Topic: Digital Communications
Given that CNDR of 256 QAM signal operating at 12.5 M Symbols/s is 100 dBHz. The value of Eb/No (Energy-per-bit to Noise Density ratio) is:
- A. 40 dB
- B. 20 dB
- C. 10 dB
- D. 24 dB
Answer: B. 20 dB
Explanation: Carrier-to-Noise Density Ratio (CNDR) = $C/N_0 = 100$ dBHz.
For 256 QAM, bits per symbol $n = \log_2(256) = 8$.
Bit rate $R_b = n \times R_s = 8 \times 12.5 \text{ Msps} = 100 \text{ Mbps} = 10^8$ bps.
In dB, $R_b = 10 \log_{10}(10^8) = 80$ dB.
$E_b/N_0 \text{ (dB)} = C/N_0 \text{ (dB)} - R_b \text{ (dB)} = 100 - 80 = 20$ dB.