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Question
An amplitude modulated wave is as shown in figure. Calculate
- the percentage modulation
- peak carrier voltage and
- peak value of information voltage.
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Solution
From the figure
`V_(max) = 100/2 = 50V, V_(min) = 20/2 = 10 V`.
i. Percentage modulation = `mu(%) = (V_(max) - V_(min))/(V_(max) + V_(min)) xx 100`
= `((50 - 10)/(50 + 10)) xx 100`
= `40/60 xx 100`
= 66.67%
ii. Peak carrier voltage = `V_c = (V_(max) + V_(min))/2`
= `(50 + 10)/2`
= 30 V
iii. Peak information voltage = `V_m = muV_c`
= `2/3 xx 30`
= 20 V.
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