Studying some control theory but having difficulty learning because my lecturer doesn't provide solutions to any of his exercises AT ALL. Below I've attached a problem I've just done and my answers and would like to know if I'm on the correct track.

enter image description here

enter image description here

For those who can't view the images:


  1. a) Consider the system

$$ \dot{x}=Ax+Bu, x(0)=x_0 \in \mathbb{R}^n ,$$

with unrestricted control u: [0,T] $ \rightarrow \mathbb{R}^m, A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n\times m} $.

(i) Define what it means for this system to be stabilizable.

(ii) State at theorem giving a sufficient condition on the matrices A,B for this system to be stabilizable.

(iii) Show that the system $$ \dot{x_1}=x_2 +u, \dot{x_2}=x_1 $$ where $ x_1, x_2, $ and u are real valued functions, is stabilizable.


(i) Consider a closed loop feedback control of the form u=kx. Then the system is stabilizable if there exists a matrix $ K \in \mathbb{R}^{m \times n} $ such that the system

$$ \dot{x} =(A+BK)x, x(0)=x_0 \in \mathbb{R}^n $$

is asymptotically stable.

(ii) If (A,B) is controllable, then it is stabilizable.

(iii) (A,B) is controllable if and only if rank(G)=n where G is the controllability matrix $ G=(B,AB,...,A^{n-1}B) \in \mathbb{R}^{n \times nm} $.

For this particular system one has $ A=\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix} B= \begin{bmatrix} 1 & 0 \end{bmatrix} $ transposed

Giving G (not writing out calculation because I'm slow at latex) $=\begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix} $.

Clearly (1,0) and (0,1) are linearly independant so rank(G)=2=n and hence (A,B) is controllable, and so is also stabilizable.

^Is this correct? Thanks.

  • $\begingroup$ Your image is unviewable on my device. Please take a moment to write it up. $\endgroup$ – Emily May 18 '15 at 14:03
  • $\begingroup$ Which image, the questions, my answers or both? $\endgroup$ – Greg May 18 '15 at 14:10
  • $\begingroup$ It looks good to me. $\endgroup$ – obareey May 25 '15 at 20:15
  • $\begingroup$ Looks good for me too. $\endgroup$ – Cybernetician Feb 16 '16 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.