泊松求和公式

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

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)

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

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

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

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

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

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

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

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

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。

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

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

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

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

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

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

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

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

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。

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${\displaystyle \sum _{n=-\infty }^{\infty }x(n)=\sum _{k=-\infty }^{\infty }X(k)}$。