flatulent said:Because the noise output is still ktB and the signal declines by the attenuation. The SNR out is worse than the SNR in by the attenuation.
rebabel said:You'd better calculate the NF of a passive components.
That is NF=SNRin/SNRout=vnout^2/(vnin^2*A^2).
It's not a complex calculation.
flatulent said:The source noise is attenuated, but the circuit adds its own noise so it is back to kTB at the output.
what if the filter is placed after the LNA or mixer??designtech said:I think u can look at it this way...the filter is usually at the beginning of the front-end..it has a certain attenuation, so it attenuates both the noise and signal, so ideally NF shud have been 1, but as the noise cannot do below kTB (physical phenomenon), noise at o/p of filter is still kTB...hence NF is equal to conversion loss
e.Horus said:Let me attempt to clarify the issue. If the passive netwrok is "ideally" noiseless, then the NF would be equal to the loss. This has nothing to do with the noise added by the filter. What the passive filter does is simply shape the noise spectral density. However, the total noise power will remain the same at the output. If the passive network is noisy, in the case of a passive mixer for example, the NF can be higher than the loss.
Element_115 said:The Noise at the I/P is the same as the O/P for Passive circuits!
But if the Signal Power is attenuated (i/p to o/p) by the I.L. Then :
F= (S/N)in / (S/N)out
You see that the signal attenuation is the only contribution to the Noise Definition.