泊松求和公式

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。

推導泊松求和公式所需的先備公式

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

${\displaystyle c_{n}}$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${\displaystyle \sum _{n=-\infty }^{\infty }c_{n}{e^{j2\pi {n{\frac {t}{T_{0}}}}}}}$

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

推導泊松求和公式

從對頻域做取樣尋找關係式

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

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

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

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

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

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

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

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

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

從對時域做取樣尋找關係式

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

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

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

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

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

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

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

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

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

週期信號的傅立葉轉換

${\displaystyle G(f)}$

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

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

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