diff function rearranges symbolic expression leading to non-vanishing terms

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I am trying to feedback linearize my non-linear system in Matlab. To that end I have to represent it in the following form:
So after I have modeled my system in the form of nonlinear differential equations, , I use the diff function to take the derrivative with respect to u to aquire g, and . This works, of course only if the system can be represented in the mentioned form. However, the diff function rewrites some of my expressions, which then after subtraction lead to non-vanishing factors multiplying u. For example, my input is still present in multiplied with a factor 5.7024e-18. I suspect, there is some quantization error introduced due to diff re-writing the derivative . Can this be somehow avoided?
g_x=[diff(dxdt,[Q_g]) diff(dxdt,[Q_b])];
for i=1:size(g_x,1)
i
f_x(i,:)=simplify(dxdt(i,:)-g_x(i,1)*Q_g-g_x(i,2)*Q_b,'steps',10,'IgnoreAnalyticConstraints',true);
end
Here Q_g and Q_b are my inputs.
An example of the re-writing of expressions:
dxdt = - (1297036692682702848*Q_b*(C_carbv - 47/20000))/15694143981460705 + ..................
g_x=1905022642377719808/9808839988412940625 - (1297036692682702848*C_carbv)/15694143981460705
f_x=Q_b*((1297036692682702848*C_carbv)/15694143981460705 - 1905022642377719808/9808839988412940625) - (1297036692682702848*Q_b*(C_carbv - 47/20000))/15694143981460705 + ..............................
simplify(Q_b*((1297036692682702848*C_carbv)/15694143981460705 - 1905022642377719808/9808839988412940625) - ...
(1297036692682702848*Q_b*(C_carbv - 47/20000))/15694143981460705)
ans =
(47*Q_b)/8242150268041428750
The .................. stands for terms that do not depend on Q_b.
  2 comentarios
Paul
Paul el 16 de Jun. de 2021
Can you provide complete code that can be replicated that illustrates the problem and is prefably as simplie as possible?
Martin Elenkov
Martin Elenkov el 16 de Jun. de 2021
Editada: Martin Elenkov el 16 de Jun. de 2021
This is a script with the model parameters and equations. The cited expression are under index 9. So, dxdt(9), f_x(9), g_x(9,1:2). It is not simple, nor clean, but I think it is fairly clear what I am doing, asside from the complicated differential equations

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Respuesta aceptada

Paul
Paul el 16 de Jun. de 2021
Why not just use collect() and coeffs() on the elements of dxdt? If the system is affine in u, then those should give the desired result directly
syms x u1 u2
dxdt = x^2 + x + x*u1 + x^3*u1 + x^2*u2 + u2;
dxdt = collect(dxdt,[u1 u2])
dxdt = 
[c,t] = coeffs(dxdt,[u1 u2])
c = 
t = 
fofx = c(find(t==1))
fofx = 
gofx = [c(find(t==u1)) c(find(t==u2))]
gofx = 
If the system is not affine in u, then extra terms will show up in t, so you can check numel(t) before doing any other processing to decide how to proceed.
dxdt = x^2 + x + x*u1 + x^3*u1 + x^2*u2 + u2 + u1*u2;
dxdt = collect(dxdt,[u1 u2])
dxdt = 
[c,t] = coeffs(dxdt,[u1 u2])
c = 
t = 
  5 comentarios
Paul
Paul el 16 de Jun. de 2021
I thought 'all' might also make it easier to determine if f(x) or an element of g(x) is zero. Alas, that appears to not be the case:
syms x u1 u2
dxdt = x*u1;
[c,t]=coeffs(dxdt,[u1 u2],'all')
c = 
t = 
So some logic will still be needed in a situation like this to set the appropriate element of g(x) to zero because u2 doesn't exist in t, even though it was explicitly requested, which is too bad for this use case.
Martin Elenkov
Martin Elenkov el 17 de Jun. de 2021
I also noticed that only the existing elements are returned, but it is an easy work around.

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Más respuestas (1)

John D'Errico
John D'Errico el 16 de Jun. de 2021
This is a floating point problem, but it comes when MATLAB takes your constants and turns them into symbolic numbers. Do you see all of those large integers? For example...
X = sym(1.34536363425474)
X = 
But that ratio is not exactly the same value as the original number. It is typically a close approximation.
vpa(X,40)
ans = 
1.345363634254739926277011363708879798651
So you get trash in there, down in digit 17 and below.
My question is, why is this a problem?
  4 comentarios
Walter Roberson
Walter Roberson el 16 de Jun. de 2021
simplify() asks to rewrite the expression to combine parts if it can figure out how. When it does that, the combined portions mostly lose the information about the number of digits the symbolic floating point numbers were evaluated at
syms Q_b C_carbv
f_x = Q_b * (vpa(83.0,2)*C_carbv - vpa(0.19,2)) - vpa(83.0,2)*Q_b*(C_carbv - vpa(2.3e-3,2))
f_x = 
expand(f_x)
ans = 
Makes sense to me, since - 0.19 is not - 2.3e-3
collect(f_x, Q_b)
ans = 
Martin Elenkov
Martin Elenkov el 16 de Jun. de 2021
Editada: Martin Elenkov el 16 de Jun. de 2021
Notice that in the first term Matlab factors only in front of the brackets, while in the second term, both the coefficient 83 and are factored out of the brackets (They both come from the same equation, Matlab just rewrites and rearranges them differently after some opperations - e.g. diff()).
83*2.3e-3
ans =
0.1909
0.19 is not 2.3e-3, but 83*2.3e-3 is aproximately equal to 0.19, depending on how you define your precision. It seems that Matlab reaproximates these coefficient after each (or at least some) opperations and that leads to this anoying problem.

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