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


1 Answer 1


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 to perform a deconvolution on $I^D(x,y)$ and try to recover $I(x,y)$ as best as we can.

This ALMA talk discusses everything I've said above and goes into some examples, as well as image weighting methods and CLEAN.

$^{\dagger}$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.

  • $\begingroup$ 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! $\endgroup$
    – Astroturf
    Jul 15, 2021 at 15:44
  • $\begingroup$ @Astroturf I'm glad to hear that! $\endgroup$
    – HDE 226868
    Jul 15, 2021 at 16:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .