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locked Bug in DFTFringe 5.0

Dale Eason
 

I discovered that if you change the mirror diameter the profile plot will not display the profile correctly until dftfringe is restarted.  I have a fix for the next release coming soon.


Re: need advice: Parabolic primary -- d = 202mm RoC = 1562mm

Joe Babendreier
 

Extremely grateful for your analysis. Thanks,

Joe


Re: need advice: Parabolic primary -- d = 202mm RoC = 1562mm

Dale Eason
 

Here it is in DFTFringe64 next release and 5%.  Yes, I'm about to release 5.1.


Re: need advice: Parabolic primary -- d = 202mm RoC = 1562mm

Dale Eason
 

I think even 5% is  more accurate for your case.  That is because using your .wft file I can adjust the Foucault image so I can see that weave pattern that shows up when many igrams with different fringe orientation are taken.   Meaning that you have averaged out most of the igram noise and are left with remnants of fringe print through.   Also because your igram looks very good with little noise.   Thus you can reduce the blur radius.       So because the surface is not as smooth as a very good mirror and with a high edge the higher blur values are artificially smoothing the surface too much at even 10%.

Your red colored ronchi I think is an actual ronchi and it too shows the surface defects well.  I think at 5% DFTFringe simulated ronchi matches your actual better.  You can see a little more noise in the 5% but it captures the edge better than the 20%.

Dale


Re: need advice: Parabolic primary -- d = 202mm RoC = 1562mm

Joe Babendreier
 

Pictures:


need advice: Parabolic primary -- d = 202mm RoC = 1562mm

Joe Babendreier
 

I've been figuring a Pyrex 8" primary for a few months now. I use DFTFringe for testing. (Plus Foucault  & Rochigrams)

My question is about the proper setting for the Gaussian blur when trying to determine the wavefront.

In my case, should I set it to 10% or 20% (or some other number) ???

I attach some pictures, including an R-gram (outside of R0C) for the mirror as it stands now.

https://www.dropbox.com/s/ss5d9jiege3mlwn/BABO%20AG%20202-1564-10pc%20385%20s883.wft?dl=0

Joe


Re: Tilt dependent astigmatism

Franz
 

Thanks Michael!
That seems very likely as my laser beam moves in a circle when I turn the fucusing element.
I will try to improve the laser and report back.


Re: Tilt dependent astigmatism

Michael Peck
 

On 5/26/2021 7:47 AM, franz.hagemann@... wrote:

It seems like astigmatism is proportional to tilt. And it is substantial.

Does anyone know if this is normal?
It makes measuring astig really hard because it only cancels out for equal tilts.
My beam separation is 6.5 mm. DCX diverger with 9 mm focal lenght.

Cheers,
Franz

Tilt dependent astigmatism isn't exactly normal but it has been observed and discussed here a number of times. It's likely due to coma in the collimating lens of your laser, which in turn is likely due to decentering.

-- 
Michael Peck
http://wildlife-pix.com
mlpeck54 -at- earthlink.net


Tilt dependent astigmatism

Franz
 
Edited

Hey folks,
for some time now I had the impression, that more fringes (or more tilt) result in more astigmatism in the wavefront.
So I made a small experiment.
The mirror is an unfinished 6 inch f 2.8.
I made some interferograms with different amounts of tilt, without touching the mirror in between.
I got this result:
x Tilt      y Tilt     Astig (polar)
12.7         11           0.055
21.9         19           0.143 
23.2        22.3         0.143
29.0        29.0         0.21

It seems like astigmatism is proportional to tilt. And it is substantial.

Does anyone know if this is normal?
It makes measuring astig really hard because it only cancels out for equal tilts.
My beam separation is 6.5 mm. DCX diverger with 9 mm focal lenght.

Cheers,
Franz


Re: Updated Wiki Page: Diverger lens Residual Spherical aberration #wiki-notice

Bruce Griffiths
 

With an equibiconvex lens the fact that the test surface brings the collimated reference beam to a focus one test surface focal length in front of the test surface whereas the test beam from the laser is collimated results in a small mismatch in SA between the test and reference beams. As long as the test surface focal length is much greater than that of the diverger then this SA mismatch is typically negligibly small.

If a PCX lens is used it contributes an SA mismatch between test and reference beams.

The formula allows this SA mismatch to be estimated.

If the refractive index is unknown using a value of 1.5 allows a ballpark estimate of the SA mismatch so one can see if the resultant SA mismatch can be neglected when testing  a particular test surface when using a PCX diverger lens (a PCX lens may be all that's immediately available). If the resultant SA mismatch is significant it can be easily measured if the diverger lens can be reversed allowing measurements to be taken with both orientations of the diverger lens.

If for example, one were testing a slow test surface a long efl diverger lens that better matches (allows sorter exposures by using more of the laser beam) the test beam to the test surface may only be available as a PCX lens. In which case an estimate of the resultant SA mismatch can be very useful.

Bruce

On 13 May 2021 at 02:24 "George Roberts (Boston)" <bb@...> wrote:

Lol.  Basically in a Bath, there if you use a biconvex lens there is a tiny contribution of that lens to the spherical aberration term.  Typically it may add less than 1/1000 of a wave of error to the DFTFringe result.  No big deal.  It depends on things like how big and fast your mirror is and what type of glass is in the thin lens and so on.  Bruce has an equation that lets you calculate the error.  I might make a web page calculator where you enter the terms but the error is so small it's probably not worth the trouble except...

If you use a PCX lens instead of a BCX lens (PCX - plano convex lens - that means one side of the diverger lens is flat) then this spherical aberration grows.  It's still typically less than 1/50th of a wave but it could be significant if you have an unusual mirror.  I have a calculator here to see how much PCX will affect your mirror here - just plug in your mirror parameters:
gr5.org/bath/

However Bruce improved the equation to include the index of refraction of your PCX lens which most people probably won't know.  So maybe best to use the older version which just assumes 1.5 which is hopefully close enough.


Re: Updated Wiki Page: Diverger lens Residual Spherical aberration #wiki-notice

George Roberts (Boston)
 

Lol.  Basically in a Bath, there if you use a biconvex lens there is a tiny contribution of that lens to the spherical aberration term.  Typically it may add less than 1/1000 of a wave of error to the DFTFringe result.  No big deal.  It depends on things like how big and fast your mirror is and what type of glass is in the thin lens and so on.  Bruce has an equation that lets you calculate the error.  I might make a web page calculator where you enter the terms but the error is so small it's probably not worth the trouble except...

If you use a PCX lens instead of a BCX lens (PCX - plano convex lens - that means one side of the diverger lens is flat) then this spherical aberration grows.  It's still typically less than 1/50th of a wave but it could be significant if you have an unusual mirror.  I have a calculator here to see how much PCX will affect your mirror here - just plug in your mirror parameters:
gr5.org/bath/

However Bruce improved the equation to include the index of refraction of your PCX lens which most people probably won't know.  So maybe best to use the older version which just assumes 1.5 which is hopefully close enough.


Re: Fourier transforms of Zernike polynomials

Bruce Griffiths
 

I foolishly assumed it would be easy to recreate and didn't note them down as I was merely establishing that the program ran at the time.

I'll see if I can recreate the issue but it may take several days.

Bruce

On 12 May 2021 at 11:38 Dale Eason <doeason@...> wrote:

Bruce,
What where the parameters of the mirror and size of the simulated wavefront?


Re: Fourier transforms of Zernike polynomials

Dale Eason
 

Bruce,
What where the parameters of the mirror and size of the simulated wavefront?


Re: Fourier transforms of Zernike polynomials

Bruce Griffiths
 

Yes, I was using the 64 bit version.

I first created a synthetic interferogram with only a defocus term and analysed it using the default size Gaussian blur and noticed the artifact near the center. Reducing the Gaussian blur size made it smaller, vanishing for a blur size of zero.


Bruce

On 12 May 2021 at 08:13 Dale Eason <doeason@...> wrote:

On Tue, May 11, 2021 at 02:53 PM, Bruce Griffiths wrote:
I had noticed that using a Gaussian blur adds a non existent hollow to the centre of the computed surface when analysing an interferogram that only  has defocus.
Bruce, Where you using DFTFringe for that experiment?   It does not sound logical too me that that should happen.   So I tried it myself and at least for the test case I chose it was not the case.   For a nulled wavefront with a defocus of 1 the 20% blur turned the edge as I expected but did nothing to the middle that I could see on a 600 x 600 surface.  MIrrors was 95mm diameter F4.6 with expected conic of -1 and a perfect wavefront with a defocus of 1.   Defocus enabled in analysis. 

Dale


Re: Fourier transforms of Zernike polynomials

Dale Eason
 

On Tue, May 11, 2021 at 02:53 PM, Bruce Griffiths wrote:
I had noticed that using a Gaussian blur adds a non existent hollow to the centre of the computed surface when analysing an interferogram that only  has defocus.
Bruce, Where you using DFTFringe for that experiment?   It does not sound logical too me that that should happen.   So I tried it myself and at least for the test case I chose it was not the case.   For a nulled wavefront with a defocus of 1 the 20% blur turned the edge as I expected but did nothing to the middle that I could see on a 600 x 600 surface.  MIrrors was 95mm diameter F4.6 with expected conic of -1 and a perfect wavefront with a defocus of 1.   Defocus enabled in analysis. 

Dale


Re: Fourier transforms of Zernike polynomials

Bruce Griffiths
 
Edited

I had noticed that using a Gaussian blur adds a non existent hollow to the centre of the computed surface when analysing an interferogram that only  has defocus. The amplitude/depth of the hole approaches zero as the width of the Gaussian blur approaches zero.


Bruce

On 12 May 2021 at 03:39 Michael Peck <mpeck1@...> wrote:

This has nothing to do directly with DFTFringe, so feel free to ignore this post. Fourier transforms of Zernike polynomials have some interesting properties and important applications -- the famous paper by Noll (1976) uses them as the starting point for an analysis of atmospheric turbulence. Zernike himself used them in the development of the phase contrast method, which is what won him the Nobel prize. His first papers on the topic actually concerned mirror testing: Zernike (2004a) (in German and still paywalled!) and Zernike (2004b) (shorter English version, but freely available). They also form the basis for the Nijboer-Zernike aberration theory, which evolved into the Extended Nijboer-Zernike theory of image formation which is still highly relevant in optics. This is mostly beyond the scope of this group though.

It turns out the FT of any Zernike polynomial has a closed form expression in terms of Bessel functions of the first kind as follows:

In polar coordinates the FT factorizes into a radial part that depends on spatial frequency and the radial Zernike order, and an azimuthal part that has the same angular dependence as the Zernikes themselves. I think the frequency k in the expression above has units of cycles per pupil diameter. Various literature sources give different values for the normalizing factors, but no matter. Here's what the radial part looks like for a few selected radial orders:

All of these except the FT of piston (top left graph) are 0 at a spatial frequency of zero. The power in each transform is always negligible up to some threshold that depends only on the radial order and they decay fairly slowly at higher spatial frequencies.

The paper I took that expression at the top from was by G. Dai (2006) titled " Zernike aberration coefficients transformed to and from Fourier series coefficients for wavefront representation," which attempts to do exactly what the title suggests. The problem is it won't work if the wavefront data is transformed using a DFT while the analytical expression is used for the FT of the Zernikes. The problem is the DFT gives a poor approximation to the FT. This can be seen for a couple of fairly high order pure aberrations below. The top panes depict the modulus and phase of the aberrations per the expression above. The bottom panes show the same values for the DFT of a 4K x 4K matrix filled with the same aberrations. The DFT gets the shape of the modulus about right although it undersamples the oscillating high frequency region. The phases are completely wrong though.


What I was actually interested in with this exercise was understanding the performance of Zernike fitting as a low pass filter, especially compared to a Gaussian blur which I think is the preferred method of smoothing in DFTFringe. Here is an example from some randomly selected real data I had laying around. First is a Zernike fit. The top left pane is the net but otherwise unfiltered wavefront (by net I mean with the software null applied and usual low order stuff removed). The two panes below it are the (modulus of) DFT of the wavefront and a zoomed in version of the same. The middle panes are the Zernike fit wavefront. The right is the residual, that is the difference between the net wavefront and the Zernike fit. Notice the dark disk in the center -- that region has virtually no power at all. The two bright specs in the residual DFT appear to be from some sinusoidal (electronic?) artifact that you can faintly see in the map.

And here's the same using a Gaussian blur. I think this used the value that Dale recommends as a default amount of smoothing. Even though this appears to have captured more detail the smoothed wavefront actually has a slightly lower RMS deviation than the Zernike fit one. Also, the former shows what looks like a complete ring at ~80% radius that's hard to pick out in the blurred map. In addition the residual from the Gaussian blur (which I think is equivalent to applying an unsharp mask) shows more power at low spatial frequencies.

So one characteristic of Zernike fitting that I think is advantageous is that it completely captures the low spatial frequency content of the measured wavefront. How much it captures can be adjusted by increasing or decreasing the number of terms in the fit (see below). The disadvantage is it's more computationally intensive and maybe prohibitively so if you really wanted to capture high frequency details. It might be worth investigating other forms of low pass filter. Something like a top hat but with a tapering edge might work.

This final graph shows the position of the first maximum of the radial part of the FT for radial orders up to 50. A fit up to order 50 using the indexing scheme of the extended Fringe set would require 676 Zernike components.


Fourier transforms of Zernike polynomials

Michael Peck
 
Edited

This has nothing to do directly with DFTFringe, so feel free to ignore this post. Fourier transforms of Zernike polynomials have some interesting properties and important applications -- the famous paper by Noll (1976) uses them as the starting point for an analysis of atmospheric turbulence. Zernike himself used them in the development of the phase contrast method, which is what won him the Nobel prize. His first papers on the topic actually concerned mirror testing: Zernike (2004a) (in German and still paywalled!) and Zernike (2004b) (shorter English version, but freely available). They also form the basis for the Nijboer-Zernike aberration theory, which evolved into the Extended Nijboer-Zernike theory of image formation which is still highly relevant in optics. This is mostly beyond the scope of this group though.

It turns out the FT of any Zernike polynomial has a closed form expression in terms of Bessel functions of the first kind as follows:

In polar coordinates the FT factorizes into a radial part that depends on spatial frequency and the radial Zernike order, and an azimuthal part that has the same angular dependence as the Zernikes themselves. I think the frequency k in the expression above has units of cycles per pupil diameter. Various literature sources give different values for the normalizing factors, but no matter. Here's what the radial part looks like for a few selected radial orders:

All of these except the FT of piston (top left graph) are 0 at a spatial frequency of zero. The power in each transform is always negligible up to some threshold that depends only on the radial order and they decay fairly slowly at higher spatial frequencies.

The paper I took that expression at the top from was by G. Dai (2006) titled "Zernike aberration coefficients transformed to and from Fourier series coefficients for wavefront representation," which attempts to do exactly what the title suggests. The problem is it won't work if the wavefront data is transformed using a DFT while the analytical expression is used for the FT of the Zernikes. The problem is the DFT gives a poor approximation to the FT. This can be seen for a couple of fairly high order pure aberrations below. The top panes depict the modulus and phase of the aberrations per the expression above. The bottom panes show the same values for the DFT of a 4K x 4K matrix filled with the same aberrations. The DFT gets the shape of the modulus about right although it undersamples the oscillating high frequency region. The phases are completely wrong though.


What I was actually interested in with this exercise was understanding the performance of Zernike fitting as a low pass filter, especially compared to a Gaussian blur which I think is the preferred method of smoothing in DFTFringe. Here is an example from some randomly selected real data I had laying around. First is a Zernike fit. The top left pane is the net but otherwise unfiltered wavefront (by net I mean with the software null applied and usual low order stuff removed). The two panes below it are the (modulus of) DFT of the wavefront and a zoomed in version of the same. The middle panes are the Zernike fit wavefront. The right is the residual, that is the difference between the net wavefront and the Zernike fit. Notice the dark disk in the center -- that region has virtually no power at all. The two bright specs in the residual DFT appear to be from some sinusoidal (electronic?) artifact that you can faintly see in the map.

And here's the same using a Gaussian blur. I think this used the value that Dale recommends as a default amount of smoothing. Even though this appears to have captured more detail the smoothed wavefront actually has a slightly lower RMS deviation than the Zernike fit one. Also, the former shows what looks like a complete ring at ~80% radius that's hard to pick out in the blurred map. In addition the residual from the Gaussian blur (which I think is equivalent to applying an unsharp mask) shows more power at low spatial frequencies.

So one characteristic of Zernike fitting that I think is advantageous is that it completely captures the low spatial frequency content of the measured wavefront. How much it captures can be adjusted by increasing or decreasing the number of terms in the fit (see below). The disadvantage is it's more computationally intensive and maybe prohibitively so if you really wanted to capture high frequency details. It might be worth investigating other forms of low pass filter. Something like a top hat but with a tapering edge might work.

This final graph shows the position of the first maximum of the radial part of the FT for radial orders up to 50. A fit up to order 50 using the indexing scheme of the extended Fringe set would require 676 Zernike components.


Re: Zernike term calculation ?

George Roberts (Boston)
 

Nice!  Now maybe watch the part of this video where I talk about Spherical Aberration and the DFTFringe null feature:
https://youtu.be/T5DNzVPwrzc?t=455


Re: Zernike term calculation ?

jeanpierre_chiappini@...
 
Edited

I used a pure sinus coma as an example.
With DFTFringe and its Zernike wavefront simulation tool I created a front with the value 0.771.
Then I saved that wavefront and turned it into an image.

     

Each point of the pupil at these coordinates and its height (h).
With the Excel spreadsheet I calculated the Wyant factor for each point I name (E).
I calculated  (Z7) = sum [E * h] / sum [E²] and found the exact term of 0.771.

To verify I have redone the calculation with these examples from other Wyants: (Z4) astigmatism, (Z6) coma cosine and (Z8) sphericity and all give 0 (in fact a very small number of powers -10 so zero).
So the principle of decomposition works well.

Thank you
@Dale Eason
@Bruce Griffiths
@Michael Peck
@George Roberts (Boston)
thanks to your help i was able to understand at my level.


Re: Zernike term calculation ?

Bruce Griffiths
 

Michael's explanation uses orthonomal Zernike polynomials which are weighted so that the integral of the square of each Zernike polynomial over the unit circle is 1. The Wyant Zernike polynomials aren't orthonormal but the scale factor required for each Wyant Zernike polynomial to convert it to orthonormal form is known.

The calculations are simpler if Orthonormal Zernike polynomials are used. However DFTFringe uses the Wyant form of the Zernike polynomials which are not orthonormal.

Bruce

On 11 May 2021 at 15:19 "George Roberts (Boston)" <bb@...> wrote:

On Mon, May 10, 2021 at 05:19 PM, <jeanpierre_chiappini@...> wrote:
Something like (Z7) is the mean of (h) / [r * (- 2 + 3 * r²) * sin (a)] ?
Yes!  Very close.  I think instead of "h / Z7" it's "h * Z7".  Instead of the "mean", it's the "integral".  Basically the same idea (for each pixel value h, find h*Z7 then divide by the surface area if radius of mirror=1 - this is basically the mean, except divide by surface area of each pixel instead of by qty of pixels).  Bruce explains it more accurately.  I guess you multiply by 8 at the end as Bruce says.

- George

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