如何利用FFT进行一维反褶积？

Question

问题

我试图用卷积定理来解卷积两个测量数据A和B。我知道，对于卷积，您应该对数据进行零点处理，以防止循环卷积。然而，我感到困惑的是，零填充是否也是反褶积所必需的。

问题

1.如何正确地根据卷积定理进行反褶积？

1. 为什么下面的示例不起作用？

逼近

因为A和B是测量的，所以我为进一步的研究创造了一个例子。其思想是在模式scipy.signal.convolve中使用same创建B。

import numpy as np 
import matplotlib.pyplot as plt 
from scipy.signal import convolve
from scipy.fftpack import next_fast_len

# A, in the description above
A = np.array([1, 1, 1, 2, 1, 1])
# The result, I want to get from the deconvolution
kernel = np.array([0, 1, 2, 1, 0, 0]) 
#B, in the description above
B = convolve(kernel, data, mode='same') 

# Using the deconvolution theorem
f_A = np.fft.fft(A)
f_B = np.fft.fft(B)
# I know that you should use a regularization here 
r = f_B / f_A

# dk should be equal to kernel
dk = np.fft.ifft(r)

dk的结果是：

dk = array([ 2.28571429-9.25185854e-18j,  1.28571429+9.25185854e-18j,
       -0.71428571-9.25185854e-18j, -0.71428571+9.25185854e-18j,
        0.28571429-9.25185854e-18j,  1.28571429+9.25185854e-18j])

预计将：

dk = array([0, 1, 2, 1, 0, 0]) 

Answer 1

实际上，由于内核是以2.0为中心的1.02.01.0(模糊和膨胀)，内核宽度为3。由于0.5上的数组A是非空的，所以在-1.6上，完全转换的数组paddedB是非空的。然而，函数scipy.signal.convolve(...,'same')返回一个集群化的卷积数组B(0..5)=paddedB(0..5)。因此，如果使用E 222 np.convolve() E 124选项(e 225)，则与paddedB(-1) 和 paddedB(6) 有关的信息丢失，恢复内核变得困难。

为了避免信息丢失，输出paddedB将被填充以包含卷积信号的支持，计算为函数A的支持和内核支持的Minkowski和。选项full of np.convolve() 直接计算 paddedB 而不丢失信息。

kernel=[1,2,1]
paddedB = convolve(kernel, A, mode='full')

为了利用卷积定理提取核，输入信号A将被填充以与函数paddedB的支持相匹配。

paddedA=np.zeros(paddedB.shape[0])
paddedA[kernel.shape[0]/2: kernel.shape[0]/2+A.shape[0]]=A[:]

# Using the deconvolution theorem
f_A = np.fft.fft(paddedA)
f_B = np.fft.fft(paddedB)
# I know that you should use a regularization here 
r = f_B / f_A

# dk should be equal to kernel
dk = np.fft.ifft(r)
# shift to get zero frequency in the middle:
dk=np.fft.fftshift(dk)

注意使用函数np.fft.fftshift()来获得中间的零频率。

import numpy as np 
import matplotlib.pyplot as plt 
from scipy.signal import convolve
from scipy.fftpack import next_fast_len

# A, in the description above
A = np.array([1, 1, 1, 2, 1, 1])

kernel=np.asarray([1,2,1])
paddedB = convolve(kernel, A, mode='full')
print paddedB

paddedA=np.zeros(paddedB.shape[0])
paddedA[kernel.shape[0]/2: kernel.shape[0]/2+A.shape[0]]=A[:]
#pad both signal and kernel. Requires the size of the kernel

# Using the deconvolution theorem
f_A = np.fft.fft(paddedA)
f_B = np.fft.fft(paddedB)
# I know that you should use a regularization here 
r = f_B / f_A

# dk should be equal to kernel
dk = np.fft.ifft(r)
# shift to get zero abscissa in the middle:
dk=np.fft.fftshift(dk)

print dk

如果不可能获得paddedB，而且B是唯一可用的数据，则可以尝试通过将B填充为零或平滑B的最后值来重建paddedB。这需要对内核大小进行一些估计。

B = convolve(A,kernel, mode='same')
paddedB=np.zeros(A.shape[0]+kernel.shape[0]-1)
paddedB[kernel.shape[0]/2: kernel.shape[0]/2+B.shape[0]]=B[:]
print paddedB

最后，窗户可以同时应用于paddedA和paddedB，这意味着在对内核进行估计时，中间的值更重要。例如，Parzen / de la Vallée Poussin窗口：

import numpy as np 
import matplotlib.pyplot as plt 
from scipy.signal import convolve
from scipy.fftpack import next_fast_len
from scipy.signal import tukey
from scipy.signal import parzen

# A, in the description above
A = np.array([1, 1, 1, 2, 1, 1])

kernel=np.asarray([1,2,1])
paddedB = convolve(kernel, A, mode='full')
print paddedB


B = convolve(A,kernel, mode='same')
estimatedkernelsize=3
paddedB=np.zeros(A.shape[0]+estimatedkernelsize-1)
paddedB[estimatedkernelsize/2: estimatedkernelsize/2+B.shape[0]]=B[:]
print paddedB

paddedA=np.zeros(paddedB.shape[0])
paddedA[estimatedkernelsize/2: estimatedkernelsize/2+A.shape[0]]=A[:]

#applying window
#window=tukey(paddedB.shape[0],alpha=0.1,sym=True) #if longer signals, should be enough.
window=parzen(paddedB.shape[0],sym=True)
windA=np.multiply(paddedA,window)
windB=np.multiply(paddedB,window)


# Using the deconvolution theorem
f_A = np.fft.fft(windA)
f_B = np.fft.fft(windB)
# I know that you should use a regularization here 
r = f_B / f_A

# dk should be equal to kernel
dk = np.fft.ifft(r)
# shift to get the zero abscissa in the middle:
dk=np.fft.fftshift(dk)

print dk

然而，由于A的大小很小，估计的内核还远远不够完善：

[ 0.08341737-6.93889390e-17j -0.2077029 +0.00000000e+00j
 -0.17500324+0.00000000e+00j  1.18941919-2.77555756e-17j
  2.40994395+6.93889390e-17j  0.66720653+0.00000000e+00j
 -0.15972098+0.00000000e+00j  0.02460791+2.77555756e-17j]

Answer 2

# I had to modify the listed code for it to work under Python3. 
# I needed to upgrade to the scipy-1.4.1 and numpy-1.18.2
# and to avoid a TypeError: slice indices must be integers
# I needed to change / to // in the line marked below
import numpy as np 
import matplotlib.pyplot as plt 
from scipy.signal import convolve 
from scipy.fftpack import next_fast_len 
# A, in the description above 
A = np.array([1, 1, 1, 2, 1, 1]) 
kernel=np.asarray([1,2,1]) 
paddedB = convolve(kernel, A, mode='full') 
print(paddedB)
paddedA=np.zeros(paddedB.shape[0]) 
# note // instead of / below
paddedA[kernel.shape[0]//2: kernel.shape[0]//2+A.shape[0]]=A[:] 
#pad both signal and kernel. Requires the size of the kernel 
# Using the deconvolution theorem 
f_A = np.fft.fft(paddedA) 
f_B = np.fft.fft(paddedB) # I know that you should use a regularization here 
r = f_B / f_A 
# dk should be equal to kernel 
dk = np.fft.ifft(r) 
# shift to get zero abscissa in the middle: 
dk=np.fft.fftshift(dk) 
print(dk)
# this gives:
#(py36) bash-3.2$ python decon.py
#[1 3 4 5 6 5 3 1]
#[ 1.11022302e-16+0.j -1.11022302e-16+0.j -9.62291355e-17+0.j
# 1.00000000e+00+0.j  2.00000000e+00+0.j  1.00000000e+00+0.j
# 9.62291355e-17+0.j -1.11022302e-16+0.j]