obsvf confused by documentation

Shouldn't (Ao,Bo) is controllable be replaced by (Ao,Co) is observable?

obsvf Observability staircase form.

[ABAR,BBAR,CBAR,T,K] = obsvf(A,B,C) returns a decomposition into the observable/unobservable subspaces.

[ABAR,BBAR,CBAR,T,K] = obsvf(A,B,C,TOL) uses tolerance TOL. If Ob=OBSV(A,C) has rank r <= n = SIZE(A,1), then there is a

similarity transformation T such that Abar = T * A * T' , Bbar = T * B , Cbar = C * T' .

and the transformed system has the form

| Ano A12| |Bno|

Abar = ---------- , Bbar = --- , Cbar = [ 0 | Co].

| 0 Ao | |Bo |

-1 -1

where (Ao,Bo) is controllable, and Co(sI-Ao) Bo = C(sI-A) B.

The last output K is a vector of length n containing the number of observable states identified at each iteration of the algorithm. The number of observable states is SUM(K).

See also obsv, ctrbf.

Answer 1

[M]

You were right, the documentation for obsvf function has been updated recently:

https://mathworks.com/help/control/ref/obsvf.html