Question:
The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ______.
Solution:
$A_{\max }=A_{c}+A_{m}=12$
$A_{\min }=A_{c}-A_{m}=3$
$\Rightarrow \mathrm{A}_{\mathrm{c}}=\frac{15}{2} \& \mathrm{~A}_{\mathrm{m}}=\frac{9}{2}$
modulation index $=\frac{\mathrm{A}_{\mathrm{m}}}{\mathrm{A}_{\mathrm{c}}}=\frac{9 / 2}{15 / 2}=0.6$
$\Rightarrow x=1$
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