GATE ECE 2004 | Question: 81

Question

 

Three analog signals, having bandwidths $1200 \mathrm{~Hz}$, $600 \mathrm{~Hz}$ and $600 \mathrm{~Hz}$, are sampled at their respective Nyquist rates, encoded with $12$ bit words, and time division multiplexed. The bit rate for the multiplexed signal is

• A. $115.2 \; \mathrm{kbps}$
• B. $28.8 \; \mathrm{kbps}$
• C. $57.6 \; \mathrm{kbps}$
• D. $38.4 \; \mathrm{kbps}$

Answer 1

$$
\begin{aligned}
f_{s} & =2(1200+600+600)=4.8 \mathrm{k} \\
R_{b} & =n f_{s}=12 \times 4.8 \mathrm{k} \\
& =57.6 \mathrm{kbps}
\end{aligned}
$$