 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/4/a/solution

Use the Fourier transform convention

f (ω) = \int_{- \infty}^{\infty} f (t) e^{- iω t} d t, f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} f (ω) e^{iω t} d ω .

Since

F [e^{i Ω t}] = 2 π δ (ω - Ω)

in distribution theory, the carrier has transform

c (ω) = π [δ (ω - Ω) + δ (ω + Ω)] .

ϕ (ω) = π [δ (ω - Ω) + δ (ω + Ω)] + \frac{1}{2} [s (ω - Ω) + s (ω + Ω)] .

The first term represents the unmodulated carrier wave; the other terms are the amplitude modulation sidebands.

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