Optical resolution in DS imaging.

[vlaiv]

5 minutes ago, Stub Mandrel said:

Would you agree that if the average noise level is close to the signal from the star (say close to 0.5) it might be visible in the first instance, but not in the second?

Well, although Andrew's example works - it is not really good example to apply to star SNR threshold. Unit impulse of DX width is not band limited - so any reconstruction of that signal is going to fail (missing high frequencies) and there will be some aliasing. If we talk about sampling of a star then sampling rate is going to be at least 3-4 pixels across the middle section to capture star shape.

Also don't mix noise with resolution - we are discussing theoretical resolution on imaging plane and that sampled by sensor. Premise of our theoretical approach is pure signal. If you have pure band limited signal and you sample it - you will be able to reconstruct it. If you have noisy samples - you will not be able to reconstruct perfect signal, due to noise, if nothing else. Noise remains part of the process but it is handled differently (longer exposure, stacking, ....).

[andrew s]

13 hours ago, Stub Mandrel said:

Would you agree that if the average noise level is close to the signal from the star (say close to 0.5) it might be visible in the first instance, but not in the second?

I think the physical reality is about the number of electrons in the well compared to the noise (read, sky bark etc.) The two examples are just about how we model the sampling process.

The physical read is of the electrons in each well.

We get one number that can be regarded as a point sample or a sample integrted over the pixel .The two are simply related in the maths. Its like measuring in cm rather than inches.

Regards Andrew

PS a better example might be measuring mass or volume when we know the density.

[andrew s]

15 hours ago, vlaiv said:

On the matter of "surface integral sampling" vs point sampling - it turns out that they are exactly the same

I tested this with a parabola 1-x^2 with integral x - ((x^3)/3) Using values x= -1 to 1 in intervals of 0.05. I then simulated 20,9 and 3 pixels covering this range using the exact definite integral. The ratio of the single point sample / surface integral sample was close to constant but the errors (measured as the ratio of the STDV to mean) were 0.2%, 1% and 2% for 20, 9, 3 pixels.

I suspect the difference is due to a parabola not being a periodic function.

Remember this is only about modeling the sampling process these error do not occur in the real world!

Regards Andrew

[vlaiv]

3 hours ago, andrew s said:

I tested this with a parabola 1-x^2 with integral x - ((x^3)/3) Using values x= -1 to 1 in intervals of 0.05. I then simulated 20,9 and 3 pixels covering this range using the exact definite integral. The ratio of the single point sample / surface integral sample was close to constant but the errors (measured as the ratio of the STDV to mean) were 0.2%, 1% and 2% for 20, 9, 3 pixels.

I suspect the difference is due to a parabola not being a periodic function.

Remember this is only about modeling the sampling process these error do not occur in the real world!

Regards Andrew

Yes, main thing with Nyquist Shannon sampling is that original signal can be reconstructed completely if samping rate is larger than twice maximum frequency of signal. This poses slight problem for any sort of practical signal - we can't have infinite sampling, and all signals that are spatially bounded are going to be unbounded in frequency domain (in order to have zeros before certain spatial value, and after certain spatial value - one needs infinite amount of sine functions - each with different frequency / phase). Even for images that are band limited by aperture, for full reconstruction of signal we need infinite number of sampling points (infinite sensor) - Airy pattern function goes off to infinity and if we want to capture perfect Airy pattern of a single star we need to sample each diffraction ring, although each successive ring being less and less in intensity they spread out to infinity.

So whenever we have spatially bounded function, or use limited number of sampling points - we are going to have some aliasing (overlap of higher harmonics in frequency domain with base frequency response) - so in real life 100% reconstruction is not possible.

One more thing to note is that many different sets of samples (although not being equal to each other) - reconstruct bandwidth limited signal perfectly. This can be easily shown if we have band limited function and sample at correct frequency, but each set of samples - we shift for small dx. So first set will be at ..., -1, 0, 1, ...., next one will be at ...., -1+dx, 0+dx, 1+dx, .... - sampling frequency does not change, and we will be able to reconstruct original signal from each sampling set, but sampling sets will have different sample values across the whole set.

All this combined makes finding correspondence between integrating and point sampling a bit hard, one can often run into examples where you don't get exact same values with both methods (as would also happen had you shifted point sampling by some constant). What is important is that we can consider integral sampling as being point sampling - thus we don't need to further examine it as a special case - everything that holds for point sampling will equally hold for integral sampling in domain where Nyquist Shannon sampling works (frequency bounded functions).

There might be a difference if we deviate from base theorem assumptions - we have frequency unbounded function, or we sample at inadequate rate.

For example - consider unitary pulse with width less than point sampling rate.

If point sampling we will get two types of responses, first type: all zeros - unitary pulse is in between of sampling points, or all zeros and single 1 at some sampling point - sampling point is "inside" unitary pulse. Here we have obvious aliasing - we don't need original function to be unitary pulse - it can have whichever shape one chooses as long as spatial extent is less than sampling wavelength. Hence term aliasing - we have one function that acts as alias for all other functions that can be sampled like that - in this case 0 function is alias for all functions of arbitrary shape that have spatial extent less than sampling wavelength and are sampled such that no sample point falls where function is non zero.

If we use integral sampling, we can have first case - all zeros - if integral width is less than sampling wavelength, and it happens that spatial extent of function is less than sampling wavelength - integral width, and there is certain alignment. We can on the other hand have 2 non zero sample values (third case). Again there will be aliasing to some other function.

So we see that there are cases, where we have function that we sample at certain rate - point sampling and integral sampling will indeed produce different results - but this is due to aliasing rather than improper sampling (again shifting point sampling will alias samples to different function and you will get different result, like in above case where in one case it aliases to 0 function and in other case to a different function).

[andrew s]

Vlaiv - I agree with most of what you say above. I also agree that you established that for periodic functions that point sampling and integral sampling were equivalent. 

I think your insight and exploration of this issue has been comprehensive.

However, for the parabola I modeled I think I established they were not quite equivalent. The two sets of data, though close, were different.

There were no gaps between samples and there was no numerical integration. This has nothing to do with phase shift as both were the same or any reconstruction issues e.g. aliasing as they were not reconstructed. 

Quite simply for this non periodic function the two were not equivalent as they produced different sets of sampled data not related by a multiplicative constant. This is not surprising as the function is not periodic.

Regards Andrew

[vlaiv]

2 minutes ago, andrew s said:

Vlaiv - I agree with most of what you say above. I also agree that you established that for periodic functions that point sampling and integral sampling were equivalent. 

I think your insight and exploration of this issue has been comprehensive.

However, for the parabola I modeled I think I established they were not quite equivalent. The two sets of data, though close, were different.

There were no gaps between samples and there was no numerical integration. This has nothing to do with phase shift as both were the same or any reconstruction issues e.g. aliasing as they were not reconstructed. 

Quite simply for this non periodic function the two were not equivalent as they produced different sets of sampled data not related by a multiplicative constant. This is not surprising as the function is not periodic.

Regards Andrew

Aliasing happens with the process of sampling - it has nothing to do with reconstruction. Sorry if I'm reiterating things that you know already. Point sampling uses pulse train and FFT of pulse train is again pulse train in frequency domain. Denser pulse train in spatial gives sparser pulse train in frequency and vice verse. So we sample by multiplying our pulse train with function that we are sampling. This in frequency domain means convolution with pulse train, and convolution with pulse train is same frequency response at each pulse overlapping. Now if frequency response is of a limited bandwidth and pulses in spatial domain are close enough (meaning they are far enough in frequency domain) - convolution of original frequency response will not overlap. No aliasing in this case - we don't have restored signal, we still have samples, but FFT of samples represents "frequency response train" in frequency domain - this is why we need filter (or signal restoration) to limit frequency response train to only one wagon.

But if we have overlap (because pulses are sparse in spatial domain, and dense in frequency domain) there is aliasing even without us trying to filter out higher "wagons" and restoring function.

No overlap / no aliasing:

Overlap / aliasing:

Note, both graphs show FFT of sampled (not restored) signal (and it continues to repeat blue frequency response both left and right over and over again).

When we want to restore original signal - then we filter out blue part, but whether or not we apply filter we have overlap and aliasing.

Now, for point sampling aliasing is obvious from above graphs, but for integral sampling, I suspect that base frequency band is the same (blue part of graph), while green graphs will be somehow different (stretched, or otherwise distorted - we need to do analysis of complete integral sampling FFT process to see what would FFT of such sampling look like) and if there is overlap then it resulting alias will be different in integral sampling than that of point sampling.

I think that we can even do that analysis rather easily - just need to fiddle around a bit with integrals and summation to do FFT of integral sampling and conclude what it produces in frequency domain.

[andrew s]

I accept that their is aliasing due to sampling, as you show above. 

However, I do not agree that this is the cause of the failure of the equivalence between point sampling and integral sampling. If the two are different in the space domain they can't be the same in the frequency domain and can't have the the same reconstruction with or without aliasing. 

Regards Andrew

[vlaiv]

4 minutes ago, andrew s said:

I accept that their is aliasing due to sampling, as you show above. 

However, I do not agree that this is the cause of the failure of the equivalence between point sampling and integral sampling. If the two are different in the space domain they can't be the same in the frequency domain and can't have the the same reconstruction with or without aliasing. 

Regards Andrew

Indeed - if two are different in space domain that means that both are different in frequency domain.

If both point sampling and integral sampling work for band limited function with proper sampling frequency - this means that only difference in frequency domain is due to higher "harmonics" of base signal.

Here is Fourier transform of square wave (to compare it to pulse train):

It differs to FFT of pulse train by having each pulse away from origin modulated by sinc function - so if it was matter of simply multiplying by square wave function (not integrating by repeating pulse function) higher images of base frequency response would be scaled and / or inverted. This means that (depending on square wave width) - aliasing part if there is aliasing - could be negative, could be positive, but it is surely smaller in intensity. With pulse train we have exact copy of base frequency response repeating each cycle (and may or may not overlap / alias).

We have shown equivalence in case of true periodic functions (by decomposition into sum of sines). In real life scenario there will be some aliasing always (not sampling with infinity sampler, and often sampling at lower than Nyquist rate in DSO imaging) - luckily we have natural low pass filter - aperture, and set of other low pass filters - seeing / guiding errors - that make aliasing artifacts for both true point and integral sampling (although different) minimal, because high frequency part of spectrum is already very attenuated.

[andrew s]

22 minutes ago, vlaiv said:

If both point sampling and integral sampling work for band limited function with proper sampling frequency - this means that only difference in frequency domain is due to higher "harmonics" of base signal.

I don't agree with this as they are not the same function in the real world we don't point sample a CCD or CMOS detector. The question was if we could model it as point sampling.

However, I think we have gone far enough with this at least for me.

Thanks for all your input.

Regards Andrew

[vlaiv]

Although I claimed that integral sampling and point sampling should yield same results - after further and deeper analysis, owing to @alan4908 for maintaining that there is indeed pixel sampling blur, and @andrew s (not trying to pull you back into discussion, just giving credit where credit is due) for providing simulation that showed that there is difference between point and integral sampling, I concluded that there is indeed additional element of blur due to pixels not being point sampling device, and here is mathematical background, as well as approximation to blur due to sampling pixel size.

First let's examine expression for sampled function when using integral sampling or pixels (expression is 1D case, but it shows important point):

Sampling function can be derived using following logic - rectangle function centered at 0 with width w/2 - P sub w in expression is shifted to sample point, multiplied with function being sampled - F and integrated over whole continuum, since P is effectively 0 outside of -w/2, w/2 range - this will be the same as integrating over that interval centered on X. Since we don't want any odd X, just those at sampling points N - we then multiply integral with delta function shifted to particular X at sampling point N. We then sum over all sampling points to give us our sampled function.

Looking at above expression it becomes evident that integral part is by definition convolution, and that summation and multiplication with delta function is standard sampling - so above formula can be interpreted - convolve F with P and then do usual point sampling. This clearly shows that result of integral sampling is blurring by kernel P sub w.

Fourier transform of rectangle function is sinc function. So in order to incorporate this blur into above equation for total blur, we need either to approximate rectangle function with gaussian, or sinc function with gaussian.

When I approximated sinc with gaussian (in frequency domain) doing simulation, conversion parameter that I've got was 3.077. Then I set off to verify that finding by using binning calculations devised by Alan and measuring sigma of binned images. I got slightly different results: best fit for binning from 0.5 to 1, for blur sigma of 0.5, 0.75, 1, 1.25, 1.5, 1.75 and 2 was coefficient of 3.35, while for x3 binning from 0.5 to 1.5, best fit was closer to 5.

This is to be expected - it is not error in process it is error in approximation of rectangle / sinc functions with respective gaussians (either in spatial or frequency domain). For most used pixel sizes (1-2"/pixel) I would say coefficient of 4.5 is good approximation. It gives 0.007 standard deviation (measured - calculated sigma) for x2 bin, and 0.008 for x3 bin.

So formula so far would be:

FWHM = 2.355 * sqrt((seeing/2.355)^2 + (airy disk * 0.367)^2 + (pixel size / 4.5)^2 + (guide RMS * unknown)^2)

Where seeing is given in FWHM arc seconds, airy disk is given in arc seconds, pixel size is given in arc seconds. We have yet to determine unknown coefficient for guide RMS measure.

 

[andrew s]

1 hour ago, vlaiv said:

Although I claimed that integral sampling and point sampling should yield same results

vilaiv - I admire your stamina and analytical powers. I found this paper https://www.radioeng.cz/fulltexts/2004/04_04_27_34.pdf which I feel confirms what you have just found - Fig 4 equation 11.

Regards Andrew

[alan4908]

3 hours ago, vlaiv said:

So formula so far would be:

FWHM = 2.355 * sqrt((seeing/2.355)^2 + (airy disk * 0.367)^2 + (pixel size / 4.5)^2 + (guide RMS * unknown)^2)

Where seeing is given in FWHM arc seconds, airy disk is given in arc seconds, pixel size is given in arc seconds. We have yet to determine unknown coefficient for guide RMS measure.

Vlaiv 

This is exactly what I was after - well done !

Alan