**泊松求和公式**（英文：）由法國數學家泊松所發現，它陳述了一個連續時間的信號，做無限多次的週期複製後，其傅立葉級數與其傅立葉轉換之間數值的關係，亦可用來求周期信號的傅立葉轉換。

公式
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设无周期函数{\displaystyle s(x)}具有傅里叶变换：

{\displaystyle S(f)\triangleq \int \_{-\infty }^{\infty }s(x)\ e^{-i2\pi fx}\,dx}
这里的{\displaystyle S(f)}也可以替代表示为{\displaystyle {\hat {s}}(f)}和 {\displaystyle {\mathcal {F}}\{s\}(f)}。有如下基本的泊松求和公式：

{\displaystyle \sum \_{n=-\infty }^{\infty }s(n)=\sum \_{k=-\infty }^{\infty }S(k)}
对于二者通过周期求和（英语：）而得到的周期函数：

{\displaystyle s\_{\_{P}}(x)\triangleq \sum \_{n=-\infty }^{\infty }s(x+nP),\quad n\in \mathbb {Z} }
{\displaystyle S\_{1/T}(f)\triangleq \sum \_{k=-\infty }^{\infty }S(f+k/T),\quad k,T\in \mathbb {Z} }
这里的参数{\displaystyle T>0}并且{\displaystyle P>0}，它们有着同{\displaystyle x}一样的单位。有如下普遍的泊松求和公式：

{\displaystyle s\_{\_{P}}(x)=\sum \_{k=-\infty }^{\infty }\underbrace {{\frac {1}{P}}\cdot S(k/P)} \_{S[k]}\ e^{i2\pi {\frac {k}{P}}x},\quad k\in \mathbb {Z} }
这是一个傅里叶级数展开，其系数是函数{\displaystyle S(f)}的采样。还有：

{\displaystyle S\_{1/T}(f)=\sum \_{n=-\infty }^{\infty }\underbrace {T\cdot s(nT)} \_{s[n]}\ e^{-i2\pi nTf},\quad n,T\in \mathbb {Z} }
这也叫做离散时间傅里叶变换。

推導泊松求和公式所需的先備公式
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考慮狄拉克δ函數{\displaystyle \delta (t)}，製作一個有無限多個{\displaystyle \delta (t)}，且間隔為{\displaystyle T\_{0}}的週期函數{\displaystyle \sum \_{n=-\infty }^{\infty }\delta (t-nT\_{0})}。

其傅立葉轉換為①{\displaystyle \sum \_{n=-\infty }^{\infty }e^{-j2\pi nT\_{0}f}=}②{\displaystyle {\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }\delta \left(f-{\frac {n}{T\_{0}}}\right)}

### 證明①轉換對

{\displaystyle {\mathcal {F}}\left\{\sum \_{n=-\infty }^{\infty }\delta (t-nT\_{0})\right\}}={\displaystyle \sum \_{n=-\infty }^{\infty }{\mathcal {F}}\left\{\delta (t-nT\_{0})\right\}}
={\displaystyle \sum \_{n=-\infty }^{\infty }e^{-j2\pi nT\_{0}f}}。

### 證明②轉換對

設{\displaystyle c\_{n}}為週期函數{\displaystyle \sum \_{n=-\infty }^{\infty }\delta (t-nT\_{0})}的傅立葉級數。

{\displaystyle \sum \_{n=-\infty }^{\infty }\delta (t-nT\_{0})}可表示為{\displaystyle \sum \_{n=-\infty }^{\infty }c\_{n}{e^{j2\pi {n{\frac {t}{T\_{0}}}}}}}。

由傅立葉級數得:

{\displaystyle c\_{n}={\frac {1}{T\_{0}}}\int \_{-{\frac {T\_{0}}{2}}}^{\frac {T\_{0}}{2}}\sum \_{m=-\infty }^{\infty }\delta (t-mT\_{0})e^{-j2\pi {n}{\frac {t}{T\_{0}}}}dt={\frac {1}{T\_{0}}}\int \_{-{\frac {T\_{0}}{2}}}^{\frac {T\_{0}}{2}}\delta (t)e^{-j2\pi {n}{\frac {t}{T\_{0}}}}dt={\frac {1}{T\_{0}}}\int \_{-{\frac {T\_{0}}{2}}}^{\frac {T\_{0}}{2}}\delta (t)e^{-j2\pi {n}{\frac {0}{T\_{0}}}}dt={\frac {1}{T\_{0}}}}。

因此，{\displaystyle {\mathcal {F}}\left\{\sum \_{n=-\infty }^{\infty }\delta (t-nT\_{0})\right\}={\mathcal {F}}\left\{\sum \_{n=-\infty }^{\infty }c\_{n}{e^{j2\pi {n{\frac {t}{T\_{0}}}}}}\right\}=\sum \_{n=-\infty }^{\infty }{\frac {1}{T\_{0}}}{\mathcal {F}}\left\{e^{j2\pi {n{\frac {t}{T\_{0}}}}}\right\}={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }\delta (f-n{\frac {1}{T\_{0}}})}。

得到等式:{\displaystyle \sum \_{n=-\infty }^{\infty }e^{-j2\pi nT\_{0}f}=}{\displaystyle {\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }\delta \left(f-{\frac {n}{T\_{0}}}\right)}，

經由適當的變量代換，{\displaystyle T\_{0}}以{\displaystyle {\frac {1}{T\_{0}}}}代換，{\displaystyle f}以{\displaystyle t}代換，得{\displaystyle \sum \_{n=-\infty }^{\infty }\delta \left(t-nT\_{0}\right)={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }e^{-j2\pi n{\frac {1}{T\_{0}}}t}={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }e^{j2\pi n{\frac {1}{T\_{0}}}t}}(因為n從負無限大到正無限大)

推導泊松求和公式
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### 從對頻域做取樣尋找關係式

{\displaystyle \sum \_{n=-\infty }^{\infty }x(t-nT\_{0})}

{\displaystyle =\sum \_{n=-\infty }^{\infty }x(t)\*\delta (t-nT\_{0})}

{\displaystyle =x(t)\*\sum \_{n=-\infty }^{\infty }\delta (t-nT\_{0})}

{\displaystyle =x(t)\*{\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }e^{j2\pi n{\frac {1}{T\_{0}}}t}}

{\displaystyle ={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }x(t)\*e^{j2\pi n{\frac {1}{T\_{0}}}t}}

{\displaystyle ={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }\int \_{-\infty }^{\infty }x(\tau )e^{j2\pi n{\frac {1}{T\_{0}}}(t-\tau )}d\tau }

{\displaystyle ={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }\left\{[\int \_{-\infty }^{\infty }x(\tau )e^{-j2\pi n{\frac {1}{T\_{0}}}\tau }d\tau ]e^{j2\pi n{\frac {1}{T\_{0}}}t}\right\}}

{\displaystyle ={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }X({\frac {n}{T\_{0}}})e^{j2\pi n{\frac {1}{T\_{0}}}t}}

當{\displaystyle t=0}時，得{\displaystyle \sum \_{n=-\infty }^{\infty }x(nT\_{0})={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }X({\frac {n}{T\_{0}}})}，

表示一個信號的在時域以{\displaystyle T\_{0}}為間隔做取樣，在頻域以{\displaystyle {\frac {1}{T\_{0}}}}為間隔做取樣，則兩者的所有取樣點的總和會有{\displaystyle T\_{0}}倍的關係。

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### 從對時域做取樣尋找關係式

{\displaystyle {\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }X(f-{\frac {n}{T\_{0}}})}

{\displaystyle ={\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }X(f)\*\delta (f-{\frac {n}{T\_{0}}})}

{\displaystyle =X(f)\*[{\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }\delta (f-{\frac {n}{T\_{0}}})]}

{\displaystyle =X(f)\*\sum \_{n=-\infty }^{\infty }e^{-j2\pi nT\_{0}f}}

{\displaystyle =\sum \_{n=-\infty }^{\infty }X(f)\*e^{-j2\pi nT\_{0}f}}

{\displaystyle =\sum \_{n=-\infty }^{\infty }\int \_{-\infty }^{\infty }X(\lambda )e^{-j2\pi nT\_{0}(f-\lambda )}d\lambda }

{\displaystyle =\sum \_{n=-\infty }^{\infty }\left\{[\int \_{-\infty }^{\infty }X(\lambda )e^{j2\pi nT\_{0}\lambda }d\lambda ]e^{-j2\pi nT\_{0}f}\right\}}

{\displaystyle =\sum \_{n=-\infty }^{\infty }x(nT\_{0})e^{-j2\pi nT\_{0}f}}

當{\displaystyle f=0}時，得{\displaystyle {\frac {1}{T\_{0}}}\sum \_{n=-\infty }^{\infty }X({\frac {n}{T\_{0}}})=\sum \_{n=-\infty }^{\infty }x(nT\_{0})}，

表示一個信號的在時域以{\displaystyle T\_{0}}為間隔做取樣，在頻域以{\displaystyle {\frac {1}{T\_{0}}}}為間隔做取樣，則兩者的所有取樣點的總和會有{\displaystyle T\_{0}}倍的關係。

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綜合上述，若時域取樣間隔{\displaystyle T\_{0}=1}時，同樣地，頻域取樣間隔{\displaystyle {\frac {1}{T\_{0}}}=1}時，得泊松求和公式{\displaystyle \sum \_{n=-\infty }^{\infty }x(n)=\sum \_{k=-\infty }^{\infty }X(k)}。

週期信號的傅立葉轉換
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考慮一個週期為{\displaystyle {T\_{0}}}的週期信號{\displaystyle g(t)}，{\displaystyle G(f)}為{\displaystyle g(t)}的傅立葉轉換，取出g(t)在區間{\displaystyle [-{\frac {T\_{0}}{2}},{\frac {T\_{0}}{2}}]}的一個完整週期{\displaystyle x(t)}，亦即{\displaystyle x(t)=g(t)rect({\frac {t}{T\_{0}}})}，{\displaystyle X(f)}是{\displaystyle x(t)}的傅立葉轉換，其中{\displaystyle rect(t)}是矩形函數。{\displaystyle a\_{n}}是{\displaystyle g(t)}的傅立葉級數。

則{\displaystyle G(f)}

{\displaystyle ={\mathcal {F}}\left\{\sum \_{n=-\infty }^{\infty }a\_{n}e^{j2\pi n{\frac {1}{T\_{0}}}t}\right\}}

{\displaystyle =\sum \_{n=-\infty }^{\infty }a\_{n}\delta (f-{\frac {n}{T\_{0}}})}

{\displaystyle =\sum \_{n=-\infty }^{\infty }{\frac {1}{T\_{0}}}X({\frac {n}{T\_{0}}})\delta (f-{\frac {n}{T\_{0}}})}

得出一週期信號的傅立葉轉換與其傅立葉級數之間的關係。

引用
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1. ↑  
Pinsky, M. Brooks Cole, 2002, ISBN 978-0-534-37660-4
2. ↑  
Zygmund, Antoni,  2nd, Cambridge University Press, 19681988, ISBN 978-0-521-35885-9

延伸阅读
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* Benedetto, J.J.; Zimmermann, G. J. Fourier Ana. App. 1997, **3** (5)  [2008-06-19], （原始内容存档于2011-05-24）
* Gasquet, Claude; Witomski, Patrick, Springer: 344–352, 1999, ISBN 0-387-98485-2
* Higgins, J.R. Bull. Amer. Math. Soc. 1985, **12** (1): 45–89  [2023-10-30], doi:10.1090/S0273-0979-1985-15293-0 可免费查阅, （原始内容存档于2020-08-12）