Subelement A: RADAR Principles – 10 Key Topics – 10 Exam Questions – 8 Drawings— Topic 4: Power, Pulse Width, PRR
Question 8-4A4
Element 8 (RADAR)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.
Related Questions
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