Question:
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:
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
- \(0.33\)
- \(0.67\)
- \(0.44\)
- \(0.63\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for modulation index.
For an amplitude modulated wave, \[ m=\frac{A_{\max}-A_{\min}}{A_{\max}+A_{\min}} \] where \[ A_{\max}=\text{maximum amplitude} \] and \[ A_{\min}=\text{minimum amplitude} \]
Step 2: Substitute the given values.
Given, \[ A_{\max}=20\ \text{V} \] \[ A_{\min}=4\ \text{V} \] Therefore, \[ m=\frac{20-4}{20+4} \] \[ m=\frac{16}{24} \] \[ m=\frac{2}{3} \] \[ m\approx0.67 \]
Step 3: Final conclusion.
Therefore, the modulation index is \[ \boxed{0.67} \]
For an amplitude modulated wave, \[ m=\frac{A_{\max}-A_{\min}}{A_{\max}+A_{\min}} \] where \[ A_{\max}=\text{maximum amplitude} \] and \[ A_{\min}=\text{minimum amplitude} \]
Step 2: Substitute the given values.
Given, \[ A_{\max}=20\ \text{V} \] \[ A_{\min}=4\ \text{V} \] Therefore, \[ m=\frac{20-4}{20+4} \] \[ m=\frac{16}{24} \] \[ m=\frac{2}{3} \] \[ m\approx0.67 \]
Step 3: Final conclusion.
Therefore, the modulation index is \[ \boxed{0.67} \]
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