Topic: Digital Communication
A system generates data at a rate of 5 Mbps. In order to provide resistance to bit errors, a rate $1/2$ error correcting code is applied. Further, the data is mapped to a 16-QAM constellation. What is the resulting symbol rate?
- A. 1.25 Msps
- B. 2.5 Msps
- C. 5 Msps
- D. 10 Msps
Answer: B. 2.5 Msps
Explanation: The raw data rate is 5 Mbps. Applying a rate $1/2$ code doubles the bit rate to $5 / 0.5 = 10$ Mbps. A 16-QAM constellation encodes $\log_2(16) = 4$ bits per symbol. Therefore, the resulting symbol rate is $10\text{ Mbps} / 4\text{ bits/symbol} = 2.5$ Msps.