 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014/6/b/solution

The profile is an even function, so its Fourier transform is real and

T (ω) = 2 \int_{0}^{2} (1 - x /2) cos (ω x) d x .

For

ω \neq = 0

\int_{0}^{2} (1 - x /2) cos (ω x) d x = \frac{1}{2 ω} \int_{0}^{2} sin (ω x) d x = \frac{1 - c o s ( 2 ω )}{2 ω ^{2}} .

Therefore

T (ω) = ⎩ ⎨ ⎧ \frac{2 sin ^{2} ω}{ω ^{2}}, 2, ω \neq = 0, ω = 0.

At zero, direct integration gives the area of the triangle, namely two; the displayed quotient has the same limit. In particular $T (ω) \geq 0$ for every real frequency. This is the Fourier transform of a triangular function.

One can also see the nonnegative square structurally:

T = \frac{1}{2} 1_{[- 1, 1]} * 1_{[- 1, 1]}

T = \frac{1}{2} (2 sin ω / ω)^{2}

 New to topics? Read the docs here!