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Exercises · 15.7

Q.For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index, μ\mu.
What would be the value of μ\mu if the minimum amplitude is zero volt?

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Given: for an AM wave, maximum amplitude Amax=10A_{max} = 10 V, minimum amplitude Amin=2A_{min} = 2 V.

Step 1 — Modulation-index formula from envelope extremes.

Since Amax=Ac+AmA_{max}=A_c+A_m and Amin=Ac−AmA_{min}=A_c-A_m, solving gives Ac=12(Amax+Amin)A_c=\tfrac{1}{2}(A_{max}+A_{min}) and Am=12(Amax−Amin)A_m=\tfrac{1}{2}(A_{max}-A_{min}), so

μ=AmAc=Amax−AminAmax+Amin\mu = \frac{A_m}{A_c} = \frac{A_{max}-A_{min}}{A_{max}+A_{min}}

Step 2 — Substitute.

μ=10−210+2=812=0.667\mu = \frac{10-2}{10+2} = \frac{8}{12} = 0.667

Step 3 — Case Amin=0A_{min}=0.

μ=Amax−0Amax+0=AmaxAmax=1\mu = \frac{A_{max}-0}{A_{max}+0} = \frac{A_{max}}{A_{max}} = 1 …

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