拉普拉斯滤波器是如何计算的？

我不太明白他们是如何得出导数方程的。有人可以详细解释一下，甚至可以提供一个有足够数学解释的链接吗？

拉普拉斯滤波器看起来像

拉普拉斯先生提出了这个方程。这就是拉普拉斯算子的简单定义：二阶导数之和（您也可以将其视为海森矩阵 https://en.wikipedia.org/wiki/Hessian_matrix).

您显示的第二个方程是有限差分近似 https://en.wikipedia.org/wiki/Numerical_differentiation到二阶导数。这是您可以对离散（采样）数据进行的最简单的近似。导数定义为斜率（方程为维基百科 https://en.wikipedia.org/wiki/Derivative#Rigorous_definition):

In a discrete grid, the smallest h is 1. Thus the derivative is f(x+1)-f(x). This derivative, because it uses the pixel at x and the one to the right, introduces a half-pixel shift (i.e. you compute the slope in between these two pixels). To get to the 2^nd order derivative, simply compute the derivative on the result of the derivative:

f'(x) = f(x+1) - f(x)
f'(x+1) = f(x+2) - f(x+1)

f"(x) = f'(x+1) - f'(x)
      = f(x+2) - f(x+1) - f(x+1) + f(x)
      = f(x+2) - 2*f(x+1) + f(x)

Because each derivative introduces a half-pixel shift, the 2^nd order derivative ends up with a 1-pixel shift. So we can shift the output left by one pixel, leading to no bias. This leads to the sequence f(x+1)-2*f(x)+f(x-1).

计算二阶导数与使用滤波器进行卷积相同[1,-2,1].

应用此过滤器及其转置，并将结果相加，相当于与内核进行卷积

[ 0, 1, 0       [ 0, 0, 0       [ 0, 1, 0
  1,-4, 1    =    1,-2, 1    +    0,-2, 0
  0, 1, 0 ]       0, 0, 0 ]       0, 1, 0 ]