observability staircase form for MIMO system

Question

atefe 2012-9-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/48514-observability-staircase-form-for-mimo-system

I have a problem with 'obsvf',I used this function to decomposes the state-space system with matrices A,C,B as below into the observability staircase form but the result isnot correct,why? A= [-1.0817,0.0477,-0.3793,-0.2781,1.0423,0.1548,-0.0242,0.4453;1.6776,-0.0384,0.7005,-0.0995,-0.5439,-0.0698,0.0345,0.4453;-1.0103,0.0448,-0.4566,-0.2778,1.1249,-0.0698,-0.0093,0.4453;3.8207,-0.5175,1.2384,0.0208,-0.5604,-0.0698,0.0601,-1.6810; 1.6725, 0.0309,0.6954,-0.0989,-361.5704,-0.0698, 0.0601,0.4453;1.6725, 0.0309,0.6954, -0.0989,-0.5604,-361.0798,0.0601,0.4453; 1.6725,0.0309,0.6954,-0.0989,-0.5604,0.0698,-360.9499, 0.4453; 1.6725, 0.0309,0.6954,202.8811,-0.5604,-0.0698,0.0601,-360.5647]; B=[0,0,0,0;0,0,0,0;0,0 ,0,0;0,0,0,0;361.0100,0,0,0;0,361.0100,0,0;0,0,361.0100,0; 0,0,0,-505.4200]; C=[0,0,0,0,3.3046,0,0,0; 0,0,0,0,0,2.2460e-01,0,0; 3.8641e-02,3.8641e-02,3.8641e-02,4.0534e-03,0,0,1.7949e-01,0; 5.4840e-01,5.4840e-01,5.4840e-01,5.7526e-02,0,0,0,2.1263; 2.8517e+01,2.8517e+01,2.8517e+01,2.9914,0,0,0,0; 0,0,0,4.0161e-01,0,0,0,0; 5.5978e-01,-4.9640e-01,-4.9640e-01,0,0,0,0,0; -1.5096e-01,9.0522e-01,-1.5096e-01,0,0,0,0,0; -4.0881e-01,-4.0881e-01,6.4737e-01,0,0,0,0,0; -9.0922e-04,-2.5509e-03,2.0460e-03,0,0,0,6.1523e-03,-5.1355e-03]; [Abar,Bbar,Cbar,T,k]= obsvf(A,B,C); sum(K)=8; but Ob = obsv(A,C); rank(Ob)=4 why????????????????

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Answer 1

Rajiv Singh 2012-10-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/48514-observability-staircase-form-for-mimo-system#answer_63346

OBSV reference page has this tip:

obsv is here for educational purposes and is not recommended for serious control design. Computing the rank of the observability matrix is not recommended for observability testing. Ob will be numerically singular for most systems with more than a handful of states. This fact is well documented in the control literature. For example, see section III in http://lawww.epfl.ch/webdav/site/la/users/105941/public/NumCompCtrl.pdf

So rank test of Ob is not going to be reliable. Note that OBSVF uses a tolerance of 10*n*norm(a,1)*eps, when not specified. This is about 1.0064e-12 for your example. Now if you check rank with this tolerance rank(Ob, 1.0064e-12), you get a value of 8 which agrees with sum(k).