There
has been a lot of
discussion on amateur astrophotography sites about the importance of f
ratio, the importance of aperture as opposed to f
ratio, the importance of both, etc. I performed this relatively
simple analysis as an exercise for myself, to work through some of the
issues and come to my own conclusions about them. The analysis
below shows that the subexposure time is proportional to the square of
the f ratio. Furthermore, once your subs are photon limited, the
pixel SNR is a constant value for a given object, camera, and imaging
site.
I. Reducing
the signal to noise equation to its basics
Let's
start by evaluating the equation for SNR in a given pixel, as a prelude
to eventually analyzing the SNR of an entire object of interest.
Much of this will reduce down to a few simple rules, but I need to go
through the derivations first.
The
signal
to noise ratio per pixel, for a single sub, is expressed as follows
(ignoring
the contribution from dark noise):
Equation 1. SNR (for single
sub) = (Obj)*tsub /
sqrt[(Sky+Obj)*tsub + R2]
where
SNR
= signal to noise ratio per pixel; Obj = object flux in
electrons/pixel/minute; Sky =
sky flux in electrons/pixel/minute; tsub = subexposure time
in
minutes; R = read noise in e RMS.
Subexposure
time (tsub) can
be calculated in two ways- one using my equation, and the
other using
John Smith's.
They are both equivalent, but I will use John's for
this analysis because most people are
familiar with it:
tsub = R2
/ [(1+p)2 - 1]*Sky, where p is the fraction of total noise
contributed by read noise, typically 0.05 (5%). Let's call 1/[(1+p)2 - 1] = Z (a constant), reducing the equation to:
Equation 2. tsub = Z * R2
/ Sky
Let's
now
consider an
expression for Sky flux, in
electrons/pixel/minute. There are several components to this,
which I will not illustrate with a diagram but which are fairly easy to
imagine. This will reduce to a constant, so don't worry too much
about the specifics, except to note where aperture and focal length are
involved:
A) Let's consider the units for photons raining down from the half
dome of sky, and falling onto the
earth's surface area. This would be a constant for a given
imaging
site. Typical units would be (photons/min per sq arcsec
of sky) / mm2
earth.
B) Multiply this by the surface area of the telescope, in mm2, to
obtain the rate of photons entering the scope. Units for this
would be
photons/min
per sq arcsec of sky. Note that this is proportional to
the
aperture squared of
the scope (surface area).
C) Multiply the value in B by the image scale squared (sq arcsec)
that the pixel covers, in order to obtain the Sky flux hitting the
pixel. Units
have now changed to photons/min
(per pixel). Note
that this is proportional to the square of [pixel
dimension/focal length (FL)]. This is because the image scale in
arcsec per pixel is proportional to pixel dimension / FL, but we need
to
express the pixel in terms of its surface area, hence this value must
be
squared.
D) Multiply the value in C by the quantum efficiency of the camera,
yielding sky flux hitting the pixel in units of electrons/pixel/minute.
The reason for
outlining steps A through D, is to show that
Sky flux for a given site is proportional to a constant (let's call it
Q) times aperture
squared
(line B above) divided by FL
squared
(line C above), or:
Equation 3. Sky flux = Q * A2 / FL2 = Q / f2,
where f is the f ratio
This is a very important result. It
says that all things being equal, the amount of sky flux hitting your
pixel will be inversely proportional to the f ratio squared. This
result will be used in the analysis that follows.
By analogy, for a given object with its associated flux, the
Obj flux hitting the pixel would be:
Equation 4. Obj flux
= T / f2
where
T is a constant. There is no need to go through steps A through D
again, because it's the identical logic.
Substituting equation 3 into equation 2, we get tsub
= (Z * R2*
f2)/Q.
Note that for a given imaging site and camera, Z, R, and Q are
constants- let's simplify further be defining Z/Q as V:
Equation 5. tsub = VR2f2
where
V is a constant related to the imaging site, R is the read noise,
and f is the f ratio. This is another important result.
This equation shows that, using the same camera and imaging site, tsub is
proportional to the focal ratio squared. Simply put, this
means
that if you are using an f4 scope
and your photon limited subexposure time is 5 minutes, if you then
switch the same camera to an f8 scope at the same imaging site, your
photon limited subexposure time will now be 5 minutes times (8/4)2, or
20 minutes.
Let's
substitute the equations in 3, 4, and 5 into the SNR equation
shown in line 1:
SNR (for single sub) = (T/f2)
*(VR2f2) /
sqrt [(Q/f2)*(VR2f2) + R2]
(I've ignored the
contribution of Obj flux to the noise, because it's usually minor)
SNR (for single sub) = TVR2/
sqrt [QVR2 + R2]
SNR (for single sub) = TVR2/
(R*sqrt [VQ + 1]), reducing to:
Equation
6. SNR (for single sub) = TVR/
sqrt [VQ + 1]
where T is a constant proportional to Obj flux; Q is a
constant proportional to Sky flux; V is a constant as defined in
equation 5; and R is read noise.
Equation 6
is very interesting. It says that the
SNR per pixel, in an individual sub, for a given camera, scope, and
imaging site, is a FIXED value
at photon limited conditions (i.e.,
assuming that you calculated your tsub
correctly). Until I did this analysis, this result would
not have been intuitively obvious to me. Let's
say that the SNR/pixel for an STL11K sub, using an FSQ106 at f5
(aperture 106 mm) at my imaging site under photon limited conditions,
is 10. What happens if I attach my camera to an 24"
RCOS at f8 (aperture 600mm), at my same imaging site? Easy- my
subexposure time will increase be a factor (8/5)^2 = 2.56 (remember
equation 5), but the SNR per pixel in an individual sub (which took
2.56 fold more
time to obtain) is still 10! We do not achieve better signal to
noise at the pixel level in
this
situation, despite using the more expensive and larger aperture scope,
assuming that we are imaging with the same
camera and at the same imaging site, as long as our subexposure time is
truly photon limited to the same degree in both cases.
II. A
few scenarios
Before we move on, I would like to reinforce the points above with a
few simple comparisons (same camera and imaging site is assumed for
both scenarios, with the only difference being the scope):
Scenario
A.
Scope 1
|
Scope 2
|
focal length = 800mm
|
focal length = 800mm
|
aperture = 200mm
|
aperture = 100mm
|
f ratio = f4
|
f ratio = f8
|
Subexposure time = 5 minutes
|
Subexposure time = 20 minutes
|
Pixel SNR = 10
|
Pixel SNR = 10
|
Note that the subexposure
times in this scenario are completely dependent upon the f ratio, not
just the aperture or
FL alone. In this particular example, the aperture happened to
dictate the f
ratio because the FL remained fixed, but it did not have to be this
way. The same results (at least with respect to subexposure times
and SNR per pixel) would have been achieved by keeping the aperture
fixed and changing the FL, to achieve the same f ratios in scope 1 and
2. So from the perspective of subexposure time,
the f ratio is the critical determinant, not the focal length in
isolation, and not the aperture in isolation. Again, based upon
equation 6 above, the SNR per pixel doesn't change once you have
achieved photon limited conditions- it is a fixed value based
upon the characteristics of your camera, the Object flux, and the Sky
flux. Note that for a given camera, the image scale (in
arcsec/pixel) is the same for Scopes 1 and 2 (because the focal length
is the same), but the achievable resolution is better for the larger
aperture scope (Scope 1). This is related to two effects:
1) Scope 1 will have a smaller Airy disk, since the
diameter of the Airy disk is proportional to the f ratio, resulting in
more pinpoint stars; and 2) The larger aperture will yield better
resolution due to the Rayleigh
Effect.
Scenario
B.
Scope 1
|
Scope 2
|
focal length = 400mm
|
focal length = 800mm
|
aperture = 100mm
|
aperture = 200mm
|
f ratio = f4
|
f ratio = f4
|
Subexposure time = 5 minutes
|
Subexposure time = 5 minutes
|
Pixel SNR = 10
|
Pixel SNR = 10
|
In
scenario B, we have two scope that differ substantially in their
apertures, but because the f ratio is the same, so is the subexposure
duration. And because we are photon limited, the SNR per pixel is
the same in both cases. So what did we gain with the larger
aperture scope in Scenario B (Scope 2)? We gained two
things: 1) a longer focal length at the same f ratio, which will
yield a superior
image scale (because image scale is proportional to the pixel dimension
divided by the FL); and 2) the larger aperture will also
yield higher resolution due
to the Rayleigh
Effect.
Note that the Airy disk will be the same in both
scopes, because the diameter of the Airy disk is proportional to the f
ratio, which is f4 in both cases.
Steve
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