傅里叶级数的推导过程

推导过程:

F m = ∑ n = 0 N − 1 f n e − 2 π i m n / N ↔ f n = 1 N ∑ m = 0 N − 1 F m e 2 π i m n / N F_m=\sum_{n=0}^{N-1}f_ne^{-2\pi imn/N}\leftrightarrow f_n=\frac{1}{N}\sum_{m=0}^{N-1}F_me^{2\pi imn/N} Fm=n=0∑N−1fne−2πimn/N↔fn=N1m=0∑N−1Fme2πimn/N

f ( x ) = ∑ n = 0 N − 1 f n δ ( x − x n ) f(x)=\sum_{n=0}^{N-1}f_n\delta(x-x_n) f(x)=n=0∑N−1fnδ(x−xn)

F m = ∫ − T T ∑ n = 0 N − 1 f n δ ( x − x n ) e − i x k m d x = ∑ n = 0 N − 1 ∫ f n δ ( x − x n ) e − i x k m d x = ∑ n = 0 N − 1 f n e − i x n k m \begin{aligned} F_m &= \int_{-T}^{T}\sum_{n=0}^{N-1}f_n\delta(x-x_n)e^{-ixk_m}dx \\&=\sum_{n=0}^{N-1}\int f_n\delta(x-x_n)e^{-ixk_m}dx \\&=\sum_{n=0}^{N-1}f_ne^{-ix_nk_m} \end{aligned} Fm=∫−TTn=0∑N−1fnδ(x−xn)e−ixkmdx=n=0∑N−1∫fnδ(x−xn)e−ixkmdx=n=0∑N−1fne−ixnkm

接下来我们假设 d x , d k dx,dk dx,dk分别是 { x n } \{x_n\} {xn}, { k n } \{k_n\} {kn}的间距，那么：

x n = n d x , k m = m d k x_n=ndx,\qquad k_m = mdk xn=ndx,km=mdk

代入上式：

F m = ∑ n = 0 N − 1 f n e − i x n k m = ∑ n = 0 N − 1 f n e − i m n d x d k \begin{aligned} F_m &=\sum_{n=0}^{N-1}f_ne^{-ix_nk_m} \\&=\sum_{n=0}^{N-1}f_ne^{-imndxdk} \end{aligned} Fm=n=0∑N−1fne−ixnkm=n=0∑N−1fne−imndxdk