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I would like to get a simple estimation of the parameters for a humidity chamber system. The humidity chamber has one constant inflow ($k_1$) and and outflow which depends on the humidity of a neighbor humidity chamber (diffusion, modeled as $k_2(RH_{ext} - RH)$. $RH$ is the humidity of the chamber, $RH_{ext}$ is the humidity of the external chamber. I have the following system description (first order): $$ \frac{dRH}{dt} = k_1 + k_2(RH_{ext} - RH) $$

Now, I have measurement data for $RH$ (output) and $RH_{ext}$ (input). $k_1$ could be seen as a second (constant) input. I would like to estimate the parameters $k_1$ and $k_2$ in a simple manner (maybe least-square based) using my (discrete) measurement data in Matlab, but I don't know how I could proceed. Use the system identification toolbox? Other ways?

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  • $\begingroup$ Hi Fabian, welcome to engineering.SE. Our site supports Latex style equation typesetting; I've edited your post to use it. $\endgroup$ Jun 9 '15 at 12:30
  • $\begingroup$ What type of measurement data do you have? Is it $RH$ vs. time with all other parameters constant? If so, you can simply integrate the equation and do a 3 parameter curve fit ($k_1$, $k_2$, and the integration constant). $\endgroup$ Jun 9 '15 at 12:33
  • $\begingroup$ Hi Chris, thanks very much for the edit and your answer. $RH_{ext}$ isn't a constant as well. Your approach seems clever, but I'm not so sure about how to do this in practice. You would simply try to fit $RH(t)$ to $f(t)=k_1 t + k2\left(\int RH_{ext} - \int RH \right)+c$, by using the numerical integrations $\int RH_{ext}$ and $\int RH$ ? $\endgroup$
    – donald
    Jun 9 '15 at 13:02
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It sounds like you have MATLAB, so it's a simple linear fit:

  • From your vector $RH(t)$, compute $\frac{d RH}{dt}$ by numerical differentiation
  • Create a matrix $A$ with ones on the first column and the values of $RH_{ext}(t) - RH(t)$ on the second column
  • Create a matrix $B$ with the values of $\frac{d RH}{dt}$ (use 0 as the initial value)
  • Your result vector is given $A$ \ $B$

    dRH = [0; diff(RH)./diff(t)];

    A = [ones(size(RH)) RH_ext-RH];

    k = A\dRH; % k(1) is k1, k(2) is k2

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  • $\begingroup$ Just to emphasize the point; $A\setminus B$ uses Matlab's left division operator which is used to solve a system of linear equations described in matrix notation. $\endgroup$ Jun 9 '15 at 16:56
  • $\begingroup$ @ChrisMueller yes, that's right. Thanks for the clarification $\endgroup$
    – am304
    Jun 9 '15 at 17:16

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