Topic: Communication Systems
A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of $p(3)=0.2$, $p(1)=0.3$, $p(-1)=0.3$, $p(-3)=0.2$ respectively. The variance for this source will be
- A. 4.2 V
- B. 3.2 V
- C. 3.6 V
- D. 4.6 V
Answer: A. 4.2 V
Explanation: Mean $E[X]=\overline{X}=0$ since the PDF of 'X' is symmetric about the origin.
$\text{Variance} = \sigma_{x}^{2} = E[(X-\overline{X})^{2}] = E[X^{2}] = \sum x_{i}^{2}p(x_{i})$
$= (-3)^{2}(0.2) + (-1)^{2}(0.3) + (1)^{2}(0.3) + (3)^{2}(0.2)$
$= 9(0.2) + 1(0.3) + 1(0.3) + 9(0.2) = 1.8 + 0.3 + 0.3 + 1.8 = 4.2~\text{V}^{2}$ (The unit is conventionally $\text{V}^{2}$, but listed as V in the options).