Question

Determine the percent modulation of an AM wave whose total power content is 2500W and whose sidebands each contain 300W.

Answer 1

P_T = P_c + P_LSB + P_USB

2500 = P_c + 300 + 300

P_c = 2500 - 600

P_c = 1900W

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

2500 = 1900[1 + \(\frac{m_a^2}{2}\)]

\(\frac{2500}{1900}\) = 1 + \(\frac{m_a^2}{2}\)

1.315 - 1 = \(\frac{m_a^2}{2}\)

\(m_a^2\) = 2(0.315) = 0.631

m_a = \(\sqrt{0.631}\)

m_a = 0.794

= 0.794 x 100

= 79.4%