Engineering Tables/Fourier Transform Table 2

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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