Correct Answer - Option 2 : 800
Concept:
QPSK is a modulation scheme that allows one symbol to transfer two bits of data. There are four possible two-bit numbers (00, 01, 10, 11).
\(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)
Analysis:
Transmission bandwidth
\(= \frac{{2{R_b}}}{{{{\log }_2}4}} = {R_b}\)
Maximum required bandwidth
\(= \frac{{{R_b}}}{2}\)
40% of \(\frac{{{R_b}}}{2} = 0.2\;{R_b}\)
Total = (0.2 + 0.5) Rb = 0.7 Rb
No. of channels required
\( = \frac{{36 \times {{10}^6}}}{{0.7 \times 64 \times {{10}^3}}} \approx 800\)