Electronic – State space modeling with disturbances using Matlab

control systemMATLABmodeling

I am doing a project which involves modeling a quadcopter in state space form using matlab. I first had to linearize the quadcopter equations of motion, and now have to implement them in a matlab state space system.

In order to do this I need to put the linearized differential equations in this form:

$$
\dot{x} = Ax + Bu
$$
$$
y = Cx
$$

The issue I am having is that one of my state equations has a constant term because of acceleration due to gravity. The equation in question is shown below, with Vz, v1^2, v2^2, v3^2, and v4^2 all being state variables, and everything else constants:

$$
\dot{V_{z}} = -\frac{K_{d}}{m}V_{z} + \frac{KC_{m}}{m}(v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}) – g
$$

How am I supposed to accompany this term (-g) into matlab's state space model?

My only thought is to use the state space model with identifiable parameters, and include the constant in the disturbance matrix K, but I am unsure of how to go about that, never having used a disturbance matrix before.

Best Answer

You have a nonzero operating point. If you had chosen an equilibrium operating point, this situation would not have occurred.

With the equation you have given, if you choose the states as \$\left\{x_1=\frac{g m}{K_d}+V_z,x_2=v_1^2,x_3=v_2^2,x_4=v_3^2,x_5=v_4^2\right\}\$ you get a linear state equation:

\$ \dot{x}_1=\frac{K \left(x_2+x_3+x_4+x_5\right) C_m}{m}-\frac{x_1 K_d}{m} \$

I am guessing this would upset some other state equation. Then you have to consider all the states and equations together. But, as I said before the best approach is to do this at the linearization step.

Related Solutions

Electronic – Brushless motor controlled via current source math model

If you are driving the motor with a known voltage, the page you linked to gives you the answer so I assume you want to drive the motor with a controlled current.

Since the motor's torque is proportional to current, you can forget the resistance, inductance and back-emf constants.

\$T = K.i\$

and

\$T=J\ddot{\theta}+b\dot{\theta}\$

this can be expressed as

\$s(Js+b)\theta=Ki\$

(from your linked article), so

\$\theta = \dfrac{Ki}{s(Js+b)}\$

and

\$\dfrac{\dot{\theta}}{i} = \dfrac{K}{Js+b}\$

Of course any current source has a finite voltage compliance, so you will also need to compute the voltage as a function of current to ensure that the current source stays under control.

Electronic – How to model this linear power system in MATLAB

To solve this problem you should look at the ODE solvers in MATLAB.

Note that your function does not describe what the response of the system is to a variation in input power Pm.

Below I have written up an implementation, based on the example in http://www.its.caltech.edu/~ae121/ae121/ode45_Ref2.pdf . In my example I have added a term Pm to the differential equations, so you can see how a disturbance might affect the dynamics.

Unfortunately I don't have MATLAB installed so I can't check whether there are any errors in this, but the idea should be clear.

function main 
% a simple example to solve ODE's
% Uses ODE45 to solve 
%    dx_dt(1) = 1*x(1)+2*x(2)+3*x(3)+4*x(4)
%    dx_dt(2) = 5*x(1)+6*x(2)+7*x(3)+8*x(4)+Pm
%    dx_dt(3) = 9*x(1)+1*x(2)+2*x(3)+3*x(4)
%    dx_dt(4) = 4*x(1)+5*x(2)+6*x(3)+7*x(4)
%set an error
options=odeset('RelTol',1e-6);

%initial conditions
X0 = [0;0;0;0]; 
Pm0=1;

%before disturbance
tspan1 = [-1,0]; 
[t1,X1] = ode45('system',tspan1,X0,options,Pm0);

%during disturbance
tspan2 = [0,1];
[t2,X2] = ode45('system',tspan2,X1(end),options,1.3*Pm0);

%after disturbance
tspan3 = [1,3];
[t3,X3] = ode45('system',tspan3,X2(end),options,Pm0);

time=[t1,t2,t3];
X=[X1,X2,X3];

%plot the results
figure 
hold on
plot(time,X)
legend('x1','x2','x3','x4');ylabel('x');xlabel('t') 
return

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [dx_dt]= system(t,x,Pm) 
%a function which returns a rate of change vector
A = [1,2,3,4;
     5,6,7,8;... 
     9,1,2,3;
     4,5,6,7]
P=[0;Pm;0;0]
dx_dt = A*x+P; 
return

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