Antennas and Noise

Question 1

Antennas definition

Answer 1

devices that turn the incident radiation into a corresponding electric signal

Question 2

Antenna signals are

Answer 2

noise-like, random fluctuations in the system

Question 3

Radiation from incoherent sources

Answer 3

have no correlation between signals from different parts of the sky and thus add powers from these parts

Question 4

Power through area A formula

Answer 4

w = SAΔv measured in watts

Question 5

power through the effective area

Answer 5

when an antenna only detects 1 polarisation

w = 1/2 S AeΔv

Question 6

for one polarisation

Answer 6

1/2 the total power

Question 7

aperture efficiency

Answer 7

ŋ = Ae/Ag

Question 8

the power through an area where the source is not on the axis of the antenna

Answer 8

w = 1/2SAe*P(θ,Φ)Δv

Question 9

signal from an extended source =

Answer 9

w = 1/2Ae∫B(θ,Φ)P(θ,Φ)dΩ*Δv

Question 10

the beam solid angle Ωa =

Answer 10

∫B(θ,Φ)P(θ,Φ)dΩ

Question 11

the reciprocity theorem

Answer 11

Antenna power patterns are the same for transmitters and recievers

Question 12

Simple antenna - the dipole diagram

Answer 12

check notes for diagram

E-field of radiation sets up currents in the antenna => voltage over the resistor

must be kept short (<λ)
only sensitive to one polarisation (E// dipole)

Question 13

a single antenna sensitive to all the radiation cannot be designed as it

Answer 13

would contradict the 2nd law of thermodynamics

Question 14

Thermal motions of electrons in a resistor generate a

Answer 14

fluctuating voltage over its ends

Question 15

Nygnist’s Noise Theorem

Answer 15

P = kT W Hz^-1

Question 16

if an antenna’s beam is filled with a blackbody source at temperature T, the power it delivers is

Answer 16

w = kTΔv measured in watts

Question 17

antenna temperature

Answer 17

Ta = w/kΔv

Question 18

Ta = true temperature of source if

Answer 18

1) the source fills the beam
2) the source is thermal, with a Planck spectrum

Question 19

expression for antenna temperature derivation

Answer 19

2kTa = S*Ae

S = ∫ B(θ,ϕ)P(θ,ϕ)dΩ

Ta = Ae/λ² * ∫ Tb(θ,ϕ)P(θ,ϕ)dΩ

if the source is not small compared with the beam , we must include P(θ,ϕ)

Question 20

antenna temperature in terms of sky brightness temperature

Answer 20

Ta = Ae/λ² * ∫ Tb(θ,ϕ)P(θ,ϕ)dΩ

Question 21

if the sky brightness is uniform then

Answer 21

Ta = Tb

Question 22

effective area formula

Answer 22

Ae = λ²/Ωa

and is true for any antenna

Question 23

directive gain

Answer 23

4π/Ωa

Question 24

system temperature

Answer 24

measure of the power per unit bandwidth contributing to the noise in a radio telescope system

Question 25

antenna temperature

Answer 25

how much noise an antenna produces in a given environment is the component of this coming from the radio source of interest

Question 26

in a phased array the signal

Answer 26

from the antennas are added.

Question 27

in a correlating interferometer the phased array of the signal

Answer 27

are multiplied

Question 28

dish surface accuracy

Answer 28

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

Question 29

observing bandwidth

Answer 29

for a blackbody thermal noise source a wider bandwidth will increase the power recieved, so give a stronger signal.

Question 30

to distinguish the noise signal from the system noise given the dish cannot move

Answer 30

chopping a secondary reflector on and off the source.

Question 31

why use a large collecting area of small dishes rather than one large dish

Answer 31

1) impractical to make a steerable single dish 2) has a very small primary beam and thus a small field of view 3) we want to spread the area out over large baselines to get good angular resolution 4) we want to preform interferometry and this can only be done with multiple recievers