JEE PYQ: Communication System Question 17
Question 17 - 2021 (26 Feb 2021 Shift 1)
The maximum and minimum amplitude of an amplitude modulated wave is 16 V and 8 V respectively. The modulation index for this amplitude modulated wave is $x \times 10^{-2}$. The value of $x$ is
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Answer: 33
Solution
mi $= \frac{A_{max} - A_{min}}{A_{max} + A_{min}} = \frac{16-8}{16+8} = \frac{8}{24} = \frac{1}{3} = 0.33 = 33 \times 10^{-2}$. So $X = 33$.
BETA
