trying to use the command hinfsyn on a rotary inverted pendulum for stabilization analysis

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Hello
I am trying to follow this paper
I have made following code for making the system looking like this
The code is given below
A=[0 0 1 0;
0 0 0 1;
0 39.32 -14.52 0;
0 81.78 -13.98 0;];
B=[0;0;25.54;24.59];
C=[1 0 0 0;
0 1 0 0];
D=[0;0;];
sys=ss(A,B,C,D);
sys.InputName={'u'};
sys.OutputName={'theta','alpha'};
s=tf('s');
Wact=0.002*(s+0.01)/(s+10);
Wact.u='u'; Wact.y='e1';
Wn1=ss(0.05);
Wn1.u='d1'; Wn1.y='wn1';
Wn2=ss(0.0275);
Wn2.u='d2'; Wn2.y='wn2';
Wx1=151.5/(s+50.5);
Wx1.u='theta'; Wx1.y='e2';
Wx2=202/(s+50.5);
Wx2.u='alpha'; Wx2.y='e3';
sdmeas=sumblk('y1 = theta+wn1');
abmeas=sumblk('y2 = alpha+wn2');
ICinputs={'d1','d2','u'};
ICoutputs={'e1','e2','e3','y1','y2'};
qsys=connect(sys,Wact,Wn1,Wn2,Wx1,Wx2,sdmeas,abmeas,ICinputs,ICoutputs);
When I am running the command hinfsyn I get following gamma
[~,~,gamma]=hinfsyn(qsys,2,1)
gamma = 0.3644
while in the paper they have
gamma=0.6951 in section 3.1 H-infinty controll. Where is my mistake or any solution to get the same answer as in the paper ?
Thanks for all answers
Tor Erik Haavik
  3 Comments
CyberRobotics
CyberRobotics on 19 Jan 2025
Edited: CyberRobotics on 19 Jan 2025
So do think that a gamma of 0.61 and if I have understand correct my code works correctly just a bad article explaining the stuff?
Paul
Paul on 19 Jan 2025
Edited: Paul on 19 Jan 2025
Based on the article, it's hard to say what is "correct" because the results from the article are not able to be replicated, at least for me. Every time I look at a different section I have more concerns. I'm really curious about how the authors developed Figure 6.

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