Question

Determine the power content of the carrier an each of side bands of an AM wave having a percent modulation of 80% and a total power of 2200W.

Answer 1

m_a = 80%

= \(\frac{80}{100}\)

= 0.8

P_T = 2200 W

P_T = P_c(1 + \(\frac{m_a^2}{2}\))

2200 = P_c[1 + \(\frac{(0.8)^2}{2}\)]

2200 = P_c[1 + 0.32]

P_c = \(\frac{2200}{1.32}\)

P_c = 1666W

P_T = P_c + Power in the sidebands

Power in two sidebands = P_T - P_c

= 2200 - 1666W

= 534W

Power in each sideband = \(\frac{534}{2}\)

= 267W