Topic: Communication Systems
At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by
- A. 6 dB
- B. 3 dB
- C. 2 dB
- D. 0 dB
Answer: B. 3 dB
Explanation: $P_e \text{ for FSK} = \frac{1}{2}\text{erfc}\sqrt{\frac{E_b}{2N_0}}$
$P_e \text{ for PSK} = \frac{1}{2}\text{erfc}\sqrt{\frac{E_b}{N_0}}$
Hence, for a given $P_e$, PSK is better than FSK by a factor of 2, which is $10 \log_{10}(2) = 3 \text{ dB}$.