Astronomical Spectroscopy - Physics - University of Cincinnati

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small signal-to-noise ratio, 10 per pixel, say (possibly 30 when one integrates over the spectral
resolution elements and spatially) then if one assumes perfect flat fielding ng = 100. Adding
in the additional noise term with p=0.008 leads to the fact that the signal-to-noise one would
achieve would be 9.97 rather than 10. In other words, if one is in the low signal-to-noise
regime flat-fielding is hardly necessary with modern detectors.
It should be further emphasized that were one to use bad flat-fields (counts comparable
to those of the object) one actually can significantly reduce the signal-to-noise of the final
spectra. How many counts does one need for a “good” flat field The rule of thumb is that
one does not want the data to be limited by the signal in the flat field. Again let n be the
number of counts per pixel in the spectrum. Let m be the number of counts per pixel in
the flat-field. Consider this purely from a propagation of errors point of view. If σ f is the
error in the flattened data, and σ n and σ m are the rms-scatter due to photon-statistics in
the spectrum and flat-field, respectively, then
σ 2 f
(ng) 2 =
σ2 n
(ng) 2 + σ2 m
(mg) 2.
(The assumption here is that the flat has been normalized to unity.) But each of those
quantities is really just the inverse of the signal-to-noise squared of the respective quantities;
i.e., the inverse of the final signal-to-noise squared will be the sum of the inverses of the
signal-to-noise squared of the raw spectrum and the inverse of the signal-to-noise squared of
the flat-field. To put another way:
1/S 2 = 1/ng + 1/mg,
where S is the final signal-to-noise. The quantity 1/ng needs to be much greater than 1/mg.
If one has just the same number of counts in one’s flat-field as one has in one’s spectrum,
one has degraded the signal-to-noise by 1/ √ 2.
Consider that one would like to achieve a signal-to-noise ratio per spectral resolution
element of 500, say. That basically requires a signal-to-noise ratio per pixel of 200 if there
are seven pixels in the product of the spatial profile with the number of pixels in a spatial
resolution element. To achieve this one expects to need something like 200 2 =40,000 e − in
the stellar spectrum. If one were not to flat-field that, one would find that the signal-tonoise
ratio was limited to about 100 per pixel, not 200. If one were to obtain only 40,000
counts in the flat field, one would achieve a signal-to-noise ratio of 140. To achieve a signalto-noise
ratio of 190 (pretty close to what was wanted) one would need about 400,000 e −
(accumulated) in all of the flat-field exposures—in other words, about 10 times what was in
the program exposure. In general, obtaining 10 times more counts in the flat than needed
for the program objects will mean that the signal-to-noise ratio will only be degraded by 5%