How do i design a polynomial system with simulink :

i have such a system
xdot=f(x)+G(x)K(x)+G(X)v , y=h(x);
with f, G, K and h polynomial functions. I am searshing for a state feedback u=K(x)+v to prove the global asymptotic stability of the equilibrium point. The goal is to find a function V(x) with : V(x)>=0 , grad(f)*V(x)+h'(x)K(x)<=0 , grad(G)V(x)=h'(x).

7 commentaires

Are you trying to model the x_dot system with Simulink? Then I would recommend the Integrator Block of the blockset Continuous and Math Operations/Polynomial
derecho
derecho le 20 Oct 2015
yep i want to model x_dot system with Simulink. f and G are Vectors of Polynomial Functions.I didn't find the desired Block in the Simulink Library Browser.
I don't know if there is a single block for your request but I think you can build it up easily: The block Simulink/Math Operations/Polynomial returns a polynomial of the input u, you can define the coefficients in the block-properties. Now if you have a small-dimensional x-vector you can take a Polynomial-block for each element of f(x), if the dimension is too big probably the For-Each block could help you. But G(x) is a Matrix, isn't it?
derecho
derecho le 21 Oct 2015
derecho
derecho le 21 Oct 2015
Modifié(e) : derecho le 21 Oct 2015
theta[1:4] are constants,S(x)=1, sigma(x)=x1 and k(x) is a polynomial function. I want to model the closed loop (see pic 2) with Simulink
Okay I was thinking of v as a vector. As G is constant, you can use a simple Matrix Gain (in this particular case a vector gain) and I would build up f in a subsystem with input x and output f(x). I hope the pictures will help you to start:
The raw system architecture.
Inside the subsystem. Of the parameters for Theta an x0 you have to declare yourself (you can also do this via a Matlab script which you run before the simulation).
derecho
derecho le 23 Oct 2015
yeess! now it is much better, I got the plots I wanted but with different G (G=[1;1;1]). For the given G=[1;0;0] occure an Error "Derivative input 1 of 'simulink1/Integrator'.." there may be a singularity..
Thanx !! it was helpful:)

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le 19 Oct 2015

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le 23 Oct 2015

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