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。

公式的形式

${\displaystyle x(t)}$ 是一個連續時間的信號，做無限次的週期複製之後，產生${\displaystyle \sum _{n=-\infty }^{\infty }x(t-nT_{0})}$ ，可由此推導出泊松求和公式。

泊松求和公式陳述${\displaystyle \sum _{n=-\infty }^{\infty }x(n)=\sum _{k=-\infty }^{\infty }X(k)}$ 。其中${\displaystyle X(f)={\mathcal {F}}\left\{x(t)\right\}}$ 。

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。