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Assume a lossless antenna connected to a lossy cable at physical temperature 300K. If this antenna is pointed into space, the antenna noise temperature will depend on the brightness temperature of objects in its beam (cosmic background radiation in this case ignoring atmospheric radiation). Practically, the noise temperature of a Yagi antenna pointed into space is around 63-68K. However, as the physical temperature of the connecting cable is much higher, is this low antenna temperature relevant for a receiver system?

From what I understand, output noise of a passive device (lossy cable) is calculated as follows. Assuming that $T_{in}$ is the input noise temperature and $A$ is the attenuation of the device (lossy cable in this case), and $T_e=(A-1)T_p$ is the equivalent temperature of the device at physical temperature $T_p$. $$ \begin{align*} T_{out} &= \frac{T_{in} + T_e}{A} \\ &= \frac{T_{in}}{A} + \frac{(A-1)T_p}{A} \\ &= \frac{T_{in} - T_p}{A} + T_p \end{align*} $$ Converting to noise power spectral densities $$ N_{out} = \frac{N_{in} - N_{p}}{A} + N_p $$ This makes sense if $N_{in} \ge N_p$. The excess noise above the thermal noise floor is attenuated by the attenuation of the passive device, and the output noise has to be atleast the thermal noise at the physical temperature of the system.

My main question is does this equation hold if $N_{in} < N_p$, as in the case of the antenna connected to a lossy cable, where the noise captured by the antenna is lower than the thermal noise floor at the physical temperature of the system.


System diagram

Assume all impedances are matched in the diagram.

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    $\begingroup$ I would really like to answer this but could you add an annotated circuit diagram? I'm always a little hesitant to answer circuit questions before I see a diagram because I want to make sure we all know exactly what we're talking about. $\endgroup$
    – DanielSank
    Oct 31, 2016 at 5:58
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    $\begingroup$ @DanielSank Added a diagram $\endgroup$
    – user80551
    Oct 31, 2016 at 6:23

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Yes, it's entirely possible for the thermal noise floor of the receiver to mask the signal you're trying to read.

This might be equivalent to what you said, but you can mentally replace the antenna + environment with a cold resistor whose:

  • resistance is the antenna's radiation resistance
  • temperature is the environment's radiation brightness temperature

(in the relevant frequency band and direction)

The noise of this "resistor" is what you're trying to measure. Then you also have a second resistor (or set of resistors) which represents the actual ohmic resistance of the antenna and cables. You can draw a simple circuit diagram with these two resistors (and any other relevant components), and use Kirchhoff's laws like usual to calculate the effects of the thermal noise voltages from each resistor, and thus figure out how each contributes to the total measured signal.

If the thermal noise is swamping the radiation noise, then yes, that makes it harder to do an accurate measurement, and you better cool your receiver or reduce the ohmic resistance of the components in series with the antenna (among other things, increase the antenna's radiation efficiency, i.e. increase the ratio of radiation resistance to ohmic resistance).

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