Imagine you have some optical spectra (intensity vs wavelength) you want to measure. You do so by measuring the intensity of the spectra hitting a detector after placing multiple long pass optical filters in front of the detector. Each successive long pass filter has a longer wavelength at which it allows light to be transmitted. If the total intensity of the optical signal is initially measured, by subtracting the measured intensity (for a particular filter) and the intensity of all previously measured bands (assuming you're measuring sequentially, say, from shorter to longer wavelengths) you can get the intensity of a "band," where the size of the band is determined by how far apart your long pass filters are. The more filters you have, the more finely you can measure an optical spectra. The only issue is, in reality, filters generally have Gaussian-like edges that are not perfect step functions. If your bands are sufficiently small, these edges will overlap. Does anyone know of some kind of transform or algorithm whereby overlapping edges could still be used to determine an optical spectra accurately? Simply subtracting intensities doesn't seem to work because of the overlap... you can't as easily attribute an intensity to a band. Any suggestions would be greatly appreciated!
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$\begingroup$ what is a "long pass" filter? $\endgroup$– hyportnexCommented Mar 28 at 23:10
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$\begingroup$ @hyportnex a long pass filter is an optical filter that allows "long" wavelengths to pass and blocks short wavelengths. In other words (in frequency) its a "low-pass" filter. Of course that its usually used in the context of your "useful"/existing wavelengths. An ideal long pass filter would be a step function that allows wavelengths longer than the specified filter's cut-off pass and blocks all shorter wavelengths. $\endgroup$– José AndradeCommented Mar 29 at 18:14
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$\begingroup$ I was going to give an answer, but I am unsure of the motivation of this question, simply because a spectrometer is a way to measure the spectrum in an incredibly easy and un-cumbersome way. The method above at first glance should work but the resolution is lower or equal to the number of filters, and needs more conditioning outside of the zone of interest, that it makes no sense to make such a cumbersome measurement. As OP is eager for suggestions on de-convolution of the signal, I am unsure if it's just curiosity or an attempt at the above, which I would recommend otherwise. $\endgroup$– José AndradeCommented Mar 29 at 18:37
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$\begingroup$ @JoséAndrade I am aware of the ease in using a spectrometer but the system I am working with is different. The application is in hyperspectral imaging. If you have any suggestions for the problem I outlined I would gladly take them. Thank you $\endgroup$– Will BraunCommented Mar 31 at 3:18
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$\begingroup$ took me literally 5 seconds of google to find this: opg.optica.org/ol/abstract.cfm?uri=ol-33-9-1023 This is the same type of answer I was going to give: your output is the multiplication of a matrix with the spectral transmissions with a vector with your source's spectrum. $\endgroup$– José AndradeCommented Apr 1 at 18:32
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