Question

An AM wave with modulation index 0.8 has total sideband power of 4.85 kW. The carrier power and the total power radiated will be nearly
1. 12.2 kW and 20 kW
2. 15.2 kW and 20 kW
3. 12.2 kW and 25 kW
4. 15.2 kW and 25 kW

Answer 1

Correct Answer - Option 2 : 15.2 kW and 20 kW

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_s = 4.85 kW and μ = 0.8.

We can write:

\({P_s} = P_c\frac{{{μ^2}}}{2}=4.85k\)

\( P_c\frac{{{(0.8)^2}}}{2}=4850\)

\(P_c=\frac{4850\times 2}{0.64}\)

P_c ≈ 15.2 kW

Now, the total power radiated will be:

\({P_t} = {P_c} + P_s\)

P_t = 15.2 + 4.85 

P_t = 20 kW