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9.3 Shutter

If we ask the data acquisition system to make an exposure of time t, and it instead makes one of time $t + \delta$ (with $\delta \neq 0$), we have to make a shutter correction, where we multiply the image with $t/(t+\delta)$. $\delta$ can be a scalar (i.e. be the same for all the pixels), or it can an image (i.e. vary across the CCD), depending on what kind of shutter the system has.

$\delta$ can be determined from two images with different ``apparent'' exposure time t1 and t2, as follows: Since the CCD is a linear detector, when it is exposed to the intensity I in the time treal, the level we read out N for a given pixel is proportional to the product of the two,

N1 $\textstyle \propto$ $\displaystyle I_1 \cdot t_{real,1}$ (9.1)
N2 $\textstyle \propto$ $\displaystyle I_2 \cdot t_{real,2}$ (9.2)

If we keep the intensity constant, and since we have $t_{real} = t + \delta$, we get

\begin{displaymath}\frac{N_1}{N_2} = \frac{t_1 + \delta}{t_2 + \delta}, \;\; t_1 \neq t_2,
\end{displaymath} (9.3)

from which we can determine $\delta$.

One can use more images to get a better determination. We used the temporal sequence of one 10 sec, ten 1 sec, and one 10 sec exposure. If we let N1' denote the sum of the levels in the ten 1 sec exposures, and N2' the sum of the levels of the two 10 sec exposures, we get

\begin{displaymath}\frac{N_1'}{N_2'} = \frac{10 \cdot ( 1 \mbox{ sec} + \delta)}
{ 2 \cdot (10 \mbox{ sec} + \delta)}
\end{displaymath} (9.4)

which yields

 \begin{displaymath}\delta = \frac{ 10 (2N_1'/N_2' - 1) } { 10 - 2N_1'/N_2' } \mbox{ sec}
\end{displaymath} (9.5)

For the determination of the shutter correction, a number of dome flats were taken, using the same lamp with constant voltage, thereby hoping to get a constant intensity. The images are listed in Table [*].

Table: The Images Used to Determine the Shutter Correction
dfsc # #frames exp. time [sec]
2233 1 10
2234 1 10
2235-2244 10 1
2245 1 10
2246 1 10
2247-2256 10 1
2257 1 10
2258 1 10
2259-2268 10 1
2269 1 10
2270 1 10
2271-2280 10 1
2281 1 10
2282 1 10
Notes: All images are from night 8. The first and the last image in the table were not used.

The images in Table [*] were corrected for bias and dark. The delta image was then computed using equation [*] for each of the four (1 $\times$ 10 sec, 10 $\times$ 1 sec, 1 $\times$ 10 sec) sequences. The mean of the four was also calculated. The results are in Table [*].

Table: Statistics of the four Individual Delta Images and the Mean Delta Image
image mean median stddev
1 0.3999 0.3997 0.01101
2 0.4093 0.4090 0.01051
3 0.4004 0.4004 0.01065
4 0.4325 0.4323 0.01073
mean 0.4105 0.4105 0.00682
Notes: ``median'' is the midpt from imstat.

A visual inspection of the four delta images showed, that they were all almost flat, as the low standard deviations also suggest. There was, however, a small gradient in the x direction of about 0.001 sec, but since it was much lower than our estimated uncertainty of 0.02 sec (see below), we decided to neglect it.

Therefore, we decided to use a scalar delta value instead of a delta image. We set this delta value to the (scalar) mean of the mean delta image, which amounts to 0.41 sec. We estimate the uncertainty on this value to be half of the maximum difference between the four individual determinations, i.e. 0.02 sec.

The differences between the four is probably due to the light not being constant. The symmetric sequence (1 $\times$ 10 sec, 10 $\times$ 1 sec, 1 $\times$ 10 sec) makes the determination insensitive to a light variation which is linear in time, so the light variations was probably more ill-behaved than that.

A cl-script b_dcorr was written to implement the scalar shutter correction.

next up previous contents
Next: 9.4 Linearity Test Up: 9. Details of the Previous: 9.2 Dark

Properties of E and S0 Galaxies in the Clusters HydraI and Coma
Master's Thesis, University of Copenhagen, July 1997

Bo Milvang-Jensen (