In amplitude modulation, the amplitude of the carrier ware is 10 V and the amplitude of one of the side…
In amplitude modulation, the amplitude of the carrier ware is 10 V and the amplitude of one of the side bands is 2 V . Then the modulation index is
- $0.8$
- $0.6$
- $0.7$
- $0.5$
Solution
In amplitude modulation (AM) wave,
$\begin{aligned}
& A_c=10 V, A_{\min }=2 V \\
& \Rightarrow A_c-A_s=2 \Rightarrow A_{s^{\prime}}=10-2=8 \mathrm{~V} \\
& \therefore \quad A_{\max }=A_s+A c=8+10=18 \mathrm{~V}
\end{aligned}$
$\therefore \quad$ Modulation index, $\mu=\frac{\mathrm{A}_{\max }-\mathrm{A}_{\min }}{\mathrm{A}_{\max }+\mathrm{A}_{\min }}$ $=\frac{18-2}{18+2}=0.8$
Asked in: AP EAMCET 2024 (20 May Shift 2)
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