离散非周期信号傅里叶变换 — Fourier Transform of Discrete-Time Aperiodic Signals

离散非周期傅里叶变换的思想就是将非周期信号拼接成为周期的离散信号来处理。如下图所示:
在这里插入图片描述
离散非周期信号 x [ n ] x[n] x[n]拼接成周期信号 x ~ [ n ] \widetilde{x}[n] x [n] n 2 − n 1 = N n2-n1=N n2n1=N。从而我们可以得到 x ~ [ n ] \widetilde{x}[n] x [n]的傅立叶级数表示:
x ~ [ n ] = ∑ k = < N > X k e i k 2 π N n \widetilde{x}[n] = \sum_{k=<N>}^{} X_{k}e^{ik\frac{2\pi}{N}n} x [n]=k=<N>XkeikN2πn

其中, X k = 1 N ∑ n = < N > x ~ [ n ] e − i k 2 π N n X_{k} = \frac{1}{N} \sum_{n=<N>}^{}\widetilde{x}[n]e^{-ik\frac{2\pi}{N}n} Xk=N1n=<N>x [n]eikN2πn。由于 n 2 − n 1 = N n2-n1=N n2n1=N,且 x [ n ] = x ~ [ n ] x[n] = \widetilde{x}[n] x[n]=x [n]在区间 [ n 1 , n 2 ] [n1,n2] [n1,n2],而对于区间 [ n 1 , n 2 ] [n1,n2] [n1,n2]之外的值, x [ n ] = 0 x[n]=0 x[n]=0。所以得:
X k = 1 N ∑ n = n 1 n 2 x [ n ] e − i k 2 π N n = 1 N ∑ n = − ∞ ∞ x [ n ] e − i k 2 π N n X_{k} = \frac{1}{N} \sum_{n=n1}^{n2}x[n]e^{-ik\frac{2\pi}{N}n} = \frac{1}{N} \sum_{n=-\infty}^{\infty}x[n]e^{-ik\frac{2\pi}{N}n} Xk=N1n=n1n2x[n]eikN2πn=N1n=x[n]eikN2πn

X ( e i w ) = ∑ n = − ∞ ∞ x [ n ] e − i w n X(e^{iw}) = \sum_{n=-\infty}^{\infty}x[n]e^{-iwn} X(eiw)=n=x[n]eiwn,其中 w = k w 0 = k 2 π N w = kw_{0} = k\frac{2\pi}{N} w=kw0=kN2π。得:
X k = 1 N X ( e i k w 0 ) X_{k} = \frac{1}{N}X(e^{ikw_{0}}) Xk=N1X(eikw0)

将上式代入 x ~ [ n ] \widetilde{x}[n] x [n]的傅立叶级数,得:
x ~ [ n ] = ∑ k = < N > 1 N X ( e i k w 0 ) e i k 2 π N n = 1 2 π ∑ k = < N > X ( e i k w 0 ) e i k w 0 n w 0 \begin{aligned} & \widetilde{x}[n] = \sum_{k=<N>}^{} \frac{1}{N}X(e^{ikw_{0}})e^{ik\frac{2\pi}{N}n} \\ & = \frac{1}{2\pi} \sum_{k=<N>}^{} X(e^{ikw_{0}})e^{ikw_{0}n}w_{0} \end{aligned} x [n]=k=<N>N1X(eikw0)eikN2πn=2π1k=<N>X(eikw0)eikw0nw0

另因为 k k k为整数, d k dk dk表示 k k k值一次的变化量,所以 d k = 1 dk=1 dk=1。那么 ∑ k = < N > … d k \sum_{k=<N>}^{} \dots dk k=<N>dk等价于 ∫ k = < N > … d k \int_{k=<N>}^{} \dots dk k=<N>dk。因此,上式继续推导:
x ~ [ n ] = 1 2 π ∑ k = < N > X ( e i k w 0 ) e i k w 0 n w 0 = 1 2 π ∫ k = < N > X ( e i k w 0 ) e i k w 0 n w 0 d k = 1 2 π ∫ 2 π X ( e i w ) e i w n d w \begin{aligned} & \widetilde{x}[n] = \frac{1}{2\pi} \sum_{k=<N>}^{} X(e^{ikw_{0}})e^{ikw_{0}n}w_{0} \\ & = \frac{1}{2\pi} \int_{k=<N>}^{} X(e^{ikw_{0}})e^{ikw_{0}n}w_{0}dk \\ & = \frac{1}{2\pi} \int_{2\pi}^{} X(e^{iw})e^{iwn}dw \end{aligned} x [n]=2π1k=<N>X(eikw0)eikw0nw0=2π1k=<N>X(eikw0)eikw0nw0dk=2π12πX(eiw)eiwndw

N → ∞ N \rightarrow \infty N时, x ~ [ n ] = x [ n ] \widetilde{x}[n] = x[n] x [n]=x[n]。至此,我们已经得到离散信号 x [ n ] x[n] x[n]的傅里叶变换 X ( e i w ) X(e^{iw}) X(eiw),且信号 x [ n ] x[n] x[n] X ( e i w ) X(e^{iw}) X(eiw)得逆变换。如下:
{ x [ n ] = 1 2 π ∫ 2 π X ( e i w ) e i w n d w X ( e i w ) = ∑ n = − ∞ ∞ x [ n ] e − i w n \begin{cases} & x[n] = \frac{1}{2\pi} \int_{2\pi}^{} X(e^{iw})e^{iwn}dw \\ & X(e^{iw}) = \sum_{n=-\infty}^{\infty}x[n]e^{-iwn} \end{cases} {x[n]=2π12πX(eiw)eiwndwX(eiw)=n=x[n]eiwn

2维离散非周期傅里叶变换

{ x [ m , n ] = 1 ( 2 π ) 2 ∫ 2 π ∫ 2 π X ( e i w 1 , e i w 2 ) e i ( w 1 m + w 2 n ) d w 1 d w 2 X ( e i w 1 , e i w 2 ) = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ x [ m , n ] e − i ( w 1 m + w 2 n ) \begin{cases} & x[m,n] = \frac{1}{(2\pi)^2} \int_{2\pi}^{}\int_{2\pi}^{} X(e^{iw_{1}},e^{iw_{2}})e^{i(w_{1}m+w_{2}n)}dw_{1}dw_{2} \\ & X(e^{iw_{1}},e^{iw_{2}}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}x[m,n]e^{-i(w_{1}m+w_{2}n)} \end{cases} {x[m,n]=(2π)212π2πX(eiw1,eiw2)ei(w1m+w2n)dw1dw2X(eiw1,eiw2)=m=n=x[m,n]ei(w1m+w2n)

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