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Show that High pass= Original - Low pass.

Mumbai University > Computer Engineering > Sem 7 > Image Processing

Marks: 10 Marks

Year: May 2016

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

$\dfrac19[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9]$

Original- Low pass

$= z_5-1/9[z_1+z2+z3+z4+z5+z6+z7+z8+z9] \\ = z_5-1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9] \\ =8/9 z_5-1/9_z1-1/9_z2-1/9_z3-1/9_z4-1/9_z6-1/9_z7-1/9_z8-1/9_z9 \\ =1/9Χ \text{high pass mask}$

-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