如何利用二维FFT计算波数域坐标

Question

我有一个二维复数数组，它代表了在真实空间中沿平面测量的势场。假设数组是128个单元格乘128个单元格，平面的总面积是500mx500m。这个数组中的每个单元格表示空间域中的一个点，坐标用x和y表示

当我在这个2d数组上使用scipy.fftpack中的2d FFT时，我得到了在波域中表示的相同信息。如何计算输出阵列点的波域坐标kx和ky？

Answer 1

这里有一些代码来充分演示这个问题和我能够找到的解决方案。

from numpy import linspace , arange , reshape ,zeros
from scipy.fftpack import fft2 , fftfreq
from cmath import pi

# create some arbitrary data
some_data = arange(0.0 , 16384.0 , dtype = complex)

# reshape it to be a 128x128 2d grid
some_data_grid = reshape(some_data , (128 , 128) )

# assign some real spatial co-ordinates to the grid points   
# first define the edge values
x_min = -250.0
x_max = 250.0
y_min = -250.0
y_max = 250

# then create some empty 2d arrays to hold the individual cell values
x_array = zeros( (128,128) , dtype = float )
y_array = zeros( (128,128) , dtype = float )

# now fill the arrays with the associated values
for row , y_value in enumerate(linspace (y_min , y_max , num = 128) ):

  for column , x_value in enumerate(linspace (x_min , x_max , num = 128) ):

    x_array[row][column] = x_value
    y_array[row][column] = y_value

# now for any row,column pair the x_array and y_array hold the spatial domain
# co-ordinates of the associated point in some_data_grid

# now use the fft to transform the data to the wavenumber domain
some_data_wavedomain = fft2(some_data_grid)

# now we can use fftfreq to give us a base for the wavenumber co-ords
# this returns [0.0 , 1.0 , 2.0 , ... , 62.0 , 63.0 , -64.0 , -63.0 , ... , -2.0 , -1.0 ]
n_value = fftfreq( 128 , (1.0 / 128.0 ) )

# now we can initialize some arrays to hold the wavenumber co-ordinates of each cell
kx_array = zeros( (128,128) , dtype = float )
ky_array = zeros( (128,128) , dtype = float )

# before we can calculate the wavenumbers we need to know the total length of the spatial
# domain data in x and y. This assumes that the spatial domain units are metres and
# will result in wavenumber domain units of radians / metre.
x_length = x_max - x_min
y_length = y_max - y_min

# now the loops to calculate the wavenumbers
for row in xrange(128):

  for column in xrange(128):

    kx_array[row][column] = ( 2.0 * pi * n_value[column] ) / x_length
    ky_array[row][column] = ( 2.0 * pi * n_value[row] ) / y_length

# now for any row,column pair kx_array , and ky_array will hold the wavedomain coordinates
# of the correspoing point in some_data_wavedomain

我知道这可能不是最有效的方法，但希望很容易理解。我希望这能帮助人们避免一点挫败感。

Answer 2

嗯，如果我使用FFT函数，则DC结束于元素0，在您的情况下，频率之间的间隔是1/500m。因此，下面（不太简洁）的片段将得到您的频率轴：

meter = 1.0
L     = 500.0 * meter
N     = 128

dF    = 1.0 / L
freqs = arange(0, N/L, dF)  # array of spatial frequencies.

自然，这些频率是以每米的周期为单位，而不是以每米的弧度为单位。如果我想让kx和ky是以rads/米为单位的空间频率阵列，我只想说

 kx = 2*pi*freqs
 ky = 2*pi*freqs

（假设我已经导入了arange和pi之类的东西）。

编辑

Stu对高于Nyquist的频率提出了一个很好的观点，你可能更愿意认为它是负的（我通常是负的，但不是在代码中）。你总是可以做到：

freqs[freqs > 0.5*N/(2*L)] -= N/L

但如果你真的想要负频率，你可能也想要玩fftshift，还有另一种蠕虫。

Answer 3

我在下面的FORTRAN代码中找到了子例程，并尝试将其作为matlab函数实现。请给我评论一下我的翻译是否正确。 这是FORTRAN子程序：

subroutine kvalue(i,j,nx,ny,dkx,dky,kx,ky)
  c Subroutine KVALUE finds the wavenumber coordinates of one
  c element of a rectangular grid from subroutine FOURN.
  c
  c Input parameters:
  c i - index in the ky direction,
  c j - index in the kx direction.
  c nx - dimension of grid in ky direction (a power of two).
  c ny - dimension of grid in kx direction (a power of two).
  c dkx - sample interval in the kx direction,
  c dky - sample interval in the ky direction,
  c
  c Output parameters:
  c kx - the wavenumber coordinate in the kx direction,
  c ky - the wavenumber coordinate in the ky direction,
  c
  real kx,ky
  nyqx=nx/2+l
  nyqy=ny/2+l
  if(j.le.nyqx)then
      kx=(j-l)*dkx
  else
      kx=(j-nx-l)*dkx
  end if
  if(i.le.nyqy)then
          ky=(i-l)*dky
  else
          ky=(i-ny-l)*dky
  end if
  return
  end

我翻译成MATLAB函数：

function [kx,ky]=kvalue(gz,nx,ny,dkx,dky)
nyq_x=nx/2+1;
nyq_y=ny/2+1;
for i=1:length(gz)
    for j=1:length(gz)
        if j <= nyq_x
            kx(j)=(j-1)*dkx;
        else
            kx(j)=(j-nx-1)*dkx;
        end
        if i <= nyq_y
            ky(i)=(i-1)*dky;
        else
            ky(i)=(i-ny-1)*dky;
        end
    end
end

谢谢你。