Question:easy

For full carrier system the transmission efficiency depends upon

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To maximize efficiency in analog broadcasting, $\mu$ is kept as close to 1 as possible without exceeding it, as $\mu \gt 1$ (over-modulation) causes severe signal distortion.
Updated On: Jul 1, 2026
  • Upper side band power
  • Lower side band power
  • Modulation index
  • Power saving
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The Correct Option is C

Solution and Explanation

1. Power Components in AM: The total power ($P_t$) in a standard AM signal consists of the carrier power ($P_c$) and the power in the two sidebands ($P_{sb}$): $$P_t = P_c + P_{sb} = P_c \left(1 + \frac{\mu^2}{2}\right)$$ where $\mu$ is the modulation index.

2. Defining Efficiency ($\eta$): Efficiency is defined as the ratio of useful power (sideband power) to the total power transmitted: $$\eta = \frac{P_{sb}}{P_t} = \frac{\frac{\mu^2}{2}}{1 + \frac{\mu^2}{2}} = \frac{\mu^2}{2 + \mu^2}$$

3. Role of the Modulation Index: The mathematical expression clearly shows that the efficiency ($\eta$) is purely a function of the

modulation index ($\mu$).

• If $\mu = 0$, efficiency is 0%.

• If $\mu = 1$ (100% modulation), efficiency is $1/(2+1) = 33.33\%$.
This indicates that even under ideal conditions, two-thirds of the power is "wasted" on the carrier, which carries no information, and the maximum efficiency is entirely determined by how deeply the carrier is modulated.
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