Question: Show that:Highpass = Original-Lowpass
0

Mumbai University > Computer Engineering > Sem 7 > Image Processing

Marks: 5 M

Year: May 2013, May 2015

 modified 3.4 years ago  • written 3.4 years ago by Juilee • 2.4k
0

When we apply the LPF on the image, the center pixel z5 changes to

$1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9]$

Original- Low pass $=z_5 - 1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9] \\ = z_5 - 1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9] \\ =8/9 z_5-1/9z_1-1/9z_2-1/9z_3-1/9z_4-1/9z_6-1/9z_7-1/9z_8-1/9z_9 \\ =1/9Χ \ high \ pass \ mask$

-1 -1 -1
-1 8 -1
-1 -1 -1

This is nothing but a high pass mask - • HPF alternates LOW frequency components and allows to pass High frequency components of the image.HPF is the sharpening Second order derivative filter. HPF image can be obtained by subtracting LPF image from original image.

Therefore, Original image = LPF image + HPF image

HPF image = Original image - LPF image