Question

For an amplitude modulated wave, the maximum amplitude is found to be $12V$ and minimum amplitude is found to be $4V$. The modulation index of this wave is __________ %.

• A. $50$
• B. $25$
• C. $75$
• D. $20$

Answer 1

The correct option is A $50$

Given,
$V_{max}=12V$
$V_{min}=4V$
The modulation index $\left(M\right)=\frac{V_{max}-V_{min}}{V_{max}+V_{min}}\times 100$
$=\frac{12-4}{12+4}\times 100=\frac{8}{16}\times 100=\frac{800}{16}$
$=50$%