Determine the percent modulation of an AM wave whose total power conte

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

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

Answer»

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%