Question

Take the z-transform of (6.2.6) and show that $\mathit{X}\left(z\right)\mathbf{=}\mathit{G}\left(z\right)\mathbf{}\mathit{Y}\left(z\right)$, where localid="1655183966907" $\mathbf{G}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}{\mathbf{(}\mathbf{zU}\mathbf{-}\mathbf{M}\mathbf{)}}^{\mathbf{-}\mathbf{1}}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}$and U is the unit matrix.

Note:localid="1655183970573" $\mathbf{G}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$is the matrix transfer function of a digital filter that represents the Jacobi or Gauss-Seidel methods. The filter poles are obtained by solvinglocalid="1655183974764" $\mathbf{det}\mathbf{(}\mathbf{zU}\mathbf{-}\mathbf{M}\mathbf{)}\mathbf{=}\mathbf{0}$. The filter is stable if and only if all the poles have magnitudes less than 1.

Short answer

The Z transform of$\left[x\left(i+1\right)\right]={\left(zU-M\right)}^{-1}{D}^{-1}Z$transformof$\left[y\left(i\right)\right]$is$X\left(z\right)=G\left(z\right)Y\left(z\right)$

Step-by-step solution

Jacobi or Gauss-Seidel matrix format formula

Write the general formula of Jacobi or Gauss-Seidel matrix format,

${\mathbf{x}}_{\mathbf{k}}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{A}}_{\mathbf{kk}}}\mathbf{[}{\mathbf{y}}_{\mathbf{k}}\mathbf{-}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{k}\mathbf{-}\mathbf{1}}{\mathbf{A}}_{\mathbf{kn}}{\mathbf{x}}_{\mathbf{n}}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{k}\mathbf{+}\mathbf{1}}^{\mathbf{N}}{\mathbf{A}}_{\mathbf{kn}}{\mathbf{x}}_{\mathbf{n}}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{]}$ ….. (1)

Here,$x_k\left(i+1\right)$ is the k^th component of x(i); x(i) is the i^thguess of the iteration; represents the coefficient of the k^th component of x(i); is other variable of k^th component.

The above equation can also be written as,

${\mathbf{x}}_{\mathbf{k}}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{Mx}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{+}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}\mathbf{y}$ ….. (2)

Here, $\mathbf{M}\mathbf{=}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{D}\mathbf{-}\mathbf{A}\mathbf{)}$and $\mathbf{D}\mathbf{=}\mathbf{[}\begin{array}{cc}\begin{array}{cccc}{\mathbf{A}}_{\mathbf{11}}& \mathbf{0}& \mathbf{\dots }& \mathbf{0}\\ \mathbf{0}& {\mathbf{A}}_{\mathbf{22}}& \mathbf{\dots }& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& & \mathbf{M}\\ \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& {\mathbf{A}}_{\mathbf{NN}}\end{array}& \end{array}\mathbf{}\mathbf{]}$

Z-transform of Jacobi or Gauss-Seidel matrix format formula

Take the Z transform of equation (2) assuming initial conditions to be zero,

$$\begin{gathered} zUX\left(z\right)=MX\left(z\right)+D^{-1}Y\left(z\right) \\ \left(zU-M\right)X\left(z\right)=D^{-1}Y\left(z\right) \end{gathered}$$

Here, U stands for unitary matrix,$X\left(z\right)$represents the Z transform of$x\left(i\right)$and$Y\left(z\right)$represents the Z transform of$Y\left(i\right)$and$zX\left(z\right)$is the Z transform of$x\left(i+1\right)$.

Prove X(z)=G(z)Y(z).

Consider the determinant of $\left(zU-M\right)$is equal to zero,$\det \left(\mathrm{zU}-\mathrm{M}\right)=0$, that is taking inverse of $\left(zU-M\right)$is possible.

Thus, dividing the above equation by $\left(zU-M\right)$.

$X\left(z\right)={\left(zU-M\right)}^{-1}{D}^{-1}Y\left(z\right)$

Also, $G\left(z\right)={\left(zU-M\right)}^{-1}{D}^{-1}$is given, therefore substitute ${\left(zU-M\right)}^{-1}{D}^{-1}\mathrm{as}G\left(z\right)$

$$\begin{gathered} X\left(z\right)={\left(zU-M\right)}^{-1}{D}^{-1}Y\left(z\right) \\ X\left(z\right)=G\left(z\right)Y\left(z\right) \end{gathered}$$

Hence, proved.