<p>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.</p>

<p>What the S/N ratio you mention means is that <em>if</em> 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.</p>

CC BY-SA 3.0
2016-09-01T20:15:22.383Z
2016-09-01T20:15:22.383Z
Peter Erwin
<p>Warning: I never worked with spectra, and I don't have a complete answer. Anyway, here's how I understand it:</p>

<p>A spectrograph has a resolution $R = \Delta \lambda / \lambda$ that tells us how well it can distinguish light of different wavelengths. <a href="http://www.eso.org/sci/facilities/paranal/instruments/vimos.html" rel="nofollow">VIMOS</a> 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.</p>

<p>Here's where I'm very unsure: giving a SNR / $(0.1$nm$)$ should be equivalent to a magnitude limit, <strong>if</strong> 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.</p>

CC BY-SA 3.0
2016-09-01T17:16:06.303Z
"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.
Peter Erwin
CC BY-SA 3.0
2016-09-01T17:29:39.273Z
I thought that's what I wrote. Low resolution -> more photons per bin.
Alex
CC BY-SA 3.0
2016-09-01T20:14:38.283Z
Oh, yes, that is what you said! My apologies...
Peter Erwin
CC BY-SA 3.0
2016-09-01T17:13:55.987Z
2016-09-01T17:13:55.987Z
Alex
<p>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? </p>

CC BY-SA 3.0
2016-08-31T20:44:56.797Z
Can you link the paper or else provide the relevant text?
zephyr
CC BY-SA 3.0
2016-08-31T21:07:36.723Z
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](https://en.wikipedia.org/wiki/Signal-to-noise_ratio#Optical_SNR).
zephyr
<data-analysis><spectroscopy>
Signal-to-noise in inverse angstrom for spectroscopy?
user1991
# Signal-to-noise in inverse angstrom for spectroscopy?

(Asked by: username_0 on 2016-08-31)

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?

**Tags:** data-analysis, spectroscopy
> **Comments:**
> **username_1**: Can you link the paper or else provide the relevant text?  
> **username_1**: 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](https://en.wikipedia.org/wiki/Signal-to-noise_ratio#Optical_SNR).  

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## 2 answers

### Answer by username_2 on 2016-09-01. Score: 2 | Accepted answer

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.


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### Answer by username_3 on 2016-09-01. Score: 1

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](http://www.eso.org/sci/facilities/paranal/instruments/vimos.html) 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.

> **Comments:**
> **username_2**: "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.  
> **username_3**: I thought that's what I wrote. Low resolution -> more photons per bin.  
> **username_2**: Oh, yes, that is what you said! My apologies...  

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