How to solve an equation system using a interp1-variable

 Hello everyone,

 

I'm trying to solve equations in a Matlab function that is part of a Simulink model.

 

One of those two equations is an interpolation of a given array.

 

% variable definiton
% T                     function argument
% p = f(T)              values are given in form of a 69x2 double
% a | b | c | h         constants / parameters (defined or calculated beforehand)

% equations
% eqn 1
% p = interp1(p(:,1),p(:,2),T)
% eqn 2
% h = a * T + p * (b + c * T)

% As I am used to solve equations with my pocket calculator I quickly found something like this while browsing:

syms T
solve (h == a * T + interp1(p(:,1),p(:,2),T) * (b + c * T),T);

% Unfortunately I do not have the required toolbox to solve equations which is why I need to find a work-around

% my next idea was to use an iteration to solve this (which I did not formulate and test until now)
% it would be something like that though:

for T = 1:100
    p = interp1(p(:,1),p(:,2),T);
    h_iter = a * T + p * (b + c * T);
    if h_iter > h - h_marge | h_iter < h + h_marge
        break
    end
end

% with h_marge being a predefined value for accuracy tolerance

I would like to avoid such an iterational process since the iteration has to be done each simulation-step which could cause long simulation times

 

depending on the accuracy and the initial condition(s) I determine.

 

Thus, I wanted to ask you if there is another method / solution to solving that problem.

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Expert Answer

 Prashant Kumar answered . 2021-10-21 06:59:13

interp1 is incompatible with symbolic parameters. You CANNOT use interp1 with solve. Period. So wherever you found that snippet of code, they were wrong.

 

You can use a tool like fzero with interp1. fzero and interp1 are compatible.

 

Next, looking at your problem, this is actually a TWO variable problem, with two equations. That is, we have known constants, a,b,c,h. And there are two unknowns. Sadly, you seem to be using p in two places at once, both as an array of known elements, and as a variable. And that is confusing as hell.

 

So I'll make the assumption that you have two variales, I'll call then T and p_T. That is p_T is given as:

 

p_T = interp1(p(:,1),p(:,2),T)

 

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