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 s ( x ) {\displaystyle s(x)} 具有傅立葉變換： S ( f ) ≜ ∫ − ∞ ∞ s ( x ) e − i 2 π f x d x {\displaystyle S(f)\triangleq \int _{-\infty }^{\infty }s(x)\ e^{-i2\pi fx}\,dx} 這裡的 S ( f ) {\displaystyle S(f)} 也可以替代表示為 s ^ ( f ) {\displaystyle {\hat {s}}(f)} 和 F { s } ( f ) {\displaystyle {\mathcal {F}}\{s\}(f)} 。有如下基本的泊松求和公式： ∑ n = − ∞ ∞ s ( n ) = ∑ k = − ∞ ∞ S ( k ) {\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： s P ( x ) ≜ ∑ n = − ∞ ∞ s ( x + n P ) , n ∈ Z {\displaystyle s_{_{P}}(x)\triangleq \sum _{n=-\infty }^{\infty }s(x+nP),\quad n\in \mathbb {Z} } S 1 / T ( f ) ≜ ∑ k = − ∞ ∞ S ( f + k / T ) , k , T ∈ 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 T > 0 {\displaystyle T>0} 並且 P > 0 {\displaystyle P>0} ，它們有著同 x {\displaystyle x} 一樣的單位。有如下普遍的泊松求和公式[1][2]： s P ( x ) = ∑ k = − ∞ ∞ 1 P ⋅ S ( k / P ) ⏟ S [ k ] e i 2 π k P x , k ∈ Z {\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 S ( f ) {\displaystyle S(f)} 的採樣。還有： S 1 / T ( f ) = ∑ n = − ∞ ∞ T ⋅ s ( n T ) ⏟ s [ n ] e − i 2 π n T f , n , T ∈ Z {\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 δ ( t ) {\displaystyle \delta (t)} ，製作一個有無限多個 δ ( t ) {\displaystyle \delta (t)} ，且間隔為 T 0 {\displaystyle T_{0}} 的週期函數 ∑ n = − ∞ ∞ δ ( t − n T 0 ) {\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT_{0})} 。 其傅立葉轉換為① ∑ n = − ∞ ∞ e − j 2 π n T 0 f = {\displaystyle \sum _{n=-\infty }^{\infty }e^{-j2\pi nT_{0}f}=} ② 1 T 0 ∑ n = − ∞ ∞ δ ( f − n T 0 ) {\displaystyle {\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }\delta \left(f-{\frac {n}{T_{0}}}\right)} 證明①轉換對 F { ∑ n = − ∞ ∞ δ ( t − n T 0 ) } {\displaystyle {\mathcal {F}}\left\{\sum _{n=-\infty }^{\infty }\delta (t-nT_{0})\right\}} = ∑ n = − ∞ ∞ F { δ ( t − n T 0 ) } {\displaystyle \sum _{n=-\infty }^{\infty }{\mathcal {F}}\left\{\delta (t-nT_{0})\right\}} = ∑ n = − ∞ ∞ e − j 2 π n T 0 f {\displaystyle \sum _{n=-\infty }^{\infty }e^{-j2\pi nT_{0}f}} 。 證明②轉換對 設 c n {\displaystyle c_{n}} 為週期函數 ∑ n = − ∞ ∞ δ ( t − n T 0 ) {\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT_{0})} 的傅立葉級數。 ∑ n = − ∞ ∞ δ ( t − n T 0 ) {\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT_{0})} 可表示為 ∑ n = − ∞ ∞ c n e j 2 π n t T 0 {\displaystyle \sum _{n=-\infty }^{\infty }c_{n}{e^{j2\pi {n{\frac {t}{T_{0}}}}}}} 。 由傅立葉級數得: c n = 1 T 0 ∫ − T 0 2 T 0 2 ∑ m = − ∞ ∞ δ ( t − m T 0 ) e − j 2 π n t T 0 d t = 1 T 0 ∫ − T 0 2 T 0 2 δ ( t ) e − j 2 π n t T 0 d t = 1 T 0 ∫ − T 0 2 T 0 2 δ ( t ) e − j 2 π n 0 T 0 d t = 1 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}}}} 。 因此， F { ∑ n = − ∞ ∞ δ ( t − n T 0 ) } = F { ∑ n = − ∞ ∞ c n e j 2 π n t T 0 } = ∑ n = − ∞ ∞ 1 T 0 F { e j 2 π n t T 0 } = 1 T 0 ∑ n = − ∞ ∞ δ ( f − n 1 T 0 ) {\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: ∑ n = − ∞ ∞ e − j 2 π n T 0 f = {\displaystyle \sum _{n=-\infty }^{\infty }e^{-j2\pi nT_{0}f}=} 1 T 0 ∑ n = − ∞ ∞ δ ( f − n T 0 ) {\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， T 0 {\displaystyle T_{0}} 以 1 T 0 {\displaystyle {\frac {1}{T_{0}}}} 代換， f {\displaystyle f} 以 t {\displaystyle t} 代換，得 ∑ n = − ∞ ∞ δ ( t − n T 0 ) = 1 T 0 ∑ n = − ∞ ∞ e − j 2 π n 1 T 0 t = 1 T 0 ∑ n = − ∞ ∞ e j 2 π n 1 T 0 t {\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 ∑ n = − ∞ ∞ x ( t − n T 0 ) {\displaystyle \sum _{n=-\infty }^{\infty }x(t-nT_{0})} = ∑ n = − ∞ ∞ x ( t ) ∗ δ ( t − n T 0 ) {\displaystyle =\sum _{n=-\infty }^{\infty }x(t)*\delta (t-nT_{0})} = x ( t ) ∗ ∑ n = − ∞ ∞ δ ( t − n T 0 ) {\displaystyle =x(t)*\sum _{n=-\infty }^{\infty }\delta (t-nT_{0})} = x ( t ) ∗ 1 T 0 ∑ n = − ∞ ∞ e j 2 π n 1 T 0 t {\displaystyle =x(t)*{\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }e^{j2\pi n{\frac {1}{T_{0}}}t}} = 1 T 0 ∑ n = − ∞ ∞ x ( t ) ∗ e j 2 π n 1 T 0 t {\displaystyle ={\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }x(t)*e^{j2\pi n{\frac {1}{T_{0}}}t}} = 1 T 0 ∑ n = − ∞ ∞ ∫ − ∞ ∞ x ( τ ) e j 2 π n 1 T 0 ( t − τ ) d τ {\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 } = 1 T 0 ∑ n = − ∞ ∞ { [ ∫ − ∞ ∞ x ( τ ) e − j 2 π n 1 T 0 τ d τ ] e j 2 π n 1 T 0 t } {\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\}} = 1 T 0 ∑ n = − ∞ ∞ X ( n T 0 ) e j 2 π n 1 T 0 t {\displaystyle ={\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }X({\frac {n}{T_{0}}})e^{j2\pi n{\frac {1}{T_{0}}}t}} 當 t = 0 {\displaystyle t=0} 時，得 ∑ n = − ∞ ∞ x ( n T 0 ) = 1 T 0 ∑ n = − ∞ ∞ X ( n T 0 ) {\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 T 0 {\displaystyle T_{0}} 為間隔做取樣，在頻域以 1 T 0 {\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 T 0 {\displaystyle T_{0}} 倍的關係。 從對時域做取樣尋找關係式 1 T 0 ∑ n = − ∞ ∞ X ( f − n T 0 ) {\displaystyle {\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }X(f-{\frac {n}{T_{0}}})} = 1 T 0 ∑ n = − ∞ ∞ X ( f ) ∗ δ ( f − n T 0 ) {\displaystyle ={\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }X(f)*\delta (f-{\frac {n}{T_{0}}})} = X ( f ) ∗ [ 1 T 0 ∑ n = − ∞ ∞ δ ( f − n T 0 ) ] {\displaystyle =X(f)*[{\frac {1}{T_{0}}}\sum _{n=-\infty }^{\infty }\delta (f-{\frac {n}{T_{0}}})]} = X ( f ) ∗ ∑ n = − ∞ ∞ e − j 2 π n T 0 f {\displaystyle =X(f)*\sum _{n=-\infty }^{\infty }e^{-j2\pi nT_{0}f}} = ∑ n = − ∞ ∞ X ( f ) ∗ e − j 2 π n T 0 f {\displaystyle =\sum _{n=-\infty }^{\infty }X(f)*e^{-j2\pi nT_{0}f}} = ∑ n = − ∞ ∞ ∫ − ∞ ∞ X ( λ ) e − j 2 π n T 0 ( f − λ ) d λ {\displaystyle =\sum _{n=-\infty }^{\infty }\int _{-\infty }^{\infty }X(\lambda )e^{-j2\pi nT_{0}(f-\lambda )}d\lambda } = ∑ n = − ∞ ∞ { [ ∫ − ∞ ∞ X ( λ ) e j 2 π n T 0 λ d λ ] e − j 2 π n T 0 f } {\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\}} = ∑ n = − ∞ ∞ x ( n T 0 ) e − j 2 π n T 0 f {\displaystyle =\sum _{n=-\infty }^{\infty }x(nT_{0})e^{-j2\pi nT_{0}f}} 當 f = 0 {\displaystyle f=0} 時，得 1 T 0 ∑ n = − ∞ ∞ X ( n T 0 ) = ∑ n = − ∞ ∞ x ( n T 0 ) {\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 T 0 {\displaystyle T_{0}} 為間隔做取樣，在頻域以 1 T 0 {\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 T 0 {\displaystyle T_{0}} 倍的關係。 綜合上述，若時域取樣間隔 T 0 = 1 {\displaystyle T_{0}=1} 時，同樣地，頻域取樣間隔 1 T 0 = 1 {\displaystyle {\frac {1}{T_{0}}}=1} 時，得泊松求和公式 ∑ n = − ∞ ∞ x ( n ) = ∑ k = − ∞ ∞ X ( k ) {\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 T 0 {\displaystyle {T_{0}}} 的週期信號 g ( t ) {\displaystyle g(t)} ， G ( f ) {\displaystyle G(f)} 為 g ( t ) {\displaystyle g(t)} 的傅立葉轉換，取出g(t)在區間 [ − T 0 2 , T 0 2 ] {\displaystyle [-{\frac {T_{0}}{2}},{\frac {T_{0}}{2}}]} 的一個完整週期 x ( t ) {\displaystyle x(t)} ，亦即 x ( t ) = g ( t ) r e c t ( t T 0 ) {\displaystyle x(t)=g(t)rect({\frac {t}{T_{0}}})} ， X ( f ) {\displaystyle X(f)} 是 x ( t ) {\displaystyle x(t)} 的傅立葉轉換，其中 r e c t ( t ) {\displaystyle rect(t)} 是矩形函數。 a n {\displaystyle a_{n}} 是 g ( t ) {\displaystyle g(t)} 的傅立葉級數。 則 G ( f ) {\displaystyle G(f)} = F { ∑ n = − ∞ ∞ a n e j 2 π n 1 T 0 t } {\displaystyle ={\mathcal {F}}\left\{\sum _{n=-\infty }^{\infty }a_{n}e^{j2\pi n{\frac {1}{T_{0}}}t}\right\}} = ∑ n = − ∞ ∞ a n δ ( f − n T 0 ) {\displaystyle =\sum _{n=-\infty }^{\infty }a_{n}\delta (f-{\frac {n}{T_{0}}})} = ∑ n = − ∞ ∞ 1 T 0 X ( n T 0 ) δ ( f − 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）Wikiwand - on Seamless Wikipedia browsing. On steroids.
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時，得
，
為間隔做取樣，在頻域以
為間隔做取樣，則兩者的所有取樣點的總和會有倍的關係。
時，同樣地，頻域取樣間隔
時，得泊松求和公式
。
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