Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 4 / g

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- Solution g

Solution

By Fourier inversion, the sinc function

sin (A t) / (π t)

has Fourier transform equal to one for

∣ ω ∣ < A

and zero for

∣ ω ∣ > A

. Thus the two bandwidths add to give

z (ω) = ⎩ ⎨ ⎧ 2, 1, 0, ∣ ω ∣ < 1, 1 < ∣ ω ∣ < 2, ∣ ω ∣ > 2.

The requested squared-magnitude spectrum therefore has heights four and one:

∣ z (ω) ∣^{2} = ⎩ ⎨ ⎧ 4, 1, 0, ∣ ω ∣ < 1, 1 < ∣ ω ∣ < 2, ∣ ω ∣ > 2.

Values at the jump points depend on the chosen Fourier convergence convention and do not affect any integral. The graph is shown below.

Figure 1.
Squared Fourier magnitude of the sum of two sinc signals
. Details.

The Parseval identity with this Fourier transform normalization gives

E = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ z (ω) ∣^{2} d ω = \frac{1}{2 π} (4 \cdot 2 + 1 \cdot 2) = \frac{5}{π} .

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