FWHM -Why do star images grow???

[Merlin66]

Using a CCD camera, each pixel records the photons and displays an ADU - the relationship between them being the gain of the camera.
Every star image produced by a telescope is basically an Airy Disk.
Due to atmospherics etc (ignore guiding errors and aberrations for a moment) a PSF (Point Spread function) curve is seen, basically the same Gaussian shape as the Airy disk but much larger in diameter.
The usual measure of this curve is the FWHM ( the width/diameter of the curve at half the peak intensity)
Still with me?
I do a 3 min exposure.....
The fainter stars appear smaller than the brighter stars ( get's even worse if the star intensity causes saturation...)
Why is this so??
The full extent of the PSF curve doesn't change.
Why doesn't the intensity just build up in each pixel within the PSF curve? Why does it appear to enlarge the PSF curve??
The reasons given say that as the exposure increases the CCD can record "fainter outer regions of the curve" and these show up as an increased diameter.....Sorry, I can't see that!
If ALL the light from the star is within a defined PSF curve and that curve, for a faint star is say 5 pixel diameter, then surely for a brighter star the total size should still be 5 pixel allbeit with a higher peak intensity.
Add 10 x 30 sec star images.....compare the size with a 5min star image????
I'm finding that if the star image is NOT saturated, then the PSF curve remains close to constant...when the star becomes a saturated image it starts to grow outwards ie 5 pixel max profile for a non-saturated star image v's 15-20 pixel for a saturated star!!!
( Ron Wodaski in his "The New CCD Astronomy", p 50-53 demonstrates this issue but gives no reasoning or answer to the possible cause.)
Can anyone add to my confusion?????

[bmitchell82]

Moooooo

:D

Sorry couldn't help myself

[Merlin66]

OK a possible lead....
Blooming and saturation?
http://learn.hamamatsu.com/articles/ccdsatandblooming.html

Interesting...the "overflow" is dumped, not transferred to nearby pixels...

[gregbradley]

Interesting thread. I have seen that effect more pronounced with sensors with a smaller well depth than with cameras with a larger well depth.

I assumed it was a case of oversaturated pixels dumping nearby but I think it is probably more the fact that there is more scatter with brighter stars and the light accumulating in nearby pixels shows that scatter more when its a bright star. Accordingly if you use shorter exposures it will have the effect of tightening up the stars but also at the expense of losing some detail in the faint areas of your image.

CCD chips have a cover slip standardly (Kodak). These can create small halos and reflections. That would add to it. Additionally I don't know how good the antireflection coatings are on these cover slips. On top of that most astro chips have a microlens layer that focuses the light to the side of the pixel that has the higher QE. So you've got a few things going on there.

Its a similar phenomena with blue filters versus other colours. Blue is often more bloated. I have read that this is because blue is more scattered. Seems to work as a theory.

Reading that article you linked it mentions anitblooming is less efficient on one side of the pixel than the other and that various antiblooming methods have different characteristics and efficiencies. The implication there is that perhaps they don't work 100% in every scenario so perhaps there still is some overflow to nearby pixels. It does not clarify that point too much.

Also I wonder if this comes into it:

http://www.paquettefamily.ca/astro/star_study/

In that colours are not registered with the same FWHM usually. Blue being worst. Sensors are often listed with QE for each colour channel and often are not equal in QE. It can vary by 10-20% more or less QE. Could that effect of bloated stars also be they were bright and bluish stars?
Greg.

[TrevorW]

maybe an article on this page can help

http://www.stark-labs.com/craig/articles/articles.html

[RickS]

I think there are a number of different factors involved here but saturation alone is enough to explain a degree of bloat on bright stars. When the top of a star profile starts to get truncated by reaching full well depth the FWHM will increase. A diagram would help to make this clear but I don't have time to draw one right now...

Cheers,
Rick.

[leinad]

Are you asking why the increase of the maximum of the FWHM with overexposure?
You could try this one.

Detecting and measuring faint point sources with a CCD
"Herbert Raab"
http://www.astrometrica.at/papers/PointSources.pdf

[irwjager]

The Airy disk does not look like a Gaussian distribution. It looks like a disk with rings around it that get ever fainter. The PSF of a single point light will affect all pixels on your CCD. E.g. the light of a point light is spread all over your image and thus will build up all over your image given a long enough exposure.
The PSF is created by diffraction and has nothing to do a priori with atmospheric turbulence. You will still see an airy disk even outside the earth's atmosphere.
The atmospheric turbulence indeed smears the disk out a little randomly over time (because it moves randomly). This random movement indeed approximates a Gaussian distribution as exposure time increases.

Does that help?

[Merlin66]

Leinad - good find!
The assumption is the PSF is maintained for all unsaturated star images...that's what I'm finding...doesn't address saturated stars?

Maybe the attached write-up will give some ideas of my data/ thoughts...
In the meantime - an analogy...
Point a laser beam at a piece of graph paper on the wall (-CCD/pixels).
The beam illuminates a patch 2 x 2 squares centred on point "A" ( - Airy disk)
After five minutes, the laser still illuminates only the 2 x 2 sq, and the intensity recorded at point "A" will have increased.
There will be no illumination recorded outside the 2 x 2 square.....
More realistically:
Same laser but supported by a shakey table, still centered on point "A" but can wobble around and illuminate up to a patch 10 x 10 squares (-Seeing conditions)
After five minutes, point "A" will still record the peak illumination and the spread will not exceed 10 x 10 squares.....

Why then would a star image appear to spread to say 20 x 20 squares??????

Other than sensor reflections etc - the feeling seems to be that the image saturation will cause the 7% residual light in the "1st Airy ring" to contribute to the image size. This doesn't seem to match....
(BTW with the Hubble...if there's no atmospherics to "smear" the Airy disk --why do brighter stars still look larger??;) )

[irwjager]

And that's exactly the case.
With what exactly? The size of the Airy disk and the spacing of its rings depend on your aperture and focal length (and CCD resolution of course).
That's because Airy disks have nothing whatsoever to do with atmospherics- diffraction is the main cause of your stars not being neat points! All telescopes will necessarily have to diffract the light, even the Hubble.

EDIT: I think your analogy is flawed as a laser beam is not a focused beam of light. All photons travel in parallel and have not been diffracted yet, thus showing no Airy disk on the paper. Also you seem to assume there is just one (or two) rings around the primary Airy disk. In fact there are many and they extend for the full surface area of the aperture.

[Merlin66]

IVO,
You need to read the document I attached to the last message...
I take into account the f ratio and the camera pixel size in the calculations...
The enlarged saturated star images appear to exceed the diameter of the Airy rings..

Re Hubble - this would then infer that the brightest star image recorded will always been no larger than the extended Airy disk...is that really the case???

[irwjager]

There is no diameter of the Airy rings. These rings extend throughout the image - e.g. all wells on the CCD (or film) get a tiny bit of light.

EDIT: I see in your document you define the 'Airy disk diameter' as 2.44 x Lambda x f ratio. This formula is an approximation of the position of the first minimum, but many, many minima (and rings) follow. The disk that you define only contains 83.3% of all the light, with the rest being spread out over an increasingly large area away from the star, concentrated in further rings. Expose this area long enough and it will start showing up, gradually 'growing' the stars until the whole image saturates.

Not sure what you're saying here, but rest assured that an incredibly bright star, which would be incredibly far away (e.g. for all intents and purposes a point light) can easily swamp Hubble's CCD wells with a long enough exposure.

[Merlin66]

IVO,
Sorry,
what I'm trying to say is that even if you add the contribution of the light from the 1st and 2nd rings ( total star energy = 94%) the apparent saturated star image still seems to exceed the linear diameter of the rings....
The background shot noise level will be sqrt max intensity, which I'm sure would exceed the residual star energy...so no additional signal to expand the image???
I probably need to use the photometry aperture feature (AA5) to better determine how the summed ADU count in each star image changes. This should be linear (for non- saturated images) - but what about a saturated image?

[irwjager]

Ok, what sort of excess energy/starlight are you observing?
You be must taking really short exposures for the shot noise to drown out the Airy disk! There are many examples of a diffracted Alnitak overwhelming the Horse Head nebula for instance (http://www.iceinspace.com.au/forum/showthread.php?t=83886), even though Alnitak was never even projected on the CCD (its light still made it through the aperture of the scope).
You hit the nail on the head - the way your CCD deals with saturation of its wells varies wildly. But it is, alas, almost never truly linear. Saturation is dealt with in various way by the anti-blooming circuits of your CCD (if any). Fitting a Gaussian curve to a saturated star is therefore not very reliable.

[leinad]

I think I see where you're getting. but confused by your analogy's.
This was also interesting.. diffraction. Wouldn't saturation cause the PSF maximum to be larger than the airy disk over x time due to the diffraction. got me glued to the screen now.
I don't know how this incorporates into your Hubble comparisons though..

http://www.willbell.com/aip/haip2_chap_1.pdf
Pg 29 - Section 1.5.4

[Merlin66]

Leinad,
That's the nub of the problem - at least for me...
If the PSF of the Airy Disk contains ALL the energy/ photons available from the star, then the maximum size of the star image can NEVER exceed this size....hence my laser pointer analogy.... there's never any illumination beyond a 10 x 10 box...so whatever the exposure the size should always be less than 10 x 10.

If you look at the profile data say for Star#1 and Star #2 (one is saturated the other is well exposed). The "max" diameter recorded for #1 is 20 pixel and for #2, 12 pixel, and the FWHM is 6 pixel v's 3.5 pixel.
Why, when the brighter star is over-exposed more does the "size" increase by 166%....Aah...the longer exposure collects the light from the outer reaches of the PSF curve...Hmmmm don't think so....

IVO,
The second comment you refer to - wasn't the Airy disk being drowned out, but the light in the extreme "wings" ie that last 6% of total energy being drowned in the background noise....

[leinad]

After going over your document; I see what you're addressing now... even more interesting. :)

/seated with popcorn.

[Merlin66]

I'm slowly getting there (I think!)
Conclusions/ Comments

For unsaturated star images, the maximum size of the star is similar to the Airy disk in diameter.
The FWHM of the star image is reasonably constant.
As the star image saturates, the “flat top” extends out to the diameter of the Airy disk.
Also, for saturated star images the “wings” include additional energy from the outer rings, and can extend, in linear diameter, to and slightly beyond the dimensions of the 1st Airy ring.

[irwjager]

The Airy Disk *is* a Point Spread Function. It describes how a single point of light (a star) is spread over (in this case) the CCD when it is shone through a circular opening (in this case the aperture of a telescope).
The sum of all the energy 'scattered' throughout the image by the PSF is indeed equal to the amount of energy received from the point light. Again, the PSF, in this case the Airy Disk, spreads the energy out over a huge area.

You seem to be confused by the 'maximum size of a star'? In a perfect world every star in your image would be a point light. But we run into two problems;


Your CCD wells saturate. They start clipping the signal and can't register a star that's brighter than 'maximum brightness', nor can your computer display display such a brightness. Therefore the signal is capped to 'pure white'. The star itself may in fact be much brighter than that but we can't record that, nor can we display that, so we cap it to the maximum brightness we can represent with our system(s).
The starlight is necessarily diffracted by the aperture of your telescope. This leads to the light being spread over a larger area than just a single point.

Now, combining the two 'problems', more starlight can be captured by your wells because the starlight is spread by the PSF over a larger area, hitting more CCD wells. No longer is the point concentrated into a single well, instead it is spread over all the wells in your CCD. Some may still saturate and be completely white, but others may not.


Again, your laser pointer analogy is flawed as the laser light is not diffracted.

Again, Gaussian Curve fitting to overexposed stars and then comparing FWHM readings is next to meaningless.

I understand that, but you can't 'drown out' a PSF. It is merely a function by which you multiply the energy of a star. There are very bright stars (ex. Alnitak) which, when you multiply them by your scope's PSF will readily show up (and ruin your image) even though their core is way off the CCD.

[irwjager]

False. The Airy Disk does not have a diameter. It is a pattern, patterns don't have diameters. You could, however, say that 83% of all starlight of the imaged star should be contained within the area that extends from the center to the first minimum.
False. The FWHM is dependent on how well you can fit a Gaussian curve to a star. You will be able to fit a Gaussian curve less well to saturated stars, depending on the response of the CCD you're using. It is also neigh impossible to fit a gaussian curve to undersampled images.

False. It can extend way beyond the first minimum.
False. They extend throughout the whole image.

[irwjager]

Ok,

http://www.startools.org/download/airy_refractor.tiff

Load this image into PhotoShop (it's important that your program supports 16-bit bit-depth).

You'll see a small star in the middle. This is a computer generated star, with an Airy Disk-like PSF applied.
Now launch a curve manipulation tool and start adjusting the white point down, way down.
Now you'll see how the star grows along the airy disk diffraction pattern as its core gets overexposed (multiplied).

Does this help?

[Merlin66]

OK
Time out!
"There will now be a short intermission - ice-cream and sweets are available in the Foyer..."

IVO, I based my wording on this:
http://en.wikipedia.org/wiki/Airy_disk

[irwjager]

:P

Mmmm... icecream...

I understand.
Your question is; "Why do star images grow?" (with longer exposures).
The answer is; because more energy is distributed according to their diffraction pattern/PSF/Airy Disk.

The wording in the Wikipedia article makes it sound like the Airy Disk is just a finite fuzzy blob around the location of a star. This is not the case (it is a simplification) and it is important to be aware of this in order to answer your question on why stars seem to 'grow'.

For purposes of Gaussian Curve fitting, the fuzzy blob gets pretty close to a Gaussian curve and we can leave it at that. And that's why Full Width Half Maximum measurements can be performed reliably and can be compare amongst eachother. Overexpose the fuzzy blob though and your Gaussian curve's top will be chopped off and/or distorted. This makes it harder to fit a Gaussian curve to and get FWHM results that you can compare with other measurements within the same image.

Now back to your question, "Why do star images grow?" (with longer exposures). You seem confused that stars can grow beyond the central fuzzy blob in the Airy Disk. The reason they can, is because the central fuzzy blob is not the end of the Airy Disk. The Airy disk has no end. Its pattern stretches out far beyond where the CCD sits.

Give the Airy Disk a bright enough star whose light needs spreading and it will saturate more and more CCD wells, further and further away from the central blob. The brighter the star, the further away from the center CCD wells get saturated. True, the central blob will be the first to completely saturate, but it doesn't stop there.

83% of all starlight falls in that central blob. But what if you got so much starlight that that 83% blob gets saturated? The other 17% all of a sudden becomes very important - that's where things start to saturate instead - the star starts to grow beyond the blob. It will take a lot of light to make it grow further and further (growing it just a single pixel will require more and more light) but it can definitely be done with enough photons. And because the spread function covers the whole CCD (and beyond), in theory, a single star can grow across the whole CCD, given enough photons.

I attached a visualisation of the underlying diffraction pattern that belongs to the tiny lonely 1-pixel star from the TIFF file I posted earlier. Diffraction is quite beautiful hey? :)
Stretch that TIFF far enough in PhotoShop and you'll see that tiny star grow according to this pattern. Try it; the data is really in there.
Note that non-linear stretching the TIFF is not what nature does. Nature just piles one more and more photons. You can do the same by adjusting the whitepoint in Photoshop, allowing pixels to gradually saturate. This is how a photograph gets exposed and how a star grows; the answer to your question.

[leinad]

Great explanations Ivo :thumbsup:
I think I was on the right idea path; just didn't have the theory to back it up.

Great read.

[Merlin66]

"Gannet on a stick...Gannet on a stick" - The cinema scene from Monty Python...
IVO...with all your smooth talking you almost had me there:hi:
To quote:
True, the central blob will be the first to completely saturate, but it doesn't stop there.

83% of all starlight falls in that central blob. But what if you got so much starlight that that 83% blob gets saturated? The other 17% all of a sudden becomes very important - that's where things start to saturate instead - the star starts to grow beyond the blob.
Humour me for a moment...
Let's call the "Blob" (which contains 83% of all the starlight) an Airy Disk, and the ongoing (decreasing) waves of light surrounding it an "Airy Pattern" In this Airy Pattern there are definite brighter/ darker rings which decrease in intensity ( #1 = 8% of the total, #2 = 3% etc etc)
By definition the image of the Airy disk for any star (not just the faint ones or the bright ones - every star, big or small) extends -
Airy disk = 1.342 x focal ratio
For a f7.5 scope this gives a size of 10 micron.
Outside this disk there's a brief gap (first dark diffraction ring) before the 1st ring....
( When you observe a star at high magnification you can always see the diffraction/ Airy disk - it never changes size between stars)

The peak intensity (for a non-saturated star image) increases with exposure..ie Brighter (or more exposed stars) have a higher intensity peak than fainter (less exposed stars)
So, if we had a peak of say 10000ADU for a bright star, the Airy disk diameter would be a constant and the sum of all the pixels in the disk would represent 83% of the total energy/photon count. Let's say the total ADU count within the Airy disk is 300,000ADU - then only 29,000 ADU would be spread into the 1st ring, and 10,800 ADU in the larger 2nd ring. Thats a very large decrease in intensity (ADU/pixel)!!!

So, back to my conclusions...
1. The Airy disk for every unsaturated star is constant
2. The peak intensity increases with brightness/ exposure.
3. You have to fill the whole Airy disk before the lower intensity of the Airy Pattern begins to contribute to the increasing area of the star image.

I couldn't get your computer generated image to show a distribution similar to that of a real star...

"Part 3 - the story so far...Mary, the dark haired divorcee ran of with Guy, the one with the bushy eyebrows......."

[leinad]

/gone to get more popcorn.

[Merlin66]

Hmmm
reading up to understand this "ever expanding star"....
I think the issue is that I'm hung up on the Airy Disk as being the manifestation of the size of the star image....and the Airy pattern of ever decreasing rings as being so low in relative intensity relative to the "core" that even if they were over-over exposed they would never reach the intensity values of the Airy disk, and would, therefore, only contribute the the "wings" of the image and not the "flat topped" over exposed, saturated area seen in and around the Airy disk..
Be gentle with me.......

[gregbradley]

I think so Rick.

A simple experiment would be to take 2 exposures of a bright star field. One is short, say 20 seconds, the other is 10 minutes and measure the FWHM's of the same stars in each. That would reveal the short image has tighter stars I am predicting.

Greg.

[irwjager]

It appears so. I don't know what else to say to persuade you otherwise. How exactly do you come to that conclusion in the face of what is on Wikipedia and what I explained and demonstrated?
The latter is just not true.

If George Bidell Airy was able to distinguish the second ring with his puny human eyes, what do you think a long exposure would show?

One last example of where you're going wrong;
Let's say my annual income is $100.000 (my CCD well depth) and I'd like to get a loan of $1000 (photons to disperse, aka star brightness). My bank offers a rate of 17% (percentage of photons dispersed outside the central fuzzy blob, or the PSF for the non-central fuzzy blob).
I'd be paying $170 per year - peanuts! It is easily serviceable with a $100.000 income and hardly noticeable when compared to my income (it's just 0.17% of my income).
By your logic I could just as easy take out a loan of $100.000.000. You seem to think the 'interest rate' (PSF) somehow is an absolute value, whereas it is a multiplication factor. Pick a high enough loan (star brightness) and it *will* start to matter. In the example of a $100.000.000 loan, I'd be paying $17.000.000 interest. I would need 170.000% my annual income (170 CCD wells) just to service the interest!
You're arguing the bump from 0.17% to 170,000% of my income isn't noticeable.

I really don't know what else to say or do. I've attached results of the TIFF multiplied by 800%, 3200%, 32000% and 320000%, just to show you it *does* grow, even though you can't see the light being spread at low energy levels.

FYI, the the initial PSF (i.e. the star in the TIFF) was obtained by creating a circular aperture (e.g. a white circle with a diameter of the 256 pixels) and then performing a forward Fourier transform and taking the magnitudes of the resulting complex numbers. Just like the math in the Wikipedia article.

[CDKPhil]

It would be the same if you were photographing the sun, with a normal SLR CCD camera and a 50mm lens ( not that it is recommended )
You might use a shutter speed of 1/64000th of a second and get a bright round circle, but use 1/2000th of a second and get your whole frame over saturated.
The only difference between the sun and the stars is the time it takes to expose them.
If you want to do a simple experiment. (not with the Sun ) get your SLR mount it on a tripod set up a LED (point Source light) in a dark room and expose it at different shutter speeds. You will see the exact same effect as what is being discussed.

Unfortunately for astronomers there are huge contrast ratios between the stars and faint fuzzy objects. So Airy disks, bloated stars what ever you want to call them, will be an issue. As CCD technology advances I am sure someone will come up with adaptive pixels, where the user could control the amount of saturation for individual pixels. It would be like the old darkroom technique of dodging and burning but on a far more advanced level.

Cheers
Phil

[irwjager]

Yes! :thumbsup:
That'd be great. And there are indeed some techniques we can use in the interim. SkyViking (Rolf Wahl Olson) used one such technique to cancel out much of Beta Pictoris diffraction, revealing the dust disc.
Another one is using Deconvolution (and knowing the exact PSF) but that only works on non-saturated stars. The future is a type of video astronomy where many short exposures are stacked to create a very high bit-depth image with a huge dynamic range (and no overexposed pixels). The only thing holding us back right now is the various types of noise that become worse (such as read noise) when using many frames.
There also the good-old HDR compositing trick we could use today (as used for M42 for example), but alas there is no catch-all formula for smoothly blending the various frames.

[mithrandir]

FWHM is commonly used as a measure of focusing and choosing stars for stacking.

Is the real point in that FWHM becomes unreliable once the central pixels come close to saturation?

Once you have a circle of saturated pixels FWHM has to be at least twice the diameter of the circle of saturated pixels.

So the moral of the story would appear to me - and to Craig Stark from the Nebulosity doco - do not use saturated stars to do any sort of measurement of focus, or alignment for stacking.

[avandonk]

Here is an animated gif of 7, 15, 30, 60, 120 and 240 seconds exposures with the last being a HDR. 6MB

http://d1355990.i49.quadrahosting.com.au/2011_11/hdr01.gif

Below are the images that made up the gif. The cluster is the Jewel Box.

Bert

[irwjager]

Comparing FWHM between different stars/frames becomes unreliable. Gaussian curve fitting (of which the FW at HM is a by product) can still be used to help with determining sub-pixel accurate location though (as used for stacking purposes) because in that case you're just interested in the location of the predicted peak of the Gaussian curve, not its width or the magnitude at its maximum.

I don't see anything wrong with using them for stacking/alignment, but Craig may have reasons I'm not aware of?

[Merlin66]

http://www.lbto.org/LBT%20Website%20General%20Public/adaptive%20optics.htm
Go down to the bottom of the page....
1. Due to the nature of light a pinpoint star produces an Airy disk, surrounded by a Airy Pattern.
2. All stars great and small produce the same linear diameter Airy disk for the same f ratio ie a 50mm f5 and a 500mm f5 produce the same sized Airy disk.

The limitations are not due to the Airy disk but the receptors ability to resolve and record the disk ( and the outlying pattern)
When you analyse a short exposure image, the star is still the Airy disk in diameter but with a much lower peak intensity. (See Star #4)
A brighter star (or more exposed star) has the same Airy disk diameter but a higher peak intensity (See Star #2)

[irwjager]

It's called diffraction. Light itself doesn't do anything by itself - it requires to encounter an aperture.

Stars don't produce an Airy disk, the Airy disk (re)produces the stars. The Airy Disk is a point spread function. A function light has to go through before it appears on the imaging plane.

Until you overexpose and the first few rings fill up (if you can even dissolve the rings at your particular magnification).

Like I said, I don't know what else I can say or do to explain better to you why stars seem to grow. You seem to have drawn your own (erroneous) conclusions, yet wonder why they don't fit reality. I think it's time I give up. :(

All the best,