Example 1: Find out the state feedback gain matrix K for the following system using two different methods such that the closed loop poles are located at 0.5 , 0.6 and 0.7.

$\displaystyle \mathbf{x}(k+1) \begin{bmatrix}0&1&0\\ 0 & 0&1\\ -1&-2&-3\end{bmatrix}\mathbf{x}(k) + \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}u(k)$

Solution:

$\displaystyle U_C=\begin{bmatrix}0&0&1\\ 0 & 1&-3\\ 1&-3&7\end{bmatrix}$

The above matrix has rank 3, so the system is controllable.

Open loop characteristic equation:

or, $\displaystyle \;\; z^3+3z^2+2z+1 =0$

Desired characteristic equation:

Since the open loop system is already in controllable canonical form, T =1.

$\displaystyle K = [\alpha_3-a_3 \;\; \alpha_2-a_2 \;\; \alpha_1-a_1 ]$

where, $\alpha_3=-0.21, \; \alpha_2=1.07, \; \alpha_1=1.8$ and $a_3=1, \; a_2=2, \; a_1 1$ . Thus

$\displaystyle K = [-1.21 \;\; -0.93 \;\; -4.8 ]$

Using Ackermann's formula:

Thus