= Solution
The <Johnson-Nyquist noise> formula gives the open-circuit mean-square <voltage> in a small <frequency> band $d\nu$. 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
$$
dP_t=\frac{\langle V^2\rangle/4}{R}=k_BT_t\,d\nu.
$$
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 $d$
$$
\frac{dS}{d\nu}=\frac{G}{4\pi d^2}k_BT_t.
$$
Multiplication by the receiving <antenna effective area> $A$ gives $dP_r=k_BT_tGA\,d\nu/(4\pi d^2)$. Define the equivalent receiving <antenna temperature> $T_A$ by the available matched-load noise $dP_r=k_BT_A\,d\nu$. Then
$$
\boxed{T_A=T_t\frac{GA}{4\pi d^2}
=(10^4\,\mathrm K)\frac{10^5(1.8\,\mathrm{m^2})}{4\pi(10^3\,\mathrm m)^2}
=143.24\,\mathrm K\simeq143\,\mathrm K.}
$$
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>.
Back to article page
