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Design >> RF Design >> noise analysis of two port
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Message started by dog1 on Jul 31^st, 2014, 3:10am

Title: noise analysis of two port
Post by dog1 on Jul 31^st, 2014, 3:10am

I am checking a book called “structured electronic design”. It is said that you can shift the noise source from the output to input through a two port.

But assuming that there is a two port connected to a resistor at its output. When using Thevenin representation for the resistor, as in the first figure, the voltage noise source is in series with the output of the two port and should be able to transfer to the input. Assume the ABCD of the two ports are all positive, then the larger R, the larger noise you get at the input.

Using Norton representation, however, there is a current source between the output ports, and the smaller the R, the larger the noise at the input. Contradicting the previous analysis.

So where is the problem, why do I have different input referred noise source when using those two ways of analysis (the former related to A and C parameter while the latter to B and D)?

Title: Re: noise analysis of two port
Post by aaron_do on Jul 31^st, 2014, 4:05am

I don't see how you reached those conclusions so easily. When you add a resistor to the output and then you want to use ABCD parameters, you need to change the ABCD parameters to include the resistor...

Aaron

Title: Re: noise analysis of two port
Post by dog1 on Jul 31^st, 2014, 7:09am

well, I think that I don't need to change the matrix.

let's say, the two port is a mosfet, and its ABCD is thus given by the character of the mosfet. whatever you do at the load or source shouldn't change the ABCD of it. because when you calculate the ABCD, the condition is the output is either shorted or opened.

Of course, you can include the resistor into the two port, so that the two port is now a mosfet with an resistor at the drain. And then again you can calculate the output noise voltage and current source of the new two port due to the resistor and put it at the end of the two port, and then shift it back to the input using the ABCD of the new two port. but that's not what I am talking about.

the way I am following is to keep the initial two port unchanged after adding the resistor, and the resistor will just be seen as a load. And my analysis follows in the way I mentioned before.

Obviously there is something wrong with that analysis, but I cannot figure out why.

Attached is the way mentioned in the book about how you can shift the noise source from the output back to the input. I think it should hold regardless of the load or source condition and only related to the ABCD of the two port itself.

Thanks!

Title: Re: noise analysis of two port
Post by dog1 on Jul 31^st, 2014, 7:32am

I think I know where the problem is.

I forgot to add the output branch in the first figure. So the noise source there isn't in series with the output branch of the two port.

Thanks for your answers anyway!

Title: Re: noise analysis of two port
Post by aaron_do on Jul 31^st, 2014, 10:20pm

You're first figure looks correct to me. When you talk about a voltage source at the output, it is not the same as a resistor with a noise voltage as the resistor will affect the actual voltage reaching the 2-port network...