second order to first order

Hello, I'm trying to convert this system to as described here:

This is my attempt but not sure...any help will be greatly appreciated. Thanks

syms x(t)
eqn = diff(x,2) + diff(x,t)*x == u;
V = odeToVectorField(eqn)

Answer 1

[Sam Chak]

Hi @Cesar Cardenas

I suspect that your 2nd-order ODE was incorrectly written. Please check. If it is a linear damped spring system, then the equation should be:

and it can be converted to the state-space form as shown below:

omega  = 2;
zeta   = sqrt(3)/4;
sympref('AbbreviateOutput', false);

syms x(t) y(t) u
eqn    = diff(x, 2) + 2*zeta*omega*diff(x) + (omega^2)*x == (omega^2)*u;
[V, S] = odeToVectorField(eqn)

V = 

S = 

From the result, and , and so, the state-space model can be constructed accordingly:

A = [0 1; -4 -sqrt(3)];
B = [0; 4];
C = [1 0];
D = 0;
sys = ss(A, B, C, D)

sys =
 
  A = 
           x1      x2
   x1       0       1
   x2      -4  -1.732
 
  B = 
       u1
   x1   0
   x2   4
 
  C = 
       x1  x2
   y1   1   0
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

Answer 2

[Cesar Cardenas]

Hi @Sam Chak, thanks much for your help. I think it is a damped spring system. I would have an additional question.

For part c (based on the image below), this is my attempt: (not sure though). and do not really know how to work out part d regarding discrete time histories. Could you please give me any clue?? Thanks much once again.

%x(t) = 0.5*x(k-1) + y(k-1)
%y(t) = -0.5*x(k) +  y(k)
%x(1) = 1, y(1) = 1

timesteps=30;
x = zeros(1, timesteps);
y = zeros(1, timesteps);
x(1) = 1;
y(1) = 1;


for k=2:timesteps
    x(k) = 0.5*x(k-1) + y(k-1);
    y(k) = -0.5*x(k) +  y(k);
end

plot(1:timesteps,x, '-b', 'Linewidth' ,3);
hold on
plot(1:timesteps,y,'--r', 'Linewidth',3);
xlabel('time');
ylabel('Quantity')
legend('x','y')

figure(2)
plot(x,y)
xlabel('x')
ylabel('y')