JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Semiconductors

The maximum amplitude wave is found to be 12 V while the minimum amplitude is found to be 3 V . The modulation index is 0.6 x where x is ____ .

Answer: 1

Solution

The modulation index follows from the extremes of the modulated envelope:

μ = \frac{A _{m a x} - A _{m i n}}{A _{m a x} + A _{m i n}} = \frac{12 - 3}{12 + 3} = \frac{9}{15} = 0.6

Comparing with

μ = 0.6 x

gives

x = 1

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer