1.1 Designing K bY thansforming the state model into controllable from

The problem is first solved for the controllable canonical form. Let us denote the controllability matrix by U_C and consider a transformation matrix T as

$\displaystyle T=U_C W$

where $\displaystyle W = \begin{bmatrix}a_{n-1} & a_{n-2} &\cdots & a_1 & 1 \\ a_{n-2}... ... \vdots \\ a_{1} & 1 &\cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix}$

's are the coefficients of the characteristic polynomial

$\vert zI - A\vert = z^n + a_1 z^{n-1} + \cdots + a_{n-1} z + a_n$ .

Define a new state vector $\mathbf{{x}}=T\mathbf{\bar x}$ . This will transform the system given by (1) into controllable canonical form, as

$\displaystyle \mathbf{\bar{x}}(k+1)=\bar{A}\mathbf{\bar{x}}(k)+\bar{B}u(k)$

(3)

You should verify that

$\displaystyle \bar{A}=T^{-1}AT= \begin{bmatrix}0 & 1 & 0 & \ldots &0 \\ 0 & 0 &... ...0 & \ldots & 1 \\ - a_n & -a_{n-1} & -a_{n-2} & \ldots & -a_1 \end{bmatrix} \;$

We first find $\bar{K}$ such that $u(k)=-\bar{K}\mathbf{\bar{x}}(k)$ places poles in desired locations. Since eigenvalues remain unaffected under similarity transformation, $u(k)=-\bar{K}T^{-1}\mathbf{{x}}(k)$ will also place the poles of the original system in desired locations.

If poles are placed at , the desired characteristic equation can be expressed as:

(4)

Since the pair $\bar{A},\bar{B}$ are in controllable-companion form then, we have

$\displaystyle \bar{A}-\bar{B}\bar{K}=\begin{bmatrix} 0 & 1 & 0 &\ldots & 0 \\ ... ... -(a_n-\bar{k}_1)&-(a_{n-1}-\bar{k}_2)& &\ldots&-(a_1-\bar{k}_n) \end{bmatrix}$

Please note that the characteristic equation of both original and canonical form is expressed as: $\vert zI-A\vert$ = $\vert zI-\bar{A}\vert$ = $z^n+a_1 z^{n-1}+\ldots+a_n$ = 0.

The characteristic equation of the closed loop system with $u=-\bar{K}\mathbf{\bar{x}}$ is given as:

$\displaystyle z^n+(a_1+\bar{k}_n)z^{n-1}+(a_2+\bar{k}_{n-1})z^{n-2}+\ldots+(a_n+\bar{k}_1)=0$

(5)

Comparing Eqs. (4) and (5), we get

$\displaystyle \bar{k}_n=\alpha_1-a_1,\;\bar{k}_{n-1}=\alpha_2-a_2,\;\bar{k}_1=\alpha_n-a_n$

(6)

We need to compute the transformation matrix T to find the actual gain matrix

$K=\bar{K}T^{-1}$ where $\bar{K}=[\bar{k}_1,\bar{k}_2,\ldots,\bar{k}_n]$ .