Question:easy

The maximum amplitude of an AM wave is found to be \(20\ \text{V}\) while its minimum amplitude is \(4\ \text{V}\). The modulation index is:

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For AM waves, \[ m=\frac{A_{\max}-A_{\min}}{A_{\max}+A_{\min}} \] A modulation index less than \(1\) indicates proper modulation without distortion.
Updated On: Jun 26, 2026
  • \(0.33\)
  • \(0.67\)
  • \(0.44\)
  • \(0.63\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Recall what modulation index means.
In an amplitude modulated (AM) wave, the amplitude of the carrier wave varies between a maximum value $A_{\text{max}}$ and a minimum value $A_{\text{min}}$. The modulation index $m$ (also called the depth of modulation) tells us how much the carrier amplitude varies relative to the carrier itself.
Step 2: Write the formula for modulation index.
The modulation index is defined as \[ m = \frac{A_{\text{max}} - A_{\text{min}}}{A_{\text{max}} + A_{\text{min}}} \] This formula directly comes from the fact that $A_{\text{max}} = A_c + A_m$ and $A_{\text{min}} = A_c - A_m$, where $A_c$ is the carrier amplitude and $A_m$ is the message signal amplitude.
Step 3: Identify the given values.
From the problem: $A_{\text{max}} = 20$ V and $A_{\text{min}} = 4$ V.
Step 4: Substitute and compute the numerator.
\[ A_{\text{max}} - A_{\text{min}} = 20 - 4 = 16\ \text{V} \]
Step 5: Compute the denominator.
\[ A_{\text{max}} + A_{\text{min}} = 20 + 4 = 24\ \text{V} \]
Step 6: Calculate the modulation index.
\[ m = \frac{16}{24} = \frac{2}{3} \approx 0.67 \] A modulation index of 0.67 means the signal is being modulated to 67% of the maximum possible depth, which is ideal for good quality transmission without distortion. \[ \boxed{m \approx 0.67} \]
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