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

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

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PT = Pc + PLSB + PUSB

2500 = Pc + 300 + 300

Pc = 2500 - 600

Pc = 1900W

PT = Pc(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

ma$\sqrt{0.631}$

ma = 0.794

= 0.794 x 100

= 79.4%