Omar,

If the linear system is not controllable, then we know that at least some of the open-loop modes are unaffected by the input signals both directly as well as indirectly.

Now if those specific uncontrollable modes happen to be unstable as well, then you are in trouble !  Because a feedback signal can not be input to influence it in any way.

Therefore either the system absolutely must have controllability of all modes to ensure stability for all of them. Or else the uncontrollable modes should be stable by default!

-Sanjay

Dear Sanjoy Roy,

Thank you for your response.

To my knowledge, controllability gives stability, the reciproque is not always true.

I have guaranteed stability of a feedback system (see example attached) and i have determined a corresponding control law which stabilize asymptotically the system even the pair (A,B) is not controllable.   for that case, has the controllability assumption a meaning ?

best regards, 

Hi Omar,

What you said "controllability gives stability, the reciproque is not always true" is essentially the same as what I explained in my message.

If you are unsure whether the system is stable or not, only then the controllability assumption has significance.

If the system is proven to be guaranteed stable, the assumption clearly loses significance because all states are inherently stable anyway.

-Sanjay

Good answers.

"Controllability" means that you can control all state-variables of your system. If your system is stable and all you care for is stability, then controllability has no meaning for you.

That's why other publications talk about "stabilizability," meaning controllability of the unstable state-variables.

However, we must be careful.

First, because controllability means that you can control all state-variables of your system, you can use the input to make any of them reach whatever values you intend to. Otherwise, the non-controllable states, even stable, may indeed reach zero, yet cannot be controlled to reach other desired value.

Moreover, a very slow pole is still stable, yet the corresponding integrator may get some noise and this will stay there for a longtime, as you have no way to affect it.

Moving from state-space to transfer function, "non-controllable" state means that a pole (s+p) in the denominator is cancelled by a corresponding zero (s+p) in the numerator. The transfer function may seem to just cancel one another, yet they may represent the real physical world.

Itzhak

Good discussion. Thanks for the thoughts.

Just to summarize what is being discussed:

"'Controllability' means that you can control all state-variables of your system." - I. Barkana.

Stabilizability means that, even though you cannot control all of your states, those you cannot control are stable.

"Can we guarantee the stability of a feedback controlled system without using controllability assumption ?"

Roughly speaking, yes, if the system is stabilizable. In other words, its non-controllable modes are stable.

Stabilizable: all unstable modes are controllable (i.e.. all uncontrollable modes are stable). As in the joined picture, you can find a matrice S, such that you can decompose the matrices of your system in two part, the first one (A11,B1) which must to be controllable. And, the poles of the second one A22, must be stables.

This can be summerazied in the second picture with a rank condition.

The assumption "The pair (A,B) is controllable" is it necessary ? - ResearchGate. Available from: https://www.researchgate.net/post/The_assumption_The_pair_A_B_is_controllable_is_it_necessary#570cd04d615e278c02068bda [accessed Apr 12, 2016].

@Omar In conclusion, since a good discussion should include also simple example for better understanding the definition provided above, you can consider the following state-space LTI system:

(1)    x1'=x1+x2+u

(2)    x2'=x2 (unstable internal dynamic)

(3)   y=[x1,x2]'

The system (1)-(3) is clearly unstable and uncontrallable since it has un unstable invariant zero due to (2), thus you could stabilize only the mode associated to (1) for example by choosing 

(4)   u=-2x1-x2

however x2 will escape for all u(x1,x2) with an exponential rate.

There are no u such that (2) can become stable!!!

Thus the system (1)-(4) is definitively unstable. For further details you could refer to the book of JJ Slotine (keyword: unstable internal dynamic) and also to the concepts of unstable invariant zero.

Assume a Linear Time-Invariant Plant, not necessary stable with m inputs and p outputs. SOLO Stabilization of Linear Time-Invariant Systems Factorization Approach • Assume that the plant can be represented by a set of n linear ordinary differential equations with constant coefficients (to assure time-invariance) or by the corresponding transfer matrix G (s)pxm. • Assume also that G (s) is proper: ( ) ∞< ∞→ sG s lim ( G(s) is analytic as s →∞ ) Therefore G (s) is in the Ring of Real-rational proper (Rp) matrices: ( ) pxm pRsG ∈ Using the fact that under those assumptions all G (s) can be factorized in two Real-rational-proper and stable (RH∞) matrices, we will obtain a parameterization of all compensators K (s)mxp that stabilize the given plant (and conversely all plants that can be stabilized by a given compensator). This parameterization will allow to define methods of optimizing the desired performances of the design over all possible stabilizing compensators.