The amplitude of upper and lower side bands of A.M.
Question:

The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency $11.21 \mathrm{MHz}$, peak voltage $15 \mathrm{~V}$ is amplitude modulated by a $7.7 \mathrm{kHz}$ sine wave of $5 \mathrm{~V}$ amplitude

are $\frac{\mathrm{a}}{10} \mathrm{~V}$ and $\frac{\mathrm{b}}{10} \mathrm{~V}$ respectively. Then the value of $\frac{\mathrm{a}}{\mathrm{b}}$ is____________

Solution:

$\frac{\mathrm{a}}{10}=\frac{\mathrm{b}}{10}=\frac{\mu \mathrm{A}_{\mathrm{C}}}{2}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=1$

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