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An audio signal $v_{m} = 15 \sin w_{1} t$ modulated on a cw of $v_{c} = 60 \sin w_{2} t$ the percentage modulation is

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50%

40%

25%

2.5%

answer is C.

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Detailed Solution

$v_{m} = 15 \sin w_{1} t A_{m} = 15 V_{c} = 60 \sin w_{2} t A_{c} = 60$

Modulation index $μ = \frac{A_{m}}{A_{c}}$

$μ = \frac{15}{60} μ = \frac{1}{4} = 0.25 μ = 25 %$

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