Question E2E11:
What is the difference between direct FSK and audio FSK?  

Choices:  
A: Direct FSK modulates the transmitter VFO  
B: Direct FSK occupies less bandwidth  
C: Direct FSK can transmit higher baud rates  
D: All these choices are correct    

Correct Answer is  A: Direct FSK modulates the transmitter VFO  

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Correct answer: Direct FSK modulates the transmitter VFO.

1) Why this is correct
- Direct FSK shifts the RF carrier frequency itself by keying the transmitter’s oscillator or synthesizer (often the VFO or a PLL/NCO). The mark and space are two RF frequencies separated by the chosen shift, created directly at RF.

2) Why the other choices are incorrect
- B: “Direct FSK occupies less bandwidth” — Not inherently. Occupied bandwidth is set primarily by the frequency shift and the keying rate (baud), plus the transmit/receive filtering, not by whether it’s direct FSK or AFSK.
- C: “Direct FSK can transmit higher baud rates” — Not inherently. Maximum practical baud rate depends on the transmitter/receiver bandwidth and filtering (and the channel), not the FSK method. Either method can support the same symbol rates if the hardware permits.
- D: “All these choices are correct” — False because B and C are not generally true.

3) Useful formulas and rules of thumb
- AFSK uses audio tones driving an SSB (or FM) transmitter. With SSB:
  $$f_{RF} = f_0 \pm f_{audio}$$
  The RF mark–space separation equals the audio tone separation:
  $$\Delta f = |f_M - f_S|$$
- Approximate occupied bandwidth for 2FSK (rule of thumb, Carson-like):
  $$B \approx 2(\Delta f + R)$$
  where:
  - $\Delta f$ = mark–space frequency shift (Hz)
  - $R$ = baud rate (symbols per second)

Key idea: Direct FSK changes the RF frequency at the source (the VFO/synthesizer). AFSK creates two audio tones that the SSB chain translates to RF; bandwidth and baud capability are determined by shift, rate, and filtering, not the FSK method itself.