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FreeTrack Forum > FreeTrack : English Forum > Support : Tracking System > Spectral analysis of home-made filters and PS3 Eye filter

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Gian92 #1 20/04/2012 - 20h33

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Hello folks, I've been reading round some interesting stuff, and I wanted to share it with you.

The following image is the result of the spectral analysis of:

_Light bulb radiation (reference)
_PS3 Eye infrared cut-off filter
_1 photographic film layer used as infrared-passing filter
_2 photographic film layers used as infrared-passing filter
_Floppy magnetic layer used as filter; it's more a cheap long-pass filter
_Infrared LED radiation

Posted Image

Credits for the single graphics go to madian of nuigroup.com: http://nuigroup.com/forums/viewthread/6458/

I stretched the graphics, coloured them, made the explicative legend and put the colour spectrum.


So, our target is to block radiation below 750 nm (~380 nm < x < ~750 nm= visible spectrum= light) and allow to pass radiation with greater wavelength (that is, infrared).

As we can see one film layer is not an absolute infrared-passing filter, since we have a peak at 600 nm. Anyway it attenuates well the radiation below ~720 nm.
Two film layers are enough to block radiation up to ~750 nm, and let pass infrared. Above 800 nm this home-made filter is almost transparent, and it is great since IR LEDs have their wavelength starting from this value. Use three layers if you have strong light coming to your tracking webcam.

The magnetic layer of a floppy disk is pretty awful compared to the film layers as it lets pass radiation from around 570 nm, although attenuated. Two floppy magnetic layers are almost opaque, i.e. useless.

On the other hand, the hot-mirror infrared cut-off filter of the PS3 Eye does great in what it was meant: it reflects almost completely the infrared and near-infrared radiation starting from 700 nm. That's why in some cases removing this filter is needed, especially when using reflective material instead of LEDs.
Edited by Gian92 on 25/05/2012 at 11h12.
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xxtraloud #2 21/04/2012 - 01h44

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Very interesting, thanks for posting this.
It is also true that the PS3 eye lets more infrared light at 850 nm go through then the two film. That's probably why people have managed to use it without removing the infrared blocking filter. I am still wondering though if this is the new or old version of the camera. That sorts of answer my question why my camera is the detecting the infrared LEDs without being modified. Also I am wondering what is the difference between unexposed and exposed film.
Gian92 #3 21/04/2012 - 02h21

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xxtraloud @ 21/04/2012 - 03h44 a dit:

It is also true that the PS3 eye lets more infrared light at 850 nm go through then the two film.


I think that it is more related to intensity of radiation as perhaps an infrared LED emits more infrared radiation than a light bulb (anyway the PS3 filter and the filter made of two photographic film layers are different: the home-made filter is transparent to infrared).

xxtraloud @ 21/04/2012 - 03h44 a dit:

I am still wondering though if this is the new or old version of the camera.


Looking at the date of the thread (2009) I guess it's the old one.

xxtraloud @ 21/04/2012 - 03h44 a dit:

Also I am wondering what is the difference between unexposed and exposed film.


If I remember correctly, an exposed film is the one that has seen light.
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Gian92 #4 21/04/2012 - 02h25

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One last thing, remember that the radiation reference is a light bulb, so you won't see here the behaviour of the home-made filters with infrared LEDs.
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xxtraloud #5 21/04/2012 - 05h38

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*I am sorry but to me the graph only makes sense if the area under the curves is the amount of intensity that a filter would allow to pass. I interpret the graph at a absolute scale so to me saying that it is based on a regular light bulb does not make much sense. The gray line supposedly depicts the intensity of the light bulb at different wavelengths.
A filter by itself or a photographic film has no intensity therefore the area under each curve must be the amount (intensity) of light at a particular wavelength.

As far as the exposed versus unexposed photographic film I just meant what would be their different responses in this particular graph.

I hope I made that clear.
Gian92 #6 21/04/2012 - 05h58

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Where does the radiation come from then?
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Ezraneut #7 24/04/2012 - 08h12

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This might solve my problem with reflectors in stead of LEDs on the head. Using a floppy disc as a filter, but im not getting enough light from the reflectors. Thanks for sharing this.
Gian92 #8 24/04/2012 - 11h30

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Ezraneut @ 24/04/2012 - 10h12 a dit:

Thanks for sharing this.



My pleasure  ^^
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xxtraloud #9 25/04/2012 - 16h48

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I reached that conclusion by looking at the axis, there is only One label for the Y axis so you can't really use different types of measures for the same curves. At least that's my understanding.

Gian92 @ 21/04/2012 - 07h58 a dit:

Where does the radiation come from then?

Gian92 #10 26/04/2012 - 14h38

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All right, I will try my best to be as clear as possible.

xxtraloud @ 21/04/2012 - 07h38 a dit:

*I am sorry but to me the graph only makes sense if the area under the curves [...]


When reading a graph, you consider the points that are in the curves, not an "area" below. Evey point has a couple of values: (x, y).

xxtraloud @ 21/04/2012 - 07h38 a dit:

I interpret the graph at a absolute scale so to me saying that it is based on a regular light bulb does not make much sense.


You are right and wrong about the scales: the x axis has its non-absolute unit, which is wavelength, while the y axis has the absolute unit (referred to as "counts" in the original thread).
Furthermore, this is not the reason why the graph is based on the radiation emitted by a light bulb, I will explain it later.

xxtraloud @ 21/04/2012 - 07h38 a dit:

The gray line supposedly depicts the intensity of the light bulb at different wavelengths.


Correct.

xxtraloud @ 21/04/2012 - 07h38 a dit:

A filter by itself or a photographic film has no intensity therefore the area under each curve must be the amount (intensity) of light at a particular wavelength.


Correct. And this returns to my question: where does the light come from?

Guess what? It comes from the light bulb. The grey curve is precisely the reference radiation, that is the amount of radiation that comes from the light bulb to the spectrometer without any alteration (i.e. with no filtering), and registered.

The other curves, except for the violet one, represent the same light of said light bulb that passes through the indicated filters which are interposed between the source (light bulb) and the spectrometer. Such filtered light is then captured and registered by the instrument. It is pretty straightforward.

xxtraloud @ 21/04/2012 - 07h38 a dit:

As far as the exposed versus unexposed photographic film I just meant what would be their different responses in this particular graph.


How can you specifically say that if they are not present in "this particular graph"? You cannot invent.

xxtraloud @ 25/04/2012 - 18h48 a dit:

I reached that conclusion by looking at the axis, there is only One label for the Y axis so you can't really use different types of measures for the same curves.


In fact, for the y axis the same unit is obviously used.

The problem is that, since we do not know how the measurements were taken - that is, for example, we do not know if the distance from the radiation source to the instrument was either maintaned or not, and other circumnstances - we are not able to say if the intensity of the whole curves varies proportionally to each other, because we do not know if the conditions at each measurement were maintained. Here is the raison d'être of the note of the graph.

It is really hard to understand how you did your own graph interpretation and your explanation about your "absolutism" of the values (the light had to come from somewhere, did not it?).

I hope I made that clear.
Edited by Gian92 on 27/04/2012 at 07h21.
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