A source transmits a symbol , taken from with equal probability, over an additive white Gaussian noise channel. The received noisy symbol is given by , where the noise is zero mean with variance 4 and is independent of . Using , the optimum symbol error probability is ______.
ANSWER
The optimum symbol error probability is , option (B).
Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)
The three symbols are equally likely and the noise is zero-mean Gaussian, so the optimum (MAP, which equals ML here) detector picks the nearest symbol. The decision thresholds lie midway between neighbouring symbols, at and .
The noise variance is , so . Each symbol is away from its nearest threshold, so every tail probability is .
Symbol has a neighbour on each side. An error occurs if or :
Symbol has only one neighbour. It is decoded wrongly only if , that is : By symmetry .
Averaging over the equal priors :
This is option (B).
KEY IDEA
With equally likely symbols the optimum detector is the minimum-distance rule. The error probability is the average of Gaussian tail probabilities, one for each decision threshold that a symbol can cross. An inner symbol can cross two thresholds and an outer symbol only one.
WATCH OUT FOR
Giving every symbol two error tails, which yields . The outer symbols can only be mistaken for their single neighbour. Another slip is using : the argument is the half-spacing divided by , which is , not . Taking instead of is a third slip.
From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 15 Digital communication questions, with notes
