Question

For an am having a power of of 500 watt and modulation index of 80%, the power content of each side band is

Answer 1

Answer:

60.61 W

Step by step explanation

The givens are:

• Power $P_t=500W$
• Modulation index $m=80\%=0.8$

The required is the power of each side band.

The total power of an AM signal is given by:

$P_t=P_{c}(1+m^2/2)\\500=P_c(1+0.8^2/2)\\P_c=378.78W$

Therefore, the power in the two side bands is 500 - 378.78 = 121.22 W

Therefore the power in each side band is 121.22/2 = 60.61 W

Answer 2

Power of each side band = 60.61 Watts

Amplitude modulated waves have a modulation index (m) and side bands. The side bands have power independent of the power of the total AM wave.

It is given that:

Modulation index (m) = 0.8 or 80 %

Total power of AM = 500 W

Now, to calculate power of the carrier band, we apply the formula:

Total power = (Carrier power) x (1 + (m^2) / 2)

Hence, carrier power = 500 / (1.32) = 378.78 W

Also, Total power = Side band power + Carrier band power

Total side bands power = 500 - 378.78 = 121.22 W

Hence, power of each side band = 121.22 / 2 = 60.61 Watts