FCC Exam Question: 6A434
If the output frequency after passing through three doubler stages is 16,800 kHz, give the crystalfrequency of a transmitter:
Explanation: A frequency doubler stage multiplies its input frequency by a factor of two. If a signal passes through three doubler stages, the original crystal frequency is multiplied by 2 x 2 x 2, resulting in a total multiplication factor of 8. To find the initial crystal frequency, you divide the final output frequency by this total multiplication factor: Crystal Frequency = Output Frequency / Total Multiplication Factor Crystal Frequency = 16,800 kHz / 8 = 2100 kHz. Comparing this calculated value to the given options: A) 2160 kHz is too high. B) 2 GHz (or 2,000,000 kHz) is orders of magnitude too high. C) 2110 kHz is very close to our calculated 2100 kHz. In multiple-choice questions, the closest plausible answer is often the intended one if there's a slight discrepancy. If the output was 16,880 kHz, this would be exact. D) 2 kHz is far too low. Therefore, 2110 kHz is the most appropriate answer, indicating the likely crystal frequency.
6A254
6A171
6A560
6A296
6A566
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Includes Elements 1, 3, 6, 7R, 8, and 9.