问题
我正在尝试使用卷积定理对两个测量数据 A 和 B 进行反卷积。我知道对于卷积，您应该对数据进行零填充以防止循环卷积。但是，对于零卷积对于反卷积也是必不可少的，我感到困惑。
问题
1.如何正确地基于卷积定理进行反卷积？
2.为什么下面的示例不起作用？
方法
因为 A 和 B 是经过测量的，所以我创建了一个示例供进一步研究。这个想法是通过在模式scipy.signal.convolve中使用same来创建 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])
预期是：
dk = array([0, 1, 2, 1, 0, 0])
答案 0 :(得分：1)
实际上，由于内核是1.0 2.0，而1.0是以2.0为中心（模糊和膨胀），因此内核宽度是3。由于数组A在[0..5]上不为null，因此全卷积数组paddedB在[-1..6]上不为null。尽管如此，函数scipy.signal.convolve(...,'same')返回了中继卷积数组B(0..5)=paddedB(0..5)。因此，与paddedB（-1）和paddedB（6）有关的信息会丢失，并且如果使用np.convolve()的选项same，则很难重新内核化内核。 。
为避免信息丢失，将对输出paddedB进行填充，以包含卷积信号的support，其计算为函数A和函数A的支持Minkowski sum。支持内核。 np.convolve()的选项full直接计算paddedB，而不会丢失信息。
kernel=[1,2,1]
paddedB = convolve(kernel, A, mode='full')
要使用卷积定理检索内核，将填充输入信号A以匹配函数paddedB的支持
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)
请注意使用功能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
如果无法获得paddedB且B是唯一可用的数据，则可以尝试通过将B填充为零或对B的最后一个值进行平滑来重建pappedB。它需要对内核大小进行一些估算。
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，window可以同时应用于paddedA和paddedB，这意味着中间的值在估计内核时更为重要。例如Parzen / de laValléePoussin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
尽管如此，由于A的大小很小，估计的内核远非完美：
[ 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]
答案 1 :(得分：0)
# 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]