Question:
For an amplitude modulated (AM) wave, the minimum amplitude is 4.0 V and the modulation index is 0.4. The maximum amplitude is:
For an amplitude modulated (AM) wave, the minimum amplitude is 4.0 V and the modulation index is 0.4. The maximum amplitude is:
Show Hint
In AM waves, the modulation index \(m\) relates max and min amplitudes:
\[
m = \frac{A_\text{max} - A_\text{min}}{A_\text{max} + A_\text{min}}
\]
and carrier amplitude \(A_c = (A_\text{max} + A_\text{min})/2\). Use these formulas for any AM amplitude problem.
Updated On: Jun 19, 2026
- 8.55 V
- 12.25 V
- 6.50 V
- 9.33 V
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Recall AM wave amplitude formula.
For an AM wave: \[ A_\text{max} = A_c (1 + m), \quad A_\text{min} = A_c (1 - m) \] where \(A_c\) is the carrier amplitude and \(m\) is the modulation index.
Step 2: Express \(A_c\) in terms of \(A_\text{min}\) and \(m\).
\[ A_\text{min} = A_c (1 - m) \Rightarrow A_c = \frac{A_\text{min}}{1 - m} \]
Step 3: Substitute values.
\[ A_c = \frac{4.0}{1 - 0.4} = \frac{4.0}{0.6} \approx 6.6667~\text{V} \]
Step 4: Calculate \(A_\text{max}\).
\[ A_\text{max} = A_c (1 + m) = 6.6667 (1 + 0.4) = 6.6667 \cdot 1.4 \approx 9.333~\text{V} \]
Step 5: Interpretation.
This shows that the maximum amplitude in AM is increased above the carrier by the modulation index factor.
Step 6: Verification.
Check using \(A_\text{min} = A_c (1 - m) = 6.6667 \cdot 0.6 = 4.0~\text{V}\). Correct.
Step 7: Conclusion.
Hence, the maximum amplitude of the AM wave is approximately 9.33 V.
For an AM wave: \[ A_\text{max} = A_c (1 + m), \quad A_\text{min} = A_c (1 - m) \] where \(A_c\) is the carrier amplitude and \(m\) is the modulation index.
Step 2: Express \(A_c\) in terms of \(A_\text{min}\) and \(m\).
\[ A_\text{min} = A_c (1 - m) \Rightarrow A_c = \frac{A_\text{min}}{1 - m} \]
Step 3: Substitute values.
\[ A_c = \frac{4.0}{1 - 0.4} = \frac{4.0}{0.6} \approx 6.6667~\text{V} \]
Step 4: Calculate \(A_\text{max}\).
\[ A_\text{max} = A_c (1 + m) = 6.6667 (1 + 0.4) = 6.6667 \cdot 1.4 \approx 9.333~\text{V} \]
Step 5: Interpretation.
This shows that the maximum amplitude in AM is increased above the carrier by the modulation index factor.
Step 6: Verification.
Check using \(A_\text{min} = A_c (1 - m) = 6.6667 \cdot 0.6 = 4.0~\text{V}\). Correct.
Step 7: Conclusion.
Hence, the maximum amplitude of the AM wave is approximately 9.33 V.
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