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Does affect the QE sentitivity curve of a camera to a filtered V or B image?

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gblasco
gblasco's picture
Does affect the QE sentitivity curve of a camera to a filtered V or B image?

Hi,

I have studied how all the photometry bands are very well defined so there are several standard type of filters like Johnsons B or V, or Cousins R, that filter the precise curve of the photometri standard.

But as far as i understand the response to the CCD or the CMOS sensor about QE is not constant along the spectrum, so I understand that this affect and should be corrected in post processing. But I don't understand how can you correct this in a Mono camera where all photons of all wavelenghts are mixed in the well.

An example of a response curve of a camera can be found here:

https://www.qhyccd.com/index.php?m=content&c=index&a=show&catid=133&id=8&cut=1#:~:text=US$99%20(Color%20Only)

And if you apply any of this this curve, you will get the effect of summing both curves:

https://www.aip.de/en/research/facilities/stella/instruments/data/johnson-ubvri-filter-curves/leadImage/image_view_fullscreen

Any help for better understanding which is the process for take in account the QE curve of response?

Thanks

--

Gonzalo

 

spp
spp's picture
Transformation

Gonzalo,

Briefly, using your camera and photometric filters you observe "standard" stars (usually a standard cluster or a Landolt standard field) which have very well established colors and magnitudes. You determine how your measurements differ from the "standard" magnitudes and colors.  From these differences you calculate transformation coefficients which can be applied to your magnitude measurements to bring them close to the "standard" system.

This process, using spreadsheets, is described in detail in Chapter 6 of the AAVSO Guide to CCD Photometry.     You can find it translated into several languages here:

https://www.aavso.org/ccd-camera-photometry-guide

Or, you can just feed your filtered images of a standard cluster to TG (AAVSO's Transform Generator software).  ... but where is the fun in that?

Transformation coefficients can be applied to your measurements using a hand calculator or one of the AAVSO's software tools.

Phil

 

 

LMA
LMA's picture
Transformation, where is the TG

Phil,

You say ``Or, you can just feed your filtered images of a standard cluster to TG (AAVSO's Transform Generator software).  ... but where is the fun in that?``

OK for the fun, but sometime one just want to be efficient and save time. And part of the fun I have problem with is dealing with spread sheet, which is always confusing to me. So, this Transformation Generator would be a welcome addition to my tool bag, where is it?

Damien

spp
spp's picture
TG

Damien,

I'm sure that TG is a godsend to people who have to calculate a lot of  transforms (i.e for the AAVSOnet telescopes).  I have no evidence for this, but I would guess that most (almost all?) AAVSO observers who transform their data now use TG for calculating the transforms for their own systems.  That said, I think that using the spreadsheet method, at least once, gives you a better understanding of the transform process.

You can find TG here:  https://www.aavso.org/tg

Phil

gblasco
gblasco's picture
As Far as I understand this

As Far as I understand this doesnt solve the problem with the non-equal response to sensitivity in different wavelenghts of the spectrum.

Sensors have a QE curve... You can see one if you click here:

https://www.qhyccd.com/uploadfile/2018/1020/20181020041802554.jpg

As an example if you use the QHY5-II-M (EXAMPLE)

and you are using a red filter. The filter for the measurement gives you a 0.2 transmission at 750nm, but your sensor is only 38% efficient in QE at this Wavelength. This mean you are understamiting the measurement. because THE SENSOR response to the wavength is not constant across all the wavelenght

If you compare with other stars with different color, you cannot correct this because in monocrome sensors all wavelnghts go to the same bucket.

If the QE is not 100% but is CONSTANT also is not a problem

But qith a non constant QE across the spectrum, I think you whould have to correct the differences in the response of the sensor... This is the subject. No idea how... 

 

gblasco
gblasco's picture
DELETED

COPIED AT THE END OF THE THREAD! I thought the post could be deleted.

https://www.aavso.org/comment/71957#comment-71957

Bernhard
Bernhard's picture
QE

Hi,

For QE, e.g. see Wikipedia:

https://en.wikipedia.org/wiki/Quantum_efficiency

Quantum efficiency of Image Sensors : Quantum efficiency (QE) is the fraction of photon flux that contributes to the photocurrent in a photodetector or a pixel. Quantum efficiency is one of the most important parameters used to evaluate the quality of a detector and is often called the spectral response to reflect its wavelength dependence. It is defined as the number of signal electrons created per incident photon. In some cases it can exceed 100% (i.e. when more than one electron is created per incident photon).

A quick brain storm leads me to the following:

  • QE is the quality of collecting an proper amount of photons at a given time.
  • CCD's or DSLR with higher QE, are needing less time to get a propper Signal to Noise.

So with on old CCD with e.g. 30 % QE, one needs more integration time to get this proper S/N. A CCD with 80% QE needs less time to get a proper or needed S/N.

The unaided eye, needs 3-5 Photons to detect photons at visual. So the eye QE ~ 3...5

(I do not think that QE is changing the response of some filters??? )

An attempt to give DSLRs a QE is here:

https://www.astrophotography.app/EOS.php

 

 

regards wbea

SFS
The QE of the human eye is

The QE of the human eye is luminance dependent, and ranges from about 0.25% to 1% per the work of Rose and Blackwell, as referenced here:  https://www.osapublishing.org/josa/abstract.cfm?uri=josa-49-7-645

It should also be noted that amplification is not included in the definition of QE; QE is normalized to unity.

TRE
TRE's picture
QE

This page says it nicely:

https://ibsen.com/technology/detector-tutorial/calculating-quantum-efficiency/

It is as simple as that for any detector. If you add a VBR ot I filter, then, for each wavelength, you multiply the detector QE times the percent filter response. Of course any piece of glass will have about a 4%/cm transmission loss, and you can really get into many more details that are not really practical. 

The equalizer is the use of transforms, ie, how does your personal optical train handle a standard field like M67 or Landolt?

Also Hamamatsu has good tech memos.

TRE

SFS
As the QE is a function of

As the QE is a function of wavelength, the answer to your question is "Yes".  It is far more complicated than inferred from the Wikipedia article; If one wants to calculate the response, it is necessary to integrate product of the star's spectral radiant luminosity, transmission loss (space, atmospheric, and intra-optical system, filter transmissivity, detector quantum efficiency and amplifier gain over the common band of all.  All but the amplifier gain are wavelength dependent.  Good luck with that.

One of the many overlooked difficulties in using DSLRs for photometry is that different camera models have different spectral responses, so the Green signal from one Canon (or Nikon or Sony or ...) for a particular target at a particular moment in time will differ from that of another.

gblasco
gblasco's picture
With the transformations

With the transformations described in chapter 6 related to TG, can you compensate this with comparison stars? I undesrtand that this is only possible with a lot of reference stars measured with all the filters in the spectrum to make a good transformation.... Or all the TG procedure doesnt really affect about the QE non-linearity in wavelenght issue im talking about.

SFS
You don't need to know

You don't need to know anything about QE to do photometry.  Just do a good job of measuring your system's transformation coefficients.

I'm not advocating ignorance here, but one of the pitfalls of amateur activities is the paralysis that results from getting ever more deeply involved in ever more minute details in the desire to buy the best possible system.  As Voltaire (and Gorshkov, later) said:  the better is the enemy of the good.

arx
arx's picture
Put simply, transformation

Put simply, transformation coefficients allow you to calculate accurate colour indices from the measurements made on images taken with (for example) your B and V filters. Comparison stars allow you to use thise results to calculate (for example) accurate B and V magnitudes for your target stars. Have a look at the DSLR and CCD observing manuals. As noted above, you do not use QE values in these calculations.

Roy

spp
spp's picture
Star colors

I think I see the misunderstanding here.

Gonzalo has been looking at the wavelength response curves of detectors, QE vs wavelength.  The rest of us have been discussing star colors.  Of course, stars don't radiate at a single wavelength. 

The color of a star, for astronomers, is the difference in magnitude or photon flux as measured in two different photometric filters, B-V or V-I, etc.  In photometry we deal in star magnitudes and star colors, not wavelengths.

Color transforms correct for the differences in detector responses to star colors.

So, Gonzalo, forget about the detector response curves and focus on understanding and calculating transforms.

Phil 

 

gblasco
gblasco's picture
Agree

I agree 100%. This is the point

gblasco
gblasco's picture
THANKS!

THANK YOU ALL! here the solution:

Chapter 6: Transforming your data

This process, using spreadsheets, is described in detail in Chapter 6 of the AAVSO Guide to CCD Photometry.     You can find it translated into several languages here:

https://www.aavso.org/ccd-camera-photometry-guide

also there are SW to make things a little easy

But the original question comes after another thought I didnt mention before

In many comments I receive in the forums, and also is remarked in the guides published that the data must be taken with Johnson or Cousins filters that are not cheap at all beacuse they give very precise curve cutoff in the wavelength.... 

Ok, the sensors after reading many spec sheets are as far as 20% or 30% again from being constant in the wavelength response against the sensitivity.... so YOU NEED TO TRANSFORM !

If the transform is needed in all cases even with the BEST - MOST expensive filters in the world, because you have to correct many other factors, Why a so such precise filters???

I mean:

  • A precise filter + non precise factors = non precise measurements -> DO at least two measures with two filters and TRANSFORM
  • A non precise filter + non precise factors = non precise measurements -> DO at least two measures with two filters and TRANSFORM

So here comes the point. Of you nee to transform in both cases, why to expend so much money in precision filters

An example: here is the 20 color filter cheap set of oriion

https://www.telescope.com/Orion-125-Premium-20-Piece-Color-Planetary-Filter-Set/p/113384.uts

Or any other 7 filter set like this one

https://www.amazon.com/Orion-5590-Stargazers-1-25-Inch-Eyepiece/dp/B0000XMXWK

If you see the curve of the UVBRI system:

https://www.researchgate.net/profile/Yamam_Tawalbeh/publication/326412740/figure/fig3/AS:648859327926282@1531711527256/3-UBVRcIc-Johnson-Cousins-Filters-Johnson-Cousins-UBVRI-filter-curves.png

You see that you can cover most of the bands with filters of similar colors, and then make measurements with adjacent colors and make the "magic" of transform your data with comparison stars.

At the paper seems legit. And if you see that the 20 filter set is priced like only one Johnson V filter. 

In other words. If you can in any way irrgular measurements because the sensor, the atmosphere, and other factors, you should measure in different color bands, and then transform. In this case, if you transform the precision of the filters will be corrected in any way with the transformation. It will be no so good as with the expensive filters, mut the data processing can correct a lot if you have data with wnough filters.

I mean . Is this a solution if you can't abborf 600USD for a serious UBVRI set?

I hope I explained

Thanks

spp
spp's picture
Ah yes, photometric filters. Here we go.

Gonzalo,

You bring up several good points.  Before we start discussing these, let me first make a few general comments.

To submit photometry measurements to the AID (AAVSO International Database) you are not required to transform your measurements, nor are you required to use photometric filters.  This may seem odd given the emphasis placed on transforms. 

The AAVSO wants you to do the best you can.  Ultimately, this would be submitting transformed measurements, but if you can't afford photometric filters there are other types of measurements which are acceptable.  In these cases it is important to describe accurately how you made your measurements.

Examples:  CCD (or CMOS) measurements made with astrophotography tricolor filters are reported as TR, TG, and TB measurements.  In these measurements you would use the RED filter with standard R magnitude values for the TR comps, the GREEN filter with V magnitude values for the TG comps, etc.

You can even submit CB, CV, CR measurements. The are CCD/CMOS measurements made without using any filters.  For CB measuresment you would use the B magnitudes for the comps, for CV the V magnitude for the comps, etc.

In the light curve plots these measurements can fall surprisingly close to the transformed measurements. 

How can this be possible?

In many cases, the effect of applying transforms is quite small.  In some cases applying transforms can add more uncertainty without actually improving the accuracy of the measurement.

If your comp star is the same color as your target variable, then your untransformed measurement is already on the standard system and there is no effect of applying transforms.

This is also the case if your target variable has a value close to 0.0 in the color band (B-V or V-I) in which you are observing.

So, you don't need to buy those expensive photometric filters to get started with photometry.  If you do decide to get serious about photometry, you don't have to buy all the filters at once.  You can spread the cost of those filters out over time.

You can start with just a V filter and submit untransformed V measurement.  This is a step better than TG.  Later you can buy a B filter.  Then you can submit transformed B and V measurements.  Later you can add an Ic filter and submit transformed B,V, I measurements.

Phil 

spp
spp's picture
Gonzalo's question

"In other words. If you can in any way irrgular measurements because the sensor, the atmosphere, and other factors, you should measure in different color bands, and then transform. In this case, if you transform the precision of the filters will be corrected in any way with the transformation. It will be no so good as with the expensive filters, mut the data processing can correct a lot if you have data with wnough filters."

I mean . Is this a solution if you can't abborf 600USD for a serious UBVRI set?"

Gonzalo,

Let me restate the question just to be sure I've got it right:

"If we are going to go to the trouble of transforming our measurements in order to adjust our results to match the standard system, do we really need to pay all that money in order to use photometric filters?  Won't the transforms also correct for the non standard filters?  Can't we just use less expensive filters, for example Green instead of V and Blue instead of B?"

 

Yes, you can do that, and I'm told that this usually works well for some filters, particularly Green for V, but for other filters it may not work well.

Let's consider transforms for B-V. 

For the thoretical ideal camera/telescope/filter system, a system which produces measurements that exactly match the standard magnitudes, the Color Transform coefficient (Tbv) would be 1.000.  The B and V magnitude transform coefficients (Tb-bv and Tv-bv) would be 0.000.  Observers who use good quality photometric filters frequently have transforms quite close to these values  Let's call these "small" transforms.

If you use Green and Blue filters your transforms will be "larger", i.e. deviate more from the ideal system's transforms.  How much larger?  You will not know until you do it.

Why is this a problem?  The "larger" your transforms, the larger the magnitude adjustments become when you apply the transforms.   I think larger magnitude adjustments result in larger magnitude effects from what might otherwise be small systmatic errors (i.e. a small Red leak in the Blue filter). 

In other words, it seems to me that "larger" transforms magnify the effect of systematic errors.  I'd like to hear other opinions on this.

I think Arne probably has some rules-of-thumbs about how far transforms can deviate from the ideal before they become unusable.  I'd like to hear his thoughts on this.

Phil

gblasco
gblasco's picture
Thank you so much

I think I was very tired that morning and with less than usual amount of coffee, and TOO much workload to read again my post and find my buggy keyboard has not worked well with my brain. 

In other words, my text is awful, but you take a time to reformulate it. and IT IS PERFECT. That was exactly what I mean, and was the point of my question.

And your reply with the help of the chapter 6 in the guide, help me a lot to find alternate ways to do value measurements with transformed values and different filters.

The key point is how much can you transform without loosing precission. "and deviate from the ideal before become unusable".

I have a little more to say, but I think it fits better in other reply of this forum, so I will go there.

Thank you very much, you hit the point and give me a valuable answer. I appreciate so much

:-)

Eric Dose
Eric Dose's picture
The real key point

The real key point about transformation suitability is this: It is the observer's own responsibility to demonstrate that his own transformations successfully correct his own magnitudes for his own equipment, targets, and likely comp stars.

No one can give you an entirely general rule for your specific system. The observer really has to verify his own case. It seems like a lot of work, but it need be done maybe 1 or 2 times per year (and it's much easier the second time).

If verified, transformations can be extended pretty far. As a perhaps extreme example: for minor planets I use a Clear filter and transform the magnitudes to Sloan r, with no more than 5-10 millimagnitude deviations. But then, this work has advantages: I apply a quadratic (not just linear) transformation, I use 50-100 comp stars, and I severely restrict the comp-star color range since the target's color is known and is very sun-like. I spent many hours verifying that this unusual transformation works well. Now I enjoy the benefit.

Ed Wiley_WEY
Ed Wiley_WEY's picture
The "Why: of photometric filters

Gonzalo,

Professional and amateur astronomers want their data to be standard and used by others. Over the years the professional community has developed standard sets of various kinds, The Johnson-Cousins system of UBVRI being one of these standards. So if you go to a catalog of strars and see that the magnitude of a star you want to measure is Johnson V = 12.4, you know that when you properly image, reduce, measure and transform that star with a V-filter you should get a value close to V = 12.4. And, if everyone using the same kind of filter then everyone should get something close to V = 12.4. Data can be shared or used by others confident that they are "playing the same game."

Perhaps a sports analogy is in order: What would happen if every time your football team showed up at the football field (and I mean real football not American football) the field was a different size or the ball was a different size? It would be confusing. 

There is nothing magic about Johnson-Cousins filters, they are simply one of several kinds of standard sets, developed over time, by the professional community. There are standard stars that were used to calibrate the behavior of these filters so that consistent results are achievable and can be shared by anyone who uses them. They are designed, as a group, to act as a crude spectrograph, sampling different areas of the spectrum (thus different colors) in a consistant manner that should be repeatable from observer to observer. They furnish all sorts of information about a star because of the vast amount of background data that are available. 

Phil's comments are on target. Although many observers start with DSLRs or other colored filters, as their interest develops, they usually move towards standard photometric filters. I know observers who only have a Johnson V.

Finally: Why expensive? They are precision made and they are "niche" items with low sales volume. But, check out some of the imaging filters, you will find the narrow band filters expensive and for the same reasons.

Ed

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