傅里叶变换推导到FFT编程（二）、由FT到DFT

（一）、傅里叶变换推导
（二）、由FT到DFT（当前）
（三）、STM32编程实现DFT和验证
（四）、由DFT到FFT
（五）、STM32编程实现FFT和验证

• DTFT推导$\to$时间离散：$t\to n\Delta t$, 采样频率为$f_s$
  $$\begin{gathered} 采样序列{\delta }_s(t)=\sum _{n=-\infty }^{\infty }\delta (t-n{T}_s),T_s=\frac{1}{f_s} \\ 对f(t)采样得f_s(t)=f(t){\delta }_s(t)=\sum _{n=-\infty }^{\infty }f(t)\delta (t-n{T}_s) \\ {f}_s(t)做傅里叶变换\mathcal{F}[f_s(t)]={\int }_{-\infty }^{\infty }\sum _{n=-\infty }^{\infty }f(t)\delta (t-n{T}_s)e^{-i\omega t}dt \\ \because {\int }_{-\infty }^{\infty }x(t)\delta (t-t_0)dt=x(t_0) \\ \therefore DTFT[f_s(t)]=\mathcal{F}[f_s(t)]=\sum _{n=-\infty }^{\infty }f(n{T}_s)e^{-i\omega n{T}_s} \end{gathered}$$
• DFT推导$\to$周期延拓：
  • $截取f(t)的N个采样点,采样频率为f_s$
  • $周期延拓\to N个采样点的周期为T=N{T}_s(T_s=\frac{1}{f_s})$
  • $采样信号表示为x(t)={\sum }_{n=0}^{N-1}f(t)\delta (t-n{T}_s)$
  • $因为周期延拓，x(t)为周期信号，代入周期函数的傅里叶级数$
    $$\begin{gathered} c_k=\frac{1}{T}{\int }_0^{T}f(t)e^{-ik\omega t}dt,(\omega =\frac{2\pi }{T}),得 \\ X[k\omega ]=\frac{1}{T}{\int }_0^T\sum _{n=0}^{N-1}f(t)\delta (t-n{T}_s)e^{-ik\omega t}dt \\ \because {\int }_{-\infty }^{\infty }x(t)\delta (t-t_0)dt=x(t_0) \\ \therefore X[k\omega ]=\frac{1}{T}\sum _{n=0}^{N-1}f(n{T}_s)e^{-ik\omega n{T}_s} \\ 又\because T=N{T}_s,\omega =\frac{2\pi }{T}=\frac{2\pi }{N{T}_s} \\ \therefore X[k\omega ]=\frac{1}{N{T}_s}\sum _{n=0}^{N-1}f(n{T}_s)e^{-i\frac{2\pi }{N}kn} \\ 令x[n]=f(n{T}_s),X[k\omega ]T_s=X[k],得 \\ DFT[x(n)]=X[k]=\sum _{n=0}^{N-1}x[n]e^{-i\frac{2\pi }{N}kn}(0\le k,n\le N-1,且k,n\in Z) \\ IDFT[X(k)]=\frac{1}{N}X[k]e^{i\frac{2\pi }{N}kn}(0\le k,n\le N-1,且k,n\in Z) \end{gathered}$$
  • $当N确定后,e^{-i\frac{2\pi }{N}}为定值，令W_N=e^{-i\frac{2\pi }{N}},则DFT为$
    $$X[k]=\sum _{n=0}^{N-1}x[n]W_N^{kn}$$