When making astronomical images with a CCD camera, the question often arises as to whether it is better to take one long exposure or to stack a number of shorter exposures
(called here sub-exposures). If the sub-exposures are accurately aligned and the background noise is random and the wanted parts of the image coherently repeatable across all images then the increase in signal to noise ratio (SNR) is proportional to the square root of the number of items stacked. However many of the noise sources in a CCD are more significant in shorter exposures than in longer exposures and thus the resultant
SNR is not as great as might be hoped for. This note examines the amount of loss.
The contri bution of the various noise sources is expressed by the well known ‘CCD equation’, which can be presented in different ways. We will use the following form:
SNR

S
STAR
* t
S
STAR
* t

N
SKY
* t

N
DARK
* t

N
RO
2
Where the symbols have the following meanings:
SNR = signal to noise ratio
Sstar = signal rate from the wanted part of the image in ADUs/second t = exposure time in seconds
Nsky = signal rate from sky background in ADUs/second
Ndark = signal rate from thermal noise in ADUS/second
Nro = readout noise in ADUs rms
We assume in the above that we are examining a single pixel in the image. The star noise, sky noise and dark noise all consist of electrons that arrive at unpredictable intervals of time, and have what is known as a Poisson distribution. They all have the nature of what is sometimes called shot noise. The readout noise has a Gaussian distribution.
If we now take a single long exposure of duration T
L
then the resulting SNR will be:
SNR

S
STAR
* T
L
------------------- (1)
S
STAR
* T
L

N
SKY
* T
L

N
DARK
* T
L

N
RO
2
Similarly the result of one short exposure of duration T
S
will be:
SNR

S
STAR
* T
S

N
SKY
S
STAR
* T
S
* T
S

N
DARK
* T
S

N
RO
2
Stacking a number M of the short exposures will give
SNR

S
STAR
* T
S
S
STAR
* T
S
*

N
SKY
* T
S

M
N
DARK
* T
S

N
RO
2
------------------- (2)

To find how many we need to stack, we equate (1) and (2) above and rearrange to give:
M

T
L
2
T
S
2


S
S
STAR
STAR
* T
S
* T
L


N
SKY
N
SKY
* T
S
* T
L


N
DARK
N
DARK
* T
S
* T
L


N
RO
2
N
RO
2


We see that if the readout noise is negligible then M reduces to T
L
/T
S
as expected.
Otherwise it may be expressed as follows:
M

T
T
S
L
*
D

N
RO
2
T
S
D

N
RO
2
T
L
, where D

S
STAR

N
SKY

N
DARK
We can see that there is no simple ‘rule of thumb’ to tell us how many sub-exposures to stack for any required result. Even for a given imaging situation, where D is constant, the extra number to stack is dependent not only on the proportions of the noise sources but also on the ratio of T
L
to T
S
: there are gains to be had from making the sub exposure period as long as conveniently possible, usually limited by tracking errors or saturation of brighter parts of the image.
To help our understanding let us look at a typical image:

This image is a single unguided 10 second subexposure of NGC2903 with a C11 telescope and f6.3 focal reducer. The camera was an Artemis 285. The temperature of the
CCD was approximately -10degC. The limiting visual magnitude was about 4.3 in this area.
This camera has the following characteristics at this temperature:
N
N
DARK

RO

0 .
0227 ADU
18 .
2 ADUs
/ sec
From measurements taken from the picture,
S
STAR

119 .
4 ADU / sec for the brightest part of the galaxy core
N
SKY

10 .
93 ADU / sec

D

119 .
4

10 .
93

0 .
0227

130 .
35 ADU / sec
To achieve the same result as one 600 second exposure:
M

600

10
130 .
35

18 .
2

18 .
2
10
18 .
2

18 .
2
130 .
35

600

74 .
92 i.e. we would need to stack 75 sub-exposures rather than the 60 expected by simple scaling.
Of more interest is to repeat the calculation for objects, such as faint outer arms of galaxies, that are at or near to the sky noise level:
S
STAR

10 .
93 ADU / sec for such objects in the above image.
And D

10 .
93

10 .
93

0 .
0227

21 .
88 ADU / sec
Now M

600

10
21 .
88

18 .
2

18 .
2
21 .
88

18 .
2
10

18 .
2
600

147 .
12
i.e. we now need to stack 147 sub-exposures. This is 2.45 times more than the expected value of 60. For objects that are below the sky noise the increase would be even greater.
Conclusions
Stacking sub-exposures does give a useful increase in SNR, but the amount of that increase is reduced by the effect of the noise sources. We see that readout noise should be small compared with sky and dark noise for the best results. Also it pays to keep the sub-exposures as long as possible, depending on other factors; not only does this increase the SNR but it also minimises the proportion of time lost downloading the images from the camera to the computer.

J C Moore
Revised 24 th May 2007