FCC Exam Question: 3-55G5

What is the output peak envelope power from a transmitter as measured on an oscilloscope showing 200 volts peak-to-peak across a 50-ohm load resistor?

A. 1,000 watts.
B. 100 watts.
C. 200 watts.
D. 400 watts.
Correct Answer: B

Explanation: To determine the Peak Envelope Power (PEP), we first need to convert the peak-to-peak voltage (Vpp) to the peak voltage (Vp). 1. **Calculate Peak Voltage (Vp):** Vp = Vpp / 2 = 200 V / 2 = 100 V 2. **Calculate PEP:** For sinusoidal RF power, the power is calculated using the peak voltage and the load resistance. The formula for power in watts, given peak voltage and resistance, is: PEP = Vp² / (2 * R) This formula accounts for the relationship between peak voltage and RMS voltage (Vrms = Vp / √2), as power is typically defined using RMS values (P = Vrms² / R). Substitute the values: PEP = (100 V)² / (2 * 50 ohms) PEP = 10000 / 100 PEP = 100 watts Therefore, the output peak envelope power is 100 watts. Option B is correct. Options A, C, and D would result from using the peak-to-peak voltage directly or incorrectly applying the power formula for peak voltage.

Pass Your FCC Exam!

Study offline, track your progress, and simulate real exams with the GMDSS Trainer app.


Includes Elements 1, 3, 6, 7R, 8, and 9.