There are a number of factors which affect the quality of photometry of star in a CCD image. One is the exposure time: the longer the exposure time, the higher the signal in a star, and so the more precise the measurements. Of course, if one exposures too long, one will saturate the CCD, which ruins the measurements. But how long an exposure time is necessary?
The answer depends on your goal. If you simply want to verify that an object is present at a particular location, you may accept a very faint, barely noticeable dot of light. If you want to measure the position of an asteroid to an arcsecond, you would need a little more signal. If you want to make a very rough estimate of magnitude, to ten percent, say, then you might need more signal. And if the goal is to look for tiny variations in light of a variable star, you might need to expose long enough that the star comes close to the non-linear regime.
So how can you figure out the appropriate exposure time for a given project? There are two approaches:
(the family car drives past a sign, "Bridge Load Limit: 10 Tons")
Calvin: Dad, how do they figure out the load limit for a bridge?
Dad: Well, they build a bridge, and then drive heavier and heavier trucks over it until the bridge collapses. Then they re-build the bridge.
You can do exactly the same thing with your astronomical observing: take a series of images with increasing exposure times, and measure the object of interest on each one. The shortest exposure which provides enough precision for your goal is the right length.
This actually isn't a bad idea if you have huge amounts of observing time. On the other hand, if you are given just a single night on a big telescope, you probably won't want to waste any time taking a series of pictures, most of which turn out to be useless.
1
fractional uncertainty = -----
S/N
So, for example, if the S/N ratio is 50, the photometry will
have an uncertainty of 0.02, which is approximately 0.02 magnitudes.
There is only one source of signal from a star: the light of the star itself. If the star causes N(star) photons to strike the CCD chip during the exposure, and all of them knock free one electron, then the image should have N(star) electrons. That's the signal.
For the standard astronomical passbands, there are equations which give the number of photons per second collected by a telescope of a particular size from a star of a particular magnitude. A good source of these zero-point fluxes is Allen's Astrophysical Quantities. Some additional references are collected at the end of today's lecture. I use these values:
Passband photons/sec/cm^2
from star of mag 0
---------------------------------
U 550,000
B 391,000
V 866,000
R 1,100,000
I 675,000
Exercise:
- How many photons should strike a CCD in a 15-second exposure with the RIT telescope through the V filter of a star with magnitude V=13?
There are three main sources of noise for simple aperture photometry on a CCD: shot noise from the star itself, shot noise from the background, and readout noise from the CCD. Let's consider each in turn.
noise(star) = sqrt of number of photons from star
= sqrt[ N(star) ]
However, just as the number of photons from a star varies randomly according to Poisson statistics, so does the number of photons from the sky. There is always some uncertainty in the amount of background light subtracted from the aperture. The noise due to this random variation is
noise(sky) = sqrt of number of photons from sky
= sqrt[ N(sky) ]
We can break this down a bit further: to calculate the sky
contribution, we determine a local sky level per pixel,
and then multiply by the number of pixels within the aperture.
N(sky) = (electrons per pixel from sky) * (number of pixels)
So we can write the noise from the sky as
noise(sky) = sqrt[ N(sky per pixel) * npix ]
electrons
readnoise R = stdev(counts) * ------------
count
In our calculations of signal-to-noise, we need to convert from the stdev to the corresponding variance:
variance = (stdev * stdev)
and then add up the variance from all the pixels which fall
inside the photometric aperture. The result is
noise from CCD = (R*R) * npix
We can add all three sources of noise together: the way to do it is to add up all the electrons they produce, and then again appeal to Poisson statistics to find the noise.
total noise = sqrt [ N(star) + N(sky)*npix + (R*R)*npix ]
In order to find the S/N ratio, we simply divide the total signal by the total noise.
signal N(star)
------ = -------------------------------------------------
noise sqrt [ N(star) + N(sky)*npix + (R*R)*npix ]
where
N(star) is the number of electrons from the star
which fall within the aperture
N(sky) is the number of electrons per pixel
due to the sky background
R is the readout noise, in electrons
npix is the number of pixels in the aperture
Exercise:
- Look in the directory richmond/apr11 for a cleaned image of Lictora called 1107b.001. Display the image.
- Using an aperture radius of 5 pixels, measure the flux of the asteroid (which is near row=122, col=274).
- Calculate the signal from the asteroid.
- Calculate the noise in the measurement.
- Calculate the signal-to-noise ratio.
- Estimate the uncertainty, in magnitudes, of a magnitude measurement of the asteroid from this image.
- Use the on-line signal-to-noise calculator to perform the same calculation. The measurements were made at an airmass of X=1.2.