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Bi-linear Interpolation 1

Bi-linear Interpolation 1. 10 10 Digital Image G(x,y). 3.6. 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. x. 0. 1. 2. 2.3. 3. 4. 5. G(3.6, 2.3) = ?. 6. 7. 8. 9. y. Bi-linear interpolation 2. G(3.6,2)=?. G(3,2)=58. G(4,2)=150. Calculate G(3.6,2) by using

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Bi-linear Interpolation 1

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  1. Bi-linear Interpolation 1 1010 Digital Image G(x,y) 3.6 0 1 2 3 4 5 6 7 8 9 x 0 1 2 2.3 3 4 5 G(3.6, 2.3) = ? 6 7 8 9 y

  2. Bi-linear interpolation 2 G(3.6,2)=? G(3,2)=58 G(4,2)=150 Calculate G(3.6,2) by using values of G(3,2) and G(4,2): G(3.6,2.3)=? Grey level 255 150 113.2 58 Horizontal position 1 2 3 4 0 3.6 G(3.6,3)=? G(3,3)=137 G(4,3)=26 G(3.6,2) = (1fx)G(3,2) + fxG(4,2), Wherefxis 0.6 (fraction part of 3.6) And G(3,2)=58, G(4,2)=150, so G(3.6,2) = 113.2 In the same way G(3.6,3) = (1fx)G(3,3) + fxG(4,3) = (1-0.6)137 + 0.626 = 70.4

  3. Bi-linear interpolation 3 G(3.6,2)=113.2 G(3,2)=58 G(4,2)=150 Calculate G(3.6,2.3) by using values of G(3.6,2) and G(3.6,3): G(3.6,2.3)=? Grey level 255 113.2 100.36 70.4 Vertical position G(3.6,3)=70.4 G(3,3)=137 G(4,3)=26 1 2 3 4 0 2.3 G(3.6,2.3) = (1fy)G(3.6,2) + fyG(3.6,3), Wherefy is 0.3 (fraction part of 2.3) And G(3.6,2)=113.2, G(3.6,3)=70.4, so G(3.6,2.3) = 100.36

  4. Bi-cubic Interpolation 1 1010 Digital Image G(x,y) 3.6 0 1 2 3 4 5 6 7 8 9 x 0 1 2 2.3 3 4 5 G(3.6, 2.3) = ? 6 7 8 9 y

  5. Bi-cubic interpolation 2 G(2,1)=34 G(3,1)=102 G(4,1)=10 G(5,1)=86 G(3.6,1)=? G(3.6,2)=? G(5,2)=205 G(2,2)=95 G(3,2)=58 G(4,2)=150 G(3.6,2.3)=? G(2,3)=220 G(3,3)=137 G(4,3)=26 G(5,3)=134 G(3.6,3)=? G(2,4)=78 G(3,4)=186 G(4,4)=18 G(5,4)=214 G(3.6,4)=?

  6. Bi-cubic interpolation 3 Calculate G(3.6,1) by using values of G(2,1), G(3,1), G(4,1) and G(5,1): Cubic curve: f(t) = a3 · t3 + a2 · t2 + a1 ·t + a0 f(2) = 34 f(3) = 102 f(4) = 10 f(5) = 86 Solve the above linear functions, we can decide the values of a3 ,a2 ,a1 and a0. Then f(3.6) can be calculated. Grey level 255 f 102 86 51.14 34 10 Horizontal position 0 4 2 3 5 1 3.6 Result: G(3.6, 1) = 51.14 Using the same strategy, we can calculate the values of G(3.6, 2) = 113.25, G(3.6, 3) = 83.37, G(3.6, 4) = 95.36

  7. Bi-cubic interpolation 4 G(2,1)=34 G(3,1)=102 G(4,1)=10 G(5,1)=86 G(3.6,1)=51.14 G(3.6,2)=113.25 G(5,2)=205 G(2,2)=95 G(3,2)=58 G(4,2)=150 G(3.6,2.3)=? G(2,3)=220 G(3,3)=137 G(4,3)=26 G(5,3)=134 G(3.6,3)=83.37 G(2,4)=78 G(3,4)=186 G(4,4)=18 G(5,4)=214 G(3.6,4)=95.36

  8. Bi-cubic interpolation 5 Calculate G(3.6,2.3) by using values of G(3.6,1), G(3.6,2), G(3.6,3) and G(3.6,4): From the previous slide we know: G(3.6, 1) = 51.14 ,G(3.6, 2) = 113.25, G(3.6, 3) = 83.37, G(3.6, 4) = 95.36. Cubic curve: f(t) = a3 · t3 + a2 · t2 + a1 ·t + a0 f(2) = 51.14 f(3) = 113.25 f(4) = 83.37 f(5) = 95.36 Solve the above linear functions, we can decide the values of a3 ,a2 ,a1 and a0. Then f(2.3) can be calculated. Grey level 255 f(2.3) = 99.21 f 113.25 95.36 83.37 51.14 Vertical position 0 4 2 3 5 1 2.3 Result: G(3.6, 2.3) = 99.21

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