单独视角：DFT公式的推导

  由博文一维连续傅里叶变换和逆变换公式的一种推导，设$f\left(t\right)$是一个连续的时间信号，其一维连续Fourier变换可取为
$$F\left(u\right)={\int }_{-\infty }^{\infty }f\left(t\right)e^{-j2\pi ut}dt$$

步骤一
  从某一时刻（记为$0$时刻）开始，每经过时间间隔$\Delta T$就对连续函数$f\left(t\right)$采样一次，共采样$N$次。对连续函数$f\left(t\right)$等间隔采样就得到一个离散序列
$$f\left(\Delta T\right),f\left(2\Delta T\right),\cdots ,f\left(N\Delta T\right)$$
记为
$$\stackrel{⌢}{f}\left(0\right),\stackrel{⌢}{f}\left(1\right),\cdots ,\stackrel{⌢}{f}\left(N-1\right)$$
而对于其他没采样到的点，即 ∀ t ∉ { Δ T ,   2 Δ T , ⋯ N Δ T } forall tnotin left{ Delta T,text{ }2Delta T,cdots NDelta T right} ∀t∈/​{ΔT, 2ΔT,⋯NΔT}，视该点处采样的函数值
$$f\left(t\right)=0$$

步骤二
  将一维连续Fourier变换公式进行拆解
$$\begin{array}{rl}& F\left(u\right)={\int }_{-\infty }^{\infty }f\left(t\right)e^{-j2\pi ut}dt\\ & ={\int }_{-\infty }^0+{\int }_0^{N\Delta T}+{\int }_{N\Delta T}^{\infty }f\left(t\right)e^{-j2\pi ut}dt\\ & ={\int }_{-\infty }^0+{\int }_{N\Delta T}^{\infty }f\left(t\right)e^{-j2\pi ut}dt+{\int }_0^{N\Delta T}f\left(t\right)e^{-j2\pi ut}dt\\ & =0+{\int }_0^{N\Delta T}f\left(t\right)e^{-j2\pi ut}dt\\ & ={\int }_0^{N\Delta T}f\left(t\right)e^{-j2\pi ut}dt\\ & =F_1\left(u\right)\end{array}$$

步骤三
  将$F_1\left(u\right)$转换为黎曼和形式$F_2\left(u\right)$，具体地说，把区间$\left[0,N\Delta T\right]$等间隔分成$N$份，第$i$个区间$k_i=\left[\left(i-1\right)\Delta T,i\Delta T\right]$，$i=1,2,\cdots ,N$，每个区间长度为$\Delta T$。在每个区间$k_i$上取右端点的函数值$f\left(i\Delta T\right)e^{-j2\pi u\cdot \left(i\Delta T\right)}$。
$$\begin{array}{rl}& F_2\left(u\right)=\sum _{i=1}^N\left[f\left(i\Delta T\right)e^{-j2\pi u\cdot \left(i\Delta T\right)}\right]\cdot \Delta T\\ & =\sum _{n=0}^{N-1}\left[f\left(\left(n+1\right)\Delta T\right)e^{-j2\pi u\cdot \left(n+1\right)\Delta T}\right]\cdot \Delta T\\ & =\sum _{n=0}^{N-1}\left[\stackrel{⌢}{f}\left(n\right)e^{-j2\pi u\cdot \left(n+1\right)\Delta T}\right]\cdot \Delta T\end{array}$$

步骤四
  若规定采样开始点和结束点之间的时长是1个单位时间，即
$$N\Delta T=1$$
则有
$$F_2\left(u\right)=\frac{1}{N}\sum _{n=0}^{N-1}\stackrel{⌢}{f}\left(n\right)e^{-j2\pi u\frac{n+1}{N}},u=0,1,\cdots ,N-1$$
这就可定义为DFT的一条公式。

其他形式
  在取相同的一维连续Fourier变换公式的基础上，更改步骤一如下。
步骤一
  从某一时刻（记为0时刻）开始，采样，然后经过时间间隔$\Delta T$才对$f\left(t\right)$采样一次，共采样$N$次。采样最后一次时再过时间$\Delta T$才终止操作。对$f\left(t\right)$等间隔采样得到一个离散序列
$$f\left(0\right),f\left(\Delta T\right),\cdots ,f\left(\left(N-1\right)\Delta T\right)$$
记为
$$\stackrel{⌢}{f}\left(0\right),\stackrel{⌢}{f}\left(1\right),\cdots ,\stackrel{⌢}{f}\left(N-1\right)$$

  原步骤二无需变动。

  原步骤三中只改动以下内容：在每个区间$k_i=\left[\left(i-1\right)\Delta T,i\Delta T\right]$上取左端点的函数值$f\left(\left(i-1\right)\Delta T\right)e^{-j2\pi u\cdot \left(i-1\right)\Delta T}$。

  在$N\Delta T=1$的限制下，就可以得到另一条公式
$$F_2\left(u\right)=\frac{1}{N}\sum _{n=0}^{N-1}\stackrel{⌢}{f}\left(n\right)e^{-j2\pi u\frac{n}{N}},u=0,1,\cdots ,N-1$$

单独视角：IDFT公式的推导

  由博文一维连续傅里叶变换和逆变换公式的一种推导，设$f\left(t\right)$是一个连续的时间信号，其一维连续Fourier逆变换可取为
$$f\left(t\right)=\frac{1}{2}{\int }_{-\infty }^{\infty }F\left(u\right)e^{j2\pi ut}du$$

  与DFT公式推导过程相似，可以得到IDFT的公式
$$f\left(n\right)=\frac{1}{2N}\sum _{u=0}^{N-1}\stackrel{⌢}{F}\left(u\right)e^{j2\pi n\frac{u+1}{N}},n=0,1,\cdots ,N-1$$
或
$$f\left(n\right)=\frac{1}{2N}\sum _{u=0}^{N-1}\stackrel{⌢}{F}\left(u\right)e^{j2\pi n\frac{u}{N}},n=0,1,\cdots ,N-1$$

综合视角：DFT, IDFT公式组的系数修正

  为了实现 对$f\left(t\right)$进行DFT，得到的结果进行IDFT还是得到$f\left(t\right)$，我们需要考虑DFT和IDFT公式的系数。令
$$\begin{array}{rl}& F\left(u\right)=c\sum _{n=0}^{N-1}f\left(n\right)e^{-j\frac{2\pi un}{N}},u=0,1,\cdots ,N-1\\ & f\left(n\right)=d\sum _{u=0}^{N-1}F\left(u\right)e^{j\frac{2\pi un}{N}},n=0,1,\cdots ,N-1\end{array}$$
设
$$\begin{array}{rl}& F={\left[F\left(0\right),F\left(1\right),\cdots ,F\left(n-1\right)\right]}^T\\ & f={\left[f\left(0\right),f\left(1\right),\cdots ,f\left(n-1\right)\right]}^T\end{array}$$
则有
$$\begin{array}{rl}& F=Af\\ & f=BF\end{array}$$
其中
$$\begin{array}{rl}& A_{N\times N}=c\left(\begin{array}{cccc}{e}^{-j\frac{2\pi }{N}\cdot 0\cdot 0}& e^{-j\frac{2\pi }{N}\cdot 0\cdot 1}& \cdots & e^{-j\frac{2\pi }{N}\cdot 0\cdot \left(N-1\right)}\\ e^{-j\frac{2\pi }{N}\cdot 1\cdot 0}& e^{-j\frac{2\pi }{N}\cdot 1\cdot 1}& \cdots & e^{-j\frac{2\pi }{N}\cdot 1\cdot \left(N-1\right)}\\ ⋮& ⋮& \ddots & ⋮\\ e^{-j\frac{2\pi }{N}\cdot \left(N-1\right)\cdot 0}& e^{-j\frac{2\pi }{N}\cdot \left(N-1\right)\cdot 1}& \cdots & e^{-j\frac{2\pi }{N}\cdot \left(N-1\right)\cdot \left(N-1\right)}\end{array}\right)\\ & B_{N\times N}=d\left(\begin{array}{cccc}{e}^{j\frac{2\pi }{N}\cdot 0\cdot 0}& e^{j\frac{2\pi }{N}\cdot 0\cdot 1}& \cdots & e^{j\frac{2\pi }{N}\cdot 0\cdot \left(N-1\right)}\\ e^{j\frac{2\pi }{N}\cdot 1\cdot 0}& e^{j\frac{2\pi }{N}\cdot 1\cdot 1}& \cdots & e^{j\frac{2\pi }{N}\cdot 1\cdot \left(N-1\right)}\\ ⋮& ⋮& \ddots & ⋮\\ e^{j\frac{2\pi }{N}\cdot \left(N-1\right)\cdot 0}& e^{j\frac{2\pi }{N}\cdot \left(N-1\right)\cdot 1}& \cdots & e^{j\frac{2\pi }{N}\cdot \left(N-1\right)\cdot \left(N-1\right)}\end{array}\right)\end{array}$$
显然有
$$\begin{array}{rl}& F=Af\\ & f=BF\end{array}\}\Rightarrow F=\left(AB\right)F$$
于是自然地想要令
  A B = I N × N ⇔ { ∀ k ∈ { 0 , 1 , ⋯   , N − 1 } ,   c d ∑ i = 0 N − 1 ( e − j 2 π N ⋅ i ⋅ k ⋅ e j 2 π N ⋅ i ⋅ k ) = c d N = 1 ∀ k 1 ≠ k 2 ,   c d ∑ i = 0 N − 1 ( e − j 2 π N ⋅ i ⋅ k 1 ⋅ e j 2 π N ⋅ i ⋅ k 2 ) = c d ∑ i = 0 N − 1 e j 2 π ( k 2 − k 1 ) N i = c d 1 − e j 2 π ( k 2 − k 1 ) N ⋅ N 1 − e j 2 π ( k 2 − k 1 ) N = 0 ⇔ c d = 1 N begin{aligned} & text{ }AB={{I}_{Ntimes N}} \ & Leftrightarrow left{ begin{aligned} & forall kin left{ 0,1,cdots ,N-1 right},text{ }cdsumlimits_{i=0}^{N-1}{left( {{e}^{-jfrac{2pi }{N}centerdot icenterdot k}}centerdot {{e}^{jfrac{2pi }{N}centerdot icenterdot k}} right)}=cdN=1 \ & forall {{k}_{1}}ne {{k}_{2}},text{ }cdsumlimits_{i=0}^{N-1}{left( {{e}^{-jfrac{2pi }{N}centerdot icenterdot {{k}_{1}}}}centerdot {{e}^{jfrac{2pi }{N}centerdot icenterdot {{k}_{2}}}} right)}=cdsumlimits_{i=0}^{N-1}{{{e}^{jfrac{2pi left( {{k}_{2}}-{{k}_{1}} right)}{N}i}}}=cdfrac{1-{{e}^{jfrac{2pi left( {{k}_{2}}-{{k}_{1}} right)}{N}centerdot N}}}{1-{{e}^{jfrac{2pi left( {{k}_{2}}-{{k}_{1}} right)}{N}}}}=0 \ end{aligned} right. \ & Leftrightarrow cd=frac{1}{N} \ end{aligned} ​ AB=IN×N​⇔⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧​​∀k∈{0,1,⋯,N−1}, cdi=0∑N−1​(e−jN2π​⋅i⋅k⋅ejN2π​⋅i⋅k)=cdN=1∀k1​​=k2​, cdi=0∑N−1​(e−jN2π​⋅i⋅k1​⋅ejN2π​⋅i⋅k2​)=cdi=0∑N−1​ejN2π(k2​−k1​)​i=cd1−ejN2π(k2​−k1​)​1−ejN2π(k2​−k1​)​⋅N​=0​⇔cd=N1​​

  因此最终我们得到DFT和IDFT的公式为
$$\begin{array}{c}F\left(u\right)=c\sum _{n=0}^{N-1}f\left(n\right)e^{-j\frac{2\pi un}{N}},u=0,1,\cdots ,N-1\\ f\left(n\right)=d\sum _{u=0}^{N-1}F\left(u\right)e^{j\frac{2\pi un}{N}},n=0,1,\cdots ,N-1\\ cd=\frac{1}{N}\end{array}$$


  此外，我们指出，对称阵$A,B$不可能为正交矩阵。事实上，有$$A{A}^T=B{B}^T=O_{N\times N}\ne I_{N\times N}$$
以$A$为例子，计算$A{A}^T$如下。
{ ∀ k ∈ { 0 , 1 , ⋯   , N − 1 } ,   c 2 ∑ i = 0 N − 1 ( e − j 2 π N ⋅ i ⋅ k ) 2 = c 2 ∑ i = 0 N − 1 e − j 4 π k N ⋅ i = c 2 1 − e − j 4 π k N ⋅ N 1 − e − j 4 π k N = 0 ∀ k 1 ≠ k 2 ,   c 2 ∑ i = 0 N − 1 ( e − j 2 π N ⋅ i ⋅ k 1 ) ( e − j 2 π N ⋅ i ⋅ k 2 ) = c 2 ∑ i = 0 N − 1 e − j 2 π ( k 1 + k 2 ) N ⋅ i = c 2 1 − e − j 2 π ( k 1 + k 2 ) N ⋅ N 1 − e − j 2 π ( k 1 + k 2 ) N = 0 ⇒   A A T = O N × N begin{aligned} & left{ begin{aligned} & forall kin left{ 0,1,cdots ,N-1 right},text{ }{{c}^{2}}sumlimits_{i=0}^{N-1}{{{left( {{e}^{-jfrac{2pi }{N}centerdot icenterdot k}} right)}^{2}}}={{c}^{2}}sumlimits_{i=0}^{N-1}{{{e}^{-jfrac{4pi k}{N}centerdot i}}}={{c}^{2}}frac{1-{{e}^{-jfrac{4pi k}{N}centerdot N}}}{1-{{e}^{-jfrac{4pi k}{N}}}}=0 \ & forall {{k}_{1}}ne {{k}_{2}},text{ }{{c}^{2}}sumlimits_{i=0}^{N-1}{left( {{e}^{-jfrac{2pi }{N}centerdot icenterdot {{k}_{1}}}} right)left( {{e}^{-jfrac{2pi }{N}centerdot icenterdot {{k}_{2}}}} right)}={{c}^{2}}sumlimits_{i=0}^{N-1}{{{e}^{-jfrac{2pi left( {{k}_{1}}+{{k}_{2}} right)}{N}centerdot i}}}={{c}^{2}}frac{1-{{e}^{-jfrac{2pi left( {{k}_{1}}+{{k}_{2}} right)}{N}centerdot N}}}{1-{{e}^{-jfrac{2pi left( {{k}_{1}}+{{k}_{2}} right)}{N}}}}=0 \ end{aligned} right. \ & \ & Rightarrow text{ }A{{A}^{T}}={{O}_{Ntimes N}} \ end{aligned} ​⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧​​∀k∈{0,1,⋯,N−1}, c2i=0∑N−1​(e−jN2π​⋅i⋅k)2=c2i=0∑N−1​e−jN4πk​⋅i=c21−e−jN4πk​1−e−jN4πk​⋅N​=0∀k1​​=k2​, c2i=0∑N−1​(e−jN2π​⋅i⋅k1​)(e−jN2π​⋅i⋅k2​)=c2i=0∑N−1​e−jN2π(k1​+k2​)​⋅i=c21−e−jN2π(k1​+k2​)​1−e−jN2π(k1​+k2​)​⋅N​=0​⇒ AAT=ON×N​​

最后

以上就是默默西装为你收集整理的一维离散傅里叶变换(DFT)和逆变换(IDFT)公式的一种推导单独视角：DFT公式的推导单独视角：IDFT公式的推导综合视角：DFT, IDFT公式组的系数修正的全部内容，希望文章能够帮你解决一维离散傅里叶变换(DFT)和逆变换(IDFT)公式的一种推导单独视角：DFT公式的推导单独视角：IDFT公式的推导综合视角：DFT, IDFT公式组的系数修正所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。