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Instruments for Optical and Radio Astronomy > Practice Questions > Flashcards

Practice Questions Flashcards

1
Q

B(v) of a blackbody is

A

B(v) = 2kT/λ^2

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2
Q

B(v) is measured in

A

W m^-2 Hz^-1 sr^-1

where Jy = Js^-1 m^-2 Hz^-1

where Js^-1 = W

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3
Q

S = ∫B(θ,φ) dΩ

A

dΩ = sinθdθdφ

and for small angles

(θ/2)^2 π

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4
Q

converting from degrees to radians

A

radians = degree π/180

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5
Q

S is measured in

A

W m^-2 Hz^1

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6
Q

converting from arc seconds to radians

A

radians = arc π/180*3600

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7
Q

L = 4πR^2S

A

where R is the distance away

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8
Q

to convert from ly to meters

A

ly = 3x10^8 * 60 * 60 * 24 * 365.25

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9
Q

gain =

A

gain = 4π/Ω(A)

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10
Q

Ω(A) =

A

λ^2/A(e)

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11
Q

w = kTΔv

A

where w is the power received per unit bandwidth

and Δv is the bandwidth

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12
Q

w = 1/2 S A(e) Δv

A

where S is the flux density and A(e) the effective area

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13
Q

derive

w/Δv = 1/2 A(e) ( ∫sky) B(θ,φ)P(θ,φ) dΩ

A

equating w = 1/2 S A(e) Δv and w = kTΔv

substituting for S

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14
Q

derive

Ta = A(e)/λ^2 ( ∫sky) Tb(θ,φ)P(θ,φ) dΩ

A

starting from

w/Δv = 1/2 A(e) ( ∫sky) B(θ,φ)P(θ,φ) dΩ

replace B for Tb

rearrange for Ta

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15
Q

For Ω(s) &laquo_space;Ω(A)

A

P ~ 1

=> S = B Ω(s)

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16
Q

Ω(A) =

A

∫ P(θ,φ) dΩ

dΩ = sinθdθdφ

for small angles

dΩ = (θ/2)^2 π

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17
Q

A(g) =

A

πD^2/4

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18
Q

A(e) =

A

ηπD^2/4

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19
Q

what is the beam efficiency of a radio telescope

A

beam efficiency = power in main lobe/total power

η(B) = Ω(M)/Ω(A)

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20
Q

phase error =

A

φ = 2π/λ 2ε = 4πε/λ

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21
Q

errors reduce the efficiency of the dish by a factor of

A

in argand plane, overall voltage response is down by ~ cosθ

=> power response down by

η = (cosθ)^2 = cos^2(4πε/λ)

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22
Q

what is the system temperature of a radio telescope

A

total noise power per unit bandwidth = kT(sys). Where T(sys) includes contributions from all noise sources

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23
Q

T(sys) =

A

T(source)+T(background) +….

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24
Q

kT(source) =

A

1/2 S A(e)

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25
Q

Noise N =

A

(Δvτ)

so

sqrt(N) = (Δvτ)^1/2

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26
Q

SNR =

A

T(source)/T(sys) (Δvτ)^1/2

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27
Q

why might the most popular choice for high sensitivity instruments be ones with a large collecting areas rather than wide bandwidths

A

SNR increases proportionally with A(e)(Δvτ)^1/2 so it is better to increase A(e) rather than Δv

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28
Q

The complex visibility measured by the interferometer can be obtained from

A

the mean conjugate product (or cross-correlation) of the signals <Ψ1Ψ2*>

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29
Q

The cross-correlation =

A

<Ψ1Ψ2*> ∝ exp(-ikr(v).θ(v))

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30
Q

the cross-correlation can be rewritten in terms of the source’s flux density S

A

<Ψ1Ψ2*> = S exp(-ikr(v).θ(v))

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31
Q

The extended source is usually

A

incoherent , the correlation between signals from two different patches of sky is zero

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32
Q

a small patch of the sky of surface brightness and solid angle will contribute a flux density. So rewriting the cross-correlation as

A

d<Ψ1Ψ2*> = B(θ) dΩ exp(-ikr(v).θ(v))

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33
Q

since the patches are incoherent the cross-correlation term can be rewritten as

A

<Ψ1Ψ2*> = ( ∫sky) B(θ) exp(-ikr(v).θ(v)) dΩ

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34
Q

the complex visibility is defined as

A

Γ(r(v)) ∝ <Ψ1Ψ2*>

Γ(0(v)) = 1

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35
Q

the constant of proportionality must be

A

1/( ∫ B(θ)dΩ) = 1/S

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36
Q

the measured complex visibility for an arbitrary extended source of surface brightness B(θ) and total flux density S is

A

Γ(r(v)) = 1/S ∫ B(θ) exp(-ikr(v).θ(v)) dΩ

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37
Q

the fringe rate is

A

the rotation rate of the complex fringe visibility Γ(D) in the argand plane.

It equals the rate of change of l measured in λ

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38
Q

if the interferometer is oriented N-S

A

D(v) = (0,D,0)
D(v) . θ(v hat) = DcosHcosδ
φ = 2π/λ cosHcosδ

taking derivative

dφ/dt = 2π/λ cosδsinH dH/dt

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39
Q

at the meridian the hour angle is

A

H = 90 degrees

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40
Q

dH/dt =

A

2π/606024

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41
Q

difference between primary beam of an interferometer and the synthesised beam

A

a primary beam is a beam of single dish and synthesised beam is effective beam of synthesised aperture

primary beam defines the field of view of the interferometer

the synthesised beam is the ‘point spreadfunction’ of the whole interferometer, and corresponds to the image produced

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42
Q

if the interferometer is oriented N-S

A

D(v) = (D,0,0)
D(v) . θ(v hat) = DcosδsinH
φ = 2π/λ cosδsinH

taking derivative

dφ/dt = 2π/λ cosδcosH dH/dt

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43
Q

max resolution =

A

λ/D(max)

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44
Q

τ =

A

1/Δv

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45
Q

max scale size =

A

λ/D(min)

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46
Q

resolution in RA =

A

λ/D

47
Q

resolution in Dec

A

λ/Dsinδ

48
Q

long baseline requires

A

imaging at high spatial frequencies. Therefore need lots of flux on small angular scales.

Surface brightness B = S/Ω

so need a high value of B

49
Q

show that the fringe rate of an interferometer can be attributed to the differential doppler shift between the signals arriving at the two ends of the baseline caused by the rotation of the Earth

A

in general let the ends of the baseline be at r1(v) + r2(v) with respect to the Earth centre

D(v) = r2(v) - r1(v)

Δv = v/c v(v).θ(v hat)
Δv = 1/λ v(v).θ(v hat)

Δv(1) - Δv(2) - 1/λ (v1(v)-v2(v)).θ(v hat)

= 1/λ(Ω(v) x D(v)).θ(v hat)

for E-W interformeter

D(v) = (D,0,0)
Ω(v) = (0,0,Ω)
θ(v hat) = (sinHcosδ,cosHcosδ,sinδ)

Δv(1) - Δv(2) = fringe rate

50
Q

the effective area of an antenna is

A

directly related to the amount of power the antenna collects from an unpolarised, white, point source of flux density S. The effective area A(e) satisfies

w = 1/2 SA(e)Δv

51
Q

the aperture efficiency is

A

simply the ratio of the effective area of the antenna to its geometrical area, when that is clearly defined

52
Q

the beam solid angle is

A

the 2-D equivalent of the Rayleigh resolution criterion for antennas

53
Q

Nyquist noise theorem

A

an antenna in an isotropic blackbody radiation field at T, over a bandwidth Δv

54
Q

The antenna temperature generated by a point source is

A

defined as the temperature corresponding to the power from a source

55
Q

If the source is not small compared to the beam (larger than the beam)

A

we must include the antenna pattern P() in the integral to account for apodisation

56
Q

CMBR is the temperature of

A

the coldest regions of the sky at any wavelength

57
Q

system temperature

A

a way of characterising the amount of noise inherent in a radio source

kTsys = w

58
Q

the system temperature has components from all the noise sources present, so that

A

T(sys) = T(background) + T(source) + … + T(LNA) + T(electronics)

at v > 500GHz Tsys&raquo_space; T(A)
at v < 300MHz T(A) often dominates

59
Q

the system temperature measures

A

the noise in the system, and therefore controls the denominator in the signal-to-noise ratio of a detected signal

60
Q

for threshold detection assume

A

SNR = 1

61
Q

Interferometer pros

A

delivers a higher angular resolution
it is easier to account for instrumental drifts
are less susceptible to small variations in front-end gain, in particular.

62
Q

I(θ) =

A

< | Ψ1+Ψ2 exp{-ikyθ} |^2 >

63
Q

|Γ(y)| =

A

visibility

I(max) - I(min) / I(max) + I(min)

of Young’s fringes

64
Q

I(max) =

A

<|Ψ1|^2> +<|Ψ2|^2> + 2<|Ψ1Ψ2*|^2>

65
Q

I(min) =

A

<|Ψ1|^2> +<|Ψ2|^2> - 2<|Ψ1Ψ2*|^2>

66
Q

a maximum of the fringe pattern occurs at the point where

A

φ+kyθ = 0

67
Q

Van cittert-zernike theorem

A

the complex fringe visibility |Γ(y)|is the Fourier transform of the normalised sky brightness

68
Q

complex fringe visibility can itself be generated by

A

computing <|Ψ1Ψ2*|> which is the correlation between the two wavefronts received at the two dishes

69
Q

The single value of Γ that is returned from one baseline measurement

A

is not sufficient to generate an image of the source as it just returns one Fourier component of the image

70
Q

Repeated measurements of Γ(y)

A

made by varying the baseline, using multiple antennas and/or using rotation synthesis will generate a sufficient number of readings of Γ(y) to give a decent map following Fourier inversion

71
Q

if the baselines do not give a good coverage of the transform plane the map will

A

contain artefacts, and will be ‘dirty’

72
Q

maps are routinely improved by

A

using reconstruction and cleaning algorithms to give the final version

73
Q

how and why would you expect the effective area to depend on wavelength

A

we would expect the effective area to decrease with wavelength because surface inaccuracies become more important as the wavelength drops

74
Q

VLBI diagram

A

see notes for answer

75
Q

uv-plane

A

is the plane perpendicular to the line of sight to the source

This is the fourier transform plane of the image

76
Q

The coordinates u and v measure

A

the projected baseline of the interferometer onto this plane in units of wavelength

77
Q

uv-tracks of 12 hour observation at the north pole

A

see notes

semi-circular track

78
Q

uv-tracks of 12 hour observation at a general declination

A

track more elliptical

79
Q

What relevance does the Van Cittert-Zernike theorem have to radio astronomical imaging

A

The VCZ theorem gives the Fourier transform of the relationship between the complex fringe visibility on a baseline x and the sky brightness distribution.

The inverse transform exists, so an image of the sky can be recovered from these observations

80
Q

a gaussian brightness profile together with a bright point at the centre

A

B(θ) = b(1)δ(θ) + b(2) exp{-θ^2/a^2}

81
Q

if a source was not overhead

A

must consider the projected baseline in the plane perpendicular to the source direction, i.e.

x becomes xcos(z)

where z is the zenith angle

path compensation may also need to be preformed

82
Q

antenna temperature

A

is the component coming from the radio source of interest

83
Q

an array of delta functions

A

F(x) = δ(x-D/2) + δ(x+D/2)

84
Q

in a phased array

A

the signal from the antennas are addded

85
Q

in a correlating interferometer

A

the signal from the antennas are multiplied

86
Q

the phase of the correlated signal can be modelled as

A

a point source at position s(0) can be modelled as a delta function so that the sky brightness distribution is

B(s) = δ(s-s(0))

therefore

Γ(s) = ∫ B(s)exp(ikxs)ds = exp{ikxs(0)}

arg(Γ) = kxs(0) = 2πxs(0)/λ

87
Q

angular radius as seen from earth

A

θ = D/d

where D is the diameter and d is the distance

88
Q

solid angle as seen from earth

A

Ω = (θr)^2π

89
Q

dish surface accuracy

A

the dish needs surface accuracy that is a small fraction of a wavelength to maintain a high efficiency

90
Q

bandwidth

A

a wider bandwidth will increase the power received so will give a stronger signal

91
Q

how to distinguish between the noise signal from the source and the system noise

A

chopping the secondary reflector on and off the source

92
Q

disadvantages to a large single dish

A

impractical to make a steerable single dish with a large collecting area

has a small primary beam -> small field of view

we want to spread the area out over a large baseline to get good angular resolution

we want to preform interferometry - two recievers

93
Q

A(g) =

A

A(e)/η

94
Q

correlation

A

process of generating the fringe visibility from two antenna signals without actually generating the fringes. For two signals
𝜓1 and 𝜓2, the correlator generates an output proportional to ⟨𝜓1𝜓2*⟩.

95
Q

Phased switched interferometer diagram

A

see notes

96
Q

why do we chop a signal

A

The gain and system temperature of a radio telescope fluctuate in time

97
Q

antenna array vs interferometer

A

An antenna array is formed when the signals from several antennas are summed to make one large single antenna.

In an interferometer, the signals from pairs of antennas are multiplied

98
Q

Antenna component

A

This turns the incident radiation into a corresponding electric signal, EM wave into a voltage.

This voltage has an approximately white spectrum. doesnt introduce much noise.

99
Q

Pre-amplifer component

A

This boosts the voltage so the antenna signal is now strong enough to not be degraded.

introduces noise

100
Q

Filter componenent

A

This reduces the range of frequencies present, defining a bandwidth and cutting out radio signals outside the band that might interfere with the observations.

101
Q

Mixer component

A

The mixer shifts the high radio frequency signals down to an intermediate frequency by mixing the signal with a local oscillator

102
Q

Square-law detector component

A

Power is proportional to v^2, so we need a block that will carry out this squaring process

103
Q

Integrator

A

This averages the fluctuating noise power output of the detector to give a signal proportion to its mean level

improves SNR ratio

104
Q

front end

A

defines sensitivity

105
Q

back end

A

preforms processing

106
Q

large antennas are

A

more sensitive and more directive

107
Q

Ω(A) is

A

the effective region of sky to which the antenna is sensitive

108
Q

the directive gain of an antenna

A

is the angular selectivity it has over an ‘isotropic’ antenna

109
Q

Beam chopping

Tsource =

A

T(on)-T(off)

110
Q

phase switch output

A

<(v1+v2)^2 -(v1-v2)^2>
=4<v1v2></v1v2>

111
Q

correlating interferometer diagram

A

see notes

112
Q

requirements for VLBI

A

the source must be very compact

timekeeping: coherence time = 1/delta v

recordings must be synchronised to better than the coherence time.

LO stability: signal phase must not wander on timescales.

delta t/T fractional stability
=> T&laquo_space;1/v 1/fractional stability

113
Q

earth rotation aperture synthesis

A

2 element interferometer at the North Pole

Instruments for Optical and Radio Astronomy (7 decks)

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