AM modulation depth requirement: For the amplitude-modulated signal s(t) = Ac cos(ωc t) + 2 cos(ωm t) cos(ωc t), what is the minimum value of carrier amplitude Ac for distortionless envelope detection?
Electronics and Communication Engineering
Exam Questions Papers
Difficulty: Medium
Choose an option
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A2
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B1
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C0.5
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D0
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ENone of these
Answer
Correct Answer: 2
Explanation
Introduction / Context:Envelope detection requires the modulation index μ ≤ 1 for distortionless demodulation. The modulation index is the ratio of the message amplitude to the carrier amplitude. If μ > 1, overmodulation occurs and the envelope detector distorts the signal.
Given Data / Assumptions:
- AM signal: s(t) = Ac cos(ωc t) + 2 cos(ωm t) cos(ωc t).
- The message term has amplitude 2.
- Carrier amplitude is Ac (unknown).
Concept / Approach:
Standard AM: s(t) = Ac cos(ωc t) [1 + (Am/Ac) cos(ωm t)]. Here, Am = 2. Modulation index μ = Am / Ac. For envelope detector to work: μ ≤ 1 ⇒ Ac ≥ Am.
Step-by-Step Solution:
Identify Am = 2 (message amplitude).Modulation index μ = Am / Ac = 2 / Ac.Condition for distortionless detection: μ ≤ 1 ⇒ 2 / Ac ≤ 1.Solve: Ac ≥ 2.Verification / Alternative check:
General rule: minimum carrier amplitude equals or exceeds the message amplitude to avoid overmodulation. Here Ac must be ≥ 2.
Why Other Options Are Wrong:
- Ac = 1 or 0.5: gives μ = 2 or 4 ⇒ overmodulation, distortion.
- Ac = 0: no carrier; signal cannot be demodulated by envelope detector.
Common Pitfalls:
- Forgetting to compare message amplitude to carrier amplitude.
- Misinterpreting the cosine product term without factoring it into AM standard form.
Final Answer:
2