Attenuators and noise

[xmo]

Assuming a matched situation with lossless interconnections and all at ‘room’ temperature such that kTB = -174 dBm/Hz, there are a few accepted facts.

▪ The noise figure of an attenuator is the same as the attenuation, thus for an example 10 dB attenuator, NF = 10 dB, loss L = 10 dB, and “gain” G = 0.1 (linear factor)
Considering signal output from the attenuator vs. signal input: SO = SI – L   or  SO = G * SI

▪ Attenuators attenuate noise as well as signals.  For the 10 dB attenuator example:
If SI = -100 dBm, then SO = -110 dBm and If NI = -120 dBm/Hz, then NO = -130 dBm/Hz

▪ Attenuators do not attenuate noise to below kTB.  Therefore, if the input to an attenuator is at -174 dBm/Hz, the output is also at -174 dBm/Hz

Now for the question and the reason for this topic: Again, for the 10 dB attenuator example, if NI = -164 dBm/Hz, what is NO =

[ejeffrey]

The output noise of an attenuator just is:

NO = G * NI + (1-G) * kTB

You can think of an attenuator as a directional coupler (itself lossless and noiseless) with both the the through and isolated ports terminated into matched loads  The coupled port sees a small copy of the input + almost all of the of the thermal noise from the termination at the isolated port.

So the answer to your example is: ~-171.2 dBm/Hz.

[xmo]

ejeffrey,

Thanks.  I did some internet research and did find one reference to the solution you posted but most noise tutorials don't explain it.  I was sure that smart people on this discussion board would confirm.

I created a drawing to show the attenuator noise contribution symbolically.