Noise Figure of an Attenuator

Question

it is known that the noise figure of an attenuator is equal to its attenuation coefficient L. A proof of this statement is provided here.

Basically, it uses the formula (Reference):

$$F = 1 + \frac{N_a}{GN_i}$$

for a general block with gain G (equal to 1/L in case of an attenuator), where Na is the output referred noise power of the attenuator.

The first source evaluates Na as the dissipated noise power in the attenuator.

I have some questions:

1. Why does it evaluate Na as the dissipated noise power in the attenuator, instead of half of it? If an amount of power is dissipated and creates noise, why should it go totally to the load (output)? I'd say it goes half to the load and half to the source (if both are 50 Ohm).

2. Why doesn't it consider the dissipated signal power to? Dissipated signal power becomes noise, I think....

Answer 1

The whole derivation of the N.F. of the perfectly matched attenuator can be done in simpler way.

Let's assume the attenuator is a transmission line hybrid which is properly terminated. It's every branch inputs and outputs noise power density kT or as well over bandwidth B there's noise power kTB going both directions in every branch assuming nothing generates new noise or if generates, it also absorbs as much noise as it generates. That lifts resistive attenuators to the same line as transmission line constructions.

The signal input power is P. The input S/N = P/kTB

The output signal power is P/L where L is the attenuation without decibels, for ex. attenuation 6dB means L=4.

The output S/N= (P/L)/kTB

The noise figure = The input S/N divided by the output S/N = L

Your linked proof of attenuator's N.F. starts from the formula of N.F. when a 2- port circuit adds its own noise. It leads to the same result when the added noise is selected well for the known result. The formally right formula for the added noise = that part of input noise which is not passed through the attenuator. That makes the input and output noise powers equal like they were in my version and so it gives the same result, too.

It's the formally right formula, but it leads readers easily to assume that what's removed from the input power in the attenuator becomes noise to the output if no heat conduction to elsewhere or storing happens. You extended that idea by assuming that also the lost input signal should be ejected as noise.

Both of these assumptions are pure nonsense. The removed part can be ejected from the attenuator through the 3rd port of the transmission line hybrid or converted to heat in resistive parts of the attenuator. The temperature of the resistive parts can slowly rise due the input signal but the resistive parts do not immediately output the absorbed input signal as noise. The resistive parts output noise power kTB like all resistors and absorb as much if they are connected to other resistive parts directly or via transmission lines and temperature is everywhere in the system =T.

The noise that resistive parts of the attenuator output is not the absorbed noise signal reflected immediately "as is", it's uncorrelated with the absorbed noise, only average powers are the same.