Problem in observibility and controlibility function ctrbf,obsvf

After doing similarity transform, transfer function of a system taking controllable modes should be same as original. But it is not showing same results.

>> clear >> A = [-.5 1 0;-1 -.5 0; 0 1 0]

A =

   -0.5000    1.0000         0
   -1.0000   -0.5000         0
         0    1.0000         0

>> b=[1;2;0]

b =

     1
     2
     0

>> c =[0 0 1]

c =

     0     0     1

>> [Abar,Bbar,Cbar,T,k]=ctrbf(A,b,c)

Abar =

         0         0         0
    0.4472   -0.5000    1.3416
    0.6667   -0.7454   -0.5000

Bbar =

         0
    0.0000
    2.2361

Cbar =

    0.7454    0.6667         0

T =

   -0.5963    0.2981    0.7454
    0.6667   -0.3333    0.6667
    0.4472    0.8944         0

k =

     1     1     0

>> c*inv([s 0 0;0 s 0;0 0 s]-A)*b ??? Undefined function or variable 's'.

>> s=sym('s'); >> c*inv([s 0 0;0 s 0;0 0 s]-A)*b

ans =

(2*(4*s + 2))/(s*(4*s^2 + 4*s + 5)) - 4/(s*(4*s^2 + 4*s + 5))

>> Cbar*inv([s 0 0;0 s 0;0 0 s]-Abar)*Bbar

ans =

(4*s + 2)/(13510798882111488*(4*s^2 + 4*s + 5)) + 40/(20*s^2 + 20*s + 25)

>> T*A*T'

ans =

         0         0         0
    0.4472   -0.5000    1.3416
    0.6667   -0.7454   -0.5000

>> Ac=[Abar(2,2) Abar(2,3);Abar(3,2) Abar(3,3)]

Ac =

   -0.5000    1.3416
   -0.7454   -0.5000

>> Bc=[Bbar(2,1);Bbar(2,3)] ??? Attempted to access Bbar(2,3); index out of bounds because size(Bbar)=[3,1].

>> Bc=[Bbar(2,1);Bbar(3,1)]

Bc =

    0.0000
    2.2361

>> Cc=[Cbar(1,2);Cbar(1,3)]

Cc =

    0.6667
         0

>> Cc=[Cbar(1,2) Cbar(1,3)]

Cc =

    0.6667         0

>> Cc*inv([s 0 ;0 s]-Ac)*Bc

ans =

(4*s + 2)/(13510798882111488*(4*s^2 + 4*s + 5)) + 8/(4*s^2 + 4*s + 5)

>>

[Walter Roberson]

It would be easier on readers if you were to cut out the places you know you made mistakes.

Answer 1

[Teja Muppirala]

This difference is just from round-off errors. Notice the second element in Bbar is 1.1102e-016. That is why you get that very large denominator (13510798882111488) in the symbolic expression

say Bbar(2) = 0, then everything works fine.

A = [-.5 1 0;-1 -.5 0; 0 1 0];
b=[1;2;0];
c =[0 0 1];
[Abar,Bbar,Cbar,T,k]=ctrbf(A,b,c)

syms s
Bbar(2) = 0;

c*( (s*eye(3)-A) \b)
Cbar*( (s*eye(3)-Abar) \Bbar)

Answer 2

[Teja Muppirala]

Just a suggestion, but when doing symbolic calculations using 's' as a transfer function variable, you might want to explicitly make s a transfer function instead of just an ordinary symbolic variable.

s = tf('s')

instead of

s = sym('s')

This will allow you to actually treat those expressions like transfer functions.

G1 = c*( (s*eye(3)-A) \b) G2 = Cbar*( (s*eye(3)-Abar) \Bbar)

bode(G1) etc...

There is also a function MINREAL that you can use to cancel out poles and zeros.

minreal(G1,1e-6)