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In radio interferometry, to get an image, the correlation of electric field observed in different antennas are Fourier transformed. This gives the "brightness function in sky coordinates".

Why do we need to Fourier transform? Does not the electric field in a direction directly give information about how the object looks? I am looking for an explanation that does not involve much mathematics.

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  • $\begingroup$ because the far field of an antenna, see, en.wikipedia.org/wiki/Fraunhofer_diffraction , is proportional to the Fourier Transform of the field in the aperture of the antenna. When reception from multiple antennas are combined coherently in the signal processor it appears as if one big antenna was used for reception, hence the improved angular resolution compared to that of a single antenna. There may also be another Fast Fourier transform just to calculate the correlation between time sequences but that is not "physics". $\endgroup$ – hyportnex Nov 29 '19 at 13:24
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Think about using a photo camera. If you simply take the photo sensor without any lens in front of it you are not getting an image on your sensor. You need to put the objective in front of the sensor to form an image. The electro-magnetic wave in the lens plane and the sensor plane are directly connected to each other by a Fourier transform. A thin lens projects the Fourier transform of the wavefield into its focus plane.

With radio waves it will be quite hard for you to build a working lens in front of your antenna array. But you have a distinct advantage over visible light, i.e. the waves are of long wavelength and slow frequency - therefore you can detect the phase of the waves, a thing that is impossible with visible light. In the consequence you can measure the full information about the radio waves and perform the function of the objective lens in your computer to obtain the image.

So in imaging you always need to perform a Fourier transform, depending on the problem it is either done optically (with a lens) or digitally on a computer.

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