Aug 13, 2018

$\cos \left({\omega}_{0} t\right) \leftrightarrow \pi \left(\delta \left(\omega - {\omega}_{0}\right) + \delta \left(\omega + {\omega}_{0}\right)\right)$

#### Explanation:

Let $f \left(t\right)$ be a real function of the proper class. By definition the Dirac distribution $\delta \left(t\right)$ is such that:

$< \delta , f > = f \left(0\right)$

so:

${\int}_{- \infty}^{+ \infty} \delta \left(t\right) {e}^{- i \omega t} \mathrm{dt} = 1$

and if we translate the distribution:

${\int}_{- \infty}^{+ \infty} \delta \left(t - {t}_{0}\right) {e}^{- i \omega t} \mathrm{dt} = {e}^{- i \omega {t}_{0}}$

So the Fourier transform of $\delta \left(t - {t}_{0}\right)$ is ${e}^{- i \omega {t}_{0}}$.

By the symmetry property of the transform if we let: ${t}_{0} = - {\omega}_{0}$ then the Fourier transform of ${e}^{- i {\omega}_{0} t}$ is $2 \pi \delta \left(\omega - {\omega}_{0}\right)$

But using the expression of the cosine as:

$\cos \left({\omega}_{0} t\right) = \frac{{e}^{i {\omega}_{0} t} + {e}^{- i {\omega}_{0} t}}{2}$

and the linearity of the transform then:

$\cos \left({\omega}_{0} t\right) \leftrightarrow \pi \left(\delta \left(\omega - {\omega}_{0}\right) + \delta \left(\omega + {\omega}_{0}\right)\right)$