I'm working on designing a radio system for CubeSats and I want to make sure that the sensitivity of my receiver is as high as it can be. I'm trying to understand what parameters are going to set the lower bound for the performance I can squeeze out of my design.

One thing I am still confused about is antenna temperature and thermal noise. The closest question I could find is here: https://physics.stackexchange.com/questions/289893/is-antenna-noise-temperature-relevant-if-the-physical-system-temperature-is-high

I don't think the top answer on that post answers the question that the original poster was asking.

Basically, the noise power for a system is given by the following formula for Johnson-Nyquist noise: \$ N = k_bT\Delta B\$. As I understand it, this is present at all points of a system and exists regardless of resistance, material, input loss, or any other parameters. Basically, this sets a lower limit on the amount of noise power you can get your system down to.

I've started learning about antenna noise temperature though and I know that this is a function of what the antenna is actually "seeing" in its radiation pattern. When pointing at the sky, the antenna noise temperature can go as low as 3 K, which is the microwave cosmic background.

For receivers at 300 K, does a low antenna noise temperature matter at all? As I understand it, any antenna at a physical temperature of 300 K or so will have a much higher thermal noise power than anything received as antenna noise.

The advantages of putting the LNA close to the antenna have to deal with avoiding resistive losses over a transmission line and not with minimizing thermal noise because thermal noise is omnipresent in equal amounts in the circuit, correct?

  • \$\begingroup\$ an antenna that's at 300 K did you mean to write receiver? An antenna at 300 K can still have a much lower noise temperature if it's low loss, as can a feeder cable. But yes, the temperature of your LNA front end will tend to dominate. But as that can be small and easily put into a dewar, and the antenna is huge, must be pointed somewhere, and connected by a potentially lossy feeder, people still have cause to worry about the latter terms. \$\endgroup\$
    – Neil_UK
    Oct 1, 2022 at 4:58
  • \$\begingroup\$ No, I meant antenna, I should clarify that I meant the physical temperature of the antenna. Basically, my confusion comes from the fact that there is an effective noise temperature of the antenna which is the same as replacing the antenna with a purely resistive element at that temperature. But, doesn't the antenna have resistance anyway that contributes thermal noise at a much higher level for a room temperature antenna pointed at the sky? Why does low loss mean that the thermal Johnson-Nyquist noise will be lower? Is it not the same at all points in the circuit? \$\endgroup\$
    – cEEa
    Oct 1, 2022 at 15:22
  • \$\begingroup\$ A lossless antenna, just like a lossless cable, doesn't add thermal noise to the signal, only losses at that temperature. That's why radio telescopes are kept clean, bird droppings at 300 K contribute a lot of noise when it's staring at cold space. An antenna will have losses. I don't recall at the moment how to equate (for instance) 0.2 dB loss at 300 K to a noise temperature or noise figure, but you should be able to google it, knowing roughly what you're looking for. \$\endgroup\$
    – Neil_UK
    Oct 1, 2022 at 15:50
  • \$\begingroup\$ There's still something I'm not understanding here and I think I'm failing to put it into words properly. Why does a lossless cable not contribute noise? The electrons inside still have random motion that gives rise to the noise voltage right? Additionally, on Wikipedia here they derive the equation for noise power in a resistive element and at the end, the formula for noise power is not a function of resistance. Does this not imply that all parts of a circuit contribute the same thermal noise power, regardless of resistance? \$\endgroup\$
    – cEEa
    Oct 1, 2022 at 21:00
  • \$\begingroup\$ Does this not imply that all parts of a circuit **contribute** the same thermal noise power, regardless of resistance? Have available into a matched load - yes - contribute - no. The low loss feeder is not matched to the LNA, so does not contribute all the available noise power to it. See my answer. Jiggling electrons is rarely a good model, and that's true in this case. We usually need to go lower, into quantum mechanics, to see how conductors behave, or higher into circuit theory or thermodynamics to understand systems. Best forget electrons, they are unnecessary and insufficient. \$\endgroup\$
    – Neil_UK
    Oct 2, 2022 at 10:00

1 Answer 1


Why does a lossless cable not contribute noise?

Let's look at the circuit, and assume a completely matched system for convenience.

R1 is the source impedance. In the case of a radio telescope, this will be the cold of space, ideally a few K or tens of K. You know how to handle this noise.

R5 is the LNA input impedance. We might be using a cryo-cooled LNA, which would make that noise contribution fairly small as well.

The lossy antenna and cable is modeled by the pi pad of R2,3,4, whose resistors are at 300 K, so are each generating their full KTB available noise. The weird value of 0.87 dB is chosen to get nice round numbers for the attenuator components. 10K and 0.5 Ω resistors would give you 0.087dB (spot a pattern?). There are any number of pi-pad calculators and formulae on the net available with a quick search.


simulate this circuit – Schematic created using CircuitLab

We know that each resistor has Johnson-Nyquist noise, which value is independent of its actual resistance. When such a resistor is loaded with its matched resistance, it will transfer KTB to the load.

I am not going to produce any formulae here. A casual reader of this question will not benefit. Somebody wanting to understand must do the sums, or at least the simulation, themselves, to drive the point home. Therefore, only hand-waving follows.

Consider the noise contribution of R4. We can model the noise, and this works fine in Spice, as a current source in parallel with R4. It also works to model the noise voltage in series though, especially in Spice, using a parallel current source means we don't have to break a wire to add it.

The R4 noise current passing through R4 would produce the normal expected noise voltage in R4. Now, when we put R5 in parallel, R5 is much, much smaller than R4, so reduces the noise voltage to roughly 1/20th of what R4 would produce without being shunted. R5 receives nothing like the noise power you would expect for a fully matched situation.

We can do a similar thing for R3's noise power. The R3 current source in parallel with it produces only a very small voltage across R3, and as a result, little of that noise current flows in R5.

Even though the resistors are at 300 K, and able to produce their full KTB noise power into a matched load, the large mismatch between them and the LNA input impedance means that only a small fraction of that possible noise power is actually manifest in the LNA input resistance.

In the limit, as the line becomes lossless, R2 and R4 tend to infinity, and R3 tends to zero, the power they can produce in R5 tends to zero.

In a very lossy line, R2 to R4 assume values more commensurate with 50 Ω. For instance, a 6dB pad uses R2,4 = 150 Ω, with R3 = 37 Ω. These are a better match to R5, and more of their noise power is transferred to it.

In the limit of a very lossy line, R2 and R4 tend to 50 Ω, R3 to infinity, and a whole KTB gets dissipated in the LNA from R4.

  • \$\begingroup\$ This is the answer I needed, I had missed the importance of the matched load when considering how that power actually gets transferred to the receiver elements or transferred from antenna elements. \$\endgroup\$
    – cEEa
    Oct 2, 2022 at 14:17
  • \$\begingroup\$ @cEEa Good. I've rather glossed over whether you can model a lossy line or antenna as an attenuator pad. You can, but some people don't grok it at first sight. There is a black box theorem which says that any 2-port resistive network can be be reduced to at most three resistors. You have three unknowns, gain and two port resistances, so only three resistors are needed. A tee-pad works as well. I can't think of what to search for to find the reference. \$\endgroup\$
    – Neil_UK
    Oct 2, 2022 at 14:37

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