The GATE Grind

GATE 2019 EC – Question 44

Communications · Digital Communications · 2 marks · Multiple choice

A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density $N_0/2$. Binary signaling, with $0\mapsto p(t)$ and $1\mapsto q(t)$, is used for the transmission, along with an optimal receiver that minimizes the bit-error probability.

Let $\varphi_1(t),\varphi_2(t)$ form an orthonormal signal set.

If we choose $p(t)=\varphi_1(t)$ and $q(t)=-\varphi_1(t)$, we would obtain a certain bit-error probability $P_b$.

If we keep $p(t)=\varphi_1(t)$, but take $q(t)=\sqrt E\,\varphi_2(t)$, for what value of $E$ would we obtain the same bit-error probability $P_b$?

  1. 0
  2. 1
  3. 2
  4. 3

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (D) 3

Explanation

The error probability is $Q\left(\frac d{\sqrt{2N_0}}\right)$ where $d$ is the distance between the two signals. For $\pm\varphi_1$, $d=2$. For $\varphi_1$ and $\sqrt E\varphi_2$ (orthogonal), $d=\sqrt{1+E}$. Equal $d$ needs $1+E=4$, so $E=3$.