Engineering Tables/Fourier Transform Table 2

	Signal	Fourier transform unitary, angular frequency	Fourier transform unitary, ordinary frequency	Remarks
	$g(t)\equiv$ $\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(\omega )e^{i\omega t}d\omega$	$G(\omega )\equiv$ $\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)e^{-i\omega t}dt$	$G(f)\equiv$ $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)e^{-i2\pi ft}dt$
10	$\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(at)$	$\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left(\frac{\omega }{2\pi a}\right)$	$\frac{1}{|a|}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left(\frac{f}{a}\right)$	The rectangular pulse and the normalized sinc function
11	$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(at)$	$\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2\pi a}\right)$	$\frac{1}{|a|}\cdot \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{f}{a}\right)$	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	${\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}}^2(at)$	$\frac{1}{\sqrt{2\pi a^2}}\cdot \mathrm{t}\mathrm{r}\mathrm{i}\left(\frac{\omega }{2\pi a}\right)$	$\frac{1}{|a|}\cdot \mathrm{t}\mathrm{r}\mathrm{i}\left(\frac{f}{a}\right)$	tri is the triangular function
13	$\mathrm{t}\mathrm{r}\mathrm{i}(at)$	$\frac{1}{\sqrt{2\pi a^2}}\cdot {\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}}^2\left(\frac{\omega }{2\pi a}\right)$	$\frac{1}{|a|}\cdot {\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}}^2\left(\frac{f}{a}\right)$	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 $\mathrm{R}\mathrm{e}(\alpha )>0$.
	$e^{ia{t}^2}={e^{-\alpha t^2}|}_{\alpha =-ia}$	$\frac{1}{\sqrt{2a}}\cdot e^{-i(\frac{{\omega }^2}{4a}-\frac{\pi }{4})}$	$\sqrt{\frac{\pi }{a}}\cdot e^{-i(\frac{{\pi }^2{f}^2}{a}-\frac{\pi }{4})}$	common in optics
	$\cos (a{t}^2)$	$\frac{1}{\sqrt{2a}}\cos (\frac{{\omega }^2}{4a}-\frac{\pi }{4})$	$\sqrt{\frac{\pi }{a}}\cos (\frac{{\pi }^2{f}^2}{a}-\frac{\pi }{4})$
	$\sin (a{t}^2)$	$\frac{-1}{\sqrt{2a}}\sin (\frac{{\omega }^2}{4a}-\frac{\pi }{4})$	$-\sqrt{\frac{\pi }{a}}\sin (\frac{{\pi }^2{f}^2}{a}-\frac{\pi }{4})$
	$e^{-a|t|}$	$\sqrt{\frac{2}{\pi }}\cdot \frac{a}{a^2+{\omega }^2}$	$\frac{2a}{a^2+4{\pi }^2{f}^2}$	a>0
	$\frac{1}{\sqrt{|t|}}$	$\frac{1}{\sqrt{|\omega |}}$	$\frac{1}{\sqrt{|f|}}$	the transform is the function itself
	$J_0(t)$	$\sqrt{\frac{2}{\pi }}\cdot \frac{\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2}\right)}{\sqrt{1-{\omega }^2}}$	$\frac{2\cdot \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(\pi f)}{\sqrt{1-4{\pi }^2{f}^2}}$	J0(t) is the Bessel function of first kind of order 0, rect is the rectangular function
	$J_n(t)$	$\sqrt{\frac{2}{\pi }}\frac{(-i{)}^n{T}_n(\omega )\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2}\right)}{\sqrt{1-{\omega }^2}}$	$\frac{2(-i{)}^n{T}_n(2\pi f)\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(\pi f)}{\sqrt{1-4{\pi }^2{f}^2}}$	it's the generalization of the previous transform; Tn (t) is the Chebyshev polynomial of the first kind.
	$\frac{J_n(t)}{t}$	$\sqrt{\frac{2}{\pi }}\frac{i}{n}(-i{)}^n\cdot U_{n-1}(\omega )$   $\cdot \sqrt{1-{\omega }^2}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2}\right)$	$\frac{2i}{n}(-i{)}^n\cdot U_{n-1}(2\pi f)$   $\cdot \sqrt{1-4{\pi }^2{f}^2}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(\pi f)$	Un (t) is the Chebyshev polynomial of the second kind