Practice Questions Flashcards
1
Q
B(v) of a blackbody is
A
B(v) = 2kT/λ^2
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
3
Q
S = ∫B(θ,φ) dΩ
A
dΩ = sinθdθdφ
and for small angles
(θ/2)^2 π
4
Q
converting from degrees to radians
A
radians = degree π/180
5
Q
S is measured in
A
W m^-2 Hz^1
6
Q
converting from arc seconds to radians
A
radians = arc π/180*3600
7
Q
L = 4πR^2S
A
where R is the distance away
8
Q
to convert from ly to meters
A
ly = 3x10^8 * 60 * 60 * 24 * 365.25
9
Q
gain =
A
gain = 4π/Ω(A)
10
Q
Ω(A) =
A
λ^2/A(e)
11
Q
w = kTΔv
A
where w is the power received per unit bandwidth
and Δv is the bandwidth
12
Q
w = 1/2 S A(e) Δv
A
where S is the flux density and A(e) the effective area
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
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
15
Q
For Ω(s) «_space;Ω(A)
A
P ~ 1
=> S = B Ω(s)
16
Q
Ω(A) =
A
∫ P(θ,φ) dΩ
dΩ = sinθdθdφ
for small angles
dΩ = (θ/2)^2 π
17
Q
A(g) =
A
πD^2/4
18
Q
A(e) =
A
ηπD^2/4
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)
20
Q
phase error =
A
φ = 2π/λ 2ε = 4πε/λ
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)
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
23
Q
T(sys) =
A
T(source)+T(background) +….
24
Q
kT(source) =
A
1/2 S A(e)
25
Noise N =
(Δvτ)
so
sqrt(N) = (Δvτ)^1/2
26
SNR =
T(source)/T(sys) (Δvτ)^1/2
27
why might the most popular choice for high sensitivity instruments be ones with a large collecting areas rather than wide bandwidths
SNR increases proportionally with A(e)(Δvτ)^1/2 so it is better to increase A(e) rather than Δv
28
The complex visibility measured by the interferometer can be obtained from
the mean conjugate product (or cross-correlation) of the signals <Ψ1Ψ2*>
29
The cross-correlation =
<Ψ1Ψ2*> ∝ exp(-ikr(v).θ(v))
30
the cross-correlation can be rewritten in terms of the source's flux density S
<Ψ1Ψ2*> = S exp(-ikr(v).θ(v))
31
The extended source is usually
incoherent , the correlation between signals from two different patches of sky is zero
32
a small patch of the sky of surface brightness and solid angle will contribute a flux density. So rewriting the cross-correlation as
d<Ψ1Ψ2*> = B(θ) dΩ exp(-ikr(v).θ(v))
33
since the patches are incoherent the cross-correlation term can be rewritten as
<Ψ1Ψ2*> = ( ∫sky) B(θ) exp(-ikr(v).θ(v)) dΩ
34
the complex visibility is defined as
Γ(r(v)) ∝ <Ψ1Ψ2*>
Γ(0(v)) = 1
35
the constant of proportionality must be
1/( ∫ B(θ)dΩ) = 1/S
36
the measured complex visibility for an arbitrary extended source of surface brightness B(θ) and total flux density S is
Γ(r(v)) = 1/S ∫ B(θ) exp(-ikr(v).θ(v)) dΩ
37
the fringe rate is
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
38
if the interferometer is oriented N-S
D(v) = (0,D,0)
D(v) . θ(v hat) = DcosHcosδ
φ = 2π/λ cosHcosδ
taking derivative
dφ/dt = 2π/λ cosδsinH dH/dt
39
at the meridian the hour angle is
H = 90 degrees
40
dH/dt =
2π/60*60*24
41
difference between primary beam of an interferometer and the synthesised beam
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
42
if the interferometer is oriented N-S
D(v) = (D,0,0)
D(v) . θ(v hat) = DcosδsinH
φ = 2π/λ cosδsinH
taking derivative
dφ/dt = 2π/λ cosδcosH dH/dt
43
max resolution =
λ/D(max)
44
τ =
1/Δv
45
max scale size =
λ/D(min)
46
resolution in RA =
λ/D
47
resolution in Dec
λ/Dsinδ
48
long baseline requires
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
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
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
the effective area of an antenna is
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
the aperture efficiency is
simply the ratio of the effective area of the antenna to its geometrical area, when that is clearly defined
52
the beam solid angle is
the 2-D equivalent of the Rayleigh resolution criterion for antennas
53
Nyquist noise theorem
an antenna in an isotropic blackbody radiation field at T, over a bandwidth Δv
54
The antenna temperature generated by a point source is
defined as the temperature corresponding to the power from a source
55
If the source is not small compared to the beam (larger than the beam)
we must include the antenna pattern P() in the integral to account for apodisation
56
CMBR is the temperature of
the coldest regions of the sky at any wavelength
57
system temperature
a way of characterising the amount of noise inherent in a radio source
kTsys = w
58
the system temperature has components from all the noise sources present, so that
T(sys) = T(background) + T(source) + ... + T(LNA) + T(electronics)
at v > 500GHz Tsys >> T(A)
at v < 300MHz T(A) often dominates
59
the system temperature measures
the noise in the system, and therefore controls the denominator in the signal-to-noise ratio of a detected signal
60
for threshold detection assume
SNR = 1
61
Interferometer pros
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
I(θ) =
< | Ψ1+Ψ2 exp{-ikyθ} |^2 >
63
|Γ(y)| =
visibility
I(max) - I(min) / I(max) + I(min)
of Young's fringes
64
I(max) =
<|Ψ1|^2> +<|Ψ2|^2> + 2<|Ψ1Ψ2*|^2>
65
I(min) =
<|Ψ1|^2> +<|Ψ2|^2> - 2<|Ψ1Ψ2*|^2>
66
a maximum of the fringe pattern occurs at the point where
φ+kyθ = 0
67
Van cittert-zernike theorem
the complex fringe visibility |Γ(y)|is the Fourier transform of the normalised sky brightness
68
complex fringe visibility can itself be generated by
computing <|Ψ1Ψ2*|> which is the correlation between the two wavefronts received at the two dishes
69
The single value of Γ that is returned from one baseline measurement
is not sufficient to generate an image of the source as it just returns one Fourier component of the image
70
Repeated measurements of Γ(y)
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
if the baselines do not give a good coverage of the transform plane the map will
contain artefacts, and will be 'dirty'
72
maps are routinely improved by
using reconstruction and cleaning algorithms to give the final version
73
how and why would you expect the effective area to depend on wavelength
we would expect the effective area to decrease with wavelength because surface inaccuracies become more important as the wavelength drops
74
VLBI diagram
see notes for answer
75
uv-plane
is the plane perpendicular to the line of sight to the source
This is the fourier transform plane of the image
76
The coordinates u and v measure
the projected baseline of the interferometer onto this plane in units of wavelength
77
uv-tracks of 12 hour observation at the north pole
see notes
semi-circular track
78
uv-tracks of 12 hour observation at a general declination
track more elliptical
79
What relevance does the Van Cittert-Zernike theorem have to radio astronomical imaging
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
a gaussian brightness profile together with a bright point at the centre
B(θ) = b(1)δ(θ) + b(2) exp{-θ^2/a^2}
81
if a source was not overhead
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
antenna temperature
is the component coming from the radio source of interest
83
an array of delta functions
F(x) = δ(x-D/2) + δ(x+D/2)
84
in a phased array
the signal from the antennas are addded
85
in a correlating interferometer
the signal from the antennas are multiplied
86
the phase of the correlated signal can be modelled as
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
angular radius as seen from earth
θ = D/d
where D is the diameter and d is the distance
88
solid angle as seen from earth
Ω = (θr)^2π
89
dish surface accuracy
the dish needs surface accuracy that is a small fraction of a wavelength to maintain a high efficiency
90
bandwidth
a wider bandwidth will increase the power received so will give a stronger signal
91
how to distinguish between the noise signal from the source and the system noise
chopping the secondary reflector on and off the source
92
disadvantages to a large single dish
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
A(g) =
A(e)/η
94
correlation
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
Phased switched interferometer diagram
see notes
96
why do we chop a signal
The gain and system temperature of a radio telescope fluctuate in time
97
antenna array vs interferometer
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
Antenna component
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
Pre-amplifer component
This boosts the voltage so the antenna signal is now strong enough to not be degraded.
introduces noise
100
Filter componenent
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
Mixer component
The mixer shifts the high radio frequency signals down to an intermediate frequency by mixing the signal with a local oscillator
102
Square-law detector component
Power is proportional to v^2, so we need a block that will carry out this squaring process
103
Integrator
This averages the fluctuating noise power output of the detector to give a signal proportion to its mean level
improves SNR ratio
104
front end
defines sensitivity
105
back end
preforms processing
106
large antennas are
more sensitive and more directive
107
Ω(A) is
the effective region of sky to which the antenna is sensitive
108
the directive gain of an antenna
is the angular selectivity it has over an 'isotropic' antenna
109
Beam chopping
Tsource =
T(on)-T(off)
110
phase switch output
<(v1+v2)^2 -(v1-v2)^2>
=4
111
correlating interferometer diagram
see notes
112
requirements for VLBI
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<< 1/v 1/fractional stability
113
earth rotation aperture synthesis
2 element interferometer at the North Pole
