# Finding the state space model of a nonlinear system

## Homework Statement

The state space model of a nonlinear system is
$$x'_1(t) = 2x^2_2(t) - 50$$
$$x'_2(t) = -x_1(t) - 3x_2(t) + u(t)$$

Where $$x_1(t)$$ and $$x_2(t)$$ are the states, and u(t) is the input. The output of the system is $$x_2(t)$$.

Find the state space model of this system linearized at the equilibrium point (-15, 5)

## Homework Equations

State space modeling, Jacobian Matrix(?)

## The Attempt at a Solution

I think my attempt lies on false assumption, but here's what I did.

Since we're linearizing at a known equilibrium point for a nonlinear system, and since $$x'_1(t) = 2x^2_2(t) - 50$$ contains non linear term, we must first take a derivative of it as denoted in Jacobian matrix. derivative of x_1(t) is

$$x''_1(t) = 4x_2(t)$$

So the state space model we get is

[x'_1] = [ 0 4 ] [x_1] + [ 0 ] u(t)
[x'_2]aa[-1 -3 ] [x_2] a[ 1 ]

(Sorry I have no clue how to put this matrix in Latex).

But then how do we get the output from this state space model?

## Answers and Replies

pasmith
Homework Helper
The linearization about (-15, 5) is obtained by evaluating the Jacobian matrix at (-15,5). Here the Jacobian matrix is $$J(x_1,x_2) = \begin{pmatrix} 0 & 4x_2 \\ -1 & -3 \end{pmatrix}$$ and the latex code is
Code:
J(x_1,x_2) = \begin{pmatrix} 0 & 4x_2 \\ -1 & -3 \end{pmatrix}

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The linearization about (-15, 5) is obtained by evaluating the Jacobian matrix at (-15,5). Here the Jacobian matrix is $$J(x_1,x_2) = \begin{pmatrix} 0 & 4x_2 \\ -1 & -3 \end{pmatrix}$$ and the latex code is
Code:
J(x_1,x_2) = \begin{pmatrix} 0 & 4x_2 \\ -1 & -3 \end{pmatrix}

So the evaluation at (-15,5) becomes

$$y'= \begin{pmatrix} 0 & 20 \\ -1 & -3 \end{pmatrix}$$

?

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