A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit is

A. The same as FDM B. The same as TDM C. A combination of FDM and TDM D. Quite different from FDM and TDM

A. C₂ ≈ 2C₁ B. C₂ ≈ C₁ + B C. C₂ ≈ C₁ + 2B D. C₂ ≈ C₁ + 0.3B

A. 1 and 2 B. 2 and 3 C. 1 and 3 D. None of the above

A. Maximum Likelihood Decoding B. Sequential Decoding C. Both A and B D. None of the above