Telescope contrast and resolution

[Ags]

3 hours ago, alpal said:

Why isn't the contrast ratio 1.0 for no obstruction?

 

Even in a refractor there is contrast loss due to diffraction. I can't recall how I got a number near 0.8 for no obstruction though - that looks too low.

[alpal]

9 minutes ago, Ags said:

Even in a refractor there is contrast loss due to diffraction. I can't recall how I got a number near 0.8 for no obstruction though - that looks too low.

Thanks - it does look wrong  - LOL.

There used to be a Newtonian calculator on the net which gave contrast ratio for the design of your choice.

Can you find that calculator for me please?

[Ags]

1 minute ago, alpal said:

Thanks - it does look wrong  - LOL.

I think I was more focused at the time on the relative performance rather than on the individual absolute values. Due to the numerical and simulation-based approach I was not able to calculate the contrast precisely (would have required hundreds of times more processing time) but I could calculate the _relative_ levels of contrast.

[alpal]

2 minutes ago, Ags said:

I think I was more focused at the time on the relative performance rather than on the individual absolute values. Due to the numerical and simulation-based approach I was not able to calculate the contrast precisely (would have required hundreds of times more processing time) but I could calculate the _relative_ levels of contrast.

OK thanks but what is the actual formula for contrast ratio as per here?

https://www.telescope-optics.net/obstruction.htm

 

[Ags]

I am afraid that telescop-optics.net is coming at this from a completely different paradigm, so I can't help you - they are using Fourier transforms for a mathematical solution, while I am modeling the waves travelling though the telescope. Because there are a lot of waves to sample, I can only approach the answer but never actually reach it. But Fourier stuff makes my brain hurt - so my approach was good enough for me.

[alpal]

6 minutes ago, Ags said:

I am afraid that telescop-optics.net is coming at this from a completely different paradigm, so I can't help you - they are using Fourier transforms for a mathematical solution, while I am modeling the waves travelling though the telescope. Because there are a lot of waves to sample, I can only approach the answer but never actually reach it. But Fourier stuff makes my brain hurt - so my approach was good enough for me.

Yes I know all about Fourier but from over 35 years ago so I forgot a lot.  84% of the starlight is  in the said Airy disk and 16% in the rings. That's due to Fourier and the wave nature of light.  here: https://www.astronomy-electronics-centre.com.au/article_resolvingpower.htm The Strehl Ratio Needless to say, the optical performance of a telescope depends on the quality of its optics, and on how they have been mounted on the telescope tube. In an imaginary “perfect” telescope the image of an out-of-focus star would appear as a perfect, well-defined circle of light. However, because of the wave nature of light, the said star will appear as a small disc surrounded by ever fainter rings, called diffraction rings. This constitutes the Airy disk and rings. A “perfect” telescope has 84% of the starlight in the said Airy disk and 16% in the rings - and it is impossible for more light to go into the disk. This 100% light distribution can be considered perfect  -  and our imaginary telescope is said to have a Strehl Ratio  =  1. Of course, less “perfect” telescopes would have more light distributed in the Airy rings, which would cause the view to become more blurred, thus making it more difficult to split the star Antares and other close double stars. To put it another way, as the Strehl Ratio drops, resolution will also drop accordingly. From what has just been said, we can see how important the Strehl Ratio (SR) really is when trying to establish the angular resolving power of a telescope!

[Ags]

27 minutes ago, alpal said:

84% of the starlight is  in the said Airy disk and 16% in the rings.

....sounds like I got quite close with a figure of 0.8ish for an unobstructed telescope! Might revisit this now i have a much more powerful computer.

[vlaiv]

May I point out that giving single number as contrast simply does not make sense in context of contrast/resolution of telescope?

I can walk you thru simple method of determining telescope MTF - which is graph of contrast vs spatial frequency in Fourier domain - without need of actually knowing how Fourier transforms work.

 

[alpal]

11 minutes ago, vlaiv said:

May I point out that giving single number as contrast simply does not make sense in context of contrast/resolution of telescope?

I can walk you thru simple method of determining telescope MTF - which is graph of contrast vs spatial frequency in Fourier domain - without need of actually knowing how Fourier transforms work.

 

Yes it's obvious from the links I have that "contrast ratio" alone does not give the full answer.

You are welcome to go through your explanation.

[vlaiv]

Here is simple version which we will elaborate.

Airy disk or more precise in this context - PSF of optical system is related to Modulation Transfer Function - MTF - which in turn explains everything you need to know about contrast.

MTF is graph of frequency attenuation - a bit like old audio equalizers - where you can set volume on different frequencies - low, medium, high. It looks like this for perfect clear aperture:

What does this mean?

Vertical axis (that goes from 0 to 1 - don't mind actual numbers above) represents attenuation factor. When line is high close to 1, that frequency remains almost as is. When line is low - that frequency is attenuated. What actually happens is that particular frequency is multiplied with height of above graph at that point.

On horizontal axis - it is spatial frequency in cycles per unit of length. At focal plane we are talking about cycles per mm (or µm), in image we are talking about cycles per pixel.

But what are those frequencies that I'm talking about? This is crucial for understanding resolution and contrast.

Image is 2d function of intensity of light over surface (or can be viewed that way). Every function can be represented as infinite sum of sine and cosine waves via Fourier analysis of that function. This image illustrates that principle well:

We have black line which is square impulse function and we think - it is square, no way it can be composed out of sine/cosine waves, but here it shows each successive step - we add finer and finer sine/cosine function to base wave and we get better and better approximation of this black function. Here we added only 5-6 terms and it's already looking pretty good - imagine having 1000 or more terms - we would not be able to tell the difference visually (there is of course still mathematical residual unless we sum all the way up to infinity ).

Yes, but what this has to do with contrast and resolution? Well, think what contrast is - it is difference between peak and valley intensity in our image - larger this intensity difference - higher the contrast.

Multiplying sine function with value less than one - reduces contrast of that sine function:

Red is the graph of 2*sin(x), black is graph of sin(x) and blue is graph of 0.5 * sin(x). Which one has the most contrast and which one has the least contrast? If you have trouble seeing it - imagine that 0 is gray and large negative is more black and large positive is more white.

Back to resolution - we now see that MTF operates across different frequencies - it attenuates low frequencies only slightly and high frequencies very much - it even cuts off all higher frequencies that certain point (this is resolution limit of the telescope - everything above that point will have 0 contrast - be just "single color" - "grey without any features").

In order to understand it better - here is example of how different frequencies are affected differently by PSF/MTF (by the way - convolution with PSF is the same as multiplication with MTF and MTF is fourier transform of PSF - if you want to know math details of the process).

Here is one image that I created from simple sin functions:

Each row has different frequency/wavelength. Rows at the top have lowest frequency / highest wavelength and rows at the bottom have highest frequencies - lowest wavelengths / distances between max and min value of sine function.

I'm going to convolve above image now with Airy PSF used to produce MTF that I posted at beginning. Look what happens:

Lower frequencies keep their contrast, but higher frequencies gradually loose their contrast - higher the frequency, more contrast loss there is - until they fade to grey / loose all contrast.

We can examine impact of Airy disk by doing the same on any image - convolution of image with airy disk.

How do we generate Airy patterns for these simulations?

Take ImageJ/Fiji, download FFTJ package.

Open new 32bit image - 512x512 pixels and run following math macro on it:

This creates our aperture. For unaberrated / clear aperture - just use this. If you want to simulate effects of central obstruction - make appropriate macro, for example:  if (d < 20) v=0; else if d(<40) v=1; else v=0;.

If you want to add aberrations - it is similar but a bit more involved - you need to make phase image as well and then decompose intensity (0 or 1 that we made) together with phase diagram into imaginary and real parts and do FFT on those. We can later do that to produce coma for example.

Next step is to launch FFTJ plugin and produce Power spectrum of this image:

Use above as real part, leave imaginary as none for this (we can use both real and imaginary for aberrations - will show later).

Then choose this options when you get calculation result:

This will produce nice Airy pattern:

Btw, size of Airy disk here is inversely proportional to diameter of aperture and both depend on size of the image (which you should keep at powers of two - so 256, 512, 1024, 2048, ... for fast and precise calculations and no scaling).

Now, if you want to do MTF of this - you then do another round of FFT on this image, but this time - select frequency spectrum instead of power spectrum.

Make sure you select right image:

Select frequency spectrum:

And you get this:

Which is 2d graph of our MTF. Select line tool and Plot profile to get actual graph:

 

[vlaiv]

Here is example of how to make airy pattern and MTF of obstructed aperture.

First make aperture pattern in the image:

Here we have 25% central obstruction.

Do FFTJ and use Power spectrum to get Airy pattern:

You can see that some of light has moved into rings (it is not linear, some rings get more light than others).

Do another round of FFTJ on this image now and use Frequency spectrum:

This time MTF looks a bit different, and in order to really see what it's shape is - we need to plot it's profile:

We've got that "obstruction belly" in our graph!

If we compare it to online graphs for central obstruction:

We see that our perfectly corresponds to 25% central obstruction one (as we used 25% central obstruction by radius).

[alpal]

Thanks Main Sequence,
I am well aware of how Fourier can reproduce a square wave.
What is a link for the calculator you are using?

There used to be a Newtonian calculator on the net which gave contrast ratio
for the design of your choice.
Have you seen it? - do you remember it? -
do you have a link?

cheers
Allan

[vlaiv]

1 minute ago, alpal said:

Thanks Main Sequence,
I am well aware of how Fourier can reproduce a square wave.
What is a link for the calculator you are using?

There used to be a Newtonian calculator on the net which gave contrast ratio
for the design of your choice.
Have you seen it? - do you remember it? -
do you have a link?

cheers
Allan

https://imagej.net/Fiji

and you should include FFTJ plugin

https://sites.google.com/site/piotrwendykier/software/parallelfftj

You can also look at Aberrator 3.0 - old software that does this pretty much automatically for you.

http://aberrator.astronomy.net/

I have no idea about Newtonian calculator that you are mentioning. I have not seen such a thing.

[alpal]

1 minute ago, vlaiv said:

https://imagej.net/Fiji

and you should include FFTJ plugin

https://sites.google.com/site/piotrwendykier/software/parallelfftj

You can also look at Aberrator 3.0 - old software that does this pretty much automatically for you.

http://aberrator.astronomy.net/

I have no idea about Newtonian calculator that you are mentioning. I have not seen such a thing.

Thanks - I'll check those out.

[vlaiv]

In the end I want to show how to use Zernike polynomials to generate different level of aberration on PSF.

First check out https://en.wikipedia.org/wiki/Zernike_polynomials

These are expressions that one can calculate on unit circle surface and are basis to describe any surface on unit circle. They are interesting to us because they are used to describe wavefront aberrations that telescope optics produces.

Let's say that we just want to see what sort of pattern will we get if we defocus our aperture. Defocus is given as:

It solely depends on rho (distance to center) and not angle.

We start by making our aperture - intensity part:

Next we need to create phase part:

You should really normalize rho and then determine how much defocus you want to do in waves, but we are going to just go with above formula and see how much defocus we get (change number multiplying d/40 in above formula to change level of defocus).

Now we have intensity and phase. We need to have real and imaginary part.

Real part = magnitude * cos(phase)

Imaginary part  = magnitude * sin(phase)

For this we can use image multiplication and math functions of sine and cosine on our magnitude and phase images.

And we get these as result:

Left is real part, right is imaginary. Now we again use FFTJ but this time we use both real and imaginary parts:

And if we apply gamma for sRGB displays, we will get image that we would visually have at a telescope:

Defocused star into two rings ...

 

 

[alpal]

Thanks Vlaiv,
the graph is here:

and the formula here:
reduction in the relative peak diffraction intensity -
and the energy content of the central maxima - by a factor    (1-ο2)2,     as per here:

https://www.telescope-optics.net/obstruction.htm

so an obstruction of say 20% is (1-ο2)2  =  (1-o.22)2  = 0.9216  or 92% contrast ratio.

so an obstruction of say 50% is (1-ο2)2  = (1-0.52)2   = 0.5625  or 56% contrast ratio.

 

Can you confirm those calculations please?
That is actually all I was after.
Whatever Newt. design program I used to use had that as part of it.
It seems that such a program no longer exists on the web.
I know that there is more to it than that because the Strehl ratio
must be taken into account for a full answer.
It also means that a mirror in any optical design can have a high Strehl ratio
which is totally ruined by having a large central obstruction in the design!

cheers
Allan

 

[vlaiv]

17 minutes ago, alpal said:

Thanks Vlaiv,
the graph is here:

and the formula here:
reduction in the relative peak diffraction intensity -
and the energy content of the central maxima - by a factor    (1-ο2)2,     as per here:

https://www.telescope-optics.net/obstruction.htm

so an obstruction of say 20% is (1-ο2)2  =  (1-o.22)2  = 0.9216  or 92% contrast ratio.

so an obstruction of say 50% is (1-ο2)2  = (1-0.52)2   = 0.5625  or 56% contrast ratio.

 

Can you confirm those calculations please?
That is actually all I was after.
Whatever Newt. design program I used to use had that as part of it.
It seems that such a program no longer exists on the web.
I know that there is more to it than that because the Strehl ratio
must be taken into account for a full answer.
It also means that a mirror in any optical design can have a high Strehl ratio
which is totally ruined by having a large central obstruction in the design!

cheers
Allan

 

All I can say is that I don't doubt the math, but:

I don't really understand how "relative peak diffraction intensity" is defined, further, I would not call it contrast ratio, regardless of its definition.

As for Strehl ratio - you can look at Strehl ratio in two different ways - both suitable for certain case.

Strehl ratio is ratio of energy contained in Airy pattern up to first minima in comparison to perfect theoretical telescope.

Choice of perfect theoretical telescope will give you Strehl ratio that is suitable for certain comparison.

For example, you have two Newtonian telescopes with same central obstruction and you want to see which one is better.

You calculate Strehl ratio of each in comparison to perfect aperture with same central obstruction and compare those.

If you want to put two scopes against each other that have same aperture size, but different or non existent central obstruction - you calculate Strehl against perfect unobstructed aperture.

However, once you start doing that - you will get misleading results. Similarly misleading results as comparing two telescopes of different aperture.

For this reason, I would suggest that Strehl ratio be used only as measure of how good optical figure is. To get that value, you should really compare it to perfect aperture with same central obstruction - regardless of central obstruction size.

If you want to compare performance of two telescopes - of different design or different size or different design and size - use MTF graphs for those two telescopes overlapped.

That will give you the best idea of how two telescopes differ in views they provide (or rather theoretical views they are capable of). MTF can be used for real telescopes as well - one just needs to asses their wavefront error and this can be measured - for example by Roddier analysis.

I'll give you example - you can't really compare performance of 4" APO refractor (we will assume it is perfect in every way) and 6" Newtonian with 25% CO (again perfect in every way) by using Strehl or above "contrast ratio".

On the other hand, making MTF diagrams is really not hard if you follow what I've written above. Here is result:

Black line is 6" with 25% CO and red line is 4" without CO. Is there any contest between the two? Can you decide which one will throw more detailed image in same conditions (same magnification and seeing good enough to show what scopes can deliver)?

We can throw 5" APO into the mix - common "wisdom" says that 6" with 25% CO will in fact perform as 4.5" clear aperture scope. This means that 5" should beat it?

Huh, tough one, isn't it? In low frequency domain, 5" unobstructed, here in blue, will in fact have slight edge over 6" with 25% CO, but in high frequency part - 6" will still show more.

This translates into following description:

"View of Jupiter thru 6" looks more washed out with colors being not as saturated as in quality 5" APO, but 6" Newtonian shows hints of those tiny ovals and ripples in belts that 5" APO just can't produce".

[alpal]

10 minutes ago, vlaiv said:

All I can say is that I don't doubt the math, but:

I don't really understand how "relative peak diffraction intensity" is defined, further, I would not call it contrast ratio, regardless of its definition.

 

Thanks Vlaiv,

Yes - I'm assuming that "relative peak diffraction intensity" is contrast ratio ?

Oh dear - we have a definition problem.

[alpal]

I found something on my computer about contrast factor not contrast ratio and I saved the pdf here with file hosting online:
http://s000.tinyupload.com/index.php?file_id=91035876064397637171

It says on page 11.
From the graphs and tables published in the referenced articles on Newtonian improvements in the Journal of the Association of Lunar and Planetary Observers (J.A.L.P.O.) a general equation can be arrived approximating the"contrast factor" value for your system (See Table I):

 

CF = 5.25 - 5.1x - 34.1x ² + 51.1x ³       where x is the obstruction ratio or secondary/primary diameters.

It says that for an unobstructed telescope the contrast factor is 5.25
and with a 20% obstruction it's  3.27.
If you divide 3.27 by 5.25 you get 62.% which doesn't agree with my calculation above of 92%.

Now I am a bit stuck here and I think it's got to do with definitions?   Any comments?

 

 

[vlaiv]

2 hours ago, alpal said:

I found something on my computer about contrast factor not contrast ratio and I saved the pdf here with file hosting online:
http://s000.tinyupload.com/index.php?file_id=91035876064397637171

It says on page 11.
From the graphs and tables published in the referenced articles on Newtonian improvements in the Journal of the Association of Lunar and Planetary Observers (J.A.L.P.O.) a general equation can be arrived approximating the"contrast factor" value for your system (See Table I):

 

CF = 5.25 - 5.1x - 34.1x ² + 51.1x ³       where x is the obstruction ratio or secondary/primary diameters.

It says that for an unobstructed telescope the contrast factor is 5.25
and with a 20% obstruction it's  3.27.
If you divide 3.27 by 5.25 you get 62.% which doesn't agree with my calculation above of 92%.

Now I am a bit stuck here and I think it's got to do with definitions?   Any comments?

 

 

I'm wondering, what would you like that number to represent?

Can you describe it's meaning - the way you would expect it to work?

Let's say for sake of argument that you have one telescope that has "contrast ratio" of 1 and another telescope that has "contrast ratio" of 0.8 or 80%.

Can you describe what would you expect differences between the two scopes to be?

[alpal]

4 minutes ago, vlaiv said:

I'm wondering, what would you like that number to represent?

Can you describe it's meaning - the way you would expect it to work?

Let's say for sake of argument that you have one telescope that has "contrast ratio" of 1 and another telescope that has "contrast ratio" of 0.8 or 80%.

Can you describe what would you expect differences between the two scopes to be?

The one which is 80% would have a more blurry view as per page 13
of the link I gave above and give again here:

http://s000.tinyupload.com/index.php?file_id=91035876064397637171

 

 

 

 

 

[vlaiv]

8 minutes ago, alpal said:

The one which is 80% would have a more blurry view as per page 13
of the link I gave above and give again here:

http://s000.tinyupload.com/index.php?file_id=91035876064397637171

 

 

 

 

 

Ah, excellent - only distinction is more blurry (or blurrier - which one is right, how does blurriness compare?), then take contrast factor to be

100% - CO%.

25% CO will have "contrast factor" of 75%

30% CO will have "contrast factor" of 70%

Hence, scope with 30% CO will give more blurry views than that with 25% CO.

Problem solved

 

[alpal]

8 hours ago, vlaiv said:

Ah, excellent - only distinction is more blurry (or blurrier - which one is right, how does blurriness compare?), then take contrast factor to be

100% - CO%.

25% CO will have "contrast factor" of 75%

30% CO will have "contrast factor" of 70%

Hence, scope with 30% CO will give more blurry views than that with 25% CO.

Problem solved

 

But those numbers don't agree with my original calculations.
It also says on page 11.   http://s000.tinyupload.com/index.php?file_id=91035876064397637171

Table I.

 

Values for obstruction and contrast factor.
The left-hand column is obstructions in percentage of secondary to primary mirror in linear diameters.
Next column gives the central spot energy percentage,
next the energy in the rings and then
the contrast factor values taken from the reference text.
 

I can't get a copy of the reference text but I understand the 5.25 number.
Hurlburt, H.W., "Improvement of the Image Contrast in a Newtonian Telescope," Journal of the
Association of Lunar and Planetary Observers (J.A.L.P.O.), Vol. 17 Nos. 7-8, July-August, 1963,pp. 153-158.

I know it's a calculation of the central spot energy divided by the Rings energy
as per the table on page 12  - here:

 

 

 

Edited December 3, 2020 by alpal

[vlaiv]

17 minutes ago, alpal said:

But those numbers don't agree with my original calculations.

I agree, but they are solution to the problem you posed:

"I need a number that will tell me if telescope is going to show more blurred image depending on size of central obstruction".

This is why I asked - what do you want a number to represent.

[alpal]

3 minutes ago, vlaiv said:

I agree, but they are solution to the problem you posed:

"I need a number that will tell me if telescope is going to show more blurred image depending on size of central obstruction".

This is why I asked - what do you want a number to represent.

It looks like a very good solution.
I think it is referring to splitting a double star?
The maximum ratio of energy between the diffraction rings is 5.25.