I have a question about RF power

Hello everyone,
I am trying to simulate in Matlab a RF recevier, but I need to understand how to "see" the receiver input power.
Should I see it how the "power avaiable from a generator" or how the "power dissipated" on a generic 50 Ohm resistor?
The two type of power have two different equations to calculate it:
"power avaiable from a generator" = Vrms^2/(4*R)
"Power dissipated" = Vrms^2/R
In order to generate the signal voltage to input to receiver, which formula should I start from?

I want to simulate the power that will effectly be generated from signal generator and measured from a power spectrum analyzer (in laboratory).

Help me to understand.
Many thanks.

Antonio L.

(V^2)/2 should be fine

Anton89 said:

Hello everyone,
I want to simulate the power that will effectly be generated from signal generator and measured from a power spectrum analyzer (in laboratory).
Antonio L.

Click to expand...

Le t me give you a hint..
As well known, the max. obtainable power from a generator that has an internal source impedance let say Zs, is defined when this generator is terminated by complex conjugate of this source resistance so conj{zs}
By this way, you can calculate how much power can be obtained from this generator.In other words, how much power is dissipated on load impedance.

Input power can be calculated like this:



just be careful voltage is not RMS here

you can replace V= Vrms x sqrt(2)

A standard RF generator has a 50 ohm output impedance. When you set it, the displayed power is referred to the nominal load of 50 ohm.
Using the schematic drawn by Ata_sa16 (Thevenin equivalent):

Vin = Vg/2 this is the voltage seen by the load

The RMS power dissipated by it will be, simply:

Pload=Vin^2/R that referred to the internal generator is Pgen=Vg^2/(4R)

Then if you know which power is sent to the nominal load of 50 ohm, let's say Prec0 then you can calculate the generator voltage:

Prec = Vin^2/R ==> Vin = sqrt(Prec0*R)
since Vin = Vg/2 ==> Vg = 2*sqrt(Prec0*R)

So, in Matlab you will have to simulate an ideal generator of voltage Vg with series resistance R=50.
Connecting it to an ideal resistor R=50 ohm you will see the Prec0 you have set. When instead you will connect it to a real receiver the dissipated power will be different, according to the actual impedance of the receiver.
Let's imagine the actual receiver has a resistive impedance Rrec, then

Vrec=Vg*Rrec/(R+Rrec) ==> Vrec= 2*sqrt(Prec0*R)*Rrec/(R+Rrec)
Prec = Vrec^2/Rrec ==> Prec = 4*Prec0*R*Rrec/(R+Rrec)^2