How do I find the the zero state diretion and the zero input direction of a transfer matrix in Matlab?

11 vues (au cours des 30 derniers jours)
I want to compute using matlab the Rosenbrock System Matrix, and in particular I am trying to do a an analysis similar to the one at the end of http://people.duke.edu/~hpgavin/ce263/zeros.pdf . So, starting from a transfer matrix I would like to show that if there is a zero at a certain position, there is a zero blocking property.
To do so, as written in the link, I have to find the zeros state direction and the zero input direction, and I am having troubles doing do.
I am considering a different system from the one in the link, so I am trying to do a similar analysis applied to another system, and I am doing this:
s = tf('s');
P = 1/(s+5);
C = 6/s;
S = 1/(1+P*C);
T = P*C/(1+P*C);
G_2 = [T S; S -S]; %transfer matrix
Now, I have that my transfer matrix is :
G_2
and for it I would like to do the analysis to find the zero state diretion and the zero input direction, but I am having troubles doing so.
Can somebody please help me? Thanks in advance.

Réponse acceptée

Star Strider
Star Strider le 11 Fév 2020
I have not done anything with Rosenbrock System Matrices since my multivariable control course in graduate school. We used the Maciejowski textbook, that I still have.
To reproduce that analysis:
s = tf('s');
P = [2/(s^2+3*s+2) 2*s/(s^2+3*s+2); -2*s/(s^2+3*s+2) -2/(s^2+3*s+2)]
S = ss(P)
Smr = minreal(S)
A = Smr.A
B = Smr.B
C = Smr.C
D = Smr.D
trz = tzero(A,B,C,D)
RSM_1 = [eye(3)-A B; -C D]
rRSM_1 = rank(RSM_1)
producing:
P =
From input 1 to output...
2
1: -------------
s^2 + 3 s + 2
-2 s
2: -------------
s^2 + 3 s + 2
From input 2 to output...
2 s
1: -------------
s^2 + 3 s + 2
-2
2: -------------
s^2 + 3 s + 2
Continuous-time transfer function.
S =
A =
x1 x2 x3 x4
x1 -3 -2 0 0
x2 1 0 0 0
x3 0 0 -3 -2
x4 0 0 1 0
B =
u1 u2
x1 2 0
x2 0 0
x3 0 2
x4 0 0
C =
x1 x2 x3 x4
y1 0 1 1 0
y2 -1 0 0 -1
D =
u1 u2
y1 0 0
y2 0 0
Continuous-time state-space model.
1 state removed.
Smr =
A =
x1 x2 x3
x1 -1.222 -0.4444 -0.5556
x2 1.556 -2.889 -1.111
x3 -1.556 0.8889 -0.8889
B =
u1 u2
x1 1 0.3333
x2 -1 1.667
x3 1 0.3333
C =
x1 x2 x3
y1 1 1 -6.939e-16
y2 -0.3333 0.3333 -1.333
D =
u1 u2
y1 0 0
y2 0 0
Continuous-time state-space model.
A =
-1.2222 -0.44444 -0.55556
1.5556 -2.8889 -1.1111
-1.5556 0.88889 -0.88889
B =
1 0.33333
-1 1.6667
1 0.33333
C =
1 1 -6.9389e-16
-0.33333 0.33333 -1.3333
D =
0 0
0 0
trz =
1
RSM_1 =
2.2222 0.44444 0.55556 1 0.33333
-1.5556 3.8889 1.1111 -1 1.6667
1.5556 -0.88889 1.8889 1 0.33333
-1 -1 6.9389e-16 0 0
0.33333 -0.33333 1.3333 0 0
rRSM_1 =
4
Verifying the analysis in the zeros PDF.
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