The Need for Deconvolution

Clearly, there has to be a better way than just reweighting the data to make the dirty beam look better, (and fatter, incidentally, since one is suppressing high spatial frequencies), But this better way has to play the dangerous game of interpolating (for short spacings and for gaps in the $u-v$ plane) and extrapolating (for values beyond the largest baseline) the visibility function which was actually measured. The standard terminology is that the imaging problem is ``underdetermined'' or ``ill-posed'' or ``ill-conditioned''. It has fewer equations than unknowns. However respectable we try to make it sound by this terminology, we are no better than someone solving $x+y=1$ for both $x$ and $y$!. Clearly, some additional criterion which selects one (or a few) solutions out of the infinite number possible has to be used. The standard terminology for this criterion is ``a priori information''. The term ``a priori'' was used by the philosopher Kant to describe things in the mind that did not seem to need sensory input, and is hence particularly appropriate here.

One general statement can be made. If one finds more than one solution to a given deconvolution problem fitting a given data set, then subtracting any two solutions should give a function whose visibility has to vanish everywhere on the data set. Such a brightness distribution, which contains only unmeasured spatial frequencies, is appropriately called an ``invisible distribution''. Our extra- /inter- polation problem consists in finding the right invisible distribution to add to the visible one!

One constraint often mentioned is the positivity of the brightness of each pixel. To see how powerful this can be, take a sky with just one point source at the field centre. The total flux and two visibilities on baselines $(D/2,0),(0,D/2)$ suffice to pin down the map completely. The only possible value for all the remaining visibilities is equal to these numbers, which are themselves equal. One cannot add any invisible distribution to this because it is bound to go negative somewhere in the vast empty spaces around our source. But this is an extreme case. The power of positivity diminishes as the field gets filled with emisssion.

Another interesting case is when the emission is known to be confined to a window in the map plane. Define a function $w(l,m)=1$ inside the window and zero outside. Let $\tilde w(u,v)$ be its Fourier transform. Multiplying the map by $w$ makes no difference. In Fourier space, this condition is quite non-trivial, viz $V(u,v)=V(u,v)*\tilde w(u,v)$. Notice how the convolution on the right transfers information from measured to unmeasured parts of the $u-v$ plane, and couples them.