The amplitude of the upper and lower sideband | vedclass

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(B) Given: Carrier amplitude $A_C = 15 \, V$,Modulating signal amplitude $A_m = 5 \, V$.The modulation index $\mu$ is defined as $\mu = \frac{A_m}{A_C

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(B) Given: Carrier amplitude $A_C = 15 \, V$,Modulating signal amplitude $A_m = 5 \, V$.The modulation index $\mu$ is defined as $\mu = \frac{A_m}{A_C} = \frac{5}{15} = \frac{1}{3}$.The amplitude of both the upper sideband $(USB)$ and lower sideband $(LSB)$ in an $A.M.$ wave is given by $\frac{\mu A_C}{2}$.Amplitude of $USB$ = $\frac{\mu A_C}{2} = \frac{(1/3) 	imes 15}{2} = \frac{5}{2} = 2.5 \, V$.Amplitude of $LSB$ = $\frac{\mu A_C}{2} = \frac{(1/3) 	imes 15}{2} = \frac{5}{2} = 2.5 \, V$.According to the problem,the amplitudes are $\frac{a}{10} \, V$ and $\frac{b}{10} \, V$.So,$\frac{a}{10} = 2.5 \Rightarrow a = 25$ and $\frac{b}{10} = 2.5 \Rightarrow b = 25$.Therefore,the value of $\frac{a}{b} = \frac{25}{25} = 1$.

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