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

3 visualizzazioni (ultimi 30 giorni)
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 Commenti
CyberRobotics
CyberRobotics il 19 Gen 2025
Modificato: CyberRobotics il 19 Gen 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 il 19 Gen 2025
Modificato: Paul il 19 Gen 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.

Accedi per commentare.

Risposte (0)

Categorie

Scopri di più su Robust Control Toolbox in Help Center e File Exchange

Prodotti


Release

R2024b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by