Sum of series (symsum) in Simulink

Hi guys,

I'm trying to solve an equation in Simulink. Solving it in Matlab works niceley but I don't know how to implement it in Simulink. It goes as follows:

k = 5;

x80 = 4;

syms mu n;

eqn = 1 - exp(-x80out/mu)*symsum((1/factorial(n))*(x80out/mu)^n, n, 0, (k-1)) == 0.8;

mu = solve(eqn, mu);

Does anybody know how to implement the sum of series in Simulink?

Alternatively I thought about using a Matlab function but it doesn't allow to introduce the variables via "syms mu n".

Thanks in advance:)

[Paul]

Hi Johannes,

What is x80out?

Are k and x80 (or x80out) to be computed (and changing) in the Simullnk model from which mu should be computed as the simulation is running?

Or are k and x80 (x80out) known a priori, and then mu only needs to be computed once before the simulation starts to execute?

[Johannes Poser]

Hi Paul,

x80out is supposed to be x80 = 4.

x80 and k are both changing during the simulation and therefore mu too.

Answer 1

[Paul]

I was able to get a workable soluion using only Simulink blocks by setting up the equation and solving for mu with an Algebraic Constraint block. The symsum part of the equation can be implemented in a For Iterator Subsystem It's much easier to use a Matlab Function block to really do the heavy lifting using @Walter Roberson code from this comment

function balance = fcn(k,x80out,mu)

n = 0 : k-1;
inner = sum( (1./factorial(n)) .* (x80out/mu).^n );
balance = 1 - exp(-x80out/mu) .* inner - 0.8;

end

[Johannes Poser]

Oh maaaaaan this is amazing I can't believe it!

Many thanks to both of you. I think I would have never figured that out without you. Im truly grateful and it works so nicely.

[Paul]

HTH.

I didn't test if that code works when k varies with time. If not, just covert the computation of inner to a simple for loop.

Answer 2

[Walter Roberson]

Nothing in the Symbolic Toolbox is supported for code generation, including not for Simulink Rapid Acceleration purposes.

You could try using coder.external coder.extrinsic -- if you do remember that symbolic variables are not permitted to be signals, so remember to double()

But you should probably instead switch to using an anonymous function and fzero, or perhaps a private function, along the lines of

mu = 0;
mu = fsolve(@mu_eqn, 1);

function balance = mu_eqn(mu)
    k = 5;
    x80out = 4;
    balance = 1 - exp(-x80out/mu)*symsum((1/factorial(n))*(x80out/mu)^n, n, 0, (k-1)) - 0.8;
end

[Johannes Poser]

Hello Walter,

thanks for the answer. This already looks more elegant. Im still confused thou how to handle the symsum "n" without the symbolic tool as I only know to use it with syms n.

[Walter Roberson]

Oh right, I missed the symsum.

Okay, the purpose of symsum() is to try to come up with a formula for the summation. For example,

syms n m
symsum(1/n, n, 1, m)

ans = 

Does your expression happen to come out to a formula?

syms x80out mu
syms n k integer
symsum((1/factorial(n))*(x80out/mu)^n, n, 0, (k-1))

ans = 

... No, it does not. And it is not an infinite summation.

So instead of using symsum(), do a vectorized calculation and a definite summation:

n = 0 : k-1;
inner = sum( (1./factorial(n)) .* (x80out/mu).^n );
balance = 1 - exp(-x80out/mu) .* inner - 0.8;

[Paul]

The following might work, I didn't try it.

Pass k and x80 into a Matlab Function block. The code would look like this

function myfunc(x80,k)

coder.extrinsic('solution')

mu = solution(x80,k)

end

Define a garden variety Matlab funtion in an m-file, call it solution.m.

function mu = solution(dx80,dk)

x80 = sym(dx80);
k   = sym(dk);

syms mu n;

eqn = 1 - exp(-x80out/mu)*symsum((1/factorial(n))*(x80out/mu)^n, n, 0, (k-1)) == 0.8;
mu = double(solve(eqn, mu));

end

If that doesn't work, or is too slow, solution .m can be modified to use fsolve. I don't see a way to call fsolve directly from a Matlab Function block in this case.

[Johannes Poser]

You guys are awesome, I can't thank you enough