State Space, Finding Characteristic equation

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Joseph
Joseph am 9 Nov. 2014
Kommentiert: fathima sugal am 20 Okt. 2021
Having trouble converting into state space. I also need to use matlab to produce the characteristic equation. here is my code
syms ks cs
A = [0 1 0 0; (ks+2000)/30 (cs+5)/30 -ks/30 -cs/30; 0 0 0 1; -ks/500 -cs/500 ks/500 cs/500];
B = [0 0 ; 66.7 .167; 0 0; 0 0];
C = [-ks/500 -cs/500 ks/500 cs/500; -1 0 0 0; -1 0 1 0];
D = [ 0 0; 1 0 ;0 0];
sys = ss(A,B,C,D);

Antworten (1)

Star Strider
Star Strider am 10 Nov. 2014
The characteristic polynomial is easy enough with the charpoly function:
CEq = charpoly(A);

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