GATE 2018 EC Q.17: solution and answer

GATE 2018 · ECQ.17 · 1 mark · NAT

A binary source generates symbols $X \in {- 1, 1}$ which are transmitted over a noisy channel. The probability of transmitting $X = 1$ is 0.5. Input to the threshold detector is $R = X + N$ . The probability density function $f_{N} (n)$ of the noise $N$ is shown below. If the detection threshold is zero, then the probability of error (correct to two decimal places) is ______.

GATE 2018 EC question 17: Plot of f_N(n) versus n: a triangular pdf. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

The probability of error is $P_{e} = 0.125 \approx 0.13$ .

Official answer: 0.12 to 0.14Source: Official answer key, GATE 2018 EC (IIT Guwahati)

The noise pdf is a symmetric triangle on $- 2 \leq n \leq 2$ with peak $0.5$ at $n = 0$ . Its area is $\frac{1}{2} \cdot 4 \cdot 0.5 = 1$ , so it is a valid pdf.
The two sides are $f_{N} (n) = 0.25 (n + 2)$ for $- 2 \leq n \leq 0$ and $f_{N} (n) = 0.25 (2 - n)$ for $0 \leq n \leq 2$ .
With threshold $0$ , the detector decides $\hat{X} = 1$ if $R \geq 0$ and $\hat{X} = - 1$ if $R < 0$ .
If $X = 1$ , then $R = 1 + N$ and an error occurs when $R < 0$ , that is $N < - 1$ . If $X = - 1$ , then $R = - 1 + N$ and an error occurs when $R > 0$ , that is $N > 1$ .
$P (N < - 1) = \int_{- 2}^{- 1} 0.25 (n + 2) d n = 0.25 [\frac{( n + 2 ) ^{2}}{2}]_{- 2}^{- 1} = 0.25 \cdot \frac{1}{2} = 0.125$ This is also the area of the small triangle with base $1$ and height $0.25$ .
The pdf is symmetric about $n = 0$ , so $P (N > 1) = 0.125$ .
$P_{e} = 0.5 (0.125) + 0.5 (0.125) = 0.125 \approx 0.13$

KEY IDEA

With a zero threshold, an error needs the noise to push the received sample across zero. For $X = 1$ that means $N < - 1$ , and for $X = - 1$ it means $N > 1$ . Each tail probability is the area of a small triangle under the noise pdf.

WATCH OUT FOR

Using the Gaussian $Q (\cdot)$ formula, or integrating from the signal level $\pm 1$ out to $\pm 2$ incorrectly. Another slip is taking the tail height at $n = \pm 1$ as $0.5$ instead of $0.25$ , which gives $0.25$ for the tail area.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 15 Digital communication questions, with notes

GATE 2018 EC Q.17: solution and answer

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