The maximum and minimum amplitude of an amplitude modulated

Question:

The maximum and minimum amplitude of an amplitude modulated wave is $16 \mathrm{~V}$ and $8 \mathrm{~V}$ respectively. The modulation index for this amplitude modulated wave is $\mathrm{x} \times 10^{-2}$. The value of $\mathrm{x}$ is

Solution:

(33)

$\mathrm{A}_{\mathrm{m}}=\frac{\mathrm{A}_{\max }-\mathrm{A}_{\min }}{2}$

$\mathrm{A}_{\mathrm{C}}=\frac{\mathrm{A}_{\max }+\mathrm{A}_{\min }}{2} \quad\left[\begin{array}{c}\mathrm{A}_{\max }=16 \mathrm{~V} \\ \mathrm{~A}_{\min }=8 \mathrm{~V}\end{array}\right]$

Modulation index $(\mathrm{mi})=\frac{A_{\mathrm{m}}}{A_{\mathrm{C}}}=\frac{\frac{A_{\max }-A_{\min }}{2}}{\frac{A_{\max }+A_{\min }}{2}}=\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}$

$\mathrm{mi}=\frac{16-8}{16+8}=\frac{8}{24}=\frac{1}{3}=0.33$

$\mathrm{mi}=33 \times 10^{-2}$

$\mathrm{X}=33$

 

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