泊松求和公式

泊松求和公式（英文：Poisson Summation Formula）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

設無周期函數${\displaystyle s(x)}$具有傅立葉變換：

${\displaystyle S(f)\triangleq \int _{-\infty }^{\infty }s(x)\ e^{-i2\pi fx}\,dx}$

這裡的${\displaystyle S(f)}$也可以替代表示為${\displaystyle {\hat {s}}(f)}$和 ${\displaystyle {\mathcal {F}}\{s\}(f)}$。有如下基本的泊松求和公式：

${\displaystyle \sum _{n=-\infty }^{\infty }s(n)=\sum _{k=-\infty }^{\infty }S(k)}$

對於二者通過周期求和（英語：Periodic summation）而得到的周期函數：

${\displaystyle s_{_{P}}(x)\triangleq \sum _{n=-\infty }^{\infty }s(x+nP),\quad n\in \mathbb {Z} }$

${\displaystyle S_{1/T}(f)\triangleq \sum _{k=-\infty }^{\infty }S(f+k/T),\quad k,T\in \mathbb {Z} }$

這裡的參數${\displaystyle T>0}$並且${\displaystyle P>0}$，它們有著同${\displaystyle x}$一樣的單位。有如下普遍的泊松求和公式^[1][2]：

${\displaystyle s_{_{P}}(x)=\sum _{k=-\infty }^{\infty }\underbrace {{\frac {1}{P}}\cdot S(k/P)} _{S[k]}\ e^{i2\pi {\frac {k}{P}}x},\quad k\in \mathbb {Z} }$

這是一個傅立葉級數展開，其係數是函數${\displaystyle S(f)}$的採樣。還有：

${\displaystyle S_{1/T}(f)=\sum _{n=-\infty }^{\infty }\underbrace {T\cdot s(nT)} _{s[n]}\ e^{-i2\pi nTf},\quad n,T\in \mathbb {Z} }$

這也叫做離散時間傅立葉變換。

推導泊松求和公式所需的先備公式

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

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

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

1. Pinsky, M., Introduction to Fourier Analysis and Wavelets., Brooks Cole, 2002, ISBN 978-0-534-37660-4
2. Zygmund, Antoni, Trigonometric Series 2nd, Cambridge University Press, 19681988, ISBN 978-0-521-35885-9

延伸閱讀

• Benedetto, J.J.; Zimmermann, G., Sampling multipliers and the Poisson summation formula, J. Fourier Ana. App., 1997, 3 (5) [2008-06-19], （原始內容存檔於2011-05-24）
• Gasquet, Claude; Witomski, Patrick, Fourier Analysis and Applications, Springer: 344–352, 1999, ISBN 0-387-98485-2
• Higgins, J.R., Five short stories about the cardinal series, Bull. Amer. Math. Soc., 1985, 12 (1): 45–89 [2023-10-30], doi:10.1090/S0273-0979-1985-15293-0 , （原始內容存檔於2020-08-12）