How to shift state space models by equilibrium points using ss and lsim?

Hi everybody,

I linearized the nonlinear ODEs describing the dynamics of the planar quadrotor at the equilibrium point: xe = [y1 z1 0 0 0 0]. Then I proceeded to put my equations in state space form x_dot = Ax + Bu and y = Cx. I obtained my A,B,C,D matrices and pass them to Matlab function ss like so:

Sys = ss(A,B,C,D). Then I pass it to lsim to simulate like so:

[Outputs, time, states] = lsim(sys, u, time, x0).

My question is: how do I make sure that all my states and my initial conditions are shifted by the equilibrium points? I understand that my state variables are now written in terms of deviation from the equilibrium point and so: y_tilde = y - ye, z_tilde = z - ze, and for the other 4 state variables x_tilde = x, since xe = 0 for all the other states.

Do I have to manually shift the initial condition x0 to make it x0_tilde before passing it to ss(A,B,C,D)? And how about the outputs from lsim do I have to shift those by xe? Please ask if I can clarify anything I understand how to define the state space system on paper but have no idea how matlab deals with it. Thanks!

Answer 1

[Paul]

The states, inputs, and outputs of Sys are all perturbations from the equilibrium solution. The intiial condition to lsim (x0) would then be an intiial perturbation from the equilibrium condition. The Outputs and States from the call to lsim are the perturbations to the equilibrium outputs and states. The input u to lsim is the perturbation to the equilibrium solution input.

[Ashwin Vadivel]

Hi Paul,

Please correct me if I am misunderstanding your answer. Does that mean the initial condition to lsim would have to be adjusted as such: x_tilde0 = x0 - xe0 --> initial perturbation from equilbrium is equal to the initial condition minus the value of the equilbrium state at time 0?

And finally the outputs from lsim (outputs_y, time, states_x), would these have to have the value of the equilibrium points added back to them to convert them from perturbations from equilibrium back to global coordinates for plotting purposes? --> so states_x_global = states_x + xe?

Thank you so much for your help, I truly appreciate it!

[Ashwin Vadivel]

So for example lets say my initial condition for a 2 state system x0 = (z0, z0_dot) = [50, 50]. The system is linearized about xe = (ze, ze_dot) = [10,10].

Then when calling lsim:

1) will the initial condition passed be x0_tilde = x0 - xe = [40, 40]?

2) and the output states from lsim lets call them x_states. To retrieve the global location of the states (x), will I have to do:

x = x_states + xe?

Sorry for the confusing syntax its just these perturbations are making my head spin :/

[Paul]

What is meant by specifying both an initial condition and an equilibrium point? I thought the context of this problem is to start with a nonlinear differential equation

xdot = f(x,u)

Now we want to find the linear approximation to this sytem around a solution to the nonlinear differential equation. One common solution around which to linearize the equation is: u(t) = 0, x(t) = xe, where xe is defined such that

f(xe,0) = 0

assuming that such a solution exists. So I don't understand how the inital condition applies. Perhaps the problem of interest is different than the problem as I understand it.

[Ashwin Vadivel]

Thanks Paul for the clarification. So in my case I am linearizing a quadrotor about it's hover configuration. In this case xdot = f(x,u) is set equal to 0 to find the equilibrium points. At the equilibrium point all derivatives of x are equal to 0 so the vehicle has no velocity in any direction in the state space. However at equilibrium, the position (states=x,y,z) are defined as non-zero since the vehicle has to be hovering above ground. Theoretically x and y can equal 0 but z cannot.

So state vector is x = [x y z phi theta psi x_dot y_dot z_dot phi_dot theta_dot psi_dot]. Where x,y,z are positions in euclidean space and phi,theta,psi are roll,pitch and yaw angles. So xdot = f(x,u) = 0 sets all velocities and accelerations (xdot ydot zdot phidot thetadot psidot + 2nd derivates of all these states) = 0. However states x,y,z are non-zero.

Actually, I did as I suggested in my previous comment(based on your advice) and the result looks good!

[Paul]

Right. You found an equilibrium point at some particular value of the state vector xe. Though unstated, presumably the input at the equilibrium is ue = 0. The output at equilibrium is ye. Now linearize around the xe and ue and ye to develop the LTI state space model sys. The state, input, and output of sys are perturbations to xe, ue, and ye respecitvely. After running lsim, the "global" state xglobal is approximated by xe + x, where x is the state variable of sys. Similarly for yglobal and uglobal. These approximations are valid for small perturbations from the equilibrium point.

[Ashwin Vadivel]

Yes makes perfect sense thanks for the help!