Relationship between f ratio, subexposure time, and pixel signal to noise ratio
Copyright Steve Cannistra

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) / mm
2 earth.
B) Multiply this by the surface area of the telescope, in mm
2, 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.

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.


 Copyright Steve Cannistra