Question

Consider the nodes of a square grid A, B, C and D (shown in figure below), where a certain parameter is measured. The distances between the points is also indicated in the same figure. For example, the value observed at point A is 120 and is indicated as A (120), and the distance between points A and B is 1.0 units. The value at point ‘X’ computed using bilinear interpolation, using the values at points A, B, C and D is _________

• A. 175
• B. 152
• C. 198
• D. 210

Hint:

Bilinear interpolation involves applying linear interpolation twice: once in the \(x\)-direction, and then in the \(y\)-direction. Use the formula: \[ f(x,y) = A(1-x)(y) + B(x)(y) + C(1-x)(1-y) + D(x)(1-y) \] with coordinates normalized over the unit square.

Answer 1

The Correct Option is A

Bilinear interpolation formula: Given points: - \( A(0,1) = 120 \) - \( B(1,1) = 200 \) - \( C(0,0) = 250 \) - \( D(1,0) = 120 \) Let: - \( x = 0.4 \) - \( y = 0.5 \) Bilinear interpolation formula is: \[ f(x,y) = A(1-x)(y) + B(x)(y) + C(1-x)(1-y) + D(x)(1-y) \] Substitute values: \[ f(0.4, 0.5) = 120(1-0.4)(0.5) + 200(0.4)(0.5) + 250(1-0.4)(1-0.5) + 120(0.4)(1-0.5) \] \[ = 120(0.6)(0.5) + 200(0.4)(0.5) + 250(0.6)(0.5) + 120(0.4)(0.5) \] \[ = 120 \times 0.3 + 200 \times 0.2 + 250 \times 0.3 + 120 \times 0.2 \] \[ = 36 + 40 + 75 + 24 = \boxed{175} \]