 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/4/a/solution

Use the Fourier transform convention

a (ω) = \int_{- \infty}^{\infty} a (t) e^{- iω t} d t

, with inverse factor

1/ (2 π)

. Direct integration over the pulse gives

a (ω) = \int_{- 1}^{1} e^{- iω t} d t = \frac{2 sin ω}{ω}, a (0) = 2.

The apparent singularity at zero is removable. Changing the values of the pulse at its two endpoints does not affect the transform.

 New to topics? Read the docs here!