如果 $$x(t)是周期为 $T$的周期函数,则该函数的连续时间指数傅立叶级数定义为,
$$\mathrm{ x(t)=\sum_{n=−\infty}^{\infty}C_{n}\:e^{jn\omega_{0} t}... (1)}$$
其中,$C_{n}$是指数傅立叶级数系数,由下式给出,
$$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T} x(t)\:e^{-jn\omega_{0} t}\ :dt... (2)}$$
设 $x_{1}(t)$和 $x_{2}(t)$两个周期为 T 且傅立叶级数系数为 $C_{n}$和 $D_{n}$的复周期函数。
如果,
$$\mathrm{x_{1}(t)\overset{FT}{\leftrightarrow}C_{n}}$$
$$\mathrm{x_{2}(t)\overset{FT}{\leftrightarrow}D_{n}}$$
然后,连续时间傅立叶级数的Parseval 定理指出
$$\mathrm{\frac{1}{T} \int_{t_{0}}^{t_{0}+T} x_{1}(t)\:x_{2}^{*}(t) \:dt =\sum_{n=−\infty}^{\infty} C_{n}\:D_{n}^{*}\:[for\:complex\: x_{1}(t)\: \& \: x_{2}(t)] ... (3)}$$
并且parseval 的傅立叶级数恒等式表明,如果
$$\mathrm{x_{1}(t)=x_{1}(t)= x(t)}$$
然后,
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}| x(t)|^{2}\:dt=\sum_{n=−\infty}^{\infty}|C_{n}|^{2}... (4)}$$
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t) \:dt=\sum_{n=−\infty}^{\infty}C_{n}\:D_{n}^{*}... (5)}$$
根据傅里叶级数的定义,我们有,
$$\mathrm{LHSof\:eq.(5)=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{ 2}^{*}(t)\:dt}$$
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}\left ( \sum_{n=−\infty}^{\infty}C_{n }e^{jn\omega_{0} t} \right )x_{2}^{*}(t)\:dt... (6)}$$
重新排列方程 (6) 的 RHS 中积分和求和的顺序,我们得到,
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t) \:dt=\sum_{n=−\infty}^{\infty}C_{n}\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T} x_{2}^{*}(t)e^{jn\omega_{0} t}\:dt \right )}$$
$$\mathrm{\Rightarrow\:\frac{1}{T}\int_{0}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t )\:dt=\sum_{n=−\infty}^{\infty}C_{n}\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T }x_{2}(t)e^{-jn\omega_{0} t}\:dt \right )^{*}... (7)}$$
将等式(7)与等式进行比较。(2),我们可以写,
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t) \:dt=\sum_{n=−\infty}^{\infty}C_{n}\:D_{n}^{*}... (8)\:\:(Hence,\:Proved)}$$
证明——帕塞瓦尔的身份
如果,
$$\mathrm{x_{1}(t)=x_{2}(t)= x(t)}$$
然后,Parseval 的关系变为,
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T} x(t)\:x^{*}(t)\:dt=\sum_{n=− \infty}^{\infty}C_{n}\:C_{n}^{*}... (9)}$$
$$\mathrm{\because\: x(t)\:x^{*}(t)=| x(t)|^{2}\:and\:C_{n}\:C_{n}^{*}=|C_{n}|^{2}}$$
现在,将这些值代入方程(9),我们得到,
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}| x(t)|^{2}\:dt=\sum_{n=−\infty}^{\infty}|C_{n}|^{2}... (10) \:\:(Hence,\:Proved)}$$