Topic: Communication Systems
What is the maximum data rate that can be transmitted using a QPSK modulation with a roll-off factor of 0.2 for a 36 MHz transponder?
- A. 7.2 Mbps
- B. 360 Mbps
- C. 43.2 Mbps
- D. 60 Mbps
Answer: D. 60 Mbps
Explanation: The bandwidth for M-ary PSK with a raised cosine filter is given by $BW = \frac{R_b}{\log_2 M}(1+\alpha)$.
For QPSK, $M = 4$, so $\log_2 4 = 2$. Thus, $BW = \frac{R_b}{2}(1+\alpha)$.
Given $BW = 36\text{ MHz}$ and $\alpha = 0.2$.
$36\text{ MHz} = \frac{R_b}{2}(1 + 0.2) = \frac{R_b}{2}(1.2) = 0.6 R_b$.
$R_b = \frac{36}{0.6} = 60\text{ Mbps}$.