In a DSB-SC system with 100% modulation, the power saving is

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In a DSB-SC system with 100% modulation, the power saving is
1. 50 %
2. 66 %
3. 75 %
4. 100 %

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Correct Answer - Option 2 : 66 %

Concept:

The generalized AM expression is represented as:

s(t) = Ac [1 + μa mn (t)] cos ωc t

The total transmitted power for an AM system is given by:

${P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)$

Pc = Carrier Power

μ = Modulation Index

The above expression can be expanded to get:

${P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}$

The total power is the sum of the carrier power and the sideband power, i.e.

${P_s} = P_c\frac{{{μ^2}}}{2}$

The power in a single sideband will be:

${P_s} = \frac{1}{2}\times P_c\frac{{{μ^2}}}{2}$

With $P_c=\frac{A^2}{2}$, the above can be written as:

${P_s} = \frac{1}{2}\times \frac{A_c^2}{2}\frac{{{μ^2}}}{2}$

${P_s} = \frac{{{A_c^2μ^2}}}{8}$

$Power Saved=\frac{P_c}{P_{total}}$  ---(1)

Power Saved = Pc in DSB - SC

$Power \ Saved=\frac{2}{2 \ + \ μ^2}$

Analysis:

When μ = 1, the transmitted power will be:

${P_t} = {P_c}\left( {1 + \frac{{{1^2}}}{2}} \right)=\frac{3}{2}P_c$

$Power Saved=\frac{P_c}{P_c(1 \ + \ \frac{μ^2}{2})}$

As μ = 1

$Power \ Saved=\frac{2}{2 \ + \ 1^2}\times 100$

Power Saved = 66 %