# Engineering Tables/Fourier Transform Table 2

Signal Fourier transform
unitary, angular frequency
Fourier transform
unitary, ordinary frequency
Remarks
$g(t)\!\equiv\!$

$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\!\!G(\omega) e^{i \omega t} d \omega \,$
$G(\omega)\!\equiv\!$

$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\!\!g(t) e^{-i \omega t} dt \,$
$G(f)\!\equiv$

$\int_{-\infty}^{\infty}\!\!g(t) e^{-i 2\pi f t} dt \,$
10 $\mathrm{rect}(a t) \,$ $\frac{1}{\sqrt{2 \pi a^2}}\cdot \mathrm{sinc}\left(\frac{\omega}{2\pi a}\right)$ $\frac{1}{|a|}\cdot \mathrm{sinc}\left(\frac{f}{a}\right)$ The rectangular pulse and the normalized sinc function
11 $\mathrm{sinc}(a t)\,$ $\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{rect}\left(\frac{\omega}{2 \pi a}\right)$ $\frac{1}{|a|}\cdot \mathrm{rect}\left(\frac{f}{a} \right)\,$ Dual of rule 10. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter.
12 $\mathrm{sinc}^2 (a t) \,$ $\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{tri} \left( \frac{\omega}{2\pi a} \right)$ $\frac{1}{|a|}\cdot \mathrm{tri} \left( \frac{f}{a} \right)$ tri is the triangular function
13 $\mathrm{tri} (a t) \,$ $\frac{1}{\sqrt{2\pi a^2}} \cdot \mathrm{sinc}^2 \left( \frac{\omega}{2\pi a} \right)$ $\frac{1}{|a|}\cdot \mathrm{sinc}^2 \left( \frac{f}{a} \right) \,$ Dual of rule 12.
14 $e^{-\alpha t^2}\,$ $\frac{1}{\sqrt{2 \alpha}}\cdot e^{-\frac{\omega^2}{4 \alpha}}$ $\sqrt{\frac{\pi}{\alpha}}\cdot e^{-\frac{(\pi f)^2}{\alpha}}$ Shows that the Gaussian function $\exp(-\alpha t^2)$ is its own Fourier transform. For this to be integrable we must have $\mathrm{Re}(\alpha)>0$.
$e^{i a t^2} = \left. e^{-\alpha t^2}\right|_{\alpha = -i a} \,$ $\frac{1}{\sqrt{2 a}} \cdot e^{-i \left(\frac{\omega^2}{4 a} -\frac{\pi}{4}\right)}$ $\sqrt{\frac{\pi}{a}} \cdot e^{-i \left(\frac{\pi^2 f^2}{a} -\frac{\pi}{4}\right)}$ common in optics
$\cos ( a t^2 ) \,$ $\frac{1}{\sqrt{2 a}} \cos \left( \frac{\omega^2}{4 a} - \frac{\pi}{4} \right)$ $\sqrt{\frac{\pi}{a}} \cos \left( \frac{\pi^2 f^2}{a} - \frac{\pi}{4} \right)$
$\sin ( a t^2 ) \,$ $\frac{-1}{\sqrt{2 a}} \sin \left( \frac{\omega^2}{4 a} - \frac{\pi}{4} \right)$ $- \sqrt{\frac{\pi}{a}} \sin \left( \frac{\pi^2 f^2}{a} - \frac{\pi}{4} \right)$
$e^{-a|t|} \,$ $\sqrt{\frac{2}{\pi}} \cdot \frac{a}{a^2 + \omega^2}$ $\frac{2 a}{a^2 + 4 \pi^2 f^2}$ a>0
$\frac{1}{\sqrt{|t|}} \,$ $\frac{1}{\sqrt{|\omega|}}$ $\frac{1}{\sqrt{|f|}}$ the transform is the function itself
$J_0 (t)\,$ $\sqrt{\frac{2}{\pi}} \cdot \frac{\mathrm{rect} \left( \frac{\omega}{2} \right)}{\sqrt{1 - \omega^2}}$ $\frac{2\cdot \mathrm{rect} (\pi f)}{\sqrt{1 - 4 \pi^2 f^2}}$ J0(t) is the Bessel function of first kind of order 0, rect is the rectangular function
$J_n (t) \,$ $\sqrt{\frac{2}{\pi}} \frac{ (-i)^n T_n (\omega) \mathrm{rect} \left( \frac{\omega}{2} \right)}{\sqrt{1 - \omega^2}}$ $\frac{2 (-i)^n T_n (2 \pi f) \mathrm{rect} (\pi f)}{\sqrt{1 - 4 \pi^2 f^2}}$ it's the generalization of the previous transform; Tn (t) is the Chebyshev polynomial of the first kind.
$\frac{J_n (t)}{t} \,$ $\sqrt{\frac{2}{\pi}} \frac{i}{n} (-i)^n \cdot U_{n-1} (\omega)\,$

$\cdot \ \sqrt{1 - \omega^2} \mathrm{rect} \left( \frac{\omega}{2} \right)$

$\frac{2 i}{n} (-i)^n \cdot U_{n-1} (2 \pi f)\,$

$\cdot \ \sqrt{1 - 4 \pi^2 f^2} \mathrm{rect} ( \pi f )$

Un (t) is the Chebyshev polynomial of the second kind