# Signal-to-noise in inverse angstrom for spectroscopy?

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?

• Can you link the paper or else provide the relevant text? Aug 31 '16 at 20:44
• 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. Aug 31 '16 at 21:07

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.

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.

• "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. Sep 1 '16 at 17:16
• I thought that's what I wrote. Low resolution -> more photons per bin.
– Alex
Sep 1 '16 at 17:29
• Oh, yes, that is what you said! My apologies... Sep 1 '16 at 20:14