First slide

Modulation

Question

For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V. The modulation index $μ$ is

Easy

A

$μ = 3 / 4$
B

$μ = 2 / 3$
C

$μ = 1 / 3$
D

$μ = 1$

Solution

since the amplitude modulated wave is given by $(A_{c} + A_{m} \sin ω_{m} t) \cos ω_{c} t$ , the maximum amplitude is $M_{1} = A_{c} + A_{m}$ .While the minimum amplitude is $M_{2} = A_{c} - A_{m}$ . Hence the modulation Index is
$μ = \frac{A_{m}}{A_{c}} = \frac{M_{1} - M_{2}}{M_{1} + M_{2}} = \frac{10 - 2}{10 + 2} = \frac{8}{12} = \frac{2}{3}$
So $μ = \frac{2}{3}$

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