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Show that High pass= Original - Low pass.
written 7.3 years ago by | • modified 7.3 years ago |
Mumbai University > Computer Engineering > Sem 7 > Image Processing
Marks: 10 Marks
Year: May 2016
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written 7.3 years ago by | • modified 7.3 years ago |
Mumbai University > Computer Engineering > Sem 7 > Image Processing
Marks: 10 Marks
Year: May 2016
written 7.3 years ago by |
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.
Therefore,
Original image = LPF image +HPF image
HPF image=Original image-LPF image