当前位置：网站首页>Second order edge detection Laplacian of Guassian Gaussian Laplacian operator

Second order edge detection Laplacian of Guassian Gaussian Laplacian operator

2022-07-19 01:55:00 【elkluh】

Catalog

One 、 Laplace operator

Two 、 Gauss Laplace operator

3、 ... and 、 First order 、 Difference of second derivative

One 、 Laplace operator

Laplace operator is equivalent to the second derivative of pixels .

First order derivation $f'(x) = \frac{f(x+\Delta x)-f(x)}{\Delta x}$ （1）

$f'(x+\Delta x) = \frac{f(x+2\Delta x)-f(x+\Delta x)} { \Delta x}$ （2）

Second order derivation $f''(x+\Delta x) = \frac {f'(x+\Delta x) - f'(x)}{\Delta x}$ （3）

$f''(x+\Delta x) =\frac{ f(x+2\Delta x) - 2f(x+\Delta x)+f(x)}{\Delta x}$ （4）

The template of the Gaussian operator is calculated as follows ：

Its advantage is simplicity , But it is easily affected by noise .

Two 、 Gauss Laplace operator

Second order edge detection can be called , Gauss' Laplace , It can also be called Mar operator (Marr Hildreth).

Because Gauss equation has smooth effect , So Gauss Laplace also has , Therefore, it is insensitive to noise .

The constant term of Gauss equation is removed as ：

$g(x,y) = e^{ \frac{ -(x^{2}+y^{2})}{ 2\sigma^{2}}}$ （5）

x The first partial derivative of is ：

$\frac{\partial g}{\partial x}=-\frac{x}{\sigma^{2}}e^{\frac{ -(x^{2}+y^{2})}{2\sigma^2}}$ （6）

x The second partial derivative of is ：

$\frac{\partial g}{\partial x^{2}}= \frac{1}{\sigma^{2}}( \frac{x_{2}}{\sigma ^{2}}-1) e^{ \frac{-(x^{2}+y^{2})}{2\sigma^2}}$ （7）

y The first partial derivative of is

$\frac{\partial g}{\partial y}=-\frac{y}{\sigma^{2}}e^{\frac{ -(x^{2}+y^{2})}{2\sigma^2}}$ （8）

y The second partial derivative of is

$\frac{\partial g}{\partial^{2}}= \frac{1}{\sigma^{2}}( \frac{y_{2}}{\sigma ^{2}}-1) e^{ \frac{-(x^{2}+y^{2})}{2\sigma^2}}$ （9）

according to （7）（9） We can get

$\triangledown g^{2}(x,y,\sigma) = g''(x) + g''(y)=\frac {e^{\frac{-(x^{2}+y^{2})}{2\sigma^{2}}}} {\sigma^{2}} (\frac{x^{2}+y^{2}}{\sigma^{2}}-2)$ (10)

hold （10） Simplification , $e^{\frac{-(x^{2}+y^{2})}{2\sigma^2}} (\frac{x^2+y^2}{\sigma^4}-\frac{2}{\sigma^2})$

$LoG=e^{\frac{-(x^{2}+y^{2})}{2\sigma^2}} \frac{(x^2+y^2-2\sigma^2)}{{\sigma^4}}{\frac{1}{2\pi \sigma^2}} = -\frac{1}{\pi\sigma^4}(1-\frac{x^2+y^2}{2\sigma^2}) e^{\frac{-(x^{2}+y^{2})}{2\sigma^2}}$

Add the constant of Gaussian function at the end 1/(2 pi sigma^2)

The shape of Gauss operator and Laplace operator

3、 ... and 、 First order 、 Difference of second derivative

For edge detection , The first derivative detects the part with the largest change rate , At this time, the greater the value represented by the derivative , Indicates that the larger the pixel change in this place , The more likely it is the edge .

The second derivative detects zero cross value , At this time, the place with the largest change value of the change rate is detected .

To judge whether the second derivative crosses the zero point, take the method shown in the figure below .

Divide the center point into the following four areas , Calculate the average value of the four regions respectively . If the average value of the midpoint of any quadrant is different from the average value of other quadrants , Then there must be a zero crossing point at the center . Calculate four averages , If the maximum is greater than zero , The smallest is less than zero , Then there must be a zero crossing point in the adjacent point .