A 100, V carrier wave is made to vary between | vedclass

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(C) The maximum amplitude of the modulated wave is given by $A_{max} = A_c + A_m = 160\, V$.The minimum amplitude of the modulated wave is given by $A

Vedclass · Accepted Answer

$0.6$

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(C) The maximum amplitude of the modulated wave is given by $A_{max} = A_c + A_m = 160\, V$.The minimum amplitude of the modulated wave is given by $A_{min} = A_c - A_m = 40\, V$.Adding these two equations: $(A_c + A_m) + (A_c - A_m) = 160 + 40$,which gives $2A_c = 200$,so $A_c = 100\, V$.Subtracting the second equation from the first: $(A_c + A_m) - (A_c - A_m) = 160 - 40$,which gives $2A_m = 120$,so $A_m = 60\, V$.The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave:$\mu = \frac{A_m}{A_c} = \frac{60}{100} = 0.6$.

$A$ $100\, V$ carrier wave is made to vary between $160\, V$ and $40\, V$ by a modulating signal. What is the modulation index?

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