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vak |
01/11/2008 09:31PM (Read 5049 times)
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Status: offline
Registered: 01/11/2008
Posts: 2
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Hi, I have a question about imcombine.
Suppose I have N images and run imcombine on them using combine=average (or median) and reject=avsigclip. Then I invert each image (multiply each image by -1) and run imcombine on them using combine=average (or median) and reject=avsigclip. Shouldn' t the resultant combined images in the two cases be inverses of each other (ie. -1 times the other)? I am finding that this is not the case for some pixels. I looked at some individual pixels, and see that this is happening because for a given pixel, a different number of images are getting rejected in the 2 cases. ( e.g. 0 images getting rejected in 1 case, and 2 getting rejected in
the inverse case). The value of nkeep is the same for both cases, and the difference arises whether I use average or median. Any ideas on what may be causing this difference? Thanks.
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valdes |
01/11/2008 09:31PM
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Status: offline
Registered: 11/11/2005
Posts: 728
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Before I could comment I would need to see a listing (lpar) or all the parameters.Yours,
Frank Valdes
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vak |
01/11/2008 09:31PM
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Status: offline
Registered: 01/11/2008
Posts: 2
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Hello,below are the parameters I used for imcobine (I also tried combine=median.)
Any suggestions on what may cause the difference I mentioned earlier would be greatly appreciated. This may be related to how imcombine with avsigclip handles rejection of pixels that are negative. Does it use absolute value of the signal at a pixel inside the square root while calculating the sigma at a given pixel? Thanks!
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>lpar imcombine
input = "@files.lst" List of images to combine
output = "comb.fits" List of output images
(headers = "") List of header files (optional)
(bpmasks = "") List of bad pixel masks (optional)
(rejmasks = "") List of rejection masks (optional)
(nrejmasks = "") List of number rejected masks (optional)
(expmasks = "") List of exposure masks (optional)
(sigmas = "") List of sigma images (optional)
(logfile = "STDOUT") Log file\n
(combine = "average") Type of combine operation
(reject = "avsigclip") Type of rejection
(project = no) Project highest dimension of input images?
(outtype = "real") Output image pixel datatype
(outlimits = "") Output limits (x1 x2 y1 y2 ...)
(offsets = "none") Input image offsets
(masktype = "none") Mask type
(maskvalue = 0.) Mask value
(blank = 0.) Value if there are no pixels\n
(scale = "none") Image scaling
(zero = "none") Image zero point offset
(weight = "none") Image weights
(statsec = "") Image section for computing statistics
(expname = "") Image header exposure time keyword\n
(lthreshold = INDEF) Lower threshold
(hthreshold = INDEF) Upper threshold
(nlow = 1) minmax: Number of low pixels to reject
(nhigh = 1) minmax: Number of high pixels to reject
(nkeep = 5) Minimum to keep (pos) or maximum to reject (neg)
(mclip = yes) Use median in sigma clipping algorithms?
(lsigma = 3.) Lower sigma clipping factor
(hsigma = 3.) Upper sigma clipping factor
(rdnoise = "rdnoise") ccdclip: CCD readout noise (electrons)
(gain = "gain") ccdclip: CCD gain (electrons/DN)
(snoise = "0.") ccdclip: Sensitivity noise (fraction)
(sigscale = 0.1) Tolerance for sigma clipping scaling corrections
(pclip = -0.5) pclip: Percentile clipping parameter
(grow = 0.) Radius (pixels) for neighbor rejection
(mode = "ql")
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valdes |
01/11/2008 09:31PM
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Status: offline
Registered: 11/11/2005
Posts: 728
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Hi,Indeed there is a dependence on the signal level with AVSIGCLIP. The idea behind this algorithm is to estimate a pseudo-gain for a line. In other words, to find a proportionality constant between the average signal and the standard deviation at each pixel in the line. Note it is the average values (excluding the minimum and maximum as possible bad points) in a line that are used. After inverting the average will be negative unless you use the scale and zero parameters. The expected Poisson variance is computed as[code:1:928b80dc9c]
max (1, scalesdaverage)
[/code:1:928b80dc9c]So if the average value is negative the expected variance will be 1 which, of course, will be different than in the case without inverting.You could use the zero or scale parameters. To do the exercise you were trying what you should do is invert about the average and not about zero; i.e. instead of mulitply by -1 you should do <image average> - <image>.Let me know if I'm not clear or you want more information.Frank
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