system unstable although its poles are negative values

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Mohamed
Mohamed 2021-10-2
回答： Star Strider 2021-10-2
Ran in:
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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回答（1 个）

Star Strider
Star Strider 2021-10-2
Ran in:
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.
.

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