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)
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 Kommentare
Paul
am 19 Jan. 2025
Bearbeitet: Paul
am 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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