Plotting response of a system with polynomial matrices - matlab

Question

How can I plot the time response of a system when the system, input, output matrices are polynomials? For example
A(x) = [0.59*x 1.67*x; 0.1 0.2]
B(x) = [2.3*x; 0.3]
C = [1 0]
Operating region of x = [-2, 2]
Initial condition x(0) = [2 0]
If the matrices were constant, I could use ss and lsim to plot it. But, how do I do it in this case? I am new to Matlab and control systems.

Answer 1

Using your example, I will run you through the basic idea of general ODE simulation in MatLab (which is very similar across programming languages).
(First though, I am assuming that the x you have written in your A and B matrices is actually x1 since you did not specify. I am also assuming from context that your state vector is [x1 x2]).
Create a function that takes in the
current time, t_now
current state vector, x_now
full input array, u
full time array, t
and uses them to compute the state derivative (xdot). The full time array (t) will only be needed for indexing the control input (u), as you will see. Basically, it will be a coded function for
xdot = f(t, x, u)
which is the general form of an ODE.
function xdot = myODE(t_now, x_now, u, t)
A = [ 0.59*x_now(1), 1.67*x_now(1);
0.1, 0.2 ] ;
B = [ 2.3*x_now(1);
0.3 ] ;
u_now = interp1(t, u, t_now) ; % get u(t=t_now)
xdot = A*x_now + B*u_now ;
end
Next, create a script that runs the simulation using an ODE solver like MatLab's ode45. If you want to know how these solvers work, read up on numerical integration. MatLab's ode45 uses a Runge-Kutta method.
%// time range over which to compute simulation
t = [0 : 0.01 : 5] ;
%// define the input signal
%// why not make it zero to look at the natural response
u = zeros(1, length(t)) ;
%// initial condition
x0 = [2, -5] ;
%// call ode45
%// first argument: anonymous call to myODE that tells it which inputs to use
%// second argument: the time range for the simulation
%// third argument: the initial condition
%// first output: the sim time, may be more detailed than the original t
%// second output: the full output including all states
[t_out, response] = ode45(#(t_now, x_now) myODE(t_now, x_now, u, t), t, x0) ;
%// The response contains two column vectors: one for x1 and one for x2 at
%// every time in t. You can extract "the true output" which based on
%// your C matrix is x1.
y = response(:,1) ;
%// Plot results
plot(t, u, 'r--', t_out, y, 'k') ;
grid on ;
legend('input', 'output')
xlabel('time')
ylabel('value')
Suppose you don't have u as a predefined input signal, but rather as a function of the current state. Then simply modify the computation of u_now in the myODE function. In the simplest case, u is not a function of time at all, in which case you don't even need to pass u or the full time array into myODE at all! For example, if
u := -k*x1
then your ODE function can be
function xdot = myODE(t_now, x_now)
A = [ 0.59*x_now(1), 1.67*x_now(1);
0.1, 0.2 ] ;
B = [ 2.3*x_now(1);
0.3 ] ;
k = 5 ;
u_now = -k*x_now(1) ;
xdot = A*x_now + B*u_now ;
end
with the call to ode45 being
[t_out, response] = ode45(myODE(t_now, x_now), t, x0) ;
and no need to define u in the script.
Hope that helps!