An AM waveform has a carrier that is 1 kV peak-to-peak. If the modulation index is 50%, what are the envelope's maximum (Emax) and minimum (Emin) voltages observed on an oscilloscope?
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A2 kV, 0.5 kV
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B1 kV, 0.5 kV
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C0.75 kV, 0.25 kV
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D0.5 kV, 1.5 kV
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E1.25 kV, 0.75 kV
Answer
Correct Answer: 0.75 kV, 0.25 kV
Explanation
Introduction / Context:For conventional amplitude modulation (AM), the oscilloscope envelope directly shows the effect of modulation index m on the carrier amplitude. Being able to compute the envelope extremes Emax and Emin is essential for avoiding overmodulation and for interpreting AM observations in lab measurements.
Given Data / Assumptions:
- Carrier peak-to-peak voltage Vpp = 1 kV ⇒ carrier peak Ec = Vpp / 2 = 0.5 kV.
- Modulation index m = 0.5 (i.e., 50%).
- AM envelope extremes: Emax = Ec(1 + m), Emin = Ec(1 − m).
Concept / Approach:In an AM signal s(t) = Ec[1 + m cos(ωm t)]cos(ωc t), the slowly varying envelope magnitudes vary between Ec(1 ± m). This directly yields the oscilloscope-measured envelope crest and trough provided the timebase is slow enough to display the envelope or a peak-detect circuit is used.
Step-by-Step Solution:
Compute carrier amplitude: Ec = 1 kV / 2 = 0.5 kV.Compute Emax: Emax = Ec(1 + m) = 0.5 × 1.5 = 0.75 kV.Compute Emin: Emin = Ec(1 − m) = 0.5 × 0.5 = 0.25 kV.Verification / Alternative check:
Check that the envelope never goes negative (m ≤ 1). With m = 0.5, there is no overmodulation, so the trough stays positive at 0.25 kV.Why Other Options Are Wrong:
2 kV, 0.5 kV and 1 kV, 0.5 kV: These misuse the peak-to-peak definition or apply m incorrectly.0.5 kV, 1.5 kV: Reverses max and min and miscomputes values.1.25 kV, 0.75 kV: Inconsistent with Ec = 0.5 kV and m = 0.5.Common Pitfalls:
Confusing peak and peak-to-peak; forgetting to multiply m by Ec, not by Vpp; swapping Emax and Emin.Final Answer:
0.75 kV, 0.25 kV