<p><em>Convert answer in comment to community wiki</em></p>

<p>You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course, the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.</p>

CC BY-SA 3.0
2017-01-03T09:48:50.460Z
2017-01-03T09:48:50.460Z
James K
<p>Both spectra you show clearly have real features. There are perturbations in the spectra at certain wavelengths that are far larger than the rms noise that can be assessed empirically (by eye, or calculation) from looking at other parts of the spectrum. </p>

<p>This would form that basis of an empirical test for the reality of spectral features. On the other hand, if your spectrum was full of (small) real features, then you would have to rely on a theoretical estimate of what the rms noise in your spectrum <em>should</em> be.</p>

<p>Either way, your questions as posed cannot really be answered in any detail. If you have a spectrum that is consistent with being featureless, there will be some amplitude of real feature below which it would be undetectable. This amplitude upper limit will increase as the signal-to-noise decreases - i.e. the better your data, the smaller the features you can rule out.</p>

<p>In the end, you can never be sure that you have a "featureless continuum", only that any features are smaller than some threshold, that must be established by calculation or simulation.</p>

CC BY-SA 3.0
2017-02-02T13:57:37.093Z
2017-02-02T13:57:37.093Z
ProfRob
<p><img src="https://i.sstatic.net/GAhCd.jpg" alt="spec"></p>

<p>A featureless spectrum and a spectrum low S/N are very similar. <br/>
Is there a way to tell the difference between them?<br/></p>

CC BY-SA 3.0
2015-04-02T11:53:55.067Z
You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.
ProfRob
CC BY-SA 3.0
2015-04-02T12:20:48.163Z
Well, it wouldnt be bad if you could tell us what are these spectra about, where they come from, what do you expect there, which telescope took the spectra, if you can plot the background flux, and such "details". But no, let's put a picture and ask a 1-line hard question.
Py-ser
CC BY-SA 3.0
2015-04-02T20:14:31.533Z
@Rob Just according to a featureless spectrum itself, we are not able to know whether it is caused by a featureless star or the spectrum is just has a low s/n, right?
questionhang
CC BY-SA 3.0
2015-04-02T22:48:54.067Z
I refer you to my previous comment. Yes, it might be difficult. You might be able to rule out spectral features that exceed a certain level of emission/absorption that depends on the S/N.
ProfRob
<spectra>
How to tell whether an optical spectrum is featureless or just with a low S/N?
questionhang
# How to tell whether an optical spectrum is featureless or just with a low S/N?

(Asked by: username_0 on 2015-04-02)

![spec](https://i.sstatic.net/GAhCd.jpg)

A featureless spectrum and a spectrum low S/N are very similar.   
Is there a way to tell the difference between them?

**Tags:** spectra
> **Comments:**
> **username_1**: You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.  
> **username_2**: Well, it wouldnt be bad if you could tell us what are these spectra about, where they come from, what do you expect there, which telescope took the spectra, if you can plot the background flux, and such "details". But no, let's put a picture and ask a 1-line hard question.  
> **username_0**: @Rob Just according to a featureless spectrum itself, we are not able to know whether it is caused by a featureless star or the spectrum is just has a low s/n, right?  
> **username_1**: I refer you to my previous comment. Yes, it might be difficult. You might be able to rule out spectral features that exceed a certain level of emission/absorption that depends on the S/N.  

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## 2 answers

### Answer by username_3 on 2017-01-03. Score: 1

*Convert answer in comment to community wiki*

You would need to do statistical tests. i.e. Is the data you have (given the known noise properties) consistent with a featureless spectrum? If not, then it has features. Of course, the lower the signal-to-noise or the weaker the features might be, then the harder it would be to rule out a featureless continuum.


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### Answer by username_1 on 2017-02-02. Score: 1

Both spectra you show clearly have real features. There are perturbations in the spectra at certain wavelengths that are far larger than the rms noise that can be assessed empirically (by eye, or calculation) from looking at other parts of the spectrum.

This would form that basis of an empirical test for the reality of spectral features. On the other hand, if your spectrum was full of (small) real features, then you would have to rely on a theoretical estimate of what the rms noise in your spectrum *should* be.

Either way, your questions as posed cannot really be answered in any detail. If you have a spectrum that is consistent with being featureless, there will be some amplitude of real feature below which it would be undetectable. This amplitude upper limit will increase as the signal-to-noise decreases - i.e. the better your data, the smaller the features you can rule out.

In the end, you can never be sure that you have a "featureless continuum", only that any features are smaller than some threshold, that must be established by calculation or simulation.


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