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?
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.
3-70K4
3-59H3
3-56G1
3-69K1
3-17B6
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Includes Elements 1, 3, 6, 7R, 8, and 9.