# Does the controllability of nominal system imply the controllability of the actual uncertain system?

Given a system dynamics $$\dot{x}=A(p)x+Bu,$$

where $A(p)$ is the uncertain system state matrix with the nominal matrix being $A(p_0)$. The uncertainty in $A(p)$ satisfies the matching condition, i.e. $$A(p)-A(p_0)=B\phi(p)$$

Now from the book "Robust control design (an optimal control approach)" I encountered an exercise problem which goes as follows:

If $(A(p_0),B)$ is controllable, show that $(A(p),B)$ is also controllable for any $\phi(p)$ where $A(p)=A(p_0)+B\phi(p)$.

My approach was to use the rank condition of the controllability matrix for the nominal case and extend that result to the original system but it does not give me the result I expected. Any hint will be much appreciated.

• I currently do not know how to proof the controllability matrix is full rank. However controllability implies that $u=K\,x$ can place the closed loop poles/eigenvalues anywhere you want. But if this holds for the nominal system, then using $u=(K-\phi(p))\,x$ for the uncertain system should also allow you to place the poles anywhere you want. – fibonatic Jul 19 '18 at 3:05
• Since $\phi(p)$ is uncertain, it cannot be used in the control implementation. – jbgujgu Jul 19 '18 at 12:17
• But that is not your question, namely you asked whether the pair $(A(p),B)$ is controllable, so if $K$ could be chosen such that the closed loop poles can be placed anywhere. In that context the value for $A(p)$ is considered known. – fibonatic Jul 19 '18 at 13:28
• So my question simplifies to the following: if the nominal system dynamics is controllable, I have to show that original uncertain system is also controllable i.e. if $(A(p_0),B)$ controllable then show that $(A(p),B)$ controllable. My last comment was on your $u=(K-\phi(p))x$ because $\phi(p)$ is uncertain and control law can not be uncertain. Additionally if it is controllable, then only the design of control law comes into context and our task is to show that the original system is controllable. – jbgujgu Jul 19 '18 at 14:53

You can prove it using the PBH test for controllability. It states that $rank \left( \begin{array}{cc} \lambda I -A & B \\ \end{array} \right) = n$ for all values of $\lambda$.
For the uncertain system this becomes $\left( \begin{array}{cc} \lambda I -A -B \phi & B \\ \end{array} \right) = \left( \begin{array}{cc} \lambda I -A & B \\ \end{array} \right).\left( \begin{array}{cc} I_n & 0 \\ -\phi & I_m \\ \end{array} \right)$ which also has full rank.
(Here $n$ is the number of states, and $m$ the number of inputs.)