FCC Exam Question: 8-4A4
If a RADAR transmitter has a pulse repetition frequency (PRF) of 900 Hz, a pulse width of 0.5 microseconds and a peak power of 15 kilowatts, what is its average power output?
Explanation: The average power output of a pulsed transmitter, like a RADAR, is determined by its peak power multiplied by its duty cycle. The duty cycle (DC) is the fraction of time the transmitter is actually emitting a pulse. It's calculated as: DC = Pulse Width (τ) × Pulse Repetition Frequency (PRF) Given: * PRF = 900 Hz * Pulse Width (τ) = 0.5 microseconds = 0.5 × 10⁻⁶ seconds * Peak Power (P_peak) = 15 kilowatts = 15,000 watts First, calculate the duty cycle: DC = (0.5 × 10⁻⁶ s) × 900 Hz = 0.00045 Next, calculate the average power (P_avg): P_avg = P_peak × DC P_avg = 15,000 watts × 0.00045 = 6.75 watts Option A is incorrect because 15 kilowatts is the *peak* power, not the average power. For a pulsed system, average power is always significantly lower than peak power because the transmitter is off for most of the time. Options B and D are incorrect based on the calculation.
8-21C3
8-50F1
8-15B4
8-38D2
8-28C3
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Includes Elements 1, 3, 6, 7R, 8, and 9.