I have just followed the trim() example in Matlab documentttion but I kind of strugling to understand how the equilibrium point is found for the second case where x = [1;1] and u=[1;1].
The example is as follow:
state space in simulink:
A = [-0.09 -0.01; 1 0];
B = [ 0 -7; 0 -2];
C = [ 0 2; 1 -5];
D = [-3 0; 1 0];
then
[x,u,y,dx,options] = trim('sys') % calling simulink model
fir example is to find the equilibrium point of the system. the answer, as expected, x = [0;0]and u = [0;0] as they make derivative of x =0
However the second example states that if another point is given, like x=[1;1] and u =[1;1]
the answer is x = [0;0] and u = [1/3;0].
I was able to calculate manually the poitn x = [0;0] using the state space equation but I CANNOT get the "u"?... How to calculate this "u"?
any clue please?
The example is as follow:
state space in simulink:
A = [-0.09 -0.01; 1 0];
B = [ 0 -7; 0 -2];
C = [ 0 2; 1 -5];
D = [-3 0; 1 0];
then
[x,u,y,dx,options] = trim('sys') % calling simulink model
fir example is to find the equilibrium point of the system. the answer, as expected, x = [0;0]and u = [0;0] as they make derivative of x =0
However the second example states that if another point is given, like x=[1;1] and u =[1;1]
the answer is x = [0;0] and u = [1/3;0].
I was able to calculate manually the poitn x = [0;0] using the state space equation but I CANNOT get the "u"?... How to calculate this "u"?
any clue please?