Mapping the Radio Universe

ICMS

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Imagine an array of N radio telescopes spread over many kilometers on the ground, each telescope directed towards the same object in the sky. This array contains ^NC₂ or [N(N-1)]/2 pairs of telescopes. For, say, an array of 30 dishes, ^NC₂ equals 435 pairs. Each pair that has a different baseline vector will measure a different value of the degree of coherence, m(p,q)_.

Hence, the problem of mapping the intensity distribution of a radio source in the sky has been reduced to the task of sampling the values of the degree of coherence, over as many different baselines as possible.

The Earths rotation also helps the radio astronomer! This is because the orientation of the physical pair of telescopes will change as seen by the source. As the Earth rotates, the same pair of telescopes appears to have different baseline vectors. Thus the pair appears as different pairs to the source at different times. Hence, by simply taking observations at different times of the day, a single pair of radio telescopes can do the job of many different pairs and sample many more points in the (p,q) plane. Typically, a single pair may be used as 1000 different pairs during a days observation.

The Fast Fourier Transform

The extremely large number of pairs (in this example: 435x1000) for which the Fourier transform must be carried out makes the computation an extremely time consuming process, even on a computer. However, an elegant shortcut method of computing discrete Fourier transforms on a computer, called the Fast Fourier Transform (F.F.T.), enables such a computation to be completed within a reasonable time.

The normal algorithm has an order: O(N²). This is because the Fourier Transform of N data samples requires N² multiplications. The F.F.T. algorithm has an order: O(Nlog₂N), as the number of multiplications required is Nlog₂N.

The tremendous difference this makes when N is very large can be seen by the following example. Suppose we want to compute a transform of N = 10⁶ data samples on a 1 MHz frequency (microsecond cycle) computer. Then what would take a week of computing time using O(N²) will take only half a minute of computing time using O(Nlog₂N)!