Complex avaible source power

[Alex_IC]

Hello!

I am reading Niknejad's lecture and wonder of p.15, why "the factor of 4 instead of 8 is used since we are now dealing with complex power"?

May be the factor 2 should be because of RMS value used in equation for Pin (V1 and I1 are RMS) and actually Pavs=Vs.rms^2/(4*Zo)?

Tnx for reply!

 

[BigBoss]

What's wrong with this ?? Available power from a source is always Vs^2/4Zs..
These are generalized equations, rms conditions are not mentioned.I think he express the signals in always with their peak values.
Remember the maximum power transfer theorem.When a source is terminated with its Zs, the maximum available power is delivered to the load.
So.. Nothing is wrong with this statement..

 

[Alex_IC]

Pavs.rms=(Vs.rms^2)/4*Zs and Pavs.rms=(Vs.peak^2)/8*Zs so is Vs peak value or RMS value in the lecture? It seems to me it should be peak value and this results in my confusing about words "the factor of 4 instead of 8 is used since we are now dealing with complex power".

Again there is an Eq. Pin=0.5*(V1*I1'+V1'*I1) (eq.1), if the V1 and I1 are the peak values the Pin is peak value too.
Because Pin is power absorbed at the input of a two port network so it is Re(V1*I1') which is exactly the eq.1. But it should be RMS value.

Let's look at p.7 there is eq. Pin=Re(Yin)*(|V1|^2)/2 from this Pin is RMS power. So now I'm confusing are Pin, Pr and Pavs RMS or peak values....?

So, my guess is that Pin, Pr, Pavs are peak values.