# What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50 ohm resistive load connected to the transmitter output?

A. 8.75 watts
B. 625 watts
C. 2500 watts
D. 5000 watts

YAY MATH

PEP or Peak Envelope Power is defined as the the highest envelope power supplied to the antenna transmission line by a transmitter during any full undistorted RF cycle or series of complete radio frequency cycles. **

Generally speaking, Power is defined as such

$P = V*I$

where V is the voltage and I is the current.  This is true in a DC circuit.  In an AC circuit, however, we need to know the RMS Voltage, or Root Mean Square Voltage of the wave.  This in effect gives us an “average” of the wave’s voltage to work with in the equation.  You find that as follows:

$V_{RMS} = 2\sqrt{2}*V_{Peak}$

Remember that the 500V given is Peak to Peak, so the wave actually only goes to positive 250VAC.

So that gives us an RMS Voltage of:

$V_{RMS} = (2\sqrt{2}*250) = 176.75V$

Now if we remember Ohm’s Law,

$V = IR$

We can use algebra to re-arrange this, we can re-write our power equation as such:

$P = (V_{RMS}^2)/R$

.

$P = (176.75^2)/50$

.

$P = 624.811 \approx 625W$

…and that’s our answer: B. 625 Watts.

[G5B14]

** https://en.wikipedia.org/wiki/Peak_envelope_power

Image: https://commons.wikimedia.org/wiki/File:RMS_voltage_average_power.svg