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

1. 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:

Then, it has another equation:

I am puzzled why they are so different?

Can you explain it to me?

Best Regards,

•

2. 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.

•

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

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

•

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

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 - - -

Originally Posted by pancho_hideboo
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.

--[[ ]]--