Mechatron
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the_emi_guy said:This is where you are getting off track, noise does not add in the manner you are describing.
We need to know the *bandwidth* of the 20KHz signal, or more specifically, the bandwidth of the receiver that will be trying to recover this signal. I'll explain why shortly.
Here you are saying that the thermal noise is both white and limited to the very small 20-1000Hz band. These are mutually exclusive statements. White means its power spectral density at 1MHz is the same as at 20Hz.
So let's say that the power spectral density is 1nW per Hz of bandwidth. This means that there will be 1nW of noise power in every 1Hz of bandwidth. If I build a perfect (doesn't create any noise of its own) receiver that tunes to 20KHz with a bandwidth of 1Hz (receives everything between 20,000Hz and 20,001Hz, rejects everything else) then I will have 1nW of noise power coming out of this receiver.
On the other hand, let's say your receiver has 1KHz of bandwidth (20KHz - 21KHz received) then we will have 1000*1nW = 1uW of noise power received.
If your 20KHz signal is 1W, then our SNR is 1W/1uW.
Hope this clarifies things, we need to know the bandwidth of the 20KHz receiver, and the noise power in whatever units (dBm/Hz, nW/Hz).
I'm sorry if things seem to be off track, but I think we're close to solving our problem.
I wanted to add that I calculated the bandwidth using the second equation in the pdf with value; 300 K, (r+R) 1000,2 Ohm, -204dBV/Hz and the Boltzmann's constant of 1,38*10^-3 which gave me a bandwidth of 2,51 *10^21 (Zetta Hz), which in the electromagnetic specter would be gamma radiation (if it was a frequency), which doesn't seem very reasonable to me. And as previously mentioned, the f=KbK/h of 6,2 THz didn't make much sense either. So I'm really, really confused.