Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section A / 2

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- Solution 2

Solution

The Johnson-Nyquist noise formula gives the open-circuit mean-square voltage in a small frequency band

d ν

. When a source electrical resistance

R

drives a load

R

with impedance matching, the load sees half the source voltage, so the available noise power is

d P_{t} = \frac{⟨ V ^{2} ⟩ /4}{R} = k_{B} T_{t} d ν .

The factor of four in the open-circuit formula is therefore not an extra factor in the delivered power.

The transmitter's antenna gain

G

makes the axial spectral energy flux at range

d

\frac{d S}{d ν} = \frac{G}{4 π d ^{2}} k_{B} T_{t} .

Multiplication by the receiving antenna effective area

A

gives

d P_{r} = k_{B} T_{t} G A d ν / (4 π d^{2})

. Define the equivalent receiving antenna temperature

T_{A}

by the available matched-load noise

d P_{r} = k_{B} T_{A} d ν

. Then

T_{A} = T_{t} \frac{G A}{4 π d ^{2}} = (1 0^{4} K) \frac{1 0 ^{5} ( 1.8 m ^{2} )}{4 π ( 1 0 ^{3} m ) ^{2}} = 143.24 K ≃ 143 K .

This is the transmitter-induced equivalent noise temperature of the receiver port. It does not determine the actual physical temperature of its internal electrical resistance, or any additional receiver and background noise. The calculation assumes the usual far-field link, negligible path loss beyond geometric spreading, matched polarization of an electromagnetic wave at the two antennas and the stated effective collecting area. No additional one-half polarization factor is needed for the matched polarization of radiation from one transmitting antenna.

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