What is the bandwidth of a 4,800-Hz frequency shift, 9,600-baud ASCII FM transmission?
15.36 kHz✓Respuesta correcta
9.6 kHz
4.8 kHz
5.76 kHz
Explicación
This is the regulatory necessary-bandwidth calculation for single-channel frequency-shift telegraphy: the baud rate plus the shift multiplied by K, where K = 1.2 is the value used for frequency modulation. (1.2 × 4800 Hz) + 9600 Hz = 5760 + 9600 = 15,360 Hz, or 15.36 kHz — neither the shift alone nor the baud rate alone accounts for the whole signal.