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Field of View for MX916 CCD and 8" 2000mm Focal Length Telescope

Setting the Telescope Up

Focussing the CCD

CCD Signal-to-Noise Ratio

Measuring the CCD Gain and Read Noise

Efficiency of Stacking Short Exposures **Powerpoint presentation added**

Determining focal ratios from CCD images

Some Useful Formulae

Here are some formulae that I have found useful.

EQ1: f/ratio = telescope focal length / telescope aperture.
e.g. My Meade LX200 is of 2000mm focal length and 200mm in aperture. Hence, its focal length is 2000/200=10 i.e. f/10.

EQ2: Magnification = telescope focal length / eyepiece focal length.
e.g. If I was to use a 9.7mm eyepiece with my scope, I would be working at a magnification of 2000/9.7 = 206

EQ3: Highest resolution of telescope = 11.6 arc seconds (") / telescope aperture in cm.
e.g. My scope has a maximum resolution of 11.6/20 = 0.58". This is it's "Dawes Limit".

EQ4: Image size at focal plane = tan(angular size of object) x telescope focal length.
e.g. Size of Jupiter in my scope with no eyepiece = tan(50/3600) x 200 = 0.05 mm. (Jupiter is roughly 50" in size, and there are 3600 arc seconds in a degree.

EQ5: Real field of view = apparent field of view / magnification.
e.g. Meade 4000 9.7mm Super Plossol eyepeice has an apparent field of view of 52^o. So, on my scope I would actually view a field of 52/206 = 0.25^o. Hence, the moon which has a diameter of 0.5^o would more than fill the field of view of a 9.7mm eyepiece used on my scope. To see the whole moon at once, I'd need a 2000x0.5/52 = 20mm eyepiece assuming it had an apparent field of view of 52^o.

EQ6: Arc-seconds per CCD pixel = pixel size in microns / (0.00485 x telescope focal length in mm). .
e.g. My MX516 CCD has rectangular pixels of 9.8x12.6 microns (equivalent to a square pixel of sqrt(9.8x12.6)= 11 microns). So, each pixel roughly covers 11/(0.00485x2000) = 1.1". For deep-sky imaging if I want to resolve 4" detail, my pixels need to span 2" (Nyquist Theorem says "sample at twice the frequency you want to resolve" i.e. 4/2 = 2, 4" is probably the best I can hope to resolve at my observing site). Therefore, I need to reduce the focal length of my scope by about a half to sample at 2". Using a 0.63 focal reducer my CCD pixels would each span 11/(0.00485x2000x0.63) = 1.8". For planetary imaging it is possible to work close to the scope's Dawes Limit as very short exposures can be used. From EQ3 this is 0.6". Hence, I need to nearly quadruple (2x1.1/0.6) my scopes focal length. A x2 Barlow lens would result in my CCD pixels spanning 11/(0.00485x2000x2) = 0.56", a x3 lens would give 0.38" per pixel.

More to come...

CCD Field of View

MX916 CCD pixel size: 11.6 x 11.2 uM
Number of pixels: 752 x 580
Telescope: F: 2000mm, f-ratio: f/10

Use Eqn. 6 to calculate "/pixel. Multiply focal length by a fraction determined by desired focal ratio / actual f-ratio e.g. to find focal length if modifed by a 0.5 focal reducer then the new focal length is f/5^*. Hence focal length in Eqn. 6 should be multiplied by 5/10.

* - actual reduction also depends on the distance from the focal reducer to the CCD chip.

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Setting the Telescope Up

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Focussing the CCD

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CCD Signal-to-Noise Ratio

CCD Signal-to-Noise Ratio

One measure of quality of an image is the amount of signal in it relative to the amount of noise. This is the signal-to-noise (SNR), and one form it may take is:

SNR = C_*t / SQRT(C_*t + C_skyt + C_darkt + R²)

Where:

t is the integration time for the image (secs)

C_* is the count rate of a star in the image (counts/sec/pixel)

C_sky is the count rate from the sky (counts/sec/pixel)

C_dark is the dark count rate from the CCD (counts/sec/pixel)

R is the readout noise from the CCD (counts/pixel)

For a bright star the equation can be simplified to:

SNR = C_*t / SQRT(C_*t)

The sections below show some consequences of this for the simple case of a bright star. Note that for the comparisons of long to stacked exposures, the individual short exposures must be long enough (or the star bright enough) for C_*t to be appreciably greater than (C_skyt + C_darkt + R²).

Some additional notes are given at the end of the document.

For a single exposure of length t₁, the SNR for a bright star is:

SNR₁ = SQRT(C_*t₁)

and for an exposure of length t₂

SNR₂ = SQRT(C_*t₂)

Comparing the SNR for both exposures:

SNR₂/SNR₁ = SQRT(C_*t₂/ C_*t₁) = SQRT(t₂/t₁)

if t₁ = 1 then

i.e. SNR increases as the square-root of the exposure time

How does the SNR compare for a single image compared to that for the average of 2 images of the same integration time?

For a single image, from above

SNR_single = SQRT(C_*t₁)

Averaging the signal in the two images gives:

Avg = (C_*t₁ + C_*t₂)/2

as the noise in one image doesn’t depend on the noise in the other, the errors in one can simply be added to the other

and the variance of this is

Var(Avg) = Var([(C_*t₁ + C_*t₂)/2] =0.25*Var[(C_*t₁ + C_*t₂)] = 0.25*(C_*t₁ + C_*t₂)

so,

SNR_stacked = [(C_*t₁ + C_*t₂)/2] / SQRT(0.25* (C_*t₁ + C_*t₂))

i.e.

SNR_stacked = (C_*t₁ + C_*t₂)/ SQRT(C_*t₁ + C_*t₂) = SQRT(C_*t₁ + C_*t₂)

and if t₂ = t₁ then

SNR_stacked = SQRT(C_*t₁ + C_*t₁) = SQRT(2C_*t₁)

Compared to the single exposure of length t₁

i.e.

SNR_stacked/ SNR_single = SQRT(2)

if k images were averaged then

So the SNR increases by the square-root of the number of images added.

Does the SNR depend on whether stacked images are averaged or summed?

From above it was seen that for averaging

SNR_averaged = (C_*t₁ + C_*t₂)/ SQRT(C_*t₁ + C_*t₂) = SQRT(C_*t₁ + C_*t₂)

Summing the signal in two images gives:

Sum = (C_*t₁ + C_*t₂)

and the variance of this is

Var(Sum) = Var([(C_*t₁ + C_*t₂)] = C_*t₁ + C_*t₂

so,

SNR_summed = [(C_*t₁ + C_*t₂)] / SQRT((C_*t₁ + C_*t₂))

i.e.

SNR_summed = SQRT(C_*t₁ + C_*t₂)

Comparing the SNR in both cases:

SNR_summed /SNR_averaged = SQRT(C_*t₁ + C_*t₂) / SQRT(C_*t₁ + C_*t₂)

So the SNR is the same for averaged images as it is for summed images.

From above it was seen that:

SNR_averaged = SQRT(C_*t₁ + C_*t₂)

i.e.

SNR_averaged = SQRT(2C_*t₁) if t₂ = t₁

For a single exposure of length t₃ , from above:

SNR_single = SQRT(C_*t₃)

and if t₃ = 2t₁ then

SNR_single = SQRT(2C_*t₁)

Comparing the SNR in both cases:

SNR_averaged / SNR_single = SQRT(2C_*t₁) / SQRT(2C_*t₁)

So the SNR of a single long exposure is the same as the SNR of stacked exposures of the same length whose individual exposure lengths are the same and sum to the length of the long exposure.

Describing the count rates:

The statistical distribution (Poisson) that describes the counts from the star, sky and dark counts, has the property that the noise is equal to the square root of the mean count. The readout noise however, is a shot noise and this is the reason it appears as a square in the SNR formula. Hence, having described the signal and the noise in an image it is possible to form the SNR equation.

Variance Formula:

Suppose x₁ and x₂ are means and multiplied by constants a₁, and a₂, so that

y = a₁x₁ + a₂x₂

then the variance (var) of y is:

var(y) = a₁² var(x₁) + a₂² var(x₂) + 2a₁a₂cov(x₁, x₂)

if x₁ and x₂ are independent (not correlated with each other) then the covariance term (cov) is equal to 0 and hence the variance formula simplifies.

Measuring the CCD Gain and Read Noise

The gain of a CCD is the amount of voltage needed to produce 1 ADU (analog-to-digital unit). ADUs are what describe the brightness levels in the CCD control software used to display the image captured by the CCD. Note that ADUs are measured in whole numbers, so if the conversion from electrons (e^-) to ADUs results in a fraction of an ADU then that fraction is ignored (this is called digitisation noise). Knowing the gain (and full-well depth - the number of electrons a pixel can hold) of the CCD is useful in working out at what value of ADU the CCD becomes saturated. Note that this may well be less than the 65,000 levels that the image downloaded to the camera control software may offer.

In reading out the captured image some noise is unavoidably added. This is called read-noise.

Howell (Handbook of CCD Astronomy) describes how to measure the gain and read-noise of a CCD. It involves taking two bias (very short dark exposures) and two flat (exposures where the CCD is illuminated by a very even light) frames.

F₁ is the mean value of the first flat frame

F₂ is the mean value of the second flat frame

B₁ is the mean value of the first bias frame

B₂ is the mean value of the second bias frame

σ²_F1-F2 is the variance of the frame produced by subtracting the second flat frame from the first flat frame

σ²_B1-B2 is the variance of the frame produced by subtracting the second bias frame from the first bias frame

The mean values of the bias and flat frames as well the variance of the frame produced by the required subtractions are all easily got from the Statistics function in Astroart (most camera software will have this function).

Using two bias and flat frames taken with my MX916 CCD I find:

Gain = 3.63 e^-/ADU

Read Noise = 14.8 e^-

The manual for the MX916 lists the gain to be ~ 5 e^-/ADU (probably for 2x2 binning mode) and the read noise as <15 e^-.

The full-well depth of the MX916 is listed as 75,000 e^-. Since we now know the gain this can be converted from electrons (e^-) into units of ADU, i.e. the saturation of the pixels occur when the ADU level reaches 75,000/3.63 = 20,673 ADU.

It is generally recommended that the exposure of flat frames should be long enough so that the CCD is at about half its saturation level. Hence, flat frames of about 10,000 ADU would be suitable for the MX916.

Efficiency of Stacking Short Exposures

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Ideally we’d like to image very faint details in deep-sky objects. To do this long exposures are required. However, it is often not practical to take long exposures due to for example a troublesome mount, bright sky, CCD saturation, or to reduce the chances of an image being spoiled by a passing cloud, satellite, or airplane.

The breaking down of a long exposure into shorter ones and then summing (or averaging) the short exposures can recover much of the benefits of long exposure frames see (CCD Signal-to-Noise Ratio) .

It would be interesting to know to what degree stacked short exposures have the same signal-to-noise ratio (SNR) as that of a single long exposure. This can be investigated by simply dividing the SNR of the long exposure by the SNR of the stacked short exposures. Lets call this new ratio the efficiency of stacked exposures (E).

Using the full CCD formula as shown in the section CCD Signal-to-Noise Ratio this gives an equation for the ratio E, if we were to stack 10 short exposures, that looks like:

y is the read noise (squared) of the CCD in ADU²

x is the sum of all of the other noise components at the level of a single CCD pixel

i.e. x = C*t + C_dark*t + C_sky*t,

C is the count rate of the star or object in ADU/s per pixel

C_dark is the dark count rate of the CCD in ADU/s per pixel

C_sky is the background sky count rate in ADU/s per pixel

and t is the exposure time of the long exposure in s.

It is easy to see that if x is much bigger than y then the efficiency is high i.e. E ≈ 1. However, if we are interested in faint details then x is not going to be much bigger than y and so the efficiency of stacking short exposures is going to be somewhat less than 1.

Using my MX916 CCD and LX200 8" SCT from my garden I find that the faintest details I can record with a clear glass filter are those with count rates of about 0.1 ADU/s. I’ve measured the dark count rate of the MX916 to be about 0.1 ADU/s and the read noise to be 14.8 e^- (see Measuring the CCD Gain and Read Noise ). The darkest part of the sky visible from my garden has a count rate of about 1 ADU/s.

Below is a set of graphs that show how the stacking efficiency varies by the count rate of the faintest detail desired to be recorded and total exposure time. The top left graph uses a sky count rate of 0.1 ADU/s, top right 0.5 ADU/s, bottom left 1.0 ADU/s, and bottom right 2.0 ADU/s. It is assumed that 10 short exposure images are stacked. If there was no loss in the SNR of stacking short exposures compared to that of a single long exposure then E would be equal to 1 (i.e. 100% efficiency).

The upper curves are the brighter (i.e. 2.0 and 1.0 ADU/s) object counts and the lower curves the fainter (i.e. 0.5 and 0.1 ADU/s) object counts. If the SNR in the single long exposure was less than 3, then that part of the curve was not plotted - this being the minimum SNR value at which one might be able to detect a faint detail. The graphs suggest, for example, in a sky count rate of 0.5 ADU/s, recording details as faint as 0.5 ADU/s in 10 stacked exposures each of duration 1 min is about 90% efficienct compared to a single long exposure of 10 min. In a bright sky x becomes much bigger than y quite quickly, see the bottom right plot, and that detecting a faint detail of 0.1 ADU/s is very difficult as the SNR did not rise above 3.

Note that due to digitisation noise a minimum exposure of 10s is required to record an object (or detail) of 0.1 ADU/s since levels of brightness are represented only by whole numbers of ADUs.

The conclusion is that stacking is an efficient way of imaging especially if either the short exposures themselves are reasonably long, the source is quite bright or the sky background is bright. Of course using a CCD camera with higher read noise would reduce the efficiency.

Ideally to get the full dynamic use of the CCD one would like to expose until the camera reaches about 80% of its saturation point. The MX916 saturates at around 20,000 ADU, so this ideal level would be 16,000 ADU. This gives the maximum length of a single exposure if the brightest features in the image are to be correctly represented. The graphs can be used to see what the stacking efficiency is for the faintest details if a long exposure was desired to record them but resulted in the saturation of the brightest features.

Determining focal ratios from CCD images

It is useful to determine the image scale of a telescope and CCD combination.  The image scale can be changed by using a focal reducer, to reduce the image scale, or a barlow lens, to increase the image scale.  This lets us determine how much sky the CCD actually sees and so we can tell beforehand if a particular object will be wholly or only fractionally seen.

By reverse if the image scale is known then the focal length of the telescope, and hence its focal ratio, can be calculated.  This can be useful because altering the distance of a focal reducer, or barlow, from the CCD can change the reducing or multiplying effect of the auxillary optic.

Figure 1 shows a triangle whose sides are the focal length of the telescope (F) and, greatly exaggerated, the width of a CCD pixel (P).  The pixel subtends (or sees) an angle on the sky of θ.   We want to find the value of  θ, since this is our image scale.

Figure 1.  The angle made on the sky by a CCD pixel of width P at the focus of a telescope of focal length F.

From figure 1:

tan(θ) = P/F,  and since θ is small this can be re-written as

θ = P/F  radians / pixel                                                                       (1)

Note that there are 57.3 degrees in a radian, and 3600 arcsecs in a degree, so equation (1) can be re-written as:

  θ = 206265 × P/F  arcsec / pixel

Now, pixel sizes are normally written in μm (10^-6m) and focal lengths in mm (10^-3m), so this can be simplified slightly to:

  θ = 206.3 × p/F  arcsec / pixel                                                            (2)

if p is simply taken to be the pixel size without regard to μm units and F is in mm.

So, for an MX916 CCD camera with 11.4 μm pixels, the image scale using a telescope of focal length 3000mm is:

  θ = 206.3 × 11.4/3000 =  0.78  arcsec / pixel            

This formula comes in useful in determining the effective focal ratio of a CCD – telescope system.

Figure 2 shows an imaginary CCD image that has 2 stars at pixel positions (x₁, y₁) and (x₂, y₂).  A image processing/camera control program such as Astroart will give pixel positions.  These have to be related to a celestial co-ordinate system such as right ascension (α) and declination (δ).  Astroart can be used to determine α and δ, since it comes with a copy of the Hubble Space Telescope Guide Star Catalogue (GSC).  It is a case of trying to identify star patterns to find the matching star.  This is most easily done by imaging a bright deep sky object and identifying field stars.

Figure 2:  CCD image showing stars at pixel positions (x₁, y₁) and (x₂, y₂).  These have celestial coordinates (α₁, δ₁) and (α₂, δ₂) respectively.

The distance between the stars can then be calculated for each of the two sets of co-ordinates.  The pixel set will of course be in pixel units, and the celestial set in arcsec.  Simply dividing one by the other will give the image scale in arcsec/pixel. 

For the pixel coordinates, the distance will be D_p:

D_p = √[((x_{1 -} x₂) ²  + (y₁ - y₂)²] pixels                                                          

For the celestial coordinates, we can use a similar formula, but we have to remember that the distance between the right ascension co-ordinates will be shortened by an amount which depends on the declination of the star – remember that lines of right ascension are all bunched together at the pole (δ=90^o) but furthest apart at the equator (δ=0^o).  So, if we take the average declination of the two stars to be δ₁₂, then the distance will be:

D_c=  3600 ×√[((α_{1 -} α₂) ²cos δ₁₂cos δ₁₂   +  (δ₁ - δ₂)²] arcsec                                      

Note that it will be easier to work with celestial coordinates in decimal form. Also, since the area of sky imaged is small, one or other of the declinations can be used.

The image scale is then simply:

D_c/D_p  arcsec/pixel

This can be plugged into equation 2, which itself can be re-written to give a value for the focal length:

F = 206.3 × p / (D_c/D_p)  mm

This will be the effective focal length of the system and it needs to be divided by the diameter of the telescope (in mm) to give the effective focal ratio.

An example:

I took an image of the Whirlpool galaxy  (M51) with an MX916 CCD, 0.63 focal reducer, and a 300mm f/10 LX200GPS.   The optical train was:

SCT – 0.63 reducer – focuser – filter unit – CCD

I wanted to know what focal ratio the system was now working at.

Startup Astroart and load image.  Go to Tools-Star Atlas to load up the GSC (you may need to insert the Astroart CD Rom).

On the Star Atlas window click on the FIND icon (hand) and type in M51.  The Star Atlas window then shows an area around M51, zoom in or out to get roughly the same field of view as the CCD image.

Click on the CCD image and then click on the Stars icon from the toolbar on the left side of the screen.  Double click on two widely space stars that can be identified in the Star Atlas window.  These will appear in the Stars window.  Now click on one of the rows in the Stars window, it will now be highlighted.  Click on the identified star in the Star Atlas window.  Do this again for the other star.  The Stars window now has all the information required.

The Xc and Yc columns contain the pixel coordinates and the R.A. ^◦  and DEC.^◦  the celestial coordinates (in decimal format). 

For this example, the pixel co-ordinates of the two stars used were found to be (57.66, 489.92) and (682.85, 374.83) and the celestial co-ordinates (202.64184, 47.2791) and (202.30553, 47.3786).  One should work with the celestial coordinates to at least 3 decimal places.

So

D_p = √[((57.66-682.85) ²  + (489.92 – 374.83)²] = 635.7 pixels

and

D_c=  3600×√[((202.6418-202.3055) ²cos(47.35)cos(47.35)   +  (47.279-47.238)²] = 833.8 arcsec

[Note:  If these are calculated using Excel, it will expect the angle to be in radians, simply multiply by 3.14/180 to do this e.g. enter cos(47.35*3.14/180)]

So the image scale = 833.8/635.7 = 1.31 arcsec / pixel

Hence the effective focal length of the telescope is (CCD pixel size = 11.4 μm):

206.3 × 11.4 / 1.31 =   1793 mm

The telescope mirror is 300mm in diameter, which gives an effective focal ratio of 793/300 = 5.98.

The native focal ratio of the telescope is f/10, so the focal reducer is working at a reduction factor of 5.98/10 = 0.6

Planets:

In trying to find the focal ratio from a planetary image, note that D_c can be got from the LX200/ASII handset, monthly astronomy magazines, or just about any planetarium program.

To measure the pixel distance it would be best to use a pixels between the 10 and 11 o’clock positions (and their opposites between the 4 and 5 o’clock positions).  The gas giants are not perfectly circular, so this method would give an average width.