[Fourier]傅里叶级数中虚数部分j去了哪里

一、本文前提

已知欧拉公式为:
ejθ=cosθ+j⋅sinθe^{j\theta}=cos\theta+j\cdot sin\thetaejθ=cosθ+j⋅sinθ
傅里叶级数公式为:
f(t)=∑n=−∞n=+∞Cn⋅ej⋅2πntTf(t)=\sum_{n=-\infty}^{n=+\infty}C_{n}\cdot e^\frac{j\cdot2\pi nt}{T}f(t)=n=−∞∑n=+∞Cn⋅eTj⋅2πnt
其中:
Cn=1T∫−T2T2f(t)e−j⋅2πntTdtC_{n}=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-\frac{j\cdot2\pi nt}{T}}d_tCn=T1∫−2T2Tf(t)e−Tj⋅2πntdt
本文不对上述公式做出证明，而是默认读者已了解上述公式，并且掌握实数、虚数的概念。
本文还用到基础的积分公式、奇偶函数积分的性质，默认读者已经掌握此类知识。

二、问题提出

~~~~    对于傅里叶级数公式f(t)=∑n=−∞n=+∞Cn⋅ej⋅2πntTf(t)=\sum_{n=-\infty}^{n=+\infty}C_{n}\cdot e^\frac{j\cdot2\pi nt}{T}f(t)=∑n=−∞n=+∞Cn⋅eTj⋅2πnt，其中左侧f(t)f(t)f(t)定义域与值域都是实数，而右侧包含ej⋅2πntTe^\frac{j\cdot2\pi nt}{T}eTj⋅2πnt，是一个虚数型式，那么为什么右侧的虚数jjj去了哪里？
~~~~    我们可以想到虚数部分很可能在公式∑\sum∑部分相互约掉了，那么是怎么约掉的呢？下面本文将推导证明虚数约去的过程。
~~~~    即证明对于任意的实数t0t_{0}t0，f(t0)=∑n=−∞n=+∞Cn⋅ej⋅2πnt0Tf(t_{0})=\sum_{n=-\infty}^{n=+\infty}C_{n}\cdot e^\frac{j\cdot2\pi nt_{0}}{T}f(t0)=∑n=−∞n=+∞Cn⋅eTj⋅2πnt0的虚部为000。

三、证明过程

对于任意的实数t0t_{0}t0，
f(t0)=∑n=−∞n=+∞Cn⋅ej⋅2πnt0Tf(t_{0})=\sum_{n=-\infty}^{n=+\infty}C_{n}\cdot e^\frac{j\cdot2\pi nt_{0}}{T}f(t0)=n=−∞∑n=+∞Cn⋅eTj⋅2πnt0
我们令
h(t0,n)=Cn⋅ej⋅2πnt0Th(t_{0},n)=C_{n}\cdot e^\frac{j\cdot2\pi nt_{0}}{T}h(t0,n)=Cn⋅eTj⋅2πnt0
那么有
f(t0)=h(t0,−∞)+...+h(t0,0)+...+h(t0,+∞)f(t_{0})=h(t_{0},-\infty)+...+h(t_{0},0)+...+h(t_{0},+\infty)f(t0)=h(t0,−∞)+...+h(t0,0)+...+h(t0,+∞)
我们只要证明h(t0,−n0)h(t_{0},-n_{0})h(t0,−n0)+h(t0,n0)h(t_{0},n_{0})h(t0,n0)的虚部为000即可以证明f(t0)f(t_{0})f(t0)的虚部为000。
A=B+C=C+D+C=2C+D\begin{aligned} A &= B + C \\ &= C + D + C \\ &= 2C + D \end{aligned}A=B+C=C+D+C=2C+D
h(t0,n0)=Cn0⋅ej⋅2πn0t0T=Cn0⋅(cos(2πn0t0T)+j⋅sin(2πn0t0T))=(1T∫−T2T2f(t)e−j⋅2πn0tTdt)⋅(cos(2πn0t0T)+j⋅sin(2πn0t0T))\begin{aligned} h(t_{0},n_{0})&=C_{n_{0}}\cdot e^\frac{j\cdot2\pi n_{0}t_{0}}{T}\\ &=C_{n_{0}}\cdot(cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot sin(\frac{2\pi n_{0}t_{0}}{T})) \\ &=(\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-\frac{j\cdot2\pi n_{0}t}{T}}d_t)\cdot(cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot sin(\frac{2\pi n_{0}t_{0}}{T})) \end{aligned} h(t0,n0)=Cn0⋅eTj⋅2πn0t0=Cn0⋅(cos(T2πn0t0)+j⋅sin(T2πn0t0))=(T1∫−2T2Tf(t)e−Tj⋅2πn0tdt)⋅(cos(T2πn0t0)+j⋅sin(T2πn0t0))
因为公式中的n0n_{0}n0，t0t_{0}t0，TTT和π\piπ都是一个确定的值，因此(cos(2πn0t0T)+j⋅sin(2πn0t0T))(cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot sin(\frac{2\pi n_{0}t_{0}}{T}))(cos(T2πn0t0)+j⋅sin(T2πn0t0))部分是一个确定的复数，可以将其写入到前面的积分中，那么h(t0,n0)h(t_{0},n_{0})h(t0,n0)可以写成:
h(t0,n0)=1T∫−T2T2f(t)e−j⋅2πn0tT⋅(cos(2πn0t0T)+j⋅sin(2πn0t0T))dt=1T∫−T2T2f(t)⋅(cos(2πn0tT)−j⋅sin(2πn0tT))⋅(cos(2πn0t0T)+j⋅sin(2πn0t0T))dt=1T∫−T2T2f(t)⋅(cos(2πn0tT)cos(2πn0t0T)−j⋅sin(2πn0tT)cos(2πn0t0T)+j⋅cos(2πn0tT)sin(2πn0t0T)−(j)2⋅sin(2πn0tT)sin(2πn0t0T))dt\begin{aligned} h(t_{0},n_{0}) &= \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-\frac{j\cdot2\pi n_{0}t}{T}}\cdot(cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot sin(\frac{2\pi n_{0}t_{0}}{T}))d_{t}\\ &= \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\cdot (cos(\frac{2\pi n_{0}t}{T})-j\cdot sin(\frac{2\pi n_{0}t}{T}))\cdot(cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot sin(\frac{2\pi n_{0}t_{0}}{T}))d_{t}\\ &= \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\cdot (cos(\frac{2\pi n_{0}t}{T})cos(\frac{2\pi n_{0}t_{0}}{T})-j\cdot sin(\frac{2\pi n_{0}t}{T})cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot cos(\frac{2\pi n_{0}t}{T})sin(\frac{2\pi n_{0}t_{0}}{T}) -(j)^{2}\cdot sin(\frac{2\pi n_{0}t}{T})sin(\frac{2\pi n_{0}t_{0}}{T}) )d_{t}\\ \end{aligned} h(t0,n0)=T1∫−2T2Tf(t)e−Tj⋅2πn0t⋅(cos(T2πn0t0)+j⋅sin(T2πn0t0))dt=T1∫−2T2Tf(t)⋅(cos(T2πn0t)−j⋅sin(T2πn0t))⋅(cos(T2πn0t0)+j⋅sin(T2πn0t0))dt=T1∫−2T2Tf(t)⋅(cos(T2πn0t)cos(T2πn0t0)−j⋅sin(T2πn0t)cos(T2πn0t0)+j⋅cos(T2πn0t)sin(T2πn0t0)−(j)2⋅sin(T2πn0t)sin(T2πn0t0))dt
因为h(t0,n0)h(t_{0},n_{0})h(t0,n0)的积分区域为∫−T2T2\int_{-\frac{T}{2}}^{\frac{T}{2}}∫−2T2T因此对于奇函数积分为000。h(t0,n0)h(t_{0},n_{0})h(t0,n0)中含有
sin(2πn0tT)sin(\frac{2\pi n_{0}t}{T})sin(T2πn0t)部分的积分后等于000，因此：
h(t0,n0)=1T∫−T2T2f(t)⋅(cos(2πn0tT)cos(2πn0t0T)−j⋅sin(2πn0tT)cos(2πn0t0T)+j⋅cos(2πn0tT)sin(2πn0t0T)−(j)2⋅sin(2πn0tT)sin(2πn0t0T))dt=1T∫−T2T2f(t)⋅(cos(2πn0tT)cos(2πn0t0T)+j⋅cos(2πn0tT)sin(2πn0t0T))dt\begin{aligned} h(t_{0},n_{0})&= \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\cdot (cos(\frac{2\pi n_{0}t}{T})cos(\frac{2\pi n_{0}t_{0}}{T})-j\cdot sin(\frac{2\pi n_{0}t}{T})cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot cos(\frac{2\pi n_{0}t}{T})sin(\frac{2\pi n_{0}t_{0}}{T}) -(j)^{2}\cdot sin(\frac{2\pi n_{0}t}{T})sin(\frac{2\pi n_{0}t_{0}}{T}) )d_{t}\\ &= \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\cdot (cos(\frac{2\pi n_{0}t}{T})cos(\frac{2\pi n_{0}t_{0}}{T})+j\cdot cos(\frac{2\pi n_{0}t}{T})sin(\frac{2\pi n_{0}t_{0}}{T}) )d_{t} \end{aligned} h(t0,n0)=T1∫−2T2Tf(t)⋅(cos(T2πn0t)cos(T2πn0t0)−j⋅sin(T2πn0t)cos(T2πn0t0)+j⋅cos(T2πn0t)sin(T2πn0t0)−(j)2⋅sin(T2πn0t)sin(T2πn0t0))dt=T1∫−2T2Tf(t)⋅(cos(T2πn0t)cos(T2πn0t0)+j⋅cos(T2πn0t)sin(T2πn0t0))dt
至此h(t0,n0)h(t_{0},n_{0})h(t0,n0)的虚部只剩下了
1T∫−T2T2f(t)⋅j⋅cos(2πn0tT)sin(2πn0t0T))dt\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t) \cdot j\cdot cos(\frac{2\pi n_{0}t}{T})sin(\frac{2\pi n_{0}t_{0}}{T}) )d_{t}T1∫−2T2Tf(t)⋅j⋅cos(T2πn0t)sin(T2πn0t0))dt
将jjj和sin(2πn0t0T)sin(\frac{2\pi n_{0}t_{0}}{T})sin(T2πn0t0)写在前面即：(sin(2πn0t0T)sin(\frac{2\pi n_{0}t_{0}}{T})sin(T2πn0t0)是个确定的数值所以可以提到积分号之外)
j⋅sin(2πn0t0T)⋅1T∫−T2T2f(t)cos(2πn0tT)dtj\cdot sin(\frac{2\pi n_{0}t_{0}}{T})\cdot \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)cos(\frac{2\pi n_{0}t}{T})d_{t}j⋅sin(T2πn0t0)⋅T1∫−2T2Tf(t)cos(T2πn0t)dt
上式即为h(t0,n0)h(t_{0},n_{0})h(t0,n0)的虚数部分。

类似的h(t0,−n0)h(t_{0}, -n_{0})h(t0,−n0)的虚数部分可以写成
j⋅sin(2π(−n0)t0T)⋅1T∫−T2T2f(t)cos(2π(−n0)tT)dtj\cdot sin(\frac{2\pi (-n_{0})t_{0}}{T})\cdot \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)cos(\frac{2\pi (-n_{0})t}{T})d_{t}j⋅sin(T2π(−n0)t0)⋅T1∫−2T2Tf(t)cos(T2π(−n0)t)dt
h(t0,n0)h(t_{0},n_{0})h(t0,n0)和h(t0,−n0)h(t_{0},-n_{0})h(t0,−n0)的虚数部分相加等于000，即得证f(t0)f(t_{0})f(t0)的虚部为000，那么即证明:
f(t)=∑n=−∞n=+∞Cn⋅ej⋅2πntTf(t)=\sum_{n=-\infty}^{n=+\infty}C_{n}\cdot e^\frac{j\cdot2\pi nt}{T}f(t)=n=−∞∑n=+∞Cn⋅eTj⋅2πnt
的虚数部分为000，求证得证。