Question

The expression for an AM voltage is

$e=5(1+0.6\cos 100\pi t)\cos 5\times {10}^6\pi tvolt.$. The percentage of modulation is :

• A. $0.6\%$
• B. $6\%$
• C. $60\%$
• D. $66\%$

Answer 1

Answer: (C)

$60\%$

Modulation Index of AM wave :

$m=\frac{a-b}{a+b}$

where $a$ is maximum amplitude and $b$ is minimum amplitude.

so $a=8V$ at $t=0$

and $b=2$

$m=\frac{8-2}{8+2}=0.6$

modulation percentage $=m\times 100=0.6\times 100=60\%$