What is the effective radiated power of a repeater with 50 watts transmitter output and specified losses?

Study for the Federal Communications Commission (FCC) Element 3 Test. Use flashcards and multiple choice questions with hints and explanations. Get prepared for your exam today!

The effective radiated power (ERP) of a repeater can be calculated by taking into account the transmitter output power and any losses in the system, such as cable losses and antenna mismatch.

In this case, if the transmitter has an output of 50 watts, and there are specified losses (commonly measured in decibels), the effective radiated power would be determined by subtracting those losses from the transmitter output. For example, if the calculated losses equal approximately 1.5 dB, this translates to a reduction in power.

In decibel terms, a loss of 1 dB corresponds to approximately a halving of the power. Therefore, if the actual loss results in the effective power being 31.5 watts after accounting for those losses, this is indeed the correct interpretation of the situation.

Thus, the effective radiated power determining that figure is why the answer provided is correct.

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