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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?

Difficulty: Easy

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
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