FCC Exam Question: 8-4A6
A search RADAR has a pulse width of 1.0 microsecond, a pulse repetition frequency (PRF) of 900 Hz, and an average power of 18 watts. The unit’s peak power is:
Explanation: In a pulsed radar system, the average power is the product of its peak power and its duty cycle. The duty cycle represents the fraction of time the transmitter is actively transmitting, calculated by multiplying the pulse width by the pulse repetition frequency (PRF). First, calculate the duty cycle (DC): DC = Pulse Width $\times$ PRF DC = $1.0 \times 10^{-6}$ seconds $\times$ 900 Hz = 0.0009 Next, use the formula relating average power and peak power: Average Power = Peak Power $\times$ Duty Cycle Rearranging to solve for Peak Power: Peak Power = Average Power / Duty Cycle Peak Power = 18 watts / 0.0009 = 20,000 watts Converting to kilowatts: 20,000 watts = 20 kilowatts.
8-21C1
8-46F4
8-48F4
8-27C4
8-43E6
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Includes Elements 1, 3, 6, 7R, 8, and 9.