Question

Consider the following amplitude modulated (AM) signal,where $f_m<B$.
$x_{AM}\left(t\right)=10(1+0.5\sin 2\pi f_{m}t)\cos 2\pi f_{c}t$
The average side-band power for the AM signal given above is

• A. $25$
• B. $12.5$
• C. $6.25$
• D. $3.125$

Answer 1

Answer: (B)

$12.5$

Average side-band power, $P_{av}=\frac{m_a}{4}{P_c}^2$
Here, $m_a=0.5$

$P_c=10$
$\therefore P_{av}=\frac{0.5\times 10\times 10}{4}=12.5$.