FCC Exam Question: 6A430

A station has an assigned frequency of 8000 kHz and a frequency tolerance of plus or minus 0.04%. Theoscillator operates at 1/8th of the input. What is the maximum permitted deviation of the oscillationfrequency in Hz, which will not exceed the tolerance?

A. Plus or minus 400 Hz
B. Plus or minus 320 Hz
C. Plus or minus 800 Hz
D. Plus or minus 640 Hz
Correct Answer: A

Explanation: A station's frequency tolerance applies to its final transmitted output frequency. The problem asks about the maximum permitted deviation of the *oscillation frequency* itself, which is then multiplied to reach the final assigned frequency. 1. **Calculate the maximum allowed deviation for the *assigned output frequency***: The assigned frequency is 8000 kHz, which is 8,000,000 Hz. The tolerance is plus or minus 0.04%. Maximum deviation at output = 8,000,000 Hz * 0.04% = 8,000,000 Hz * (0.04 / 100) = 8,000,000 Hz * 0.0004 = 3200 Hz. 2. **Relate oscillator deviation to output deviation**: The problem states the oscillator operates at 1/8th of the input (meaning, 1/8th of the final assigned frequency). This implies the final output frequency is 8 times the oscillator frequency. In a frequency multiplier circuit, any deviation in the original oscillator frequency is also multiplied by the same factor. To keep the final output within the calculated 3200 Hz tolerance, the oscillator's initial deviation must be 1/8th of that amount. 3. **Calculate the maximum allowed deviation for the *oscillation frequency***: Maximum oscillator deviation = (Maximum output deviation) / (Multiplication factor) Maximum oscillator deviation = 3200 Hz / 8 = 400 Hz. Therefore, the maximum permitted deviation of the oscillation frequency is plus or minus 400 Hz.

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Includes Elements 1, 3, 6, 7R, 8, and 9.