Mathematics - The MathWorks - #30

t = [0 .3 .8 1.1 1.6 2.3]';

y = [.82 .72 .63 .60 .55 .50]';

Try modeling the data with a decaying exponential function:

The preceding equation says that the vector y should be approximated by a linear combination of two other vectors, one the constant vector containing all ones and the other the vector with components e"*. The unknown coefficients, c₁ and c₂, can be computed by doing a least squares fit, which minimizes the sum of the squares of the deviations of the data from the model. There are six equations in two unknowns, represented by the 6-by-2 matrix:

E = [ones(size(t)) exp(-t)] E =

1.0000 1.0000

1.0000 0.7408

1.0000 0.4493

1.0000 0.3329

1.0000 0.2019

1.0000 0.1003

Usethebackslashoperatortoget theleast squaressolution:

c = E\y

c =

0.4760 0.3413

In other words, the least squares fit to the data is

y(f)= 0.4760 + 0.3412 e⁴

The following statements evaluate the model at regularly spaced increments in t, and then plot the result, together with the original data:

T = (0:0.1:2.5)';