Johnson–Nyquist noise contribution Antenna noise temperatures?

Question

I understand that for an antenna the greatest contribution to the noise comes from the noise that the antenna captures from the environment, not the "internal" thermal noise of the antenna. But I do not know how to calculate this myself (and prove to myself that the Johnson-Nyquist noise is negligible).

I am imagining a 50 Ohm antenna in a room at temperature \$ T \$, receiving \$ P_{noise} \$ over bandwidth \$ B \$, matched to a 50 Ohm receiver. Hence the noise temperature will be \$ T_n = \frac{P_{noise}}{k_bB} \ne T \$.

However, if the wire making up the antenna has resistance \$ R=1\Omega\$, what happens to the thermal noise power \$P_{Johnson}=k_bTB\$? Is it just mismatched because \$ R \ll 50\Omega\$, and hence not delivered to the receiver?

Similarly, is there a way to express the antenna noise temperature as the sum of a component due to noise received from the environment and intrinsic thermal Johnson-Nyquist noise?

Answer 1

It depends on the ambient noise & stray signal in each band but I know LNA's are essential for Sat. Rx and GPS Rx as well for VLF global Rx. The Tx levels, Friis path loss, antennae gains and noise figure and BW of Preamp all are included in C/N ratios and with demodulation gains to S/R ratios.

So it is not universally true that Johnson-Nyquist Noise is negligible and depends on background interference and Rx threshold & acceptable BER.

\$P_\mathrm{dBm} = 10\ \log_{10}(k_\text{B} T \times 1000) + 10\ \log_{10}(\Delta f)\$

which is more commonly seen approximated for room temperature (T = 300K) as:

\$P_\mathrm{dBm} = -174 + 10\ \mathrm\log_{10} \ (\Delta f)\$

• for 0dBm= 1mW

e.g. for \$(\Delta f)\$

• 1 MHz −114 dBm Bluetooth channel ( well below ambient noise)
• 20 MHz −101 dBm WLAN 802.11 channel ( >-80 dBm signal min. typical needed)
• 80 MHz −95 dBm WLAN 802.11ac 80 MHz channel (>-65 dBm often needed )

Even small Capacitors have thermal noise due to V and C and those not using NP0 material are even microphonic.

\$v_{n} = \sqrt{ k_\text{B} T / C }\$

• e.g. 1pF 64µV, 1nF 2µV

Then we have \$1/f\$ pink solid-state noise for the spectrum of pink noise in one-dimensional signals and for 2D signals (e.g., images) the power spectrum is \$1/f^2\$.

The most common units of measure for noise is \$dB/\sqrt{Hz}\$.

The understanding of threshold noise depends on many factors including Shannon's Law for SNR vs BER and non-"matched receivers" that match the signal BW and non-ideal discriminators and Ricean Fading loss (phase cancelling reflections) and many other factors.

Answer 2

While the question in the title is fine, there were some misconceptions in the detailed description of the question. An antenna has both radiation resistance related to the EM radiation it generates \$R_{an}\$ (what we usually talk about) and dissipative resistance leading to thermal losses \$R_{th}\$ (due to the material of the wire making up the antenna, a resistance we usually neglect). Typically \$R_{th}\ll R_{an}\$.

The antenna will be receiving from background EM bath of temperature \$T_{an}\$ (290K if pointing to the warm Earth, 4K if pointing to deep space), hence the noise temperature of the antenna will be \$T_{an}\$ if we neglect \$R_{th}\$.

However, if we take both into account the circuit will look as follows (next to each resistor I have placed the corresponding noise source and I have included the transmission line that would be necessary for the computation of delivered noise power):

^{simulate this circuit – Schematic created using CircuitLab}

For a small bandwidth \$\Delta\nu\$:

• \$RMS(V_{th}) = \sqrt{4kT_{th}R_{th}\Delta\nu}\$
• \$RMS(V_{an}) = \sqrt{4kT_{an}R_{an}\Delta\nu}\$

Hence the total voltage (given that the two sources are not correlated) is \$RMS(V_{total\ noise}) = \sqrt{4k(T_{an}R_{an}+T_{th}R_{th})\Delta\nu}\$.

Hence the Johnson-Nyquist noise temperature is suppressed in comparison to the background EM noise temperature by a factor of \$\frac{R_{th}}{R_{an}}\$ which is typically less than a hundredth.

Now we can attempt to derive noise power delivered in the transmission line:

\$P = V_{delivered\ noise}^2/R_{line} =\left(\frac{R_{line}}{R_{line}+R_{an}+R{th}}\right)^2 4k(T_{an}R_{an}+T_{th}R_{th})\Delta\nu/R_{line}\$

For a matched antenna \$R_{an}=R_{line}\$ hence:

\$P = \left(\frac{R_{an}}{2R_{an}+R{th}}\right)^2 4k(T_{an}R_{an}+T_{th}R_{th})\Delta\nu/R_{an}\$

and given that \$R_{th}\ll R{an}\$:

\$P \approx k(T_{an}+T_{th}\frac{R_{th}}{R_{an}})(1-\frac{R_{th}}{R_{an}}) \Delta\nu\$.

Therefore the noise temperature is to first order:

\$T = T_{an}+(T_{th}-T_{an})\frac{R_{th}}{R_{an}}\$