Forward euler calculation of linearized non-linear induction motor model

Hello,

I am desperately trying to simulate an induction motor model (non-linear) by linearizing it (via calcualtion of Jacobians) and then discretizing it via Euler forward method. You can find the code in the file ASM_discreteModelMatlab.m

I somehow can not get any correct results. I can simulate the non-linear model with ode45 and get correct results.

I calculate the the linearized state system matrizes with the file TOOLS_jacobian.m and update it at each time step during simulation in ASM_discreteModelMatlab.m

I do not know if I do something wrong when I am linearizing the model because linearization is defined as (from https://de.mathworks.com/help/slcontrol/ug/linearizing-nonlinear-models.html):

""

Extending the concept of linearization to dynamic systems, you can write continuous-time nonlinear differential equations in this form:

x'(t)=f(x(t),u(t),t)y(t)=g(x(t),u(t),t).

In these equations, x(t) represents the system states, u(t) represents the inputs to the system, and y(t) represents the outputs of the system.

A linearized model of this system is valid in a small region around the operating point t=t0, x(t0)=x0, u(t0)=u0, and y(t0)=g(x0,u0,t0)=y0.

To represent the linearized model, define new variables centered about the operating point:

δx(t)=x(t)−x0

δu(t)=u(t)−u0

δy(t)=y(t)−y0

The linearized model in terms of δx, δu, and δy is valid when the values of these variables are small:

δx'(t)=Aδx(t)+Bδu(t)δy(t)=Cδx(t)+Dδu(t)

""

For me it is not clear how to handle δx(t) because I linearize at x0 and x0 is equal to x(k). When I want to set δx(k) = x(k) - x0 it results to 0. Thats why I just set δx(k) = x(k). As you can see I do have some struggles understanding the linearization approach.

[Siddharth Jawahar]

Hi Henning,

It would be good if you can share your model and explain the reproduction steps and the end result which you want after linearizing the model.

Sid