z-transforms of a system of coupled difference equations

Question

I have the following system of difference equations $$\mathbf{x}_{n+1}=A\mathbf{x}_{n},\qquad A=\frac{1}{3}\begin{bmatrix} 1&5&2\\ 2&0&0\\ 0&2&0 \end{bmatrix}.$$ $L$ is diagonalisable, and so via the spectral decomposition determining a general $\mathbf{x}_{n}$ is easy. Although I am required to implement z-transforms on the original coupled system, this is what I'm having issues with. Some hints on where to start with this would be great. (I am given $\mathbf{x}_0=\begin{bmatrix} 6 & 0& 0\end{bmatrix}^T$). Thank you

Answer 1

We start from $$ \left\{ \matrix{ {\bf x}_{\,n} = 0\quad \left| {\;n < 0} \right. \hfill \cr {\bf x}_{\,0} = {\bf x}_{\,0} \hfill \cr {\bf x}_{\,n + 1} = {\bf x}_{\,n} \,A \hfill \cr} \right. $$

we put 2nd and 3rd into a single identity $$ {\bf x}_{\,n} I = {\bf x}_{\,n - 1} \,A + [0 = n]\,{\bf x}_{\,0} I $$ where $[P]$ denotes the Iverson bracket $$ \left[ P \right] = \left\{ {\begin{array}{*{20}c} 1 & {P = TRUE} \\ 0 & {P = FALSE} \\ \end{array} } \right. $$

Define the vectorial function of $z$ $$ {\bf F}(z) = \left( {f_{\,1} (z),f_{\,2} (z), \cdots ,f_{\,q} (z)} \right) = \sum\limits_{0\, \le \,n} {{\bf x}_{\,n} \,z^{\,n} } $$ then the recurrence becomes $$ \eqalign{ & {\bf F}(z)I = \sum\limits_{0\, \le \,n} {{\bf x}_{\,n} \,z^{\,n} I} = \sum\limits_{0\, \le \,n} {{\bf x}_{\,n - 1} z^{\,n} } \,A + \sum\limits_{0\, \le \,n} {[0 = n]\,z^{\,n} {\bf x}_{\,0} } I = \cr & = {\bf x}_{\, - 1} z^{\,0} + \sum\limits_{1\, \le \,n} {{\bf x}_{\,n - 1} z^{\,n} } \,A + z^{\,0} {\bf x}_{\,0} I = \cr & = \sum\limits_{1\, \le \,n} {{\bf x}_{\,n - 1} z^{\,n} } \,A + {\bf x}_{\,0} I = \cr & = \sum\limits_{0\, \le \,n - 1} {{\bf x}_{\,n - 1} z^{\,n - 1 + 1} } \,A + {\bf x}_{\,0} I = \cr & = z\sum\limits_{0\, \le \,n - 1} {{\bf x}_{\,n - 1} z^{\,n - 1} } \,A + {\bf x}_{\,0} I = \cr & = z{\bf F}(z)\,A + {\bf x}_{\,0} I \cr} $$

i.e. $$ \eqalign{ & {\bf F}(z)I = z{\bf F}(z)\,A + {\bf x}_{\,0} I\quad \Rightarrow \quad {\bf F}(z)\left( {I - zA} \right) = {\bf x}_{\,0} I\quad \Rightarrow \cr & \Rightarrow \quad {\bf F}(z) = {\bf x}_{\,0} \left( {I - zA} \right)^{\, - 1} \cr} $$