https://obs-web.rs.umbc.edu/ 2026-09-05T13:31:34+00:00 https://obs-web.rs.umbc.edu/ https://obs-web.rs.umbc.edu/lib/tpl/bootstrap3/images/favicon.ico text/html 2023-11-09T12:54:43+00:00 Anonymous (anonymous@undisclosed.example.com) https://obs-web.rs.umbc.edu/doku.php?id=wiki:astronomy:observational_astronomy:calibration_frames:bias_signal&rev=1699552483&do=diff Measuring Bias The BIAS calibration frame is perhaps the easiest (and fastest!) to take. Consider taking a $0s$ integration time. What is the resulting signal? $$L_{ij}(\lambda) = \Bigl[\epsilon_{ij}(\lambda)I_{ij}(\lambda) + T_{ij}(\lambda)\Bigr]t + B_{ij}(\lambda)$$ $$L_{ij}(\lambda) = \Bigl[\epsilon_{ij}(\lambda)I_{ij}(\lambda) + T_{ij}(\lambda)\Bigr]0 + B_{ij}(\lambda)$$ $$L_{ij}(\lambda) = B_{ij}(\lambda)$$ Yep. That's it. text/html 2023-11-11T20:18:43+00:00 Anonymous (anonymous@undisclosed.example.com) https://obs-web.rs.umbc.edu/doku.php?id=wiki:astronomy:observational_astronomy:calibration_frames:thermal_signal&rev=1699751923&do=diff Measuring Thermal Signal The THERMAL calibration frame is trickier than collecting a BIAS frame. Due to the thermal signal being a time-integrated term, we cannot simply take a $0s$ exposure as we did with the BIAS--there would be no thermal signal in such a frame!$$L_{ij}(\lambda) = \Bigl[\epsilon_{ij}(\lambda)I_{ij}(\lambda) + T_{ij}(\lambda)\Bigr]t + B_{ij}(\lambda) ~~|~~ I_{ij}(\lambda)=0 ~ \forall ~ i,j$$$$L_{ij}(\lambda) = T_{ij}(\lambda)t + B_{ij}(\lambda)$$$$L_{ij}(\lambda) = T_{ij}(\… text/html 2023-11-09T12:55:13+00:00 Anonymous (anonymous@undisclosed.example.com) https://obs-web.rs.umbc.edu/doku.php?id=wiki:astronomy:observational_astronomy:calibration_frames:uniformity&rev=1699552513&do=diff Measuring Uniformity To measure the uniformity across the detector and optical system, we take a frame of a uniformly illuminated field. It is important that the field covers the entirety of the detector. Also important is that the integration time is chosen to not saturate any pixels in the frame.$C(\lambda)$$$L_{ij}(\lambda) = \Bigl[\epsilon_{ij}(\lambda)I_{ij}(\lambda) + T_{ij}(\lambda)\Bigr]t + B_{ij}(\lambda) ~~|~~ I_{ij}(\lambda)=C(\lambda) ~ \forall ~ i,j$$$$L_{ij}(\lambda) = \Bigl[\epsi…