Question

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: \[ Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-\frac{t^2}{2}} dt, \] the optimum symbol error probability is \(\_\_\_\_\).

• A. \(\frac{2}{3}Q(2)\)
• B. \(\frac{4}{3}Q(1)\)
• C. \(\frac{2}{3}Q(1)\)
• D. \(\frac{4}{3}Q(2)\)

Hint:

For symbol error calculations in Gaussian noise channels, identify the decision boundaries and use the \(Q\)-function with \(\frac{|s|}{\sigma}\) as the argument.

Answer 1

The Correct Option is B

Step 1: Understand the symbols and noise distribution.
The given symbols are equally spaced, and the additive noise is Gaussian with variance \(\sigma^2 = 4\), giving a standard deviation of \(\sigma = 2\). The decision boundaries lie halfway between adjacent symbols. Step 2: Symbol error probability in a Gaussian channel.
The probability of error for one symbol interval is determined using the \(Q\)-function: \[ P_e = 2Q\left(\frac{|s|}{\sigma}\right), \] where \(|s|\) is the spacing between symbols divided by 2. Here, the distance between symbols is \(4\), so: \[ \frac{|s|}{\sigma} = \frac{4}{3}. \] Substituting this into the \(Q\)-function expression, we have: \[ P_e = \frac{4}{3}Q(1). \] Final Answer: \[ \boxed{{(2) } \frac{4}{3}Q(1)} \]