我有一个由复数组成的二维数组,表示在现实空间中沿平面测量的潜力场。假设该数组为128个单元格乘以128个单元格,平面的总面积为500米乘以500米。该数组中的每个单元格都代表空间域中具有x和y坐标的点。
当我对这个二维数组使用scipy.fftpack中的2d FFT时,我得到了相同信息在波域中的表示。如何计算输出数组中的点的波域坐标kx和ky?
当我对这个二维数组使用scipy.fftpack中的2d FFT时,我得到了相同信息在波域中的表示。如何计算输出数组中的点的波域坐标kx和ky?
下面是一些代码,完整演示了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,但希望它易于理解。我希望这能帮助某人避免一些挫败感。
meter = 1.0
L = 500.0 * meter
N = 128
dF = 1.0 / L
freqs = arange(0, N/L, dF) # array of spatial frequencies.
自然地,这些频率以每米循环为单位,而不是弧度/米。如果我想让kx和ky成为弧度/米的空间频率数组,我只需要这样说:
kx = 2*pi*freqs
ky = 2*pi*freqs
(假设我已经导入了arange和pi等内容)。
编辑
Stu提出了一个关于超过奈奎斯特频率的频率的好观点,你可能更喜欢将其视为负数(我通常这样做,但代码中不是这样)。你可以随时执行以下操作:
freqs[freqs > 0.5*N/(2*L)] -= N/L
但是如果你真的想要负频率,你可能也想尝试一下fftshift -- 另一个棘手的问题。
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
我将其翻译为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。
fftfreq()
函数时,感觉非常幸福! - PhilMacKay