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March 18th, 2016, 07:29 AM   #1
Joined: Mar 2016
From: South Africa

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Exclamation log linear weighted least squares application


I am trying to fit a curve to a set of data using a weighted least squares approach. The reason I am using the weighted approach is to bias my solution to my more reliable data. I am however having a problem trying to derive the analytical solution to the problem.

T curve I am trying to fit is a creep strain equation of the following form, where beta are the coefficients I am trying to solve for:
$\displaystyle \epsilon(t,\sigma, T) = 10^{\beta_1}\sigma^{\beta_2}te^{\beta_3/T}$

I then log the equation in order to linearise it as follows:
$\displaystyle log(\epsilon) = \beta_1 + \beta_2log(\sigma) + log(t) - \beta_3log(e^{1/T})$

I have experimental results for time t to reach strain $\displaystyle \epsilon$ at a constant test stress $\displaystyle \sigma$ and temperature T. My problem is when I try to implement the analytical solution for the weighted least squares problem which has the form, where W is a diagonal matrix of my chosen weightings:
$\displaystyle \beta = (X^TWX)^{-1}X^TWY $

As I understand I have the following for n data points:
$\displaystyle \beta = [\beta_1,\ \beta_2,\ \beta_3]$
$\displaystyle Y^T = [log(\epsilon_1),\ log(\epsilon_2,\ ...\ ,log(\epsilon_n))]$
$\displaystyle X^T = [1,\ log(\sigma_i),\ log(e^{1/T_i})]$ with i=n rows

My confusion is what do I do with the $\displaystyle \log(t)$ expression. As I see it, this term is an "x" data point with a coefficient of 1. However, I cannot see how to include this and solve for my $\displaystyle \beta$ coefficients. I know that I could go the optimisation route and solve it that way. I am however looking for an analytical solution to the least squares problem.

Any assistance would be greatly appreciated. Thanks in advance.
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March 21st, 2016, 09:59 PM   #2
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The given equation says that strain is proportional to time and the constant of proportionality depends on stress $\displaystyle \sigma$ and temperature T.

$\displaystyle \epsilon (t ,\sigma ,T)=f(\sigma ,T)t\\$

The objective is to determine f by measuring the slope of strain/time curves at different values of stress and temperature.

If the form of f given in OP were exactly correct and repeatable, then two experiments would suffice to determine f:

$\displaystyle f(\sigma _{1},T_{1})=K_{1}\\$

$\displaystyle f(\sigma _{2},T_{2})=K_{2}\\$

where $\displaystyle K_{1}$ and $\displaystyle K_2$ are the slopes of the strain\time curves under given conditions of stress and temperature. Solve for $\displaystyle \beta _{1}$ and $\displaystyle \beta_{2}$.

In reality, f is an approximation and you have to use curve fitting for $\displaystyle \beta _{1}$ and $\displaystyle \beta_{2}$. t doesn't enter into it.
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