| example problems |
|
| 1. |
Suppose that we detect 1000
photo-electrons from an
astronomical source in a certain time interval. In the absence of
other noise sources, what is the likely error in the measurement and
what signal-to-noise ratio (SNR) is obtained?
From Poisson statistics, an estimate of the error is simply the square root of the number of detected photo-electrons, σ = (1000)0.5 = 31.6. Thus we can write that the number of photo-electrons = 1000 ± 32. The SNR is simply the signal divided by the noise, which in this case is again the square root of the number of detected photo-electrons: SNR = S / N = S / S0.5 = S0.5 = (1000)0.5 = 31.6. |
| 2. |
A CCD detects a total of 600 counts/s from an astronomical
source, and 1200 counts/s/pixel from the sky. If the
exposure time is 1000 s, the readout noise of the CCD
is 5 e-/pixel and the dark current is 2
e-/pixel/s, what is the error in the measurement of the
source and what is the SNR? You may assume that the
gain of the CCD
is 1.2 e-/ADU and that the light from the source is spread over
50 pixels.
The error in the measurement, i.e. the noise N, is given by the denominator of the CCD equation: N = [ (Sobj . t . g) + (Ssky . t . g . npix) + (Sdark . t . npix) + (R2 . npix) ]0.5. Note that we have chosen the form here that includes the CCD gain, g, as this is required to convert the object and sky signals, which are in units of counts, into units of photo-electrons. Hence N = [ (600 x 1000 x 1.2) + (1200 x 1000 x 1.2 x 50) + (2 x 1000 x 50) + (52 x 50) ]0.5 = 8534 e-. The SNR is then SNR = (Sobj . t . g) / N = (600 x 1000 x 1.2) / 8534 = 84. |
| 3. |
A detector has incident upon it a total of 1 photon/s from an
astronomical source and a total of 2 photons/s from the sky underneath
the source. If the dark current and readout noise are negligible, how
long an exposure is required to achieve a SNR of 30?
In this question, the QE of the detector is not given and so we are unable to convert the incident photons into detected photo-electrons. So, we must assume that it is a perfect detector, i.e. 100% QE. Since the dark current and readout noise are negligible, we can write for the SNR that: SNR = (Sobj . t) / [ (Sobj . t) + (Ssky . t) ]0.5. Rearranging for t, we obtain: t = SNR2 . [(Sobj + Ssky) / (Sobj)2]. Hence t = (302) . 3 / 1 = 2700 seconds. |