Fourier transform

Shifting theorem

$$ \begin{align*} \left( \mathcal{F} x(t-t_0) \right) (s) &= \left( \mathcal{F} x(t)*\delta(t-t_0) \right) (s) \\ &= \mathcal{F}x(s) \cdot  \left( \mathcal{F} \delta(t-t_0) \right) (s) \\ &= e^{-2\pi i st_0} \mathcal{F}x(s) \end{align*} $$

$$ \therefore \left( \mathcal{F} x(t-t_0) \right) (s) = e^{-2\pi i st_0} \mathcal x(s) $$

 

Duality

$$ \mathcal{F}f(-s) = \int_{-\infty}^\infty e^{-2 \pi i (-s)t} f(t)dt = \int_{-\infty}^\infty e^{2\pi i st} f(t)dt = \mathcal{F}^{-1}f(s) $$

$$ \mathcal{F}^{-1}f(-t) = \int_{-\infty}^\infty e^{2\pi i s(-t)} f(s)ds = \int_{-\infty}^\infty e^{-2\pi i st} f(s)ds = \mathcal{F}f(t)$$

 

Shifted impulse

$$ \left( \mathcal{F} \delta(t-t_0) \right) (s) = \int_{-\infty}^\infty e^{-2\pi i st}\delta(t-t_0) dt = e^{-2\pi i s t_0} $$

$$ \begin{align*} e^{2\pi i s_0 t} &= \int_{-\infty}^\infty e^{2\pi i st} \delta(s-s_0) ds \\ &= \int_{-\infty}^\infty e^{2\pi i st} \left( \mathcal{F} e^{2\pi i s_0 t} \right) (s) ds \end{align*}$$

$$ \therefore \begin{align*} \left( \mathcal{F} \delta (t-t_0) \right) (s) = e^{-2\pi i st_0} \\ \left(\mathcal{F} e^{2\pi i s_0 t} \right) (s) = \delta(s-s_0) \end{align*} $$

Hence, the Fourier transform of the shifted impulse is a complex exponential, and the Fourier transform of the complex exponential is the shifted impulse in the frequency domain.