Question E8C06:
What is the bandwidth of an FT8 signal?  

Choices:  
A: 10 Hz  
B: 50 Hz  
C: 600 Hz  
D: 2.4 kHz    

Correct Answer is  B: 50 Hz  

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FT8 uses 8-tone FSK (MFSK) with a tone spacing of about 6.25 Hz and a symbol rate of about 6.25 baud. The occupied bandwidth is therefore on the order of 50 Hz.

Why 50 Hz is correct:
- FT8 tones are spaced by $\Delta f \approx 6.25\ \text{Hz}$.
- With $M = 8$ tones, the total span from the lowest to highest tone is approximately $(M - 1)\Delta f = 7 \times 6.25 \approx 43.75\ \text{Hz}$.
- Allowing for shaping and practical occupied bandwidth, this is rounded to about 50 Hz.

Why the other choices are incorrect:
- 10 Hz: Too narrow. Modes like WSPR (~6 Hz) or JT9 (~15 Hz) are in this range, not FT8.
- 600 Hz: Far wider than a single FT8 signal; closer to typical RTTY/CW filter widths, not the signal itself.
- 2.4 kHz: Typical SSB voice bandwidth, much wider than any single FT8 signal.

Useful rules of thumb:
- For narrow MFSK, a quick estimate is:
  $$BW \approx (M - 1)\Delta f$$
  Then account for shaping/filters, giving FT8 ≈ 50 Hz.
- Memorize common narrow digital mode widths:
  - FT8 ≈ 50 Hz
  - FT4 ≈ 90 Hz
  - JT9 ≈ 15 Hz
  - WSPR ≈ 6 Hz
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