Q.Define transmission efficiency and state its relation with m_a.

Transmission efficiency of an AM wave tells us what fraction of the total transmitted power is actually carried by the sidebands, since only the sidebands carry information. …

[!TLDR]

Transmission efficiency is the ratio of useful (sideband) power to total AM power, $\eta = m_a^2/(2 + m_a^2)$, with a maximum of about 33.3 % at $m_a = 1$.

In an AM wave the total power is the sum of the carrier power and the power in the two sidebands, while only the sidebands carry the message. Transmission efficiency is therefore defined as the ratio of the sideband power to the total transmitted power:

Using $P_{total} = P_c\left(1 + \dfrac{m_a^2}{2}\right)$ and sideband power $P_{SB} = P_c\dfrac{m_a^2}{2}$, we get

$\eta = \frac{\dfrac{m_a^2}{2}}{1 + \dfrac{m_a^2}{2}} = \frac{m_a^2}{2 + m_a^2}$ …