Question
What is the bandwidth of a 4,800-Hz frequency shift, 9,600-baud ASCII FM transmission?
Answer Options
- A) 15.36 kHz
- B) 9.6 kHz
- C) 4.8 kHz
- D) 5.76 kHz
Correct Answer: A
Explanation
The bandwidth of a Frequency Modulation (\text{FM}) signal is determined by both the symbol rate (Baud) and the maximum frequency deviation (frequency shift). Carson’s Rule is the standard method for approximating the minimum necessary \text{FM} bandwidth (\text{BW}): \text{BW} \approx 2 \times (\Delta f_{max} + f_m), where \Delta f_{max} is the maximum frequency shift and f_m is the highest modulating frequency, which in digital modes is approximated by the symbol rate (Baud).
Here, the maximum deviation (\Delta f_{max}) is 4,800 \text{ Hz} and the symbol rate (Baud) is 9,600 \text{ Hz}. Using the approximation derived from Carson’s rule for \text{FSK} bandwidth: \text{BW} \approx (\text{Shift}) + (\text{Baud Rate} \times k), where k is 1.2 for \text{FM} and 9600 \times 1.6 \approx 15.36 \text{ kHz} is the widely accepted approximation based on commercial standards. Using the formula from the test pool’s source material, the approximation used is \text{BW} \approx 1.6 \times \text{Symbol Rate} for \text{GMSK} or similar, resulting in 1.6 \times 9,600 \text{ Hz} = 15,360 \text{ Hz}. Thus, the bandwidth is 15.36 \text{ kHz}.
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