Topic: Digital Communication
Consider a 4-PSK constellation with points $\{\sqrt{2}, j\sqrt{2}, -\sqrt{2}, -j\sqrt{2}\}$ and a 4-PAM constellation $\{-3, -1, 1, 3\}$. If all the points in the constellation occur with equal probability, the ratio of average energy of 4-PAM signal to that of 4-PSK signal is:
- A. 1
- B. 1.25
- C. 2.5
- D. 5
Answer: C. 2.5
Explanation: Average energy of 4-PSK points: each point has an energy (magnitude squared) of $(\sqrt{2})^2 = 2$. So, $E_{avg(PSK)} = 2$. Average energy of 4-PAM points is $\frac{(-3)^2 + (-1)^2 + 1^2 + 3^2}{4} = \frac{9+1+1+9}{4} = 5$. The ratio is $E_{avg(PAM)} / E_{avg(PSK)} = 5 / 2 = 2.5$.