Texas Instruments TI-89 Tip List page 203

Bilinear interpolation
If we are given four (x,y) points such that
f(x ,y ) = z
1 1 1
f(x ,y ) = z
1 2 2
f(x ,y ) = z
2 1 3
f(x ,y ) = z
2 2 4
then we can solve for a function in four unknown coefficients. There are an infinite number of such
equations, but the most simple candidate is
z = ax + bx + cxy + d
so the system to solve for a, b, c and d is
ax + by + cx y
1 1 1 1
ax + by + cx y
1 2 1 2
ax + by + cx y
2 1 2 1
ax + by + cx y
2 2 2 2
We could solve this directly for a, b, c and d, but we can get a more simple solution by scaling x and y,
like this:
x−x 1
t =
x 2 −x 1 and
With this scaling, t = 0 when x = x
= y . This equation system to solve simplifies to this:
2
d = z
1
b + d = z
2
a + d = z
3
a + b + c + d = z
4
So the equation to estimate z in terms of t and u is
z = (z - z )t + (z
3 1 2
We can expand this equation, collect on z
z = z (1-u)(1-t) + z
1
The function bilinint() does the interpolation with this formula
bilinint(xa,xb,ya,yb,zaa,zab,zba,zbb,x,y)
Func
©(xa,xb,ya,yb,zaa,zab,zba,zbb,x,y) Lin interpolate z=f(x,y)
©8oct00 dburkett@infinet.com
local t,u
(x-xa)/(xb-xa)→t
and
+ d = z
1
+ d = z
2
+ d = z
3
+ d = z
4
y−y 1
u =
y 2 −y 1
and t = 1 when x = x
1
or
- z )u + (z - z - z + z
1 1 2 3
, z
1
u(1-t) + z t(1-u) + z ut
2 3 4
x > x
2 1
y > y
2 1
. Similarly, u = 0 when y = y
2
a = z - z
3 1
b = z - z
2 1
c = z - z - z
1 2 3
d = z
1
)ut + z
4 1
, z and z , and further factor that result to get
2 3 4
and u = 1 when y
1
+ z
4
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