Question E8C07:
What is the bandwidth of a 4,800-Hz frequency shift, 9,600-baud ASCII FM transmission?  

Choices:  
A: 15.36 kHz  
B: 9.6 kHz  
C: 4.8 kHz  
D: 5.76 kHz    

Correct Answer is  A: 15.36 kHz  

To estimate the occupied bandwidth of an ASCII FM (binary FSK) signal, use the MSK/FSK rule of thumb when the frequency shift equals half the bit rate. Here:

- Baud rate: $R_b = 9600$ baud
- Frequency shift (mark–space separation): $f_s = 4800$ Hz
- Peak deviation is half the shift: $\Delta f = f_s/2 = 2400$ Hz
- Modulation index for CPFSK: $h = \dfrac{2\Delta f}{R_b} = \dfrac{2 \cdot 2400}{9600} = 0.5$ (MSK condition)

For $h \approx 0.5$ (MSK), a common rule of thumb for occupied bandwidth is:
$$BW \approx 1.6\,R_b$$

Compute:
$$BW \approx 1.6 \times 9600 = 15{,}360\ \text{Hz} = 15.36\ \text{kHz}$$

Why the correct answer (A) is correct:
- The given shift and baud rate produce $h=0.5$ (MSK), for which $BW \approx 1.6R_b$ is used. That yields 15.36 kHz.

Why the other choices are incorrect:
- B: 9.6 kHz equals the baud rate and ignores FM deviation and sidebands; too narrow.
- C: 4.8 kHz equals the shift only; far too narrow for the FM spectrum.
- D: 5.76 kHz has no basis in standard FM/FSK bandwidth estimates; too narrow.

Useful formulas and rules of thumb:
- Carson’s Rule for FM: 
  $$BW \approx 2(\Delta f + f_m)$$ 
  For NRZ data, the highest significant modulating frequency is approximately $f_m \approx R_b/2$.
  With $\Delta f = 2400$ Hz and $R_b = 9600$ baud: 
  $$BW \approx 2(2400 + 4800) = 14.4\ \text{kHz}$$
  This is close, but the exam convention for $h=0.5$ (MSK-like) uses $BW \approx 1.6R_b = 15.36$ kHz to allow for the data spectrum’s sidelobes.
- Modulation index for binary CPFSK:
  $$h = \frac{2\Delta f}{R_b}$$
  If $h=0.5$ (MSK), use $BW \approx 1.6R_b$.