system unstable although its poles are negative values

Question

Mohamed 2021-10-2

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1465124-system-unstable-although-its-poles-are-negative-values

回答： Star Strider 2021-10-2

Hello friends

anyone has an idea why my system is ustable although its poles are located on the left hand side with negative values as you can see?

I am not be able to design an observer or controller for it.

rs=0.84; Vin=40; Vfd=0; rL=.695; C=46.27e-6; Ra=1.73; La=1.17e-3; B=0.000138; Jq=0.000115; 
kt=0.0663; ke=0.0663; Tfr=0.0284;  L=2.473e-3; D=0.6; Tl=20e-3; 
%   syms rs Vin Vfd rL C Ra La B Jq kt ke Tfr L D Tl R
 A =[   (B*(D - 1))/Jq - (B*D)/Jq, (D*kt)/Jq - (kt*(D - 1))/Jq,  0,  0;...
 (ke*(D - 1))/La - (D*ke)/La, (Ra*(D - 1))/La - (D*Ra)/La, D/La - (D - 1)/La,0;...
   0,  (D - 1)/C - D/C, 0, D/C - (D - 1)/C;...
   0,       0,   (D - 1)/L - D/L, (rL*(D - 1))/L - (D*(rL))/L]
A = 4×4
	1.0e+04 *

   -0.0001    0.0577         0         0
   -0.0057   -0.1479    0.0855         0
         0   -2.1612         0    2.1612
         0         0   -0.0404   -0.0281
     
     BB=[((Tfr + Tl)*(D - 1))/Jq - (D*(Tfr + Tl))/Jq 0;...
                                           0 0;...         
                                           0 0;...
                 0 (Vfd*(D - 1))/L + (D*Vin)/L]
BB = 4×2
	1.0e+03 *

   -0.4209         0
         0         0
         0         0
         0    9.7048
     Egg=eig(A)
Egg = 
   1.0e+03 *

  -0.5434 + 5.1597i
  -0.5434 - 5.1597i
  -0.0174 + 0.0000i
  -0.6567 + 0.0000i
     stablity= istable(A)
stablity = logical
   0
   

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Answer 1

Star Strider 2021-10-2

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https://ww2.mathworks.cn/matlabcentral/answers/1465124-system-unstable-although-its-poles-are-negative-values#answer_799849

在 MATLAB Online 中打开

You nare not giving the isstable function the correct input.

Try this instead —

rs=0.84; Vin=40; Vfd=0; rL=.695; C=46.27e-6; Ra=1.73; La=1.17e-3; B=0.000138; Jq=0.000115;

kt=0.0663; ke=0.0663; Tfr=0.0284; L=2.473e-3; D=0.6; Tl=20e-3;

% syms rs Vin Vfd rL C Ra La B Jq kt ke Tfr L D Tl R

A =[ (B*(D - 1))/Jq - (B*D)/Jq, (D*kt)/Jq - (kt*(D - 1))/Jq, 0, 0;...

(ke*(D - 1))/La - (D*ke)/La, (Ra*(D - 1))/La - (D*Ra)/La, D/La - (D - 1)/La,0;...

0, (D - 1)/C - D/C, 0, D/C - (D - 1)/C;...

0, 0, (D - 1)/L - D/L, (rL*(D - 1))/L - (D*(rL))/L]

A = 4×4

1.0e+04 * -0.0001 0.0577 0 0 -0.0057 -0.1479 0.0855 0 0 -2.1612 0 2.1612 0 0 -0.0404 -0.0281

BB=[((Tfr + Tl)*(D - 1))/Jq - (D*(Tfr + Tl))/Jq 0;...

0 0;...

0 (Vfd*(D - 1))/L + (D*Vin)/L]

BB = 4×2

1.0e+03 * -0.4209 0 0 0 0 0 0 9.7048

Egg=eig(A)

Egg =

1.0e+03 * -0.5434 + 5.1597i -0.5434 - 5.1597i -0.0174 + 0.0000i -0.6567 + 0.0000i

sys = ss(A,BB,ones(1,size(A,2)),0)

sys = A = x1 x2 x3 x4 x1 -1.2 576.5 0 0 x2 -56.67 -1479 854.7 0 x3 0 -2.161e+04 0 2.161e+04 x4 0 0 -404.4 -281 B = u1 u2 x1 -420.9 0 x2 0 0 x3 0 0 x4 0 9705 C = x1 x2 x3 x4 y1 1 1 1 1 D = u1 u2 y1 0 0 Continuous-time state-space model.

figure

pzmap(sys)

figure

stepplot(sys)

Test = isstable(sys)

Test = logical

1

And as expected from the pole-zero plot and the step response, the system is stable.

.