3
$\begingroup$

In a paper I came across the description of the signal-to-noise ratio (SNR) for observations with a spectograph. This was reported as $10\:1/\mathring{A}$. I am rather new to spectroscopy, so could someone clarify these units for me? Why is it used and how should I read it?

$\endgroup$
  • $\begingroup$ Can you link the paper or else provide the relevant text? $\endgroup$ – zephyr Aug 31 '16 at 20:44
  • $\begingroup$ I can't provide a full answer, but it seems to me the source of the inverse angstrom is because the noise is calculated over a band or wavelength regime. See for example this description on wiki. $\endgroup$ – zephyr Aug 31 '16 at 21:07
2
$\begingroup$

A spectrum imaged onto a CCD or other detector consists of photons spread out along a wavelength axis. Signal-to-noise (S/N) depends on how many photons you have from the source, how many from the background, intrinsic noise from the electronics, etc. -- and what kind of bin you're adding up the photons in. If you summed up all the photons across the spectrum (over all wavelengths), you'd get one number with a very high S/N, but no spectral information. If you divided the spectrum into N different wavelength bins, you'd get a lower S/N per bin, but more spectral information.

What the S/N ratio you mention means is that if you divided that spectrum into bins that were 1 Angstrom wide, you'd have enough photons in each bin for a S/N of 10 in each bin (at least for some region of the spectrum, probably near the central wavelength). You can then have some idea how the S/N would improve for bins larger than 1 Angstrom in size, or how it would get worse for smaller bins. S/N per Angstrom is conventional for optical spectroscopy, both because Angstroms are traditional and because moderate resolutions spectrographs often have scales of a few tenths to several Angstroms per pixel along the wavelength direction.

| improve this answer | |
$\endgroup$
1
$\begingroup$

Warning: I never worked with spectra, and I don't have a complete answer. Anyway, here's how I understand it:

A spectrograph has a resolution $R = \Delta \lambda / \lambda$ that tells us how well it can distinguish light of different wavelengths. VIMOS from the paper has a resolution from 200 to 2500. With the same flux from the source and for a low-resolution spectrograph, you get more photons per wavelength bin than for a high-resolution one. This means you'll also get better SNR. So, giving SNR / wavelength makes inherent sense.

Here's where I'm very unsure: giving a SNR / $(0.1$nm$)$ should be equivalent to a magnitude limit, if the spectral energy distribution is the same. Or it could be a minimum quality criterion on the spectrum - stop taking data as soon as the SNR is reached.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "With the same flux from the source and for a low-resolution spectrograph, you get more photons per wavelength bin than for a high-resolution one." -- Actually, it's the opposite way around: low-spectral resolution means more photons per wavelength bin. $\endgroup$ – Peter Erwin Sep 1 '16 at 17:16
  • $\begingroup$ I thought that's what I wrote. Low resolution -> more photons per bin. $\endgroup$ – Alex Sep 1 '16 at 17:29
  • $\begingroup$ Oh, yes, that is what you said! My apologies... $\endgroup$ – Peter Erwin Sep 1 '16 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.