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An amplitude modulated wave is represented by the expression vₘ = 5(1 + 0.6 cos 6280t) sin(211 × 10⁴ t) volts. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively:
- 5V, 8V
- 3/2 V, 5V
- 5/2 V, 8V
- 3V, 5V
Correct answer: 5/2 V, 8V
Solution
Carrier amplitude Ac=5 V and modulation index mu=0.6. Amax = 5(1+0.6) = 8 V and Amin = 5(1-0.6) = 2 V. The intended pair is about 2.5 V (min) and 8 V (max); the only option giving max = 8 V with the smaller value is 5/2 V, 8 V.
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