Question

Determine the poles of the Jacobi and Gauss-Seidel digital filters for the general two-dimensional problem$\left(N=2\right)$:

$\left[\begin{array}{ccc}{A}_{11}& A_{12}& \\ A_{21}& A_{22}& \end{array}\right]\left[\begin{array}{c}{X}_2\\ X_2\end{array}\right]\mathbf{=}\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]$

Then determine a necessary and sufficient condition for convergence of these filters when $\mathbf{N}\mathbf{=}\mathbf{2}$.

Short answer

The poleof the Jacobi and Gauss-Seidel digital filter for$\left[\begin{array}{ccc}{\mathrm{A}}_{11}& {\mathrm{A}}_{12}& \\ {\mathrm{A}}_{21}& {\mathrm{A}}_{22}& \end{array}\right]\left[\begin{array}{c}{\mathrm{X}}_2\\ {\mathrm{X}}_2\end{array}\right]=\left[\begin{array}{c}{\mathrm{y}}_1\\ {\mathrm{y}}_2\end{array}\right]$when$N=2$is$z=\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}$.

Step-by-step solution

Write the given data

The general two-dimensional problem given in the question is as below:

$\mathbf{\left[}\begin{array}{ccc}{\mathbf{A}}_{\mathbf{11}}& {\mathbf{A}}_{\mathbf{12}}& \\ {\mathbf{A}}_{\mathbf{21}}& {\mathbf{A}}_{\mathbf{22}}& \end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{X}}_{\mathbf{2}}\\ {\mathbf{X}}_{\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{c}{\mathbf{y}}_{\mathbf{1}}\\ {\mathbf{y}}_{\mathbf{2}}\end{array}\mathbf{\right]}$

Determine the poles of Jacobi digital filter

The general formula to solve the given matrix for Jacobi digital filter is

$\mathbf{M}\mathbf{=}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}\left(D-A\right)$

Here, A is the general matrix of $N\times N$, given that $N=2$and D is the matrix having only diagonal elements of matrix A

Therefore, $A=\left[\begin{array}{c}{A}_{11}{A}_{12}\\ A_{21}{A}_{22}\end{array}\right]\mathrm{and}\mathrm{D}=\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ 0{\mathrm{A}}_{22}\end{array}\right]$

Calculate and substitute the values of A and D respectively,

$$\begin{gathered} M=D^{-1}\left(D-A\right) \\ ={\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ 0{\mathrm{A}}_{22}\end{array}\right]}^{-1}\left(\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ 0{\mathrm{A}}_{22}\end{array}\right]-\left[\begin{array}{c}{\mathrm{A}}_{11}{\mathrm{A}}_{12}\\ {\mathrm{A}}_{21}{\mathrm{A}}_{22}\end{array}\right]\right) \\ ={\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ 0{\mathrm{A}}_{22}\end{array}\right]}^{-1}\left[\begin{array}{c}0{\mathrm{A}}_{12}\\ {\mathrm{A}}_{21}0\end{array}\right] \end{gathered}$$

Further solving the inverse matrix,

$$\begin{gathered} M={\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ 0{\mathrm{A}}_{22}\end{array}\right]}^{-1}\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ -{\mathrm{A}}_{21}0\end{array}\right] \\ =\left[\begin{array}{cc}\frac{A_{22}}{A_{11}{A}_{22}}& 0\\ 0& \frac{A_{11}}{A_{11}{A}_{22}}\end{array}\right]\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ -{\mathrm{A}}_{21}0\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{A_{11}}& 0\\ 0& \frac{1}{A_{22}}\end{array}\right]\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ -{\mathrm{A}}_{21}0\end{array}\right] \\ =\left[\begin{array}{cc}0& \frac{-A_{12}}{A_{11}}\\ \frac{-A_{21}}{A_{22}}& 0\end{array}\right] \end{gathered}$$

Calculate the determinant of $\left[\begin{array}{cc}0& \frac{-A_{12}}{A_{11}}\\ \frac{-A_{21}}{A_{22}}& 0\end{array}\right]$

$$\begin{gathered} det\left(zU-M\right)=0 \\ det\left(z\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& \frac{-A_{12}}{A_{11}}\\ \frac{-A_{21}}{A_{22}}& 0\end{array}\right]\right)=0 \\ det\left(\left[\begin{array}{cc}z& 0\\ 0& z\end{array}\right]-\left[\begin{array}{cc}0& \frac{-A_{12}}{A_{11}}\\ \frac{-A_{21}}{A_{22}}& 0\end{array}\right]\right)=0 \\ det\left(\left[\begin{array}{cc}z& \frac{-A_{12}}{A_{11}}\\ \frac{-A_{21}}{A_{22}}& z\end{array}\right]\right)=0 \end{gathered}$$

Further solving the above matrix,

$$\begin{gathered} z^2-\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}=0 \\ z=\pm \sqrt{\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}} \end{gathered}$$

Determine the poles of Gauss-Seidel digital filter

The general formula to solve the given matrix for Gauss-Seidel digital filter is

$\mathbf{M}\mathbf{=}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{D}\mathbf{-}\mathbf{A}\mathbf{)}$

Here, A is the general matrix of $N\times N$, given that $N=2$and D is the matrix having only lower triangular elements of matrix A

Therefore,$A=\left[\begin{array}{c}{A}_{11}{A}_{12}\\ A_{21}{A}_{22}\end{array}\right]\mathrm{and}\mathrm{D}=\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ 0\mathrm{A}{}_{22}\end{array}\right]$

Calculate $M$and substitute the values of A and D respectively,

$$\begin{gathered} M=D^{-1}\left(D-A\right) \\ ={\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ A_{21}{\mathrm{A}}_{22}\end{array}\right]}^{-1}\left(\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ A_{21}{\mathrm{A}}_{22}\end{array}\right]-\left[\begin{array}{c}{\mathrm{A}}_{11}{\mathrm{A}}_{12}\\ {\mathrm{A}}_{21}{\mathrm{A}}_{22}\end{array}\right]\right) \\ ={\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ A_{21}{\mathrm{A}}_{22}\end{array}\right]}^{-1}\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ 00\end{array}\right] \end{gathered}$$

Further solving the inverse matrix,

$$\begin{gathered} M={\left[\begin{array}{c}{\mathrm{A}}_{11}0\\ A_{21}{\mathrm{A}}_{22}\end{array}\right]}^{-1}\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ 00\end{array}\right] \\ =\left[\begin{array}{cc}\frac{A_{22}}{A_{11}{A}_{22}}& 0\\ \frac{-A_{11}}{A_{11}{A}_{22}}& \frac{A_{11}}{A_{11}{A}_{22}}\end{array}\right]\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ 00\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{A_{11}}& 0\\ \frac{-A_{11}}{A_{11}{A}_{22}}& \frac{1}{A_{22}}\end{array}\right]\left[\begin{array}{c}0-{\mathrm{A}}_{12}\\ 00\end{array}\right] \\ =\left[\begin{array}{cc}0& \frac{-A_{12}}{-A_{11}}\\ 0& \frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\end{array}\right] \end{gathered}$$

Calculate the determinant of $\left[\begin{array}{cc}0& \frac{-A_{12}}{-A_{11}}\\ 0& \frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\end{array}\right]$

$$\begin{gathered} det\left(zU-M\right)=0 \\ det\left(z\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& \frac{-A_{12}}{A_{11}}\\ 0& \frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\end{array}\right]\right)=0 \\ det\left(\left[\begin{array}{cc}z& 0\\ 0& z\end{array}\right]-\left[\begin{array}{cc}0& \frac{-A_{12}}{A_{11}}\\ 0& \frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\end{array}\right]\right)=0 \\ det\left(\left[\begin{array}{cc}z& \frac{-A_{12}}{A_{11}}\\ 0& z-\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\end{array}\right]\right)=0 \end{gathered}$$

Further solving the above matrix,

$z\left(z-\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\right)=0$

Now, as z cannot be equal to zero.

Thus

$$\begin{gathered} \left(z-\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}\right)=0 \\ z=\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}} \end{gathered}$$

Hence, the root for both Jacobi and Gauss-Seidel digital filters converges at$\frac{A_{12}{A}_{21}}{A_{11}{A}_{22}}$.