Question E2E13:
Which of these digital modes has the highest data throughput under clear communication conditions?  

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

Correct Answer is  D: PACTOR IV  

Correct answer: D — PACTOR IV

Why D is correct
- PACTOR IV is an adaptive, wideband, multi-carrier QAM mode that achieves the highest HF data throughput under good SNR. In a ~2.4 kHz channel it can deliver up to about 10 kbps (often 5–10 kbps depending on conditions), far exceeding weak-signal or narrowband modes.

Why the others are incorrect
- A: MFSK16  
  - 16 tones at about 15.625 baud. Raw rate:  
    $$R_b \approx R_s \times \log_2(16) = 15.625 \times 4 \approx 62.5\ \text{bps}$$  
  - With FEC/overhead, net is on the order of a few tens of bps. Much slower than PACTOR IV.

- B: 170 Hz shift, 45 baud RTTY  
  - Baud rate: 45.45 baud with 5-bit Baudot plus start/stop (7.5 bits/char).  
    Characters per second:  
    $$CPS = \frac{45.45}{7.5} \approx 6\ \text{chars/s}$$  
    User data rate (5 bits/char):  
    $$R_b \approx 6 \times 5 \approx 30\ \text{bps}$$  
  - Extremely low compared to PACTOR IV.

- C: FT8  
  - Highly robust but very low throughput. Roughly one short message (~75–80 info bits) every ~12.6 s:  
    $$R_b \approx \frac{77}{12.6} \approx 6\ \text{bps}$$  
  - Optimized for weak-signal DX, not speed.

Useful formulas / rules of thumb
- Bit rate:  
  $$R_b = R_s \times \log_2(M) \times \eta$$  
  where $R_s$ = symbol rate, $M$ = modulation order, $\eta$ = coding/overhead efficiency (0–1).
- Throughput increases with bandwidth, symbol rate, and higher-order modulation under clear conditions. Thus, a wideband, adaptive, high-order scheme like PACTOR IV wins on throughput.