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== Cubic Spline Interpolation == | == Cubic Spline Interpolation == |
Revision as of 16:49, 1 September 2008
Contents
Linear Interpolation
A linear function may be written as Given any two data points, , and , we can determine a linear function that exactly passes through these points. We can do this by solving for and . Since there are two unknowns we require two equations. They are:
These may be written in matrix form as
Given values for , and , these equations may be easily solved in MATLAB. But since they are so simple, we can easily by hand to obtain a general solution,
This gives the coefficients, which may then be substituted into the original equation and simplified to obtain
This is a very convenient equation for performing linear interpolation.
Example
The density of air at various temperatures and atmospheric pressure is given in the following table
Temperature (°F) | 1.39 | 1.16 | 0.99 | 0.87 | 0.78 | 0.69 | 0.63 | 0.58 | 0.54 | 0.50 | 0.46 | 0.44 | 0.41 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Density (kg/m3) | -10 | 80 | 170 | 260 | 350 | 440 | 530 | 620 | 710 | 800 | 890 | 980 | 1070 |
Estimate the density of air at 32 °F. Using linear interpolation, we have . We define and . Now we can calculate
Implementation in Matlab
There are three ways to do interpolation in MATLAB.
- Solve the linear system.
T1=-9.7; T2=80; rho1=1.39; rho2=1.16; T = 32; A = [ 1 T1; 1 T2 ]; b = [ rho1; rho2 ]; a = A\b; rho = a(1) + a(2)*T;
- Use the general equation we derived above:
T1=-9.7; T2-80; rho1=1.39; rho2=1.16; T = 32; rho = (rho2-rho1)/(T2-T1) * (T-T1) + rho1;
- Use MATLAB's built-in interpolation tool
Ti = [-9.7 80]; rhoi = [1.39 1.16]; rho = interp1( Ti, rhoi, 32 );
All three of these methods provide exactly the same answer. However, using MATLAB's interp1 function allows you to use a vector for the points that you want to interpolate to. For example, if we wanted the density at 30, 50, 70, 10, and 110 °F, we could accomplish this by:
Ti = [-9.7 80 170 ];
rhoi = [1.39 1.16 0.995];
T = 30:20:110;
rho = interp1( Ti, rhoi, T );
Implementation in Excel
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Polynomial Interpolation
Polynomial interpolation is a simple extension of linear interpolation. In general, an nth degree polynomial is given asIf n=1 then we recover a first-degree polynomial, which is linear. The formula gives .
In general, if we want to interpolate a set of data using an nth degree polynomial, then we must determine n+1 coefficients, . Therefore, we require n+1 points to interpolate using an nth degree polynomial.
The equations that must be solved are given as
Solving these equations provides the values for the coefficients, . Once we know these coefficients, we can evaluate the polynomial at any point.
Example
Using a third degree polynomial and the data in the table above, approximate the density of air at 50 °F and 300 °F.
We require 4 points to fit a third degree polynomial, . We will choose the 4 points closest to the temperature that we wish to interpolate to.
Implementation in Matlab
Our procedure is:
- Determine which points we need to use.
- Set up and solve the linear system to obtain the polynomial coefficients .
- Evaluate the polynomial to obtain the temperature.
The MATLAB code to obtain the density at 50 °F is
% Set the data points from the table.
Ti = [ -9.7 80 170 260 350 440 530 ...
620 710 800 890 980 1070 ]';
rhoi=[ 1.39 1.16 0.99 0.87 0.78 0.69 0.63 ...
0.58 0.54 0.50 0.46 0.44 0.41 ]';
% Set the temperature we want to evaluate the density at.
T = 50;
% determine the set of points that we will use to construct the interpolant.
points = 1:4;
% set up the linear system
A = [ ones(4,1), ones(4,1).*Ti(points), ones(4,1)*Ti(points).^2, ones(4,1)*Ti(points).^3 ];
b = rhoi(points);
% solve for the polynomial coefficients
a = A\b;
% evaluate the density.
rho = a(0) + a(1)*T + a(2)*T^2 + a(3)*T^3;
For T=300 we would repeat the above with the only change being:
T=300;
points=4:8;
Cubic Spline Interpolation
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Example
Implementation in Matlab
Lagrange Polynomial Interpolation
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Given points the order Lagrange polynomial that interpolates these function values, are expressed as
where is the Lagrange polynomial given by
and . In other words, for an order interpolation, we require points.
The operator represents the continued product, and is analogous to the operator for summations. For example, .