GATE 2024 EC – Question 45
A source transmits a symbol $s$, taken from $\{-4,0,4\}$ with equal probability, over an additive white Gaussian noise channel. The received noisy symbol $r$ is given by $r=s+w$, where the noise $w$ is zero mean with variance 4 and is independent of $s$. Using $Q(x)=\dfrac1{\sqrt{2\pi}}\int_x^\infty e^{-t^2/2}dt$, the optimum symbol error probability is ______.
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Correct answer: (B) $\dfrac43Q(1)$
Explanation
The optimum (maximum-likelihood) thresholds are the midpoints $\pm2$, at distance 2 from each symbol, and the noise standard deviation is $\sigma=2$, so each threshold crossing has probability $Q(2/2)=Q(1)$. The symbols $\pm4$ have one neighbouring threshold each (error probability $Q(1)$) and the symbol 0 has two ($2Q(1)$). $P_e=\frac13\left[Q(1)+2Q(1)+Q(1)\right]=\frac43Q(1)$.