The carrier of an AM signal has power of 1,000 Watts. If the percentage of modulation is 80, what is the power in the upper sideband?

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ALP CBT 2 Electronic Mechanic Previous Paper: Held on 21 Jan 2019 Shift 2

Answer 1

Concept:

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.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

Calculation:

Given: P_c = 1000 W and μ = 0.8.

We can write:

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

\(= 1000 \times \frac{0.8^2}{2}\)

\(P_s=320~W\)

The total sideband power = 320 W

power in upper sideband + power in lower side band = 320 W

power in upper sideband = power in lower sideband = 160 W