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    Why does the available power equal abs(b)^2-abs(a)^2?

    Hi,
    I read a S-parameter tutorial. It first has such an equation:

    Click image for larger version. 

Name:	Pav0.PNG 
Views:	18 
Size:	23.2 KB 
ID:	147879

    Then, it has another equation:

    Click image for larger version. 

Name:	Pav.PNG 
Views:	20 
Size:	46.8 KB 
ID:	147878


    I am puzzled why they are so different?

    Can you explain it to me?

    Best Regards,
    Last edited by ruwan2; 15th July 2018 at 06:20.

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    Re: Why is (b_nXa_n^*b_n^*Xa_n) pure imaginary?

    Pin is the Power which is entered into circuit.Pavs is Available Power from Source.
    They are obviously different.If Pr is zero ( conjugate matching ) all power available from the source enters into circuit.



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    Re: Why does the available power equal abs(b)^2-abs(a)^2?

    Quote Originally Posted by ruwan2 View Post
    Then, it has another equation:
    Equations are wrong.
    Pavs has to be Pin.



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    Re: Why does the available power equal abs(b)^2-abs(a)^2?

    To start with we should clarify some terminology:
    Pavs (aka the available, incident, exchangeable power from the source)=|a|˛
    Pr (reflected power)=|b|˛
    Pin (absorbed, transmitted, or actual power) =|a|˛-|b|˛.

    So the equation in the first attachment is fine: Pin=Pavs-Pr.

    But the second attachment is completely off. I'm not even sure why the condition ΓLS* is relevant. That doesn't have anything to do with conjugate match condition...

    - - - Updated - - -

    Quote Originally Posted by pancho_hideboo View Post
    Equations are wrong.
    Pavs has to be Pin.
    I'm guessing that here "Pin" doesn't mean "incident" power, but rather power absorbed by the load... I know, very sloppy.
    Last edited by mtwieg; 16th July 2018 at 18:26.



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