# Sliding mode observer gain -- How do I find it?

Hi,
I have a 2nd order of sliding mode observer of the form:
$$\dot{\hat{x}} = \hat{f}(x,t) + \delta f + \Psi(u,y) +[ \frac{d \Omega}{dx}]^{-1} \upsilon$$
where ##\upsilon##:
$$\upsilon_1= \alpha_1 \lambda_1^{1/2} | y_1 -\hat{x}_1|^{1/2}*sign(y_1 -\hat{x}_1)$$
$$\upsilon_2= \alpha_2 \lambda_1*sign(\upsilon_1)$$
$$\upsilon_3= \alpha_3 \lambda_2^{1/2} | y_2 -\hat{x}_2|^{1/2}*sign(y_2 -\hat{x}_2)$$
$$\upsilon_4= \alpha_4 \lambda_2*sign(\upsilon_3)$$
....
where ##y_{1,2}## is the output and and ##x_{1,2}## are the estimations of measurable outputs. How do I find the correct gain of sliding mode observer?
##\delta f## is the uncertainties and ##[ \frac{d \Omega}{dx}]^{-1} ## is some mathematical stuff based on transformations. The main question is how do you find the proper gain of an observer by linearizing a system?
Do I linearize the plant and observer? and then what graphs should I look at to find the appripriate gains?

donpacino
Gold Member
Can you convert your equations into state space format? It would be MUCH easier to solve if you can do that.
Is this for a class on non-linear controls, or just in general?

If this is a nonlinear controls problem, and the sign function implies, then yes you'll have to linear to get the function into state space format.

You would solve for an observer pretty much the same way you solve for a controller. You determine your desired eigenvalues and work backwards to determine the gains required. Ackermans formula for example is one way to do it.

donpacino
Gold Member
Do you know how to compute your Jacobin matrix?