**Question:** Compute the Fourier transform of the signal x(t) equal to:

$ x(t) = cos(2\pi t) \, $

**Student's Answer:**
The Fourier Transform of a signal in Continuous Time is defined by:

$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \, $

Using this, we obtain:

$ X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \, $

Knowing that cos(t) is equal to: $ \frac{e^{jt}+e^{-jt}}{2} $:

$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{j2\pi t}+e^{-j2\pi t})e^{-j\omega t}dt \, $

$ X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \, $

**Mimi's grading and comments**

- The last step of this computation is highly mysterious: how exactly do you know that there is a factor $ \pi $ in front of the deltas? And how is it exactly that the answer is zero when $ \omega \neq 2\pi, -2\pi $? I would give about 2/10 for this answer. --Mboutin 15:09, 17 October 2008 (UTC)