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

Because the received signal-to-noise ratio depends on signal power, the power carried by an AM wave is a key design quantity. An AM wave is a carrier plus two sidebands, so the total radiated power is their sum. Working with rms values in antenna resistance $R$:

Adding these gives the central result

Two important conclusions follow. First, the carrier power is exactly the same in the modulated wave as in the unmodulated carrier — the extra power comes entirely from the sidebands, which carry all the useful information. Second, at 100% modulation ($m_a = 1$), $P_T = 1.5\,P_C$.

The total sideband power is $P_{SB} = (m_a^2/2)P_C$, so the transmission efficiency — useful (sideband) power divided by total power — is

$\eta = \frac{P_{SB}}{P_T} = \frac{m_a^2}{2 + m_a^2}$ …