How close is a 2D Gaussian to an Airy disk?

by omoplata
Tags: airy, disk, gaussian
omoplata is offline
Mar9-12, 08:18 AM
P: 315
So we were taking measurements for an experiment in our radio astronomy lab. For the first part of the experiment, we recorded the intensity of a far away point source ( the signal from a TV satellite was used for the point source ) detected by commercial satellite dish and receiver.

When we plotted the intensity against the angle ( we traversed the dish across the satellite, so the curve we had was a 1D cross section of the 2D intensity distribution ), it looked like a Gaussian ( ). I was actually able to fit a Gaussian to it.

But then I read about what we were supposed to get, and found that we were supposed to get an Airy disk from a point source ( ).

So I'm curious. How close is an Airy disk to a Gaussian?

Can I get a Gaussian from an Airy disk by making some parameter close to zero or infinty or something?

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Andy Resnick
Andy Resnick is offline
Mar9-12, 03:57 PM
Sci Advisor
P: 5,468
The central peak of an Airy function can probably be approximated as a Gaussian function, so the issue is why didn't you measure the secondary maxima? I can think of a few reasons, but without knowing any details it's hard to say. How big is your receiver, in terms of wavelength?
omoplata is offline
Mar9-12, 10:16 PM
P: 315
The detector ( LNBF ) that we are using has a bandpass centered at around 2.7 cm ( 11 GHz ). It's used with an off-axis parabolic satellite dish ( the width of the dish is about 1 m, if that matters ).

we didn't measure the secondary maxima ( if what you mean are the secondary peaks of the Airy disk ) specifically.

We just used a an interface instrument ( Vernier Lab Pro ) with a PC to record the voltage coming off the detector 10 times a second, and moved the satellite dish with constant speed, using a motor, so that the direction it's pointing towards traced a line across the position of the satellite in the sky.

So we were taking data of Voltage vs. time, but the dish was moving with constant angular velocity that we know, so we can convert the data to Voltage vs. angle.

If there were supposed to be secondary peaks in the plot of Voltage vs. angle ( which there should be if it was the cross section of an Airy disk ), they are not visible.

Andy Resnick
Andy Resnick is offline
Mar10-12, 08:40 AM
Sci Advisor
P: 5,468

How close is a 2D Gaussian to an Airy disk?

Here are a few things to consider:

0) if the diffraction pattern is indeed an Airy disc, how large should it be at your receiver (assuming none of the below effects)?
1) the satellite dish/detector is not sensitive to a singe direction- there is an angular acceptance angle.
2) Propagation of light through the atmosphere introduces scattering (and attenuation), blurring the diffraction pattern
3) your source could be moving during signal acquisition
4) Your measured data is the convolution of the 'ideal' signal and all these effects, similar to blurring in any other optical imaging system.

What do other members of your group think?
omoplata is offline
Mar10-12, 11:08 AM
P: 315
0) If by "how large" you mean the angular diameter of the Airy disk, looking at the wikipedia article, the intensity of the Airy disk [itex]I(\theta)[/itex] is related to the radius of the aperture [itex]a[/itex] ( in this case the radius of the satellite dish? ), using the same notation as the wikipedia article. ( I'm using the term Airy "disk" and not "function" here, because it seems different from an Airy function )
1) If detector (LBNF) has full sensitivity to an angle larger than the angle of the dish edge, I think we can take ignore the angular sensitivity distribution of the detector, because then the aperture is going to be the whole dish ( if the background behind the dish emits little radio signals of the relevent frequency) . If the sensitivity of the detector falls off with increasing angle from the central axis with such a rapid rate that it drops off before it reaches the dish edge, then I guess we have to take that in to account.
2) This is certainly an issue.
3) We were detecting a geocentric satellite. So it did not move in relation to our position.
4) So I guess considering 1) and 2), it is possible that our plot can NOT be the cross section of an Airy disk.

When I told the professor in charge of the experiment that I fitted a Gaussian, he said "Why did you fit a Gaussian? It's supposed to be an Airy disk." The reason why did not fit one yet is because the fitting software I am using right now does not have Bessel functions ( the wikipedia Airy disk equation is given in terms of Bessel functions ). I'm going to find another fitting software which has Bessel functions and fit it. So this observation might be premature, but I can't see any secondary peaks.

The other two members in our group have the same problem. How to find a fitting software that can fit an Airy disk. But we think we found one ( IDL ), and we are going to try it out.

This link lists the functions in the NASA IDL Astronomy User's Library for IDL (which is very widely used in the Astronomy community) , and even they seem to be fitting Point Spread Functions with Gaussians and not Airy disks ( functions FIND and GETPSF ).
Andy Resnick
Andy Resnick is offline
Mar11-12, 07:09 PM
Sci Advisor
P: 5,468
Ok- I think you are on the right track. However, answer (0) should be more quantitative- I mean you should calculate how large a diameter (in meters/cm/feet/cubits..) you expect the Airy disk to be at your detector, and compare that to the size of your detector. Yes, the dish is the aperture.

And don't forget, since you are detecting the intensity and not the field, the fit function is actually [J_0(kr)/kr]^2.

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