# 傅里叶变换

（重定向自傅立葉轉換

“傅里叶变换”一词既指变换操作本身（将函数 ${\displaystyle f}$ 进行傅里叶变换），又指该操作所生成的复数函数（${\displaystyle {\hat {f}}}$${\displaystyle f}$ 的傅里叶变换）。

## 定义

${\displaystyle {\hat {f}}(\xi )=\int _{-\infty }^{\infty }f(x)\ e^{-2\pi ix\xi }\,dx}$ ξ为任意实数

${\displaystyle f(x)=\int _{-\infty }^{\infty }{\hat {f}}(\xi )\ e^{2\pi i\xi x}\,d\xi }$ x为任意实数。

## 傅里叶变换的不同变种

${\displaystyle {\hat {f}}(\omega )=\int _{\mathbf {R} ^{n}}f(x)e^{-i\omega \cdot x}\,dx.}$

${\displaystyle f(x)={\frac {1}{(2\pi )^{n}}}\int _{\mathbf {R} ^{n}}{\hat {f}}(\omega )e^{i\omega \cdot x}\,d\omega .}$

${\displaystyle {\hat {f}}(\omega )={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbf {R} ^{n}}f(x)e^{-i\omega \cdot x}\,dx}$
${\displaystyle f(x)={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbf {R} ^{n}}{\hat {f}}(\omega )e^{i\omega \cdot x}\,d\omega .}$

普通频率ξ（ 赫兹） ${\displaystyle \displaystyle {\hat {f}}_{1}(\xi )\ {\stackrel {\mathrm {def} }{=}}\ \int _{\mathbf {R} ^{n}}f(x)e^{-2\pi ix\cdot \xi }\,dx={\hat {f}}_{2}(2\pi \xi )=(2\pi )^{n/2}{\hat {f}}_{3}(2\pi \xi )}$ ${\displaystyle \displaystyle f(x)=\int _{\mathbf {R} ^{n}}{\hat {f}}_{1}(\xi )e^{2\pi ix\cdot \xi }\,d\xi \ }$ ${\displaystyle \displaystyle {\hat {f}}_{2}(\omega )\ {\stackrel {\mathrm {def} }{=}}\int _{\mathbf {R} ^{n}}f(x)e^{-i\omega \cdot x}\,dx\ ={\hat {f}}_{1}\left({\frac {\omega }{2\pi }}\right)=(2\pi )^{n/2}\ {\hat {f}}_{3}(\omega )}$  ${\displaystyle \displaystyle f(x)={\frac {1}{(2\pi )^{n}}}\int _{\mathbf {R} ^{n}}{\hat {f}}_{2}(\omega )e^{i\omega \cdot x}\,d\omega \ }$ ${\displaystyle \displaystyle {\hat {f}}_{3}(\omega )\ {\stackrel {\mathrm {def} }{=}}\ {\frac {1}{(2\pi )^{n/2}}}\int _{\mathbf {R} ^{n}}f(x)\ e^{-i\omega \cdot x}\,dx={\frac {1}{(2\pi )^{n/2}}}{\hat {f}}_{1}\left({\frac {\omega }{2\pi }}\right)={\frac {1}{(2\pi )^{n/2}}}{\hat {f}}_{2}(\omega )}$  ${\displaystyle \displaystyle f(x)={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbf {R} ^{n}}{\hat {f}}_{3}(\omega )e^{i\omega \cdot x}\,d\omega \ }$

### 傅里叶级数

${\displaystyle f(x)=\sum _{n=-\infty }^{\infty }F_{n}\,e^{inx},}$

${\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left[a_{n}\cos(nx)+b_{n}\sin(nx)\right]}$

### 离散傅里叶变换

${\displaystyle X_{k}=\sum _{n=0}^{N-1}x_{n}e^{-i{\frac {2\pi }{N}}kn}\qquad k=0,\dots ,N-1}$

## 常用傅里叶变换表

### 函数关系

${\displaystyle \displaystyle f(x)\,}$  ${\displaystyle \displaystyle {\hat {f}}(\xi )=}$

${\displaystyle \displaystyle \int _{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }\,dx}$

${\displaystyle \displaystyle {\hat {f}}(\omega )=}$

${\displaystyle \displaystyle {\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(x)e^{-i\omega x}\,dx}$

${\displaystyle \displaystyle {\hat {f}}(\nu )=}$

${\displaystyle \displaystyle \int _{-\infty }^{\infty }f(x)e^{-i\nu x}\,dx}$

101 ${\displaystyle \displaystyle a\cdot f(x)+b\cdot g(x)\,}$  ${\displaystyle \displaystyle a\cdot {\hat {f}}(\xi )+b\cdot {\hat {g}}(\xi )\,}$  ${\displaystyle \displaystyle a\cdot {\hat {f}}(\omega )+b\cdot {\hat {g}}(\omega )\,}$  ${\displaystyle \displaystyle a\cdot {\hat {f}}(\nu )+b\cdot {\hat {g}}(\nu )\,}$  线性
102 ${\displaystyle \displaystyle f(x-a)\,}$  ${\displaystyle \displaystyle e^{-2\pi ia\xi }{\hat {f}}(\xi )\,}$  ${\displaystyle \displaystyle e^{-ia\omega }{\hat {f}}(\omega )\,}$  ${\displaystyle \displaystyle e^{-ia\nu }{\hat {f}}(\nu )\,}$  时域平移
103 ${\displaystyle \displaystyle e^{2\pi iax}f(x)\,}$  ${\displaystyle \displaystyle {\hat {f}}\left(\xi -a\right)\,}$  ${\displaystyle \displaystyle {\hat {f}}(\omega -2\pi a)\,}$  ${\displaystyle \displaystyle {\hat {f}}(\nu -2\pi a)\,}$  频域平移，变换102的频域对应
104 ${\displaystyle \displaystyle f(ax)\,}$  ${\displaystyle \displaystyle {\frac {1}{|a|}}{\hat {f}}\left({\frac {\xi }{a}}\right)\,}$  ${\displaystyle \displaystyle {\frac {1}{|a|}}{\hat {f}}\left({\frac {\omega }{a}}\right)\,}$  ${\displaystyle \displaystyle {\frac {1}{|a|}}{\hat {f}}\left({\frac {\nu }{a}}\right)\,}$  在时域中定标。如果${\displaystyle \displaystyle |a|\,}$ 值较大，则${\displaystyle \displaystyle f(ax)\,}$ 会收缩到原点附近，而${\displaystyle \displaystyle {\frac {1}{|a|}}{\hat {f}}\left({\frac {\omega }{a}}\right)\,}$ 会扩散并变得扁平。当${\displaystyle \displaystyle |a|\,}$ 趋向无穷时，${\displaystyle \displaystyle f(ax)\,}$ 成为狄拉克δ函数
105 ${\displaystyle \displaystyle {\hat {f}}(x)\,}$  ${\displaystyle \displaystyle f(-\xi )\,}$  ${\displaystyle \displaystyle f(-\omega )\,}$  ${\displaystyle \displaystyle 2\pi f(-\nu )\,}$  傅里叶变换的二元性性质。这里${\displaystyle {\hat {f}}}$ 的计算需要运用与傅里叶变换那一列同样的方法。通过交换变量${\displaystyle x}$ ${\displaystyle \xi }$ ${\displaystyle \omega }$ ${\displaystyle \nu }$ 得到。
106 ${\displaystyle \displaystyle {\frac {d^{n}f(x)}{dx^{n}}}\,}$  ${\displaystyle \displaystyle (2\pi i\xi )^{n}{\hat {f}}(\xi )\,}$  ${\displaystyle \displaystyle (i\omega )^{n}{\hat {f}}(\omega )\,}$  ${\displaystyle \displaystyle (i\nu )^{n}{\hat {f}}(\nu )\,}$  傅里叶变换的微分性质
107 ${\displaystyle \displaystyle x^{n}f(x)\,}$  ${\displaystyle \displaystyle \left({\frac {i}{2\pi }}\right)^{n}{\frac {d^{n}{\hat {f}}(\xi )}{d\xi ^{n}}}\,}$  ${\displaystyle \displaystyle i^{n}{\frac {d^{n}{\hat {f}}(\omega )}{d\omega ^{n}}}}$  ${\displaystyle \displaystyle i^{n}{\frac {d^{n}{\hat {f}}(\nu )}{d\nu ^{n}}}}$  变换106的频域对应
108 ${\displaystyle \displaystyle (f*g)(x)\,}$  ${\displaystyle \displaystyle {\hat {f}}(\xi ){\hat {g}}(\xi )\,}$  ${\displaystyle \displaystyle {\sqrt {2\pi }}{\hat {f}}(\omega ){\hat {g}}(\omega )\,}$  ${\displaystyle \displaystyle {\hat {f}}(\nu ){\hat {g}}(\nu )\,}$  记号${\displaystyle \displaystyle f*g\,}$ 表示${\displaystyle f}$ ${\displaystyle g}$ 的卷积—这就是卷积定理
109 ${\displaystyle \displaystyle f(x)g(x)\,}$  ${\displaystyle \displaystyle ({\hat {f}}*{\hat {g}})(\xi )\,}$  ${\displaystyle \displaystyle ({\hat {f}}*{\hat {g}})(\omega ) \over {\sqrt {2\pi }}\,}$  ${\displaystyle \displaystyle {\frac {1}{2\pi }}({\hat {f}}*{\hat {g}})(\nu )\,}$  变换108的频域对应。
110 ${\displaystyle \displaystyle f(x)}$ 是实变函数 ${\displaystyle \displaystyle {\hat {f}}(-\xi )={\overline {{\hat {f}}(\xi )}}\,}$  ${\displaystyle \displaystyle {\hat {f}}(-\omega )={\overline {{\hat {f}}(\omega )}}\,}$  ${\displaystyle \displaystyle {\hat {f}}(-\nu )={\overline {{\hat {f}}(\nu )}}\,}$  埃尔米特对称。${\displaystyle \displaystyle {\overline {z}}\,}$ 表示复共轭
111 ${\displaystyle \displaystyle f(x)}$ 是实偶函数 ${\displaystyle \displaystyle {\hat {f}}(\omega )}$ , ${\displaystyle \displaystyle {\hat {f}}(\xi )}$ ${\displaystyle \displaystyle {\hat {f}}(\nu )\,}$ 都是实偶函数
112 ${\displaystyle \displaystyle f(x)}$ 是实奇函数 ${\displaystyle \displaystyle {\hat {f}}(\omega )}$ , ${\displaystyle \displaystyle {\hat {f}}(\xi )}$ ${\displaystyle \displaystyle {\hat {f}}(\nu )}$ 都是奇函数
113 ${\displaystyle \displaystyle {\overline {f(x)}}}$  ${\displaystyle \displaystyle {\overline {{\hat {f}}(-\xi )}}}$  ${\displaystyle \displaystyle {\overline {{\hat {f}}(-\omega )}}}$  ${\displaystyle \displaystyle {\overline {{\hat {f}}(-\nu )}}}$  复共轭，110的一般化

### 平方可积函数

${\displaystyle g(t)\!\equiv \!}$

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

${\displaystyle {\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\!\!g(t)e^{-i\omega t}\mathrm {d} t\,}$
${\displaystyle G(f)\!\equiv }$

${\displaystyle \int _{-\infty }^{\infty }\!\!g(t)e^{-i2\pi ft}\mathrm {d} t\,}$
10 ${\displaystyle \mathrm {rect} (at)\,}$  ${\displaystyle {\frac {1}{\sqrt {2\pi a^{2}}}}\cdot \mathrm {sinc} \left({\frac {\omega }{2\pi a}}\right)}$  ${\displaystyle {\frac {1}{|a|}}\cdot \mathrm {sinc} \left({\frac {f}{a}}\right)}$  矩形脉冲和归一化的sinc函数
11 ${\displaystyle \mathrm {sinc} (at)\,}$  ${\displaystyle {\frac {1}{\sqrt {2\pi a^{2}}}}\cdot \mathrm {rect} \left({\frac {\omega }{2\pi a}}\right)}$  ${\displaystyle {\frac {1}{|a|}}\cdot \mathrm {rect} \left({\frac {f}{a}}\right)\,}$  变换10的频域对应。矩形函数是理想的低通滤波器，sinc函数是这类滤波器对反因果冲击的响应。
12 ${\displaystyle \mathrm {sinc} ^{2}(at)\,}$  ${\displaystyle {\frac {1}{\sqrt {2\pi a^{2}}}}\cdot \mathrm {tri} \left({\frac {\omega }{2\pi a}}\right)}$  ${\displaystyle {\frac {1}{|a|}}\cdot \mathrm {tri} \left({\frac {f}{a}}\right)}$  tri三角形函数
13 ${\displaystyle \mathrm {tri} (at)\,}$  ${\displaystyle {\frac {1}{\sqrt {2\pi a^{2}}}}\cdot \mathrm {sinc} ^{2}\left({\frac {\omega }{2\pi a}}\right)}$  ${\displaystyle {\frac {1}{|a|}}\cdot \mathrm {sinc} ^{2}\left({\frac {f}{a}}\right)\,}$  变换12的频域对应
14 ${\displaystyle e^{-\alpha t^{2}}\,}$  ${\displaystyle {\frac {1}{\sqrt {2\alpha }}}\cdot e^{-{\frac {\omega ^{2}}{4\alpha }}}}$  ${\displaystyle {\sqrt {\frac {\pi }{\alpha }}}\cdot e^{-{\frac {(\pi f)^{2}}{\alpha }}}}$  高斯函数${\displaystyle \exp(-\alpha t^{2})}$ 的傅里叶变换是他本身.只有当${\displaystyle \mathrm {Re} (\alpha )>0}$ 时，这是可积的。
15 ${\displaystyle e^{iat^{2}}=\left.e^{-\alpha t^{2}}\right|_{\alpha =-ia}\,}$  ${\displaystyle {\frac {1}{\sqrt {2a}}}\cdot e^{-i\left({\frac {\omega ^{2}}{4a}}-{\frac {\pi }{4}}\right)}}$  ${\displaystyle {\sqrt {\frac {\pi }{a}}}\cdot e^{-i\left({\frac {\pi ^{2}f^{2}}{a}}-{\frac {\pi }{4}}\right)}}$  光学领域应用较多
16 ${\displaystyle \cos(at^{2})\,}$  ${\displaystyle {\frac {1}{\sqrt {2a}}}\cos \left({\frac {\omega ^{2}}{4a}}-{\frac {\pi }{4}}\right)}$  ${\displaystyle {\sqrt {\frac {\pi }{a}}}\cos \left({\frac {\pi ^{2}f^{2}}{a}}-{\frac {\pi }{4}}\right)}$
17 ${\displaystyle \sin(at^{2})\,}$  ${\displaystyle {\frac {-1}{\sqrt {2a}}}\sin \left({\frac {\omega ^{2}}{4a}}-{\frac {\pi }{4}}\right)}$  ${\displaystyle -{\sqrt {\frac {\pi }{a}}}\sin \left({\frac {\pi ^{2}f^{2}}{a}}-{\frac {\pi }{4}}\right)}$
18 ${\displaystyle \mathrm {e} ^{-a|t|}\,}$  ${\displaystyle {\sqrt {\frac {2}{\pi }}}\cdot {\frac {a}{a^{2}+\omega ^{2}}}}$  ${\displaystyle {\frac {2a}{a^{2}+4\pi ^{2}f^{2}}}}$  a>0
19 ${\displaystyle {\frac {1}{\sqrt {|t|}}}\,}$  ${\displaystyle {\frac {1}{\sqrt {|\omega |}}}}$  ${\displaystyle {\frac {1}{\sqrt {|f|}}}}$  变换本身就是一个公式
20 ${\displaystyle J_{0}(t)\,}$  ${\displaystyle {\sqrt {\frac {2}{\pi }}}\cdot {\frac {\mathrm {rect} \left({\frac {\omega }{2}}\right)}{\sqrt {1-\omega ^{2}}}}}$  ${\displaystyle {\frac {2\cdot \mathrm {rect} (\pi f)}{\sqrt {1-4\pi ^{2}f^{2}}}}}$  J0(t)0阶第一类贝塞尔函数
21 ${\displaystyle J_{n}(t)\,}$  ${\displaystyle {\sqrt {\frac {2}{\pi }}}{\frac {(-i)^{n}T_{n}(\omega )\mathrm {rect} \left({\frac {\omega }{2}}\right)}{\sqrt {1-\omega ^{2}}}}}$  ${\displaystyle {\frac {2(-i)^{n}T_{n}(2\pi f)\mathrm {rect} (\pi f)}{\sqrt {1-4\pi ^{2}f^{2}}}}}$  上一个变换的推广形式; Tn (t)第一类切比雪夫多项式
22 ${\displaystyle {\frac {J_{n}(t)}{t}}\,}$  ${\displaystyle {\sqrt {\frac {2}{\pi }}}{\frac {i}{n}}(-i)^{n}\cdot U_{n-1}(\omega )\,}$

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

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

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

Un (t)第二类切比雪夫多项式

### 分布

${\displaystyle g(t)\!\equiv \!}$

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

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

${\displaystyle \int _{-\infty }^{\infty }\!\!g(t)e^{-i2\pi ft}dt\,}$
23 ${\displaystyle 1\,}$  ${\displaystyle {\sqrt {2\pi }}\cdot \delta (\omega )\,}$  ${\displaystyle \delta (f)\,}$  ${\displaystyle \delta (\omega )}$ 代表狄拉克δ函数分布.这个变换展示了狄拉克δ函数的重要性：该函数是常函数的傅立叶变换
24 ${\displaystyle \delta (t)\,}$  ${\displaystyle {\frac {1}{\sqrt {2\pi }}}\,}$  ${\displaystyle 1\,}$  变换23的频域对应
25 ${\displaystyle e^{iat}\,}$  ${\displaystyle {\sqrt {2\pi }}\cdot \delta (\omega -a)\,}$  ${\displaystyle \delta (f-{\frac {a}{2\pi }})\,}$  由变换3和24得到.
26 ${\displaystyle \cos(at)\,}$  ${\displaystyle {\sqrt {2\pi }}{\frac {\delta (\omega \!-\!a)\!+\!\delta (\omega \!+\!a)}{2}}\,}$  ${\displaystyle {\frac {\delta (f\!-\!{\begin{matrix}{\frac {a}{2\pi }}\end{matrix}})\!+\!\delta (f\!+\!{\begin{matrix}{\frac {a}{2\pi }}\end{matrix}})}{2}}\,}$  由变换1和25得到，应用了欧拉公式${\displaystyle \cos(at)=(e^{iat}+e^{-iat})/2.}$
27 ${\displaystyle \sin(at)\,}$  ${\displaystyle {\sqrt {2\pi }}{\frac {\delta (\omega \!-\!a)\!-\!\delta (\omega \!+\!a)}{2i}}\,}$  ${\displaystyle {\frac {\delta (f\!-\!{\begin{matrix}{\frac {a}{2\pi }}\end{matrix}})\!-\!\delta (f\!+\!{\begin{matrix}{\frac {a}{2\pi }}\end{matrix}})}{2i}}\,}$  由变换1和25得到
28 ${\displaystyle t^{n}\,}$  ${\displaystyle i^{n}{\sqrt {2\pi }}\delta ^{(n)}(\omega )\,}$  ${\displaystyle \left({\frac {i}{2\pi }}\right)^{n}\delta ^{(n)}(f)\,}$  这里, ${\displaystyle n}$ 是一个自然数. ${\displaystyle \delta ^{(n)}(\omega )}$ 是狄拉克δ函数分布的${\displaystyle n}$ 阶微分。这个变换是根据变换7和24得到的。将此变换与1结合使用，我们可以变换所有多項式
29 ${\displaystyle {\frac {1}{t}}\,}$  ${\displaystyle -i{\sqrt {\frac {\pi }{2}}}\operatorname {sgn}(\omega )\,}$  ${\displaystyle -i\pi \cdot \operatorname {sgn}(f)\,}$  此处${\displaystyle \operatorname {sgn}(\omega )}$ 符号函数；注意此变换与变换7和24是一致的.
30 ${\displaystyle {\frac {1}{t^{n}}}\,}$  ${\displaystyle -i{\begin{matrix}{\sqrt {\frac {\pi }{2}}}\cdot {\frac {(-i\omega )^{n-1}}{(n-1)!}}\end{matrix}}\operatorname {sgn}(\omega )\,}$  ${\displaystyle -i\pi {\begin{matrix}{\frac {(-i2\pi f)^{n-1}}{(n-1)!}}\end{matrix}}\operatorname {sgn}(f)\,}$  变换29的推广.
31 ${\displaystyle \operatorname {sgn}(t)\,}$  ${\displaystyle {\sqrt {\frac {2}{\pi }}}\cdot {\frac {1}{i\ \omega }}\,}$  ${\displaystyle {\frac {1}{i\pi f}}\,}$  变换29的频域对应.
32 ${\displaystyle u(t)\,}$  ${\displaystyle {\sqrt {\frac {\pi }{2}}}\left({\frac {1}{i\pi \omega }}+\delta (\omega )\right)\,}$  ${\displaystyle {\frac {1}{2}}\left({\frac {1}{i\pi f}}+\delta (f)\right)\,}$  此处${\displaystyle u(t)}$ 单位阶跃函数;此变换根据变换1和31得到.
33 ${\displaystyle e^{-at}u(t)\,}$  ${\displaystyle {\frac {1}{{\sqrt {2\pi }}(a+i\omega )}}}$  ${\displaystyle {\frac {1}{a+i2\pi f}}}$  ${\displaystyle u(t)}$ 单位阶跃函数，且${\displaystyle a>0}$ .
34 ${\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT)\,}$  ${\displaystyle {\begin{matrix}{\frac {\sqrt {2\pi }}{T}}\end{matrix}}\sum _{k=-\infty }^{\infty }\delta \left(\omega -k{\begin{matrix}{\frac {2\pi }{T}}\end{matrix}}\right)\,}$  ${\displaystyle {\frac {1}{T}}\sum _{k=-\infty }^{\infty }\delta \left(f-{\frac {k}{T}}\right)\,}$  狄拉克梳状函数英语Dirac comb——有助于解释或理解从连续到离散时间的转变.

### 二元函数

400 ${\displaystyle \displaystyle f(x,y)}$  ${\displaystyle \displaystyle {\hat {f}}(\xi _{x},\xi _{y})=}$
${\displaystyle \displaystyle \iint f(x,y)e^{-2\pi i(\xi _{x}x+\xi _{y}y)}\,dx\,dy}$
${\displaystyle \displaystyle {\hat {f}}(\omega _{x},\omega _{y})=}$
${\displaystyle \displaystyle {\frac {1}{2\pi }}\iint f(x,y)e^{-i(\omega _{x}x+\omega _{y}y)}\,dx\,dy}$
${\displaystyle \displaystyle {\hat {f}}(\nu _{x},\nu _{y})=}$
${\displaystyle \displaystyle \iint f(x,y)e^{-i(\nu _{x}x+\nu _{y}y)}\,dx\,dy}$
401 ${\displaystyle \displaystyle e^{-\pi \left(a^{2}x^{2}+b^{2}y^{2}\right)}}$  ${\displaystyle \displaystyle {\frac {1}{|ab|}}e^{-\pi \left(\xi _{x}^{2}/a^{2}+\xi _{y}^{2}/b^{2}\right)}}$  ${\displaystyle \displaystyle {\frac {1}{2\pi \cdot |ab|}}e^{\frac {-\left(\omega _{x}^{2}/a^{2}+\omega _{y}^{2}/b^{2}\right)}{4\pi }}}$  ${\displaystyle \displaystyle {\frac {1}{|ab|}}e^{\frac {-\left(\nu _{x}^{2}/a^{2}+\nu _{y}^{2}/b^{2}\right)}{4\pi }}}$
402 ${\displaystyle \displaystyle \mathrm {circ} ({\sqrt {x^{2}+y^{2}}})}$  ${\displaystyle \displaystyle {\frac {J_{1}\left(2\pi {\sqrt {\xi _{x}^{2}+\xi _{y}^{2}}}\right)}{\sqrt {\xi _{x}^{2}+\xi _{y}^{2}}}}}$  ${\displaystyle \displaystyle {\frac {J_{1}\left({\sqrt {\omega _{x}^{2}+\omega _{y}^{2}}}\right)}{\sqrt {\omega _{x}^{2}+\omega _{y}^{2}}}}}$  ${\displaystyle \displaystyle {\frac {2\pi J_{1}\left({\sqrt {\nu _{x}^{2}+\nu _{y}^{2}}}\right)}{\sqrt {\nu _{x}^{2}+\nu _{y}^{2}}}}}$

400： 变量ξxξyωxωyνxνy为实数。 对整个平面积分。

401： 这两个函数都是高斯分布，而且可能不具有单位体积。

402： 此圆有单位半径，如果把circ（t）认作阶梯函数u(1-t); Airy分布用J1（1阶第一类贝塞尔函数）表达。（Stein & Weiss 1971，Thm. IV.3.3）

### 三元函数

${\displaystyle \mathrm {circ} ({\sqrt {x^{2}+y^{2}+z^{2}}})}$  ${\displaystyle 4\pi {\frac {\sin[2\pi f_{r}]-2\pi f_{r}\cos[2\pi f_{r}]}{(2\pi f_{r})^{3}}}}$  此球有单位半径；fr是频率矢量的量值{fx,fy,fz}.

## 參考資料

### 文内资料引用

1. ^ 杨毅明. 数字信号处理（第2版）. 北京: 机械工业出版社. 2017年: 第25、29页. ISBN 9787111576235.