MCQ
The maximum amplitude for an amplitude modulated wave is found to be $12\, {V}$ while the minimum amplitude is found to be $3\, {V}$. The modulation index is $0.6\, {x}$ where ${x}$ is $....\, .$
- A$2$
- ✓$1$
- C$3$
- D$4$
✓
Answer
Correct option: B.
$1$
b
As we know
As we know
$A_{\max }=A_{c}+A_{m}=12$
$A_{\max }=A_{c}-A_{m}=3$
$\Rightarrow A_{c}=\frac{15}{2}\, \& \,A_{m}=\frac{9}{2}$
$\text { Modulation index }=\frac{A_{m}}{A_{c}}=\frac{9 / 2}{15 / 2}=0.6$
$\Rightarrow x=1$
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