Solve state space equation and initial conditions

Hello everybody.

I am trying to implement an adaptive filter identification system. The mass is variable so I have to calculate with Matlab (no Simulink) each transfer function for each mass. The problem is the continuity: I have all the state spaces but I can not solve de last "position" in order to introduce it as initial condition in the next state space.

I have state space equation like:

xdot=A.x+B.u(t) where u=sen(t)

y= C.x where C is [0 C2] so I can solve x2 because I know the value of the amplitude y(t).

The problem is that I can not solve x1, I dont know how to use ode45 for this.

The purpose is to use lsim(sys,u,t,x0) where x0=[x1 x2]

How could I resolve the value of x1 in the last position of my last state space?

A =                                         
        x1     x2
 x1     -4  -12.5
 x2      8      0

B = 
     u1
 x1   4
 x2   0

C = 
        x1     x2
 y1      0  3.125

D = 
     u1
 y1   0

As aspicted in the attached figure, the next state space starts in the correct position (last state space position) because I solved x2 but not with the proper velocity and acceleration because I didnt solve x1. The mass changes in each vertical line, the mass change point is marked with a circle

[Aquatris]

Is it not possible for you to change the C such that it gives you both x1 and x2?

[Eloy Pena Asensio]

No, I have already tried. It doesnt work. Changing the C changes the responses.I think that only works for first order.

[Aquatris]

Wrong. It works for any order. If changing the C matrix (adding another row) changes the response of your system (assuming you compare the same output channel), then you are doing something wrong.

Also are you combining 2 different simulation results together instead of running a single simulation?

[Eloy Pena Asensio]

What do you mean with "adding another row"?

[M]

Adding another row in the matrix C such that you get both x1 and x2, something like:

C=[0 3.125;
   1 0];

This should not change the first output, but it will add a second one, corresponding to the evolution of x1.

[Eloy Pena Asensio]

And how could I solve this equation? y= [0 1;3.125 0][x1 x2]

[Aquatris]

You use the same method as the one you use when C only had 1 row. Why are you confused?

[Eloy Pena Asensio]

And how could I solve this equation? y= [0 1;3.125 0][x1 x2]

Because in the firt example I only have to solve the equation and the result is a value. If i add a row to [C] the solution would be a differential equation.

[Aquatris]

First, the thing you solve with ode is xd = Ax+Bu, which is the differential equation part of state space. The part y = Cx is not a differential equation. It is a simple linear equation. The only purpose of y = Cx is to extract the output value using the states. I recommend you review state space again.

[Eloy Pena Asensio]

But the initial condiction x0, x1 express the velocity and the position. That is why i am confusse about to extrat from y=Cx. What parameter should I uso in y to extract x1?

[Aquatris]

That depends on your system. How did you find the "3.125" number in your C matrix to get the position? You will use the same method to get C value for velocity.

Answer 1

[Akshay Khadse]

For using “ode45”, you will need to create a MATLAB function independent of “u” by substituting “sin(t)” in its place. Then, the “ode45” could be used as

[t,x] = ode45(@functionName,tspan,y0);

where “tspan = [t0 tf]” is the vector with start and end time of the solution and “y0” is the initial condition vector

Examples regarding these can be viewed in documentation of ode45 .