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Matched two-port, noise figure = loss

SummerSilk

Apr 14, 2014
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Apr 14, 2014
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Hello,

This isnt homework, but past paper revision (we will, annoying, not be given answers).

Can anyone give more detail on what this means by 'loss', and how the equation on the second page of this document prove that the stated fact is true?
The equation is:

F =( Psig * (P(Noise in Signal) + (P(Noise from Amp)))/Psig*P((P(Noise in Signal))

THats fairly nasty to read, so I am linking.


http://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=1&cad=rja&uact=8&ved=0CC4QFjAA&url=http%3A%2F%2Frfic.eecs.berkeley.edu%2F~niknejad%2Fee242%2Fpdf%2Feecs242_lect8_twoportnoise.pdf&ei=XvJLU9ilGO-u7Aadk4CYBw&usg=AFQjCNFB8wFso98qA__zs8R-B9RsS8wbAg&sig2=i9bkHps2R5ry1dVO5raD9A
I hope that link works, if not a quick google for
"

If a matched two-port has loss, then the noise figure is equal to the loss. To see this, we use the following equivalent formulation" Yields the PDF as the top result.
 
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