Formulas

Question 1

X(ij) =

Answer 1

P(ij) + T(ij) + E(ij) + C(ij)

P(ij) = contribution from pixel charge due to photons from the astronomical source

T(ij) = contribution from pixel charge due to thermal effects

E(ij) = contribution from readout process

C(ij) = contribution from pixel charge produced by cosmic ray hits

where X(ij) is in DN

Question 2

B(ij) =

Answer 2

T(ij) + E(ij)

Question 3

N(ij) =

Answer 3

[X(ij) -B(ij)]/F(ij)

X(ij) - ‘raw’ image
N(ij) - DN per pixel due to astronomical sources
B(ij) - background
F(ij) - flat field

Question 4

N =

Answer 4

n/g ± σ(0)

N is the DN read out
g is the gain
n is the number of photo-electrons
σ(0) is the readout noise

Question 5

σ(n) =

Answer 5

√Ng in e-

√N/g in DN

Question 6

σ^2(N) =

Answer 6

σ(0)^2 + N/g

Question 7

n(ij) =

Answer 7

N(ij) g

where n(ij) is the number of photo-electrons produced by a pixel

Question 8

n(ph) =

Answer 8

n(ij)/Q

where Q is the quantum efficiency

Question 9

λ(mean) =

Answer 9

[ ∫ P(λ)s(λ)dλ]/[ ∫P(λ)s(λ)dλ]

s(λ) is the normalised source spectrum
P(λ) is the probability of detecting a photon of wavelength λ

Question 10

J =

Answer 10

[n(ph) h v(mean)]/[A(pix) Δt]

=

F (D/θf)^2

where θ = y/f

y is the pixel size
A is the pixel area
F is the flux
f the focal length
θ the angle subtended by the object being imaged
Δt is the exposure time

Question 11

dynamic range =

Answer 11

full well/total noise

Question 12

total noise^2 =

Answer 12

sum of all noise sources squared

i.e.

σ^2(dc) + σ^2(r) + …

Question 13

max CTE =

Answer 13

CTE^(charge transfers)

Question 14

charge transfers =

Answer 14

for a 2048 x 2048 pixel format

2047 + 2047 = 4094

Question 15

max fraction of CTE lost =

Answer 15

1 - max CTE

Question 16

O(x(2)) =

of a single source

Answer 16

I(x(1)) PSF(x(2)-x(1))

Question 17

O(x(2)) =

source distribution

Answer 17

I*PSF = (∞ ∫ 0) I(x(1)) PSF(x(2)-x(1))dx(1)

Question 18

I =

Answer 18

F^-1[F(0)/F(PSF)]

Question 19

O(λ(2)) =

single line

Answer 19

I(λ(1)) ILP(λ(2)-λ(1))

Question 20

O(λ(2)) =

a spectrum

Answer 20

I*ILP = (∞ ∫ 0) I(λ(1)) ILP(λ(2)-λ(1))dλ(1)

Question 21

G(u) ∝

Answer 21

(∞ ∫ -∞) F(v(LOS)) S(u-v(LOS))dv(LOS) = F*S

Question 22

CCF(v(LOS)) =

Answer 22

(∞ ∫ -∞) G(u) S(u-v(LOS))du

Question 23

CCF(τ) =

Answer 23

(∞ ∫ -∞) f(t) g(t-τ)dt

τ is a variable time lag/shift

Question 24

ACF(τ) =

Answer 24

(∞ ∫ -∞) f(t) f(t-τ)dt

Question 25

τ = nT

Answer 25

max ACF

Question 26

τ = (2n+1)T/2

Answer 26

get a min ACF

Question 27

PSD(v) ∝

Answer 27

|F[f(t)]|^2

Question 28

PSD(v)^2 =

Answer 28

C(v)^2 + S(v)^2

Question 29

Wavelet transform W(a,b)

Answer 29

see formula sheet changing a -> 1/freq change changing b -> time shift

Question 30

Resolving power =

Answer 30

R = λ/Δλ = Nn N = total number of lines on the grating n = order number of spectrum

Question 31

Angular dispersion

Answer 31

dθ/dλ = n/acosθ a = spacing between lines of grating θ = angle at which spectral feature is formed

Question 32

Grating response width

Answer 32

W = λ/Na

Question 33

Interpolation

Answer 33

N(i') = N(i(1)') + m(1-Δi') m = [N(i(2)') - N(i(1)')]/[i(2)'-i(1)'] Δi' = i(2)'-i(1)'

Question 34

sky background

Answer 34

N(b)(λ) = mλ + c

Question 35

total real counts in the source

Answer 35

(λ2 Σ λ=λ1) [N(λ) - N(b)(λ)]

Question 36

Significance =

Answer 36

SNR in σ = Ns/√[Ns+2Nb] ~ √Ns

Question 37

line profile =

Answer 37

φ(λ) = [Bc - Bλ]/Bc Bc = continuum intensity Bλ = line intensity

Question 38

equivalent width =

Answer 38

W =(∞ ∫ 0) [Bc - Bλ]/Bc dλ

Question 39

Optical depth

Answer 39

Bλ/Bc = e^(-τ(λ)) -> W = (∞ ∫ 0) 1-e^(-τ(λ)) dλ

Question 40

Weak spectral line

Answer 40

τ(λ) << 1 -> W = (∞ ∫ 0) τλ dλ

Question 41

a narrow slab of absorbing gas has optical depth

Answer 41

τ(λ) = nσ(λ) Δs = σ(λ)N Δs = thickness column density N = nΔs σ(λ) = absorption cross-section

Question 42

Boltzmann factor

Answer 42

p(i) = p(0)g(i) exp[-Ei/kT] E(i) is the energy of this state g(i) is the degeneracy p(0) is a constant found by normalising the probability over all possible states of energy E(i)

Question 43

the average number of transitions per second per unit volume

Answer 43

n(ij) = n(i) n(e) C(ij) C(ij) is the collisional rate coefficient and depends on the electron speed and the atomic properties n(e) is the number of electrons per unit volume n(i) is the number of atoms in state i per unit volume

Question 44

in the case where upwards transitions are collision-dominated and downward transitions are due to spontaneous radiation coronal approximation

Answer 44

n(i) n(e) C(ij) = n(j) A(ji) = I(ij)

Question 45

detailed balance

Answer 45

n(i) n(e) C(ij) = n(j) A(ji) + n(j) n(e) C(ji)

Question 46

emission lines from a hot gas are broadened due to random particle motions

Answer 46

[λ - λ(0)]/λ(0) = v/c

Question 47

turbulent broadening

Answer 47

(Δλ(turb)/λ(0))^2 = (v(turb)/c)^2