transformation of a difference equation

Question

How can I translate the difference equation $$x_{k+3}+4x_{k+2}+3x_{k+1}+x_k=2u_{k+2}$$

into a state-space representation of the following form (A and B are matrices)

$$x_{k+1}=Ax_k+Bu_k$$

Answer 1

We can shift back to time by two samples i.e. $$x_{k+1}+4x_{k}+3x_{k-1}+x_{k-2}=2u_{k}$$ We only need one input and three states. Define $$X_k:=[x_{k-2}, x_{k-1}, x_{k}]^T$$ then $$X_{k+1}=\left[\matrix{ 0 & 1 & 0\\0 & 0 & 1\\ -1 & -3 & -4}\right]X_k+\left[\matrix{0\\0\\2}\right]u_k$$

Answer 2

As always in such cases. Let $$ X_k := \begin{pmatrix} x_k \\ x_{k+1}\\ x_{k+2} \end{pmatrix} $$ and $$ U_k := \begin{pmatrix} u_k \\ u_{k+1} \\ u_{k+2}\end{pmatrix} $$ Then the above can be written as \begin{align*} X_{k+1} &= \begin{pmatrix} x_{k+1} \\ x_{k+2}\\ x_{k+3} \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & -3 & -4\end{pmatrix} \begin{pmatrix} x_k \\ x_{k+1}\\ x_{k+2} \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 2\end{pmatrix} \begin{pmatrix} u_k \\ u_{k+1}\\ u_{k+2} \end{pmatrix}\\ &=: AX_k + BU_k \end{align*}

Answer 3

One way is $$ \begin{bmatrix} x_{k+3}\\x_{k+2}\\x_{k+1} \end{bmatrix} = \begin{bmatrix} 4&3&1\\ 1&0&0\\ 0&1&0 \end{bmatrix} \begin{bmatrix} x_{k+2}\\x_{k+1}\\x_k \end{bmatrix} + \begin{bmatrix} 2&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix} \begin{bmatrix} u_{k+2}\\u_{k+1}\\u_k \end{bmatrix} $$