Designer’s Guide Community

Multicathode New Member Offline Posts: 5	Effective Noise Figure of this block Feb 05^th, 2012, 10:46am Hi guys, Please help me find the effective noise figure of this system. The parallel blocks are identical amplifiers I have attached the file. I dont know how to add an image to the post. I would appreciate your help Thanks
Back to top	noise_figure.JPG IP Logged

aaron_do Senior Fellow Offline Posts: 1398	Re: Effective Noise Figure of this block Reply #1 - Feb 5^th, 2012, 5:23pm Hi, I think the answer is just "F". Its not difficult to calculate from F=total output noise/output noise due to source. Assume some input-referred noise of the amplifier (Namp for example). regards, Aaron
Back to top	there is no energy in matter other than that received from the environment - Nikola Tesla IP Logged

Multicathode

New Member

Offline

Posts: 5

Re: Effective Noise Figure of this block
Reply #2 - Feb 5^th, 2012, 6:16pm

So you mean to say that I assume that Namp be the Noise input �for the amplifier and then calculate like we do for cascaded stages. I got it.

I have a doubt here.

In splitter,

Will the input noise get splited? �Wat about the inout signal?
What I am assuming is that

Namp(Noise at ip of 1 amp)=(G1Ni)/4 + Na

where Na is noise added by splitter
Ni=input noise and G1 is the Gain of splitter

and for the signal

Samp(signal at ip of 1 amp)=(G1Si)/4

This is the thing that I am doing.

Am I doing it correctly?

Back to top

IP Logged

aaron_do

Senior Fellow

Offline

Posts: 1398

Re: Effective Noise Figure of this block
Reply #3 - Feb 5^th, 2012, 7:49pm

Hi,

The input noise power will get split in the same way that the signal power is split. The splitter itself has no knowledge of what is being fed into it.

The splitter is a passive component, so the noise that it adds is only related to its insertion loss. This could be small depending on the splitter you are using. For an ideal power splitter, you can assume that no noise is added.

If you are talking about a non-ideal power splitter, then I would have to think about it a bit more. Your equations look to be correct, but how they really work out is a bit unclear since in reality, G1 and Na are related as noise factor is normally equal to insertion loss (for a passive component). I would start with that fact and then simply divide the noise power and signal power 4 ways.

regards,
Aaron

Back to top

there is no energy in matter other than that received from the environment - Nikola Tesla

IP Logged