore, if we Fourier transform this quantity into image space, we won't get the true image $I(x,y)$ but rather another quantity $I^{D}(x,y)$, which is what we call the dirty image. We can also write this as a convolution:
$$I^D(x,y)=B(x,y)\*I(x,y)$$
where $B(x,y)$ is the dirty beam, the inverse Fourier transform of $S(u,v)$, and $I(x,y)$ is the true sky brightness.$^{\dagger}$

In short: if we were able to sample the $(u,v)$ plane perfectly - meaning that $S(u,v)=1$ everywhere in the $(u,v)$ plane - then this wouldn't be an issue. But our coverage will always be incomplete, so we have to take into account the sampling function and the dirty beam, and then use an algorithm like [CLEAN](https://archive.lib.msu.edu/crcmath/math/math/c/c368.htm) to perform a deconvolution on $I^D(x,y)$ and try to recover $I(x,y)$ as best as we can.

[This ALMA talk](http://www.alma.inaf.it/images/Imaging_feb16.pdf) discusses everything I've said above and goes into some examples, as well as image weighting methods and CLEAN.

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$^{\dagger}$Minor note: In the impossible (but ideal) case where we sample the entire $(u,v)$ plane, the dirty beam should be a two-dimensional delta function, since $S(u,v)=1$ and the Fourier transform of a delta function yields 1. Then, since the convolution of any function with a delta function yields the original function, we have that
$$I^D(x,y)=\delta(x,y)\*I(x,y)=I(x,y)$$
as we would expect; perfect sampling means the dirty image is the same as the true sky image.

> **Comments:**
> **username_0**: I just wanted to comment that this explanation has been tremendously helpful. There were some points that were subtle to me that I was unable to reconcile, so thanks again!  
> **username_1**: @username_0 I'm glad to hear that!  

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