FCC Exam Question: 6A431
A transmitter operating on 5000 kHz uses a 1000 kHz crystal with a tempered coefficient of - 4Hz/MHz/0 degrees centigrade. What is the change in the output frequency of the transmitter if thetemperature increases 6 degrees centigrade?
Explanation: A crystal's frequency changes with temperature according to its temperature coefficient. First, determine the crystal's frequency shift: 1. **Crystal Frequency Shift:** The crystal is 1000 kHz (1 MHz). With a coefficient of -4 Hz/MHz/°C, the crystal drifts by -4 Hz for every degree Celsius increase. For a 6°C increase, the crystal frequency shifts by (-4 Hz/°C) * 6°C = -24 Hz. 2. **Transmitter Multiplier:** The transmitter operates on 5000 kHz using a 1000 kHz crystal, meaning it has a frequency multiplier of 5 (5000 kHz / 1000 kHz). 3. **Output Frequency Shift:** Any frequency drift in the crystal is also multiplied by this factor. So, the output frequency shifts by -24 Hz * 5 = -120 Hz. 4. **Convert to kHz:** -120 Hz is equivalent to -0.12 kHz. 5. **New Output Frequency:** The original output frequency was 5000 kHz. With a -0.12 kHz shift, the new output frequency would be 5000 kHz - 0.12 kHz = 4999.88 kHz. The question asks for the "change in the output frequency," but the options provide specific frequencies. Given that the correct answer is A) 499.88 kHz, it implies that the original transmitter operating frequency in the question was likely intended to be 500 kHz (not 5000 kHz). If the transmitter operated on 500 kHz and experienced a -0.12 kHz shift (as calculated for a 5x multiplier and a 1000 kHz crystal), the new frequency would be 500 kHz - 0.12 kHz = 499.88 kHz.
6A63
6A128
6A592
6A50
6A574
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Includes Elements 1, 3, 6, 7R, 8, and 9.