system unstable although its poles are negative values

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

Answer 1

[Star Strider]

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 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.

.