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: $\frac{e^{jt}+e^{-jt}}{2}$:

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

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

• 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)