GATE 2020 EC Q.48: solution and answer

GATE 2020 · ECQ.48 · 2 marks · NAT

In a digital communication system, a symbol $S$ randomly chosen from the set ${s_{1}, s_{2}, s_{3}, s_{4}}$ is transmitted. It is given that $s_{1} = - 3$ , $s_{2} = - 1$ , $s_{3} = + 1$ and $s_{4} = + 2$ . The received symbol is $Y = S + W$ . $W$ is a zero-mean unit-variance Gaussian random variable and is independent of $S$ . $P_{i}$ is the conditional probability of symbol error for the maximum likelihood (ML) decoding when the transmitted symbol $S = s_{i}$ . The index $i$ for which the conditional symbol error probability $P_{i}$ is the highest is ______.

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Answer and solution

ANSWER

The conditional symbol error probability is highest for $i = 3$ , with $P_{3} \approx 0.467$ .

Official answer: 3Source: Official answer key, GATE 2020 EC (IIT Delhi)

The noise is Gaussian with the same variance for every symbol, so ML decoding is minimum-distance decoding. The decision thresholds are the midpoints of adjacent symbols: $- 2$ (between $- 3$ and $- 1$ ), $0$ (between $- 1$ and $+ 1$ ) and $+ 1.5$ (between $+ 1$ and $+ 2$ ).
Here $W \sim N (0, 1)$ , so $P (W > a) = P (W < - a) = Q (a)$ .
$s_{1} = - 3$ : decoded correctly if $Y < - 2$ , that is $W < 1$ . An error needs $W > 1$ , so $P_{1} = Q (1) \approx 0.159$ .
$s_{2} = - 1$ : decoded correctly if $- 2 < Y < 0$ , that is $- 1 < W < 1$ . An error occurs on either side, so $P_{2} = 2 Q (1) \approx 0.317$ .
$s_{3} = + 1$ : decoded correctly if $0 < Y < 1.5$ , that is $- 1 < W < 0.5$ . An error needs $W < - 1$ or $W > 0.5$ , so $P_{3} = Q (1) + Q (0.5) \approx 0.159 + 0.309 = 0.467$ .
$s_{4} = + 2$ : decoded correctly if $Y > 1.5$ , that is $W > - 0.5$ . An error needs $W < - 0.5$ , so $P_{4} = Q (0.5) \approx 0.309$ .
Comparing, $P_{3} \approx 0.467 > P_{2} \approx 0.317 > P_{4} \approx 0.309 > P_{1} \approx 0.159$ . Symbol $s_{3}$ has a decision region of width only $1.5$ with a threshold on each side (at distances $1$ and $0.5$ from the symbol), so errors are possible in both directions and it is the worst case.
Hence the index with the highest conditional symbol error probability is $i = 3$ .

KEY IDEA

With equal-variance Gaussian noise, ML decoding is minimum-distance decoding. Put thresholds at the midpoints between adjacent symbols, then find each symbol's error probability from its distance to the thresholds with the $Q$ function.

WATCH OUT FOR

Assuming the error probability is largest for the symbol with the nearest neighbour, or for an outer symbol. Outer symbols have only one threshold. Symbols $s_{2}$ and $s_{3}$ have thresholds on both sides, and $s_{3}$ has the closer pair, at distances $1$ and $0.5$ . Also, a wrong placement of the threshold between $+ 1$ and $+ 2$ (it is at $1.5$ , not $1$ or $2$ ) changes the result.

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GATE 2020 EC Q.48: solution and answer

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