written 8.4 years ago by | modified 2.8 years ago by |
(i) Digital Negative (ii) Bit Plane Slicing (iii) Thresholding
Mumbai University > Electronics > Sem 7 > Digital image processing
Marks: 10 M
Year: Dec 2012
written 8.4 years ago by | modified 2.8 years ago by |
(i) Digital Negative (ii) Bit Plane Slicing (iii) Thresholding
Mumbai University > Electronics > Sem 7 > Digital image processing
Marks: 10 M
Year: Dec 2012
written 8.4 years ago by |
i) Digital Negative is calculated by following formula:
$s = intensity_{max} – r$
In given image, $intensity_{max}$ = 7
Therefore, Digital negative = | 5 | 6 | 4 | 0 | 3 | |---|---|---|---|---| | 3 | 2 | 5 | 7 | 6 | | 4 | 2 | 6 | 3 | 1 | | 7 | 3 | 7 | 5 | 4 | | 5 | 6 | 1 | 6 | 3 |
ii) Bit plane Slicing
The given image has maximum grey level of 7, hence it is a 3 bit image.
2 | 1 | 3 | 7 | 4 |
---|---|---|---|---|
4 | 5 | 2 | 0 | 1 |
3 | 5 | 1 | 4 | 6 |
0 | 4 | 0 | 2 | 3 |
2 | 1 | 6 | 1 | 4 |
Converting image into binary
010 | 001 | 011 | 111 | 100 |
---|---|---|---|---|
100 | 101 | 010 | 000 | 001 |
011 | 101 | 001 | 100 | 110 |
000 | 010 | 000 | 010 | 011 |
010 | 001 | 110 | 001 | 100 |
Separating the bit planes, we obtain
iii) Thresholding
Thresholding of image is obtained using following formula.
s = 1 (r > threshold)
s= 0 (r <= threshold)
Let’s assume threshold to be 2
Therefore, image obtained after Thresholding is –
0 | 0 | 3 | 7 | 4 |
---|---|---|---|---|
4 | 5 | 0 | 0 | 0 |
3 | 5 | 0 | 4 | 6 |
0 | 4 | 0 | 0 | 3 |
0 | 0 | 6 | 0 | 4 |