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Design >> RF Design >> Effective Noise Figure of this block
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Message started by Multicathode on Feb 5^th, 2012, 10:46am

Title: Effective Noise Figure of this block
Post by Multicathode on Feb 5^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

Title: Re: Effective Noise Figure of this block
Post by aaron_do on 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

Title: Re: Effective Noise Figure of this block
Post by Multicathode on 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?

Title: Re: Effective Noise Figure of this block
Post by aaron_do on 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