Separation of H alpha and NII

[pmlogg]

I've read posts suggesting subtracting H alpha 3nm from H alpha 5nm to separate the H Alpha and NII data.

I apologise if this a silly suggestion but in the absence of having both bandwidth H alpha filters would a viable option be to instead subtract 3nm NII from 5nm H alpha? 

[wimvb]

If the Nii source is bright enough, this may work. But the 5 nm filter will also contain red background, more than the 3 nm filter. So, it depends how much each filter will let through outside the Ha and Nii peak as well as outside its passband. Plus you will have to consider the transmissivity at peak for both filters. All in all a non trivial exercise in mathematics.

[pmlogg]

Wim

Thanks for that.  A far from simple process to get it right.

Peter

[vlaiv]

On 18/06/2020 at 17:15, pmlogg said:

I've read posts suggesting subtracting H alpha 3nm from H alpha 5nm to separate the H Alpha and NII data.

I apologise if this a silly suggestion but in the absence of having both bandwidth H alpha filters would a viable option be to instead subtract 3nm NII from 5nm H alpha? 

If you already have NII filter - why would you want to extract it from Ha? You have means to record it as is.

I suppose that original post suggested that you can get NII data by using 5nm Ha and 3nm Ha.

5nm Ha is wide enough to have both Ha and NII data combined as their wavelengths are 656.461 and 658.527, so 656.5 +/- 2.5 => 654nm to 659nm and that includes 658.5, while Ha 3nm will not do that 656.5 +/- 1.5 => 655nm to 658nm does not include 658.5nm.

5nm will be Ha signal + NII signal, while 3nm will be only Ha signal and therefore 5nm image - 3nm image = Ha signal + NII signal  - Ha signal = NII signal.

I would not worry much about red component of LP - as it will be offset or maybe very slight gradient - but that thing can be wiped out from the image.

[pmlogg]

The reason is because rather than having two H alpha filters I have a H alpha 5nm and a NII 3nm.

The suggested method was what you wrote i.e. H alpha 3nm = H alpha and H alpha 5nm - H alpha 3nm = NII.

What I was asking about doing the reverse NII = NII (of course) but then H alpha 5nm - NII 3nm = H alpha.

[vlaiv]

1 hour ago, pmlogg said:

The reason is because rather than having two H alpha filters I have a H alpha 5nm and a NII 3nm.

The suggested method was what you wrote i.e. H alpha 3nm = H alpha and H alpha 5nm - H alpha 3nm = NII.

What I was asking about doing the reverse NII = NII (of course) but then H alpha 5nm - NII 3nm = H alpha.

Ok, I get it now.

No, that won't work or be beneficial. In fact - it will work in one particular case, but it won't be beneficial.

Let me explain.

Consider two cases - H alpha 5nm band contains both Ha signal and NII signal would be the first case. Second case would be - it contains only Ha signal. We won't consider third case - only NII signal as it makes no sense as you would end up with "blank image" (NII - NII = 0).

First case:

From SNR perspective, capturing both signals at the same time is beneficial. It improves SNR as there is more signal that you capture. Now you could take another image of NII and remove NII signal from first image, but results would be far worse than doing pure Ha 3nm.

First - 5nm passes more LP than 3nm, that means 5nm Ha + NII will have more LP and associated noise. When you remove NII signal - you won't remove noise associated with it. In fact you'll add a bit more shot noise associated with it - one from "standalone" NII image. You'll also add all other noise in NII image.

In the end, you will end up with Ha signal / (Ha signal shot noise + LP noise 5nm + LP noise 3nm + NII shot noise + NII shot noise + read and dark noises from both images)

vs

Ha signal / (Ha signal shot noise + 3nm LP noise + single dark and read noise)

if you use 3nm Ha signal.

Compare that to original

Ha signal + NII signal / (Ha signal shot noise + NII shot noise + 5nm LP noise + single dark and read noise)

If NII is strong, 5nm filter could produce better SNR than 3nm that captures only Ha

Second case:

There is not much to this - because 3nm will not capture any signal - it will just inject read noise, lp noise and dark current noise into your 5nm image - nothing gained, quite a bit of noise injected.

 

[pmlogg]

Thanks for that explanation.  I had surmised from Wim that it wasn't a good idea but you have filled in the detail.

I'll just use the filters separately, and as appropriate, use the NII as an additional channel.  Perhaps as some stage I'll make the change to 3nm Ha.