Surface contrast in CaK for various filter bandpasses.
Posted: Fri Sep 14, 2018 8:14 pm
Hello Everyone.
I recently posted an animated spectral sweep through the CaK assembled from images taken with my spectroheliograph on Aug08.
viewtopic.php?f=8&t=25028
It brought up some interesting questions from Valery and Christian that I thought I'd look into. The basic question was: What filter bandpass would be required to see filaments in a CaK image? The bandpass of my SHG (in the configuration used) was about 180 milliAngstroms. This is narrow enough to clearly see filaments with very good contrast. Here's a line centre image:
To answer the question above, I simulated the performance of a filter of given bandpass by averaging over a specific number of frames used in the spectral sweep animation. Here is a montage of 25 out of 40 of these frames around line centre.
A large size version of this image with frames labelled in wavelength offsets can be found here:
https://www.astrobin.com/full/366718/0/?nc=user
The wavelength separation between frames is 95.8 milliAngstroms and I've taken line centre to be located at frame 20. To give a reasonable simulation of filter performance, I used a weighted average of frames, weighted in such a way to mimic a Gaussian filter profile. For example, here is a set of weighting coefficients for a filter fwhm bandpass of about 1.1 Angstroms, shown in graphical form. To make my job simpler, I stopped averaging frames whose contribution to the average was determined by 1% or less (weighting coefficient of 0.01 or less). The image frame number corresponds to the number shown in the montage above and the corresponding wavelength offsets from line centre are also shown in the plot.
Here are some results, showing simulated CaK images for fwhm filter bandpasses of 1113, 860, 593, 500, 400 and 300 milliAngstroms. I'll leave it to you to judge at what point filament contrast becomes good. To me, it looks like a fwhm of less than 400 milliAngstroms would be very desirable. By the way, there was a bit of processing applied: levels and sharpening. Note that the Lunt CaK module has a bandpass 0f ~2400 milliAngstroms, making filaments essentially invisible.
There is a fine point to the analysis I used. To take into account the native bandpass of the SHG, I used the following equation to relate the desired filter bandpass to the SHG bandpass and the fwhm of the Gaussian curve used to give the weighting coefficients for the average ("SIM" in the equation):
This equation basically relates the bandpass of the modelled filter, the SHG and the simulation curve (SIM) through convolution. So, for example, a filter bandpass of 500 milliAngstroms involved a "SIM" fwhm of 466.5 milliAngstroms convoluted with the 180 milliAngstrom SHG bandpass.
So, the conclusion is that a CaK filter bandpass of less than 0.4 Angstroms would do a very good job, giving very contrasty views of the Ca network and filaments. I don't know much about filter manufacturing fine points, but perhaps this is not too difficult to achieve?
Cheers.
Peter
I recently posted an animated spectral sweep through the CaK assembled from images taken with my spectroheliograph on Aug08.
viewtopic.php?f=8&t=25028
It brought up some interesting questions from Valery and Christian that I thought I'd look into. The basic question was: What filter bandpass would be required to see filaments in a CaK image? The bandpass of my SHG (in the configuration used) was about 180 milliAngstroms. This is narrow enough to clearly see filaments with very good contrast. Here's a line centre image:
To answer the question above, I simulated the performance of a filter of given bandpass by averaging over a specific number of frames used in the spectral sweep animation. Here is a montage of 25 out of 40 of these frames around line centre.
A large size version of this image with frames labelled in wavelength offsets can be found here:
https://www.astrobin.com/full/366718/0/?nc=user
The wavelength separation between frames is 95.8 milliAngstroms and I've taken line centre to be located at frame 20. To give a reasonable simulation of filter performance, I used a weighted average of frames, weighted in such a way to mimic a Gaussian filter profile. For example, here is a set of weighting coefficients for a filter fwhm bandpass of about 1.1 Angstroms, shown in graphical form. To make my job simpler, I stopped averaging frames whose contribution to the average was determined by 1% or less (weighting coefficient of 0.01 or less). The image frame number corresponds to the number shown in the montage above and the corresponding wavelength offsets from line centre are also shown in the plot.
Here are some results, showing simulated CaK images for fwhm filter bandpasses of 1113, 860, 593, 500, 400 and 300 milliAngstroms. I'll leave it to you to judge at what point filament contrast becomes good. To me, it looks like a fwhm of less than 400 milliAngstroms would be very desirable. By the way, there was a bit of processing applied: levels and sharpening. Note that the Lunt CaK module has a bandpass 0f ~2400 milliAngstroms, making filaments essentially invisible.
There is a fine point to the analysis I used. To take into account the native bandpass of the SHG, I used the following equation to relate the desired filter bandpass to the SHG bandpass and the fwhm of the Gaussian curve used to give the weighting coefficients for the average ("SIM" in the equation):
This equation basically relates the bandpass of the modelled filter, the SHG and the simulation curve (SIM) through convolution. So, for example, a filter bandpass of 500 milliAngstroms involved a "SIM" fwhm of 466.5 milliAngstroms convoluted with the 180 milliAngstrom SHG bandpass.
So, the conclusion is that a CaK filter bandpass of less than 0.4 Angstroms would do a very good job, giving very contrasty views of the Ca network and filaments. I don't know much about filter manufacturing fine points, but perhaps this is not too difficult to achieve?
Cheers.
Peter