# How to transform a second order ODE to the format xddot=A.X+B.U

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### Answers (3)

Star Strider
on 30 Jan 2023

Those should work if you want to solve them using the MATLAB ordinary differential equation integrators.

If you want to use them with the Control System Toolbox, that will require a different approach.

##### 2 Comments

Star Strider
on 6 Feb 2023

Edited: Star Strider
on 7 Feb 2023

The linear control systems assume that all the derivatives are first-degree. The odeToVectorField function will put them in that form, and the second ‘Subs’ output will give the row order of the variables.

Adding:

sympref('AbbreviateOutput',false);

after the initial syms call might also be helpful.

EDIT — (7 Feb 2023 at 11:54)

Another option (that I do not often use and so forgot about) is the odeFunction function. That might be more appropriate here.

.

Paul
on 6 Feb 2023

Edited: Paul
on 6 Feb 2023

Hi Ítalo,

Let p = x and q = xdot. Then the standard, first order state space model is

[pdot;qdot] = [zeros(n) eye(n); A zeros(n)] * [p ; q] + [0*B; B] * u;

For example, let A = 2 and B = 3. The output of the system is x.

A = 2; B = 3;

hss = ss([0 1;A 0],[0*B;B],[1 0],0)

htf = tf(hss)

We see from the transfer function that the input/ouput ode is: ydddot = 2*y + 3*u, exactly as it should be because y = x.

##### 0 Comments

Sam Chak
on 7 Feb 2023

The odeToVectorField() should work. But since you already have the linear 2nd-order system in this ODE form , then you should be able to readily transform it to the State-space form through some basic matrix operations:

, where and

Then, the optimal gain can be obtained from the lqr() function.

Check this example if this is something you are look for. You can also try applying the sparse 2nd-order state-space model as recommended by @Paul.

A = magic(4)

B = diag([2, 3, 5, 7])

n = size(A)

Aa = [zeros(n) eye(n);

A zeros(n)]

Bb = [zeros(n); B]

K = lqr(Aa, Bb, eye(2*n), eye(n))

##### 2 Comments

Sam Chak
on 8 Feb 2023

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