# What is the total power carried by sidebands of the AM wave (DSB) for tone modulation of μ = 0.4 ?

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What is the total power carried by sidebands of the AM wave (DSB) for tone modulation of μ = 0.4 ?

1. 7.4%
2. 11.11%
3. 4.3%
4. 8.3%

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Correct Answer - Option 1 : 7.4%

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.

${P_s} = P_c\frac{{{μ^2}}}{2}$

Analysis:

Total sideband power is calculated as:

${P_s} = P_c\frac{{{μ^2}}}{2}$

% of sideband power is given as:

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

⇒ $\frac{μ^2}{μ^2+2}$

putting μ = 0.4, we get

⇒ $\frac{0.4^2}{0.4^2\;+\;2} \times 100 = 0.074 \times 100$

= 7.4 %