JediWill |
12/02/2006 02:12AM (Read 6315 times)
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Registered: 06/23/2006
Posts: 13
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Hello, I have a question about CCDPROC and Dark images. During my summer project I was using dark frames that were exposed for much longer than my object frames. Normally they are about 1.5 to 2 times as long. My advisor told me that when I use CCDPROC, it will automatically scale the dark frames to the same time as my object frames. I need to know if this is true or not.
Also, if not, is there a way to correct for this time difference without gathering new darks?Thanks, William
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fitz |
12/02/2006 02:12AM
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Registered: 09/30/2005
Posts: 4040
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William,From the CCDPROC help page description of the dark correction:[quote:47e0128749]Dark counts are subtracted by scaling a dark count calibration image to the same exposure time as the input image and subtracting.[/quote:47e0128749]-Mike
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JediWill |
12/02/2006 02:12AM
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Status: offline
Registered: 06/23/2006
Posts: 13
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Thank you for pointing that out. But, I am wondering how it is actually scaling the darks. (i.e if it is linear, quadratic ect...) Thanks William
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fitz |
12/02/2006 02:12AM
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AFAIK, it is the ratio of the exposure times as a scaling factor, so....linear scaling.-Mike
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knine |
12/02/2006 02:12AM
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Registered: 11/29/2005
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Make sure darkcombine.scale = exposure so it uses the exposure time.Now, if it REALLY works on your CCD is a different question. I have a reference in front of me that says their experience is that dark current seldom scales linearly. So before you assume it works for your CCD, I've been told to try a few things. For example, take 180 second darks and 90 second darks. Divide by the 180 second by 2, then subtract from the 90 second dark and see if you have anything significant left.I have yet to do this myself, but I will probably this week, possibily with a chip that was used on data for your summer project. :wink:
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JediWill |
12/02/2006 02:12AM
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Status: offline
Registered: 06/23/2006
Posts: 13
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Thanks Knine,But, what is the definition of significant are we using. I found a 150 sec and 80 sec dark, multiplyed the 150 s by .533 (conversion to 80 sec) and subracted the result from the 80 s dark.
By displaying the result, I am seeing the values go between -50 to 50 in ds9. Imstat of my new image gives a Mean of 38.2 and a STD of 64.67Let me know what you think,Thanks to everyone
William
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fitz |
12/02/2006 02:12AM
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Status: offline
Registered: 09/30/2005
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[quote:e2c18108e7]By displaying the result, I am seeing the values go between -50 to 50 in ds9. ....Imstat of my new image gives a Mean of 38.2 and a STD of 64.67[/quote:e2c18108e7]If "By displaying..." you mean you used the DISPLAY task, remember that the pixel readouts you'll get from DS9 are an approximation based on the z1/z2 scaling and not real values. This is a limitation in DS9, but if you load the image standalone you'll see real values. Note that IMEXAM can also be used to access the real pixel values if you DISPLAY the image. Knine may still be right, don't let the display values fool you though.-Mike
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