泊松求和公式

${\displaystyle x(t)}$ 是一个连续时间的信号，做无限次的周期复制之后，产生${\displaystyle \sum _{n=-\infty }^{\infty }x(t-nT_{0})}$ ，可由此推导出泊松求和公式。

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

考虑狄拉克δ函数${\displaystyle \delta (t)}$ ，制作一个有无限多个${\displaystyle \delta (t)}$ ，且间隔为${\displaystyle T_{0}}$ 的周期函数${\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT_{0})}$ 。

其傅立叶转换为①${\displaystyle \sum _{n=-\infty }^{\infty }e^{-j2\pi nT_{0}f}=}$ ②${\displaystyle {\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }\delta \left(f-{\frac {n}{T_{0}}}\right)}$ 

证明①转换对

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

证明②转换对

设${\displaystyle c_{n}}$ 为周期函数${\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT_{0})}$ 的傅立叶级数。

${\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT_{0})}$ 可表示为${\displaystyle \sum _{n=-\infty }^{\infty }c_{n}{e^{j2\pi {n{\frac {t}{T_{0}}}}}}}$ 。

由傅立叶级数得:

${\displaystyle c_{n}={\frac {1}{T_{0}}}\int _{-{\frac {T_{0}}{2}}}^{\frac {T_{0}}{2}}\sum _{m=-\infty }^{\infty }\delta (t-mT_{0})e^{-j2\pi {n}{\frac {t}{T_{0}}}}dt={\frac {1}{T_{0}}}\int _{-{\frac {T_{0}}{2}}}^{\frac {T_{0}}{2}}\delta (t)e^{-j2\pi {n}{\frac {t}{T_{0}}}}dt={\frac {1}{T_{0}}}\int _{-{\frac {T_{0}}{2}}}^{\frac {T_{0}}{2}}\delta (t)e^{-j2\pi {n}{\frac {0}{T_{0}}}}dt={\frac {1}{T_{0}}}}$ 。

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:${\displaystyle \sum _{n=-\infty }^{\infty }e^{-j2\pi nT_{0}f}=}$ ${\displaystyle {\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }\delta \left(f-{\frac {n}{T_{0}}}\right)}$ ，

经由适当的变量代换，${\displaystyle T_{0}}$ 以${\displaystyle {\frac {1}{T_{0}}}}$ 代换，${\displaystyle f}$ 以${\displaystyle t}$ 代换，得${\displaystyle \sum _{n=-\infty }^{\infty }\delta \left(t-nT_{0}\right)={\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }e^{-j2\pi n{\frac {1}{T_{0}}}t}={\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }e^{j2\pi n{\frac {1}{T_{0}}}t}}$ (因为n从负无限大到正无限大)

从对频域做取样寻找关系式

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

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

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

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

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

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

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

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

当${\displaystyle t=0}$ 时，得${\displaystyle \sum _{n=-\infty }^{\infty }x(nT_{0})={\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }X({\frac {n}{T_{0}}})}$ ，

表示一个信号的在时域以${\displaystyle T_{0}}$ 为间隔做取样，在频域以${\displaystyle {\frac {1}{T_{0}}}}$ 为间隔做取样，则两者的所有取样点的总和会有${\displaystyle T_{0}}$ 倍的关系。

从对时域做取样寻找关系式

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

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

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

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

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

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

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

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

当${\displaystyle f=0}$ 时，得${\displaystyle {\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }X({\frac {n}{T_{0}}})=\sum _{n=-\infty }^{\infty }x(nT_{0})}$ ，

表示一个信号的在时域以${\displaystyle T_{0}}$ 为间隔做取样，在频域以${\displaystyle {\frac {1}{T_{0}}}}$ 为间隔做取样，则两者的所有取样点的总和会有${\displaystyle T_{0}}$ 倍的关系。

综合上述，若时域取样间隔${\displaystyle T_{0}=1}$ 时，同样地，频域取样间隔${\displaystyle {\frac {1}{T_{0}}}=1}$ 时，得泊松求和公式${\displaystyle \sum _{n=-\infty }^{\infty }x(n)=\sum _{k=-\infty }^{\infty }X(k)}$ 。

考虑一个周期为${\displaystyle {T_{0}}}$ 的周期信号${\displaystyle g(t)}$ ，${\displaystyle G(f)}$ 为${\displaystyle g(t)}$ 的傅立叶转换，取出g(t)在区间${\displaystyle [-{\frac {T_{0}}{2}},{\frac {T_{0}}{2}}]}$ 的一个完整周期${\displaystyle x(t)}$ ，亦即${\displaystyle x(t)=g(t)rect({\frac {t}{T_{0}}})}$ ，${\displaystyle X(f)}$ 是${\displaystyle x(t)}$ 的傅立叶转换，其中${\displaystyle rect(t)}$ 是矩形函数。${\displaystyle a_{n}}$ 是${\displaystyle g(t)}$ 的傅立叶级数。

则${\displaystyle G(f)}$ 

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

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

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

得出一周期信号的傅立叶转换与其傅立叶级数之间的关系。