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I am doing some work that involves dividing two stellar spectra from the same star. Those stellar spectra are constructed by summing random samples of multiple spectra from the same star to improve the signal to noise ratio (SNR). I know that for a sum, the $SNR$ will vary with $\sqrt N$ where $N$ is the number of spectra I am adding to construct each one of my spectra.

My problem is: Assuming $SNR_1$ and $SNR_2$ are the signal to noise ratios of the spectras I am dividing by each other, how can I obtain the $SNR$ of the resulting spectrum?

Do you recommend any documentation where thse operations are explained in detail?


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up vote 1 down vote accepted

The usual approximation for arithmatic of quantities with uncorrelated uncertainties that for small(ish) uncertainties $\Delta x_i$ or measurements $x_i$ let's us write for multiplicative operation $$ \begin{array}~ y &=& \left(\frac{x_1}{x_2}\right) \, \text{ or }\, \left(x_1 x_2\right) \\ \frac{\Delta y}{y} &=& \sqrt{ \left(\frac{\Delta x_1}{x_1}\right)^2 + \left(\frac{\Delta x_2}{x_2}\right)^2 } \end{array} $$ (i.e. add relative uncertainties in quadrature).

You can get this kind of result from a bastardization of the chain the rule. Given $y = f(x_1, x_2 , \dots)$ or each input measurement $x_i$, compute $\left(\frac{\partial f(x_1, x_2 , \dots)}{\partial x_i}\right) \Delta x_i$ and add all the resulting terms in quadrature.

You have also been a little free with your nomenclature here. Call $s$ the underling signal and $n$ the random noise (this can be counting statistics or any other random process such as shot noise in the detector, but not a constant bias which must be subtracted off--our noise is assumed to have a mean of zero) with a distribution whose width is characterized by $\sigma$. A single measurement is then $m_i = s_i + n_i$, and the population has a signal to noise ration of $\frac{s}{\sigma}$.

The win from addition is that the sum of $N$ such measurements is $$ M = \sum_{i=1}^N m_i = Ns + \sqrt{N}\sigma $$ meaning that the signal to noise ratio of the sum is $\frac{Ns}{\sqrt{N}\sigma} = \sqrt{N}\left(\frac{s}{\sigma}\right)$, an improvement.

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So, would I be correct if I said that: $ \frac{s}{\sigma} = \left(\sqrt{\left(\frac{s_{1}}{\sigma_{1}}\right)^{-2}+\left(\frac{s_{2}}{\sigma‌​_{2}}\right)^{-2}}\right)^{-1}$ Thanks! – jorgehumberto Feb 21 '13 at 17:33
Also, is there any book that you would recomend on this type of spectra statistics? – jorgehumberto Feb 21 '13 at 17:40
@jorgehumberto It's not astronomy-specific, but John R. Taylor's "An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements" is pretty much the canonical book for college-level error analysis. – Thriveth Sep 15 '13 at 21:08
@Thriveth Cool, thanks! It will help for sure refreshing my concepts as was out of the academic medium for almost 10 years :) – jorgehumberto Sep 17 '13 at 11:00

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