2
$\begingroup$

This question is related to an other question I've asked yesterday here. Havin not received a reply, I believe there isn't a ready-made function for non-linear coordinate transformation in order to obtain a normal form of an affine state system as in nonlinear control theory. I wish to address here some of the problems in implementing such a function.

We would have the following non-linear control system:

x' = f(x) + g(x) u

y = h(x)

First step As is well explained in this book Isidori, what is firstly necessary is to define a nonlinear coordinate transformation. In order to do this, we ought to firstly define a Lie derivative. I've implemented the following code to do this:

QDim[a_, b_] := TrueQ[Dimensions[a] == Dimensions[b]]
Lie[l_, f_, x_] := 
 With[{}, If[QDim[f, x] == True, 
   dl = Table[D[l, x[[i]]], {i, 1, Dimensions[x][[1]]}];
   Sum[dl[[i]] f[[i]], {i, 1, Dimensions[x][[1]]}], 
   Print["Error: dimensional mismatch."]]]
DLie[l_, f_, x_, k_] := 
 With[{}, For[i = 0; t = l, i < k, i++, t = Lie[t, f, x]]; t]

Another necessary function to be defined is a function that would calculate the relative degree of the nonlinear control system. This was quite easy as Mathematica has such a function:

RelativeDegree[f_, g_, h_, x_] := 
 With[{}, sys = AffineStateSpaceModel[{f, g, {h}}, x]; 
  r = SystemsModelVectorRelativeOrders[sys]]

Second step. What we ought to do now is define an invertible nonlinear coordinate transformation that satisfies the following conditions. If we call r the relative degree defined above,we must have:

phi_1 = DLie[h,f,x,0]

phi_2 = DLie[h,f,x,1]

...

phi_{r} = DLie[h,f,x,r-1]

As for phi_{k} with r < k <= n = Dimensions[x], these coordinate transformations must fulfill the following condition:

DLie[phi_{k},g,x,1]=0

In order that the transformation:

phi={phi_1, phi_2, ... , phi_n}

has a non-singular Jacobian.

How can I achieve this result? The main issues involve the fact that the condition for phi_{k} involves DSolve and thus I would have "constants" like C[1]. The point would then be to find C[1] such that phi has a non-singular Jacobian.

Example

h = x3;
f = {-x1, x1 x2, x2};
x = {x1, x2, x3};
g = {{Exp[x2]}, {1}, {0}};
r = RelativeDegree[f, g, h, x][[1]];

The first r elements of phi would be:

phir = Table[DLie[l, f, x, i], {i, 0, r - 1}];

As for the remaining n-r elements have to satisfy:

DLie[phi3[x], Flatten[g], x, 1]

This would correctly output:

Derivative[{0, 1, 0}][phi3][{x1, x2, x3}] + E^x2*Derivative[{1, 0, 0}][phi3][{x1, x2, x3}]

I would then attempt to solve this equation with DSolve. I should obtain something like this:

C[1][E^x2 - x1]

Yet I wouldn't know how to ask Mathematica to choose a C1 that satisfies the non-singularity of the Jacobian of phi.

A correct answer to this example is:

phi = {x3, x2, 1+x1-E^x2}
$\endgroup$
2
$\begingroup$

First find the desired transformation:

u[x1, x2, x3] /. 
DSolve[{D[u[x1, x2, x3], x1] Exp[x2] + D[u[x1, x2, x3], x2] == 0} , 
       u[x1, x2, x3], {x1, x2, x3}][[1]]
trans = % /. C[1][x3] :> (1 - # &)

Then use StateSpaceTransform:

StateSpaceTransform[AffineStateSpaceModel[{f, g, h}, x], 
                   {Automatic, {z1 -> x3, z2 -> x2, z3 -> trans}}]

enter image description here

$\endgroup$
  • $\begingroup$ Thanks for your answer. I would like to ask you how may I achieve this directly from my Lie Derivative: 'DLie[phi3[x], Flatten[g], x, 1]' The problem seems to be that the argument of the phi3 is {x1,x2,x3}: phi3[{x1,x2,x3}]... this messes up DSolve. How can get around this? $\endgroup$ – Mirko Aveta Nov 7 '16 at 16:56
  • $\begingroup$ I don't understand what you are asking. 'DLie[phi3[x], Flatten[g], x, 1]' evaluates to $\text{x1}+e^{\text{x2}} \text{x2}$. $\endgroup$ – Suba Thomas Nov 7 '16 at 17:26
  • $\begingroup$ It shouldn't. It should keep phi3 as an unknown to be found with DSolve as you did with u. @Suba Thomas. Since my final goal is to implement a whole function that does all of this, I wanted to obtain the transformed state space form by simply assigning f,g,h,x. $\endgroup$ – Mirko Aveta Nov 7 '16 at 18:53
  • $\begingroup$ I apologize, I have realized your answer perfectly fulfills my request. Thanks again for helping. $\endgroup$ – Mirko Aveta Nov 10 '16 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.