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Four messages band limited to $\text{W, W, 2W}$ and $\text{3W}$ respectively are to be multiplexed using $\text{Time Division Multiplexing (TDM)}.$ The minimum bandwidth required for transmission of this $\text{TDM}$ signal is

  1. $\mathrm{W}$
  2. $3 \mathrm{W}$
  3. $6 \mathrm{W}$
  4. $7 \mathrm{W}$

1 Answer

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$$
\begin{aligned}
f_{s_{1}} & =2 \times \mathrm{W}=2 \mathrm{~W} \\
f_{s_{2}} & =2 \times \mathrm{W}=2 \mathrm{~W} \\
f_{s_{3}} & =2 \times 2 \mathrm{~W}=4 \mathrm{~W} \\
f_{s_{4}} & =2 \times 3 \mathrm{~W}=2 \mathrm{~W} \\
f_{s} & =f_{s_{1}}+f_{s_{2}}+f_{s_{3}}+f_{s_{4}}
\end{aligned}
$$

For minimum bandwidth, $n=1$
$$
\begin{aligned}
R_{b} & =n f_{s} \\
R_{b} & =1 \times 14 \mathrm{~W}=14 \mathrm{~W} \\
(\text { B.W. })_{\min } & =\frac{R_{b}}{2} \\
(\text { (B.W. })_{\min } & =\frac{14 \mathrm{~W}}{2}=7 \mathrm{~W}
\end{aligned}
$$
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