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A bit about CCD imaging Paul McGale. Signal-to-noise ratio SNR = C star T / √(C star T + C sky T+ C dark T + R 2 ) where: T is the total integration time.

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Presentation on theme: "A bit about CCD imaging Paul McGale. Signal-to-noise ratio SNR = C star T / √(C star T + C sky T+ C dark T + R 2 ) where: T is the total integration time."— Presentation transcript:

1 A bit about CCD imaging Paul McGale

2 Signal-to-noise ratio SNR = C star T / √(C star T + C sky T+ C dark T + R 2 ) where: T is the total integration time for the image (secs) C star is the count rate of a star in the image (ADU/sec/pixel) C sky is the count rate from the sky (ADU/sec/pixel) C dark is the dark count rate from the CCD (ADU/sec/pixel) R is the readout noise from the CCD (ADU/pixel)

3 Sub-exposure stacking efficiency (1) e.g. 1 long exposure vs. average of 10 short ones E = √[(x + y) / (x + 10y)] where x is C star T+ C sky T+ C dark T y is R 2

4 Sub-exposure stacking efficiency (2) ** Dark sky **

5 Sub-exposure stacking efficiency (3) ** Light-polluted sky **

6 Number of sub-exposures (1) For a bright object SNR ≈ √(C star T) ≡ √(N, the number of sub exposures) How does SNR change with increasing N? Rate of change in SNR with N is 1/√(4N)  stack 4, rate = 1/4, SNR decreasing quickly  stack 25, rate = 1/10, SNR decreasing slowly

7 Number of sub-exposures (2) Faintest part of object visible has SNR=3 i.e. C star T/ √(C star T+ C sky T) = 3 Solve for C star(snr=3): C star(snr=3) = 9 + √(81 + 36C sky T)/(2T) or for N sub-exposures, length t C star(snr=3) = 9 + √(81 + 36C sky Nt)/(2Nt)

8 Number of sub-exposures (3)


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