What is the output PEP from a transmitter, if an oscilloscope measures 400 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?

To calculate Peak Envelope Power (PEP) from an oscilloscope's peak-to-peak voltage trace, you can use the standard shortcut formula: P = (Vp-p)² / (8 * R). Substitute the given values into the formula: P = (400)² / (8 * 50). This equals 160,000 / 400, which simplifies exactly to 400 watts. You can verify this by finding the RMS voltage first. The peak voltage is half of 400V, which is 200V. The RMS voltage is 200V * 0.707 = 141.4V. Using P = E² / R, (141.4)² / 50 = 20,000 / 50 = 400 watts. In this specific question, the PEP in watts happens to equal the peak-to-peak voltage, but this is merely a mathematical coincidence for a 400Vp-p signal on a 50-ohm load.