출처 : https://ena.etsmtl.ca/pluginfile.php/137982/mod_resource/content/8/Fourier-table.pdf

 

 

 

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출처 : https://en.wikipedia.org/wiki/Fourier_transform

 

Functional relationships, one-dimensional

 

The Fourier transforms in this table may be found in [Erdélyi 1954] or [Kammler 2000, appendix].

	Function	Fourier transform unitary, ordinary frequency	Fourier transform unitary, angular frequency	Fourier transform non-unitary, angular frequency	Remarks
	$$f(x)$$	$$\begin{array}{rl}& \hat{f}(\xi )\triangleq \widehat{f_1}(\xi )\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i2\pi \xi x}dx\end{array}$$	$$\begin{array}{rl}& \hat{f}(\omega )\triangleq \widehat{f_2}(\omega )\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\omega x}dx\end{array}$$	$$\begin{array}{rl}& \hat{f}(\omega )\triangleq \widehat{f_3}(\omega )\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\omega x}dx\end{array}$$	Definitions
101	$$af(x)+bg(x)$$	$$a\hat{f}(\xi )+b\hat{g}(\xi )$$	$$a\hat{f}(\omega )+b\hat{g}(\omega )$$	$$a\hat{f}(\omega )+b\hat{g}(\omega )$$	Linearity
102	$$f(x-a)$$	$$e^{-i2\pi \xi a}\hat{f}(\xi )$$	$$e^{-ia\omega }\hat{f}(\omega )$$	$$e^{-ia\omega }\hat{f}(\omega )$$	Shift in time domain
103	$$f(x)e^{iax}$$	$$\hat{f}(\xi -\frac{a}{2\pi })$$	$$\hat{f}(\omega -a)$$	$$\hat{f}(\omega -a)$$	Shift in frequency domain, dual of 102
104	$$f(ax)$$	$$\frac{1}{|a|}\hat{f}\left(\frac{\xi }{a}\right)$$	$$\frac{1}{|a|}\hat{f}\left(\frac{\omega }{a}\right)$$	$$\frac{1}{|a|}\hat{f}\left(\frac{\omega }{a}\right)$$	Scaling in the time domain. If $$abs|a$$ is large, then $$f(ax)$$ is concentrated around 0 and $$\frac{1}{|a|}\hat{f}\left(\frac{\omega }{a}\right)$$ spreads out and flattens.
105	$$\widehat{f_n}(x)$$	$$\widehat{f_1}(x)\stackrel{{\mathcal{F}}_1}{⟷}f(-\xi )$$	$$\widehat{f_2}(x)\stackrel{{\mathcal{F}}_2}{⟷}f(-\omega )$$	$$\widehat{f_3}(x)\stackrel{{\mathcal{F}}_3}{⟷}2\pi f(-\omega )$$	The same transform is applied twice, but x replaces the frequency variable (ξ or ω) after the first transform.
106	$$\frac{d^{n}f(x)}{d{x}^n}$$	$$(i2\pi \xi {)}^n\hat{f}(\xi )$$	$$(i\omega {)}^n\hat{f}(\omega )$$	$$(i\omega {)}^n\hat{f}(\omega )$$	nth-order derivative. As $$f$$ is a [[Schwartz space|Schwartz function]]
106.5	$${\int }_{-\mathrm{\infty }}^{x}f(\tau )d\tau$$	$$\frac{\hat{f}(\xi )}{i2\pi \xi }+C\delta (\xi )$$	$$\frac{\hat{f}(\omega )}{i\omega }+\sqrt{2\pi }C\delta (\omega )$$	$$\frac{\hat{f}(\omega )}{i\omega }+2\pi C\delta (\omega )$$	Integration.[1] Note: $$\delta$$ is the [[Dirac delta function]] and $$C$$ is the average ([[DC component|DC]]) value of $$f(x)$$ such that $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(f(x)-C)dx=0$$
107	$$x^{n}f(x)$$	$${\left(\frac{i}{2\pi }\right)}^n\frac{d^n\hat{f}(\xi )}{d{\xi }^n}$$	$$i^n\frac{d^n\hat{f}(\omega )}{d{\omega }^n}$$	$$i^n\frac{d^n\hat{f}(\omega )}{d{\omega }^n}$$	This is the dual of 106
108	$$(f\ast g)(x)$$	$$\hat{f}(\xi )\hat{g}(\xi )$$	$$\sqrt{2\pi }\hat{f}(\omega )\hat{g}(\omega )$$	$$\hat{f}(\omega )\hat{g}(\omega )$$	The notation $$f\ast g$$ denotes the [[convolution]] of $$f$$ and $$g$$ — this rule is the [[convolution theorem]]
109	$$f(x)g(x)$$	$$(\hat{f}\ast \hat{g})(\xi )$$	$$\frac{1}{\sqrt{2\pi }}(\hat{f}\ast \hat{g})(\omega )$$	$$\frac{1}{2\pi }(\hat{f}\ast \hat{g})(\omega )$$	This is the dual of 108
110	For $$f(x)$$ purely real	$$\hat{f}(-\xi )=\overline{\hat{f}(\xi )}$$	$$\hat{f}(-\omega )=\overline{\hat{f}(\omega )}$$	$$\hat{f}(-\omega )=\overline{\hat{f}(\omega )}$$	Hermitian symmetry. $$\bar{z}$$ indicates the [[complex conjugate]].
113	For $$f(x)$$ purely imaginary	$$\hat{f}(-\xi )=-\overline{\hat{f}(\xi )}$$	$$\hat{f}(-\omega )=-\overline{\hat{f}(\omega )}$$	$$\hat{f}(-\omega )=-\overline{\hat{f}(\omega )}$$	$$\bar{z}$$ indicates the [[complex conjugate]].
114	$$\overline{f(x)}$$	$$\overline{\hat{f}(-\xi )}$$	$$\overline{\hat{f}(-\omega )}$$	$$\overline{\hat{f}(-\omega )}$$	[[Complex conjugation]], generalization of 110 and 113
115	$$f(x)\cos (ax)$$	$$\frac{\hat{f}(\xi -\frac{a}{2\pi })+\hat{f}(\xi +\frac{a}{2\pi })}{2}$$	$$\frac{\hat{f}(\omega -a)+\hat{f}(\omega +a)}{2}$$	$$\frac{\hat{f}(\omega -a)+\hat{f}(\omega +a)}{2}$$	This follows from rules 101 and 103 using [[Euler's formula]]: $$\cos (ax)=\frac{e^{iax}+e^{-iax}}{2}.$$
116	$$f(x)\sin (ax)$$	$$\frac{\hat{f}(\xi -\frac{a}{2\pi })-\hat{f}(\xi +\frac{a}{2\pi })}{2i}$$	$$\frac{\hat{f}(\omega -a)-\hat{f}(\omega +a)}{2i}$$	$$\frac{\hat{f}(\omega -a)-\hat{f}(\omega +a)}{2i}$$	This follows from 101 and 103 using [[Euler's formula]]: $$\sin (ax)=\frac{e^{iax}-e^{-iax}}{2i}.$$

^[1] The Integration Property of the Fourier Transform

	Function	Fourier transform unitary, ordinary frequency	Fourier transform unitary, angular frequency	Fourier transform non-unitary, angular frequency	Remarks
	$$f(x)$$	$$\begin{array}{rl}& \hat{f}(\xi )\triangleq {\hat{f}}_1(\xi )\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i2\pi \xi x}dx\end{array}$$	$$\begin{array}{rl}& \hat{f}(\omega )\triangleq {\hat{f}}_2(\omega )\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\omega x}dx\end{array}$$	$$\begin{array}{rl}& \hat{f}(\omega )\triangleq {\hat{f}}_3(\omega )\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\omega x}dx\end{array}$$	Definitions
201	$$\mathrm{rect}(ax)$$	$$\frac{1}{|a|}\mathrm{sinc}\left(\frac{\xi }{a}\right)$$	$$\frac{1}{\sqrt{2\pi a^2}}\mathrm{sinc}\left(\frac{\omega }{2\pi a}\right)$$	$$\frac{1}{|a|}\mathrm{sinc}\left(\frac{\omega }{2\pi a}\right)$$	The rectangular pulse and the normalized sinc function, here defined as $$1=\text{sinc}(x)=\frac{\sin (\pi x)}{\pi x}$$
202	$$\mathrm{sinc}(ax)$$	$$\frac{1}{|a|}\mathrm{rect}\left(\frac{\xi }{a}\right)$$	$$\frac{1}{\sqrt{2\pi a^2}}\mathrm{rect}\left(\frac{\omega }{2\pi a}\right)$$	$$\frac{1}{|a|}\mathrm{rect}\left(\frac{\omega }{2\pi a}\right)$$	Dual of rule 201. The rectangular function is an ideal low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The sinc function is defined here as $$1=\text{sinc}(x)=\frac{\sin (\pi x)}{\pi x}$$
203	$${\mathrm{sinc}}^2(ax)$$	$$\frac{1}{|a|}\mathrm{tri}\left(\frac{\xi }{a}\right)$$	$$\frac{1}{\sqrt{2\pi a^2}}\mathrm{tri}\left(\frac{\omega }{2\pi a}\right)$$	$$\frac{1}{|a|}\mathrm{tri}\left(\frac{\omega }{2\pi a}\right)$$	The function $$\text{tri}(x)$$ is the triangular function
204	$$\mathrm{tri}(ax)$$	$$\frac{1}{|a|}{\mathrm{sinc}}^2\left(\frac{\xi }{a}\right)$$	$$\frac{1}{\sqrt{2\pi a^2}}{\mathrm{sinc}}^2\left(\frac{\omega }{2\pi a}\right)$$	$$\frac{1}{|a|}{\mathrm{sinc}}^2\left(\frac{\omega }{2\pi a}\right)$$	Dual of rule 203.
205	$$e^{-ax}u(x)$$	$$\frac{1}{a+i2\pi \xi }$$	$$\frac{1}{\sqrt{2\pi }(a+i\omega )}$$	$$\frac{1}{a+i\omega }$$	The function $$u(x)$$ is the Heaviside unit step function and $$a>0$$.
206	$$e^{-\alpha x^2}$$	$$\sqrt{\frac{\pi }{\alpha }}{e}^{-\frac{(\pi \xi {)}^2}{\alpha }}$$	$$\frac{1}{\sqrt{2\alpha }}{e}^{-\frac{{\omega }^2}{4\alpha }}$$	$$\sqrt{\frac{\pi }{\alpha }}{e}^{-\frac{{\omega }^2}{4\alpha }}$$	This shows that, for the unitary Fourier transforms, the Gaussian function $$e^{-\alpha x^2}$$ is its own Fourier transform for some choice of $$\alpha$$. For this to be integrable we must have $$Re(\alpha )>0$$.
208	$$e^{-a|x|}$$	$$\frac{2a}{a^2+4{\pi }^2{\xi }^2}$$	$$\sqrt{\frac{2}{\pi }}\frac{a}{a^2+{\omega }^2}$$	$$\frac{2a}{a^2+{\omega }^2}$$	For $$Re(a)>0$$. That is, the Fourier transform of a two-sided decaying exponential function is a Lorentzian function.
209	$$\mathrm{sech}(ax)$$	$$\frac{\pi }{a}\mathrm{sech}\left(\frac{{\pi }^2}{a}\xi \right)$$	$$\frac{1}{a}\sqrt{\frac{\pi }{2}}\mathrm{sech}\left(\frac{\pi }{2a}\omega \right)$$	$$\frac{\pi }{a}\mathrm{sech}\left(\frac{\pi }{2a}\omega \right)$$	Hyperbolic secant is its own Fourier transform
210	$$e^{-\frac{a^2{x}^2}{2}}{H}_n(ax)$$	$$\frac{\sqrt{2\pi }(-i{)}^n}{a}{e}^{-\frac{2{\pi }^2{\xi }^2}{a^2}}{H}_n\left(\frac{2\pi \xi }{a}\right)$$	$$\frac{(-i{)}^n}{a}{e}^{-\frac{{\omega }^2}{2a^2}}{H}_n\left(\frac{\omega }{a}\right)$$	$$\frac{(-i{)}^n\sqrt{2\pi }}{a}{e}^{-\frac{{\omega }^2}{2a^2}}{H}_n\left(\frac{\omega }{a}\right)$$	$$H_n$$ is the $$n$$th-order Hermite polynomial. If $$a=1$$ then the Gauss–Hermite functions are eigenfunctions of the Fourier transform operator. For a derivation, see Hermite polynomial. The formula reduces to 206 for $$n=0$$.

	Function	Fourier transform unitary, ordinary frequency	Fourier transform unitary, angular frequency	Fourier transform non-unitary, angular frequency	Remarks
	$$f(x)$$	$$\begin{array}{rl}& \hat{f}(\xi )\triangleq {\hat{f}}_1(\xi )\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i2\pi \xi x}dx\end{array}$$	$$\begin{array}{rl}& \hat{f}(\omega )\triangleq {\hat{f}}_2(\omega )\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\omega x}dx\end{array}$$	$$\begin{array}{rl}& \hat{f}(\omega )\triangleq {\hat{f}}_3(\omega )\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)e^{-i\omega x}dx\end{array}$$	Definitions
301	$$1$$	$$\delta (\xi )$$	$$\sqrt{2\pi }\delta (\omega )$$	$$2\pi \delta (\omega )$$	The distribution $${}^{″}{\delta }^{″}{(}^{″}{\xi }^{″})$$ denotes the [[Dirac delta function]].
302	$$\delta (x)$$	$$1$$	$$\frac{1}{\sqrt{2\pi }}$$	$$1$$	Dual of rule 301.
303	$$e^{iax}$$	$$\delta (\xi -\frac{a}{2\pi })$$	$$\sqrt{2\pi }\delta (\omega -a)$$	$$2\pi \delta (\omega -a)$$	This follows from 103 and 301.
304	$$\cos (ax)$$	$$\frac{\delta (\xi -\frac{a}{2\pi })+\delta (\xi +\frac{a}{2\pi })}{2}$$	$$\sqrt{2\pi }\frac{\delta (\omega -a)+\delta (\omega +a)}{2}$$	$$\pi (\delta (\omega -a)+\delta (\omega +a))$$	This follows from rules 101 and 303 using [[Euler's formula]]: $$\cos (ax)=\frac{e^{iax}+e^{-iax}}{2}.$$
305	$$\sin (ax)$$	$$\frac{\delta (\xi -\frac{a}{2\pi })-\delta (\xi +\frac{a}{2\pi })}{2i}$$	$$\sqrt{2\pi }\frac{\delta (\omega -a)-\delta (\omega +a)}{2i}$$	$$-i\pi (\delta (\omega -a)-\delta (\omega +a))$$	This follows from 101 and 303 using $$\sin (ax)=\frac{e^{iax}-e^{-iax}}{2i}.$$
306	$$\cos \left(a{x}^2\right)$$	$$\sqrt{\frac{\pi }{a}}\cos (\frac{{\pi }^2{\xi }^2}{a}-\frac{\pi }{4})$$	$$\frac{1}{\sqrt{2a}}\cos (\frac{{\omega }^2}{4a}-\frac{\pi }{4})$$	$$\sqrt{\frac{\pi }{a}}\cos (\frac{{\omega }^2}{4a}-\frac{\pi }{4})$$	This follows from 101 and 207 using $$\cos (a{x}^2)=\frac{e^{ia{x}^2}+e^{-ia{x}^2}}{2}.$$
307	$$\sin \left(a{x}^2\right)$$	$$-\sqrt{\frac{\pi }{a}}\sin (\frac{{\pi }^2{\xi }^2}{a}-\frac{\pi }{4})$$	$$\frac{-1}{\sqrt{2a}}\sin (\frac{{\omega }^2}{4a}-\frac{\pi }{4})$$	$$-\sqrt{\frac{\pi }{a}}\sin (\frac{{\omega }^2}{4a}-\frac{\pi }{4})$$	This follows from 101 and 207 using $$\sin (a{x}^2)=\frac{e^{ia{x}^2}-e^{-ia{x}^2}}{2i}.$$
308	$$e^{-\pi i\alpha x^2}$$	$$\frac{1}{\sqrt{\alpha }}{e}^{-i\frac{\pi }{4}}{e}^{i\frac{\pi {\xi }^2}{\alpha }}$$	$$\frac{1}{\sqrt{2\pi \alpha }}{e}^{-i\frac{\pi }{4}}{e}^{i\frac{{\omega }^2}{4\pi \alpha }}$$	$$\frac{1}{\sqrt{\alpha }}{e}^{-i\frac{\pi }{4}}{e}^{i\frac{{\omega }^2}{4\pi \alpha }}$$	Here it is assumed $$\alpha$$ is real. For the case that alpha is complex see table entry 206 above.
309	$$x^n$$	$${\left(\frac{i}{2\pi }\right)}^n{\delta }^{(n)}(\xi )$$	$$i^n\sqrt{2\pi }{\delta }^{(n)}(\omega )$$	$$2\pi i^n{\delta }^{(n)}(\omega )$$	Here, $$n$$ is a [[natural number]] and $${}^{″}{\delta }^{″}isup|{(}^{″}{n}^{″})\left(\xi \right)$$ is the $$n$$th distribution derivative of the Dirac delta function. This rule follows from rules 107 and 301. Combining this rule with 101, we can transform all [[polynomial]]s.
310	$${\delta }^{(n)}(x)$$	$$(i2\pi \xi {)}^n$$	$$\frac{(i\omega {)}^n}{\sqrt{2\pi }}$$	$$(i\omega {)}^n$$	Dual of rule 309. $${}^{″}{\delta }^{″}isup|{(}^{″}{n}^{″})\left(\xi \right)$$ is the $$n$$th distribution derivative of the Dirac delta function. This rule follows from 106 and 302.
311	$$\frac{1}{x}$$	$$-i\pi \text{sgn}(\xi )$$	$$-i\sqrt{\frac{\pi }{2}}\text{sgn}(\omega )$$	$$-i\pi \text{sgn}(\omega )$$	Here $$sgn(\xi )$$ is the [[sign function]]. Note that $$\frac{1}{x}$$ is not a distribution. It is necessary to use the [[Cauchy principal value]] when testing against [[Schwartz functions]]. This rule is useful in studying the [[Hilbert transform]].
312	$$\begin{array}{rl}& \frac{1}{x^n}\\ & :=\frac{(-1{)}^{n-1}}{(n-1)!}\frac{d^n}{d{x}^n}\log |x|\end{array}$$	$$-i\pi \frac{(-i2\pi \xi {)}^{n-1}}{(n-1)!}\text{sgn}(\xi )$$	$$-i\sqrt{\frac{\pi }{2}}\frac{(-i\omega {)}^{n-1}}{(n-1)!}\text{sgn}(\omega )$$	$$-i\pi \frac{(-i\omega {)}^{n-1}}{(n-1)!}\text{sgn}(\omega )$$	$$\frac{1}{x^n}$$ is the [[homogeneous distribution]] defined by the distributional derivative $$\frac{(-1{)}^{n-1}}{(n-1)!}\frac{d^n}{d{x}^n}\log |x|$$
313	$$|x{|}^{\alpha }$$	$$-\frac{2\sin \left(\frac{\pi \alpha }{2}\right)\mathrm{\Gamma }(\alpha +1)}{|2\pi \xi {|}^{\alpha +1}}$$	$$\frac{-2}{\sqrt{2\pi }}\frac{\sin \left(\frac{\pi \alpha }{2}\right)\mathrm{\Gamma }(\alpha +1)}{|\omega {|}^{\alpha +1}}$$	$$-\frac{2\sin \left(\frac{\pi \alpha }{2}\right)\mathrm{\Gamma }(\alpha +1)}{|\omega {|}^{\alpha +1}}$$	This formula is valid for $$0>\alpha >-1$$. For $$\alpha >0$$ some singular terms arise at the origin that can be found by differentiating 320. If $$Re\alpha >-1$$, then $$|x{|}^{\alpha }$$ is a locally integrable function, and so a tempered distribution. The function $$\alpha \mapsto |x{|}^{\alpha }$$ is a holomorphic function from the right half-plane to the space of tempered distributions. It admits a unique meromorphic extension to a tempered distribution, also denoted $$|x{|}^{\alpha }$$ for $$\alpha \ne -1,-3,...$$ (See [[homogeneous distribution]].)
	$$\frac{1}{\sqrt{|x|}}$$	$$\frac{1}{\sqrt{|\xi |}}$$	$$\frac{1}{\sqrt{|\omega |}}$$	$$\frac{\sqrt{2\pi }}{\sqrt{|\omega |}}$$	Special case of 313.
314	$$\text{sgn}(x)$$	$$\frac{1}{i\pi \xi }$$	$$\sqrt{\frac{2}{\pi }}\frac{1}{i\omega }$$	$$\frac{2}{i\omega }$$	The dual of rule 311. This time the Fourier transforms need to be considered as a [[Cauchy principal value]].
315	$$u(x)$$	$$\frac{1}{2}(\frac{1}{i\pi \xi }+\delta (\xi ))$$	$$\sqrt{\frac{\pi }{2}}(\frac{1}{i\pi \omega }+\delta (\omega ))$$	$$\pi (\frac{1}{i\pi \omega }+\delta (\omega ))$$	The function $$u(x)$$ is the Heaviside [[Heaviside step function|unit step function]]; this follows from rules 101, 301, and 314.
316	$$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (x-nT)$$	$$\frac{1}{T}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (\xi -\frac{k}{T})$$	$$\frac{\sqrt{2\pi }}{T}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (\omega -\frac{2\pi k}{T})$$	$$\frac{2\pi }{T}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (\omega -\frac{2\pi k}{T})$$	This function is known as the [[Dirac comb]] function. This result can be derived from 302 and 102, together with the fact that $$\begin{array}{rl}& \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{inx}\\ =& 2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (x+2\pi k)\end{array}$$ as distributions.
317	$$J_0(x)$$	$$\frac{2\mathrm{rect}(\pi \xi )}{\sqrt{1-4{\pi }^2{\xi }^2}}$$	$$\sqrt{\frac{2}{\pi }}\frac{\mathrm{rect}\left(\frac{\omega }{2}\right)}{\sqrt{1-{\omega }^2}}$$	$$\frac{2\mathrm{rect}\left(\frac{\omega }{2}\right)}{\sqrt{1-{\omega }^2}}$$	The function $$J_0(x)$$ is the zeroth order [[Bessel function]] of first kind.