Question E2E10:
Which of these digital modes has the narrowest bandwidth?  

Choices:  
A: MFSK16  
B: 170 Hz shift, 45-baud RTTY  
C: FT8  
D: PACTOR IV    

Correct Answer is  C: FT8  

FT8 is correct because it is designed as a very narrow, weak-signal mode. Its occupied bandwidth is about 50 Hz, much less than the others listed.

Why FT8 is correct
- FT8 uses 8-FSK with tone spacing of about 6.25 Hz and a low symbol rate.
- Approximate width: $B \approx (M-1)\,\Delta f \approx 7 \times 6.25 \approx 43.8\ \text{Hz}$, which with filtering is about 50 Hz on the air.

Why the other choices are incorrect
- MFSK16: Uses 16 tones with relatively wider spacing and rate; typical occupied bandwidth is about 300 Hz (often quoted ≈316 Hz).
- 170 Hz shift, 45-baud RTTY: A common rule of thumb for FSK bandwidth is
  $$B \approx 2(\Delta f + R)$$
  where $\Delta f$ is frequency deviation (half the shift) and $R$ is the baud rate.
  For 170 Hz shift RTTY: $\Delta f = 85\ \text{Hz}$, $R \approx 45.45\ \text{Bd}$,
  $$B \approx 2(85 + 45.45) \approx 2 \times 130.45 \approx 261\ \text{Hz}$$
  Typically observed ≈250–300 Hz, much wider than FT8.
- PACTOR IV: Adaptive, high-throughput HF mode; occupied bandwidth up to about 2.4 kHz.

Useful rules of thumb
- Narrower bandwidth comes from lower symbol rates and smaller tone spacing/shift.
- For simple FSK: $B \approx 2(\Delta f + R)$.
- For multi-tone FSK: $B$ grows roughly with the number of tones and their spacing, i.e., $B \sim (M-1)\,\Delta f$ (plus filtering overhead).

Bottom line: FT8 (~50 Hz) is significantly narrower than MFSK16 (~300 Hz), 45-baud 170 Hz RTTY (~250–300 Hz), and PACTOR IV (up to ~2.4 kHz).