Continued Fraction Expansions of Stable Discrete-Time Systems of Difference Equations

Abstract

We provide a systematic procedure for generating the coefficients of the continued fraction expansion of the test function associated with the characteristic polynomial of a stable system of difference equations. We illustrate the feasibility of the procedure, and we provide an application on the stability of two-dimensional digital filters.

1. Introduction

The subject of continued fractions and its applications to both continuous and discrete systems have a long history and continue to impact research in various stability contexts [1,2,3,4,5,6,7,8,9]. The discovery of interesting relationships between continued fraction expansions and Schur polynomials, i.e., polynomials having their zeros inside the unit circle [1], led to several investigations into the stability of polynomials via continued fractions.

In a recent work [10], the symmetric properties between Routh–Hurwitz and Schur–Cohn stability types were highlighted. In [10], the test functions of each of these stability types were expanded in continued fraction forms satisfying specific conditions.

In the Routh–Hurwitz case; Theorem 3 of [10], the coefficients of the continued fraction expansion of the test function associated with the characteristic polynomial of the system are relatively easy to obtain using sequential long division. In the Schur–Cohn case; Theorem 4 of [10], generating such coefficients is far from trivial. The aim of this paper is to develop a systematic procedure for generating the coefficients of the continued fraction expansion in the Schur case and to illustrate the procedure through an engineering application to the stability of two-dimensional digital filters.

It should be noted that assessing the stability of 2-D digital filters requires one to check the location of the zeros of complex polynomials [11]. Hence, all polynomials considered in this paper have complex coefficients.

In Section 2, we lay out some definitions, notations, and the required results from [10]. In Section 3, we provide a systematic procedure to generate the coefficients of the continued fraction expansion associated with a Schur stable polynomial. The feasibility of the procedure is illustrated in Section 4. An application to the stability of 2-D digital filters is advanced in Section 5 to verify the proposed method.

2. Definitions and Notations

A reminder of the required definitions and results established in [10].

Definition 1.

A linear discrete-time system of difference equations is stable if and only if all its eigenvalues lie inside the unit disc. If

$$g\left(z\right)=a_0+a_{1}z+a_2{z}^2+\cdots +a_{n-1}{z}^{n-1}+a_n{z}^n$$

(1)

is the characteristic polynomial of the system, then the system is stable if all zeros of g(z) lie inside the unit disc. Such polynomials are said to be Schur stable.

Definition 2.

The reciprocal of g is defined by $g^{\tau }(z)=z^n\overline{g(1/\overline{z})}$. Then, Then, g^τ can be written as $g^{\tau }(z)={\overline{a}}_n+{\overline{a}}_{n-1}z+{\overline{a}}_{n-2}{z}^2+\cdots +{\overline{a}}_0{z}^n$ where ${\overline{a}}_k$ denotes the complex conjugate of $a_k$ for $k=0,1,\dots ,n.$

Definition 3.

The test function of the given discrete-time system is defined by

$$\Psi \left(z\right)=\frac{g\left(z\right)-g^{\tau }\left(z\right)}{g\left(z\right)+g^{\tau }\left(z\right)}$$

(2)

Theorem 1

([10] Theorem 4). The linear discrete-time system of difference equations characterized by (1) is stable if and only if the test function Ψ(z) defined by (2) can be written in the continued fraction expansion

$$\Psi \left(z\right)=h_0\frac{z-1}{z+1}+k_0+\frac{1}{\begin{array}{c}{h}_1\frac{z-1}{z+1}+k_1+\\ ⋮\\ +\frac{1}{h_n\frac{z-1}{z+1}+k_n}\end{array}}$$

(3)

$\mathit{where}{h}_0\ge 0,h_1>0,\dots ,h_n>0\mathit{and}{k}_j\mathit{are}\mathit{imaginary}\mathit{or}\mathit{zero}\mathit{for}0\le j\le n.$ This expansion in$\frac{z-1}{z+1}$ is known as the bilinear transformation.

3. A Procedure to Generate the Coefficients of $\Psi \left(z\right)$

Motivated by [12], we would like to substitute the variable $\frac{z-1}{z+1}\mathrm{in}\left(3\right)\mathrm{by}\frac{s-1}{s}.\mathrm{Solving}\frac{z-1}{z+1}=\frac{s-1}{s}\mathrm{for}\mathrm{z}\mathrm{leads}\mathrm{to}z=2s-1.$

Therefore, we define the function T (s) in the following way:

$$T\left(s\right)=\Psi \left(2s-1\right).$$

$T\left(s\right)$ can now be written as

$$T\left(s\right)=h_0\frac{s-1}{s}+k_0+\frac{1}{\begin{array}{c}{h}_1\frac{s-1}{s}+k_1+\\ ⋮\\ +\frac{1}{h_n\frac{s-1}{s}+k_n}\end{array}}$$

$\mathrm{This}\mathrm{expansion}\mathrm{in}\frac{s-1}{s}\mathrm{is}\mathrm{known}\mathrm{as}\mathrm{the}\mathrm{backward}\mathrm{difference}\mathrm{transform}.$ We begin by breaking up the above form of $T\left(s\right)$ in the following way. Define

$$T\left(s\right)=T_0\left(s\right)=h_0\frac{s-1}{s}+k_0+\frac{1}{T_1\left(s\right)},$$

$$T_1\left(s\right)=h_1\frac{s-1}{s}+k_1+\frac{1}{T_2\left(s\right)}.$$

In general, define

$$T_j\left(s\right)=h_j\frac{s-1}{s}+k_j+\frac{1}{T_{j+1}\left(s\right)}\mathrm{for}0\le j\le n-1$$

(4)

Finally,

$$T_n\left(s\right)=h_n\frac{s-1}{s}+k_n.$$

$\mathrm{Since}\mathrm{each}{T}_j\left(s\right)\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{function},\mathrm{write}\mathrm{it}\mathrm{in}\mathrm{the}\mathrm{form}:$

$$T_j\left(s\right)=\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^l}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l}$$

(5)

The following theorem determines the values of the coefficients $h_j$ of (3).

Theorem 2.

$$h_j=-\frac{a_{j,j}}{b_{j,j+1}}for0\le j\le n.$$

Proof.

$$\mathrm{Multiply}{T}_j\left(s\right)\mathrm{of}\left(5\right)\mathrm{by}\frac{s}{s-1}.$$

$$\begin{array}{c}{T}_j\left(s\right)·\frac{s}{s-1}=\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^l}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l}·\frac{s}{s-1}\\ =\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^{l+1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l·\left(s-1\right)}\\ =\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^{l+1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}-{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l}.\end{array}$$

By changing the indices, the last form can be written as:

$$T_j\left(s\right)·\frac{s}{s-1}=\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^{l+1}}{-b_{j,j+1}s+{{\displaystyle \sum }}_{l=1}^{n-j-1}(b_{j,j+l}-b_{j,j+l+1})s^{l+1}+b_{j,n}{s}^{n-j+1}}.$$

By a simple application of L’Hopital rule, we get:

$$\underset{s\to 0}{\lim }\left[T_j\left(s\right)·\frac{s}{s-1}\right]=-\frac{a_{j,j}}{b_{j,j+1}}.$$

Going back to the expression of $T_j\left(s\right)$ in (4), $T_j\left(s\right)=h_j\frac{s-1}{s}+k_j+\frac{1}{T_{j+1}\left(s\right)}$

$$\mathrm{Multiply}\mathrm{both}\mathrm{sides}\mathrm{by}\frac{s}{s-1}\mathrm{to}\mathrm{get}$$

$$\begin{array}{c}{T}_j\left(s\right)·\frac{s}{s-1}=h_j\frac{s-1}{s}·\frac{s}{s-1}+k_j·\frac{s}{s-1}+\frac{1}{T_{j+1}\left(s\right)}·\frac{s}{s-1}\\ =h_j+k_j·\frac{s}{s-1}+\frac{1}{T_{j+1}\left(s\right)}·\frac{s}{s-1}.\end{array}$$

Hence,

$$\underset{s\to 0}{\lim }\left[T_j\left(s\right)·\frac{s}{s-1}\right]=\underset{s\to 0}{\lim }\left[h_j+k_j·\frac{s}{s-1}+\frac{1}{T_{j+1}\left(s\right)}·\frac{s}{s-1}\right]=h_j.$$

That leads to the desired conclusion:

$$h_j=-\frac{a_{j,j}}{b_{j,j+1}}\mathrm{for}0\le j\le n.\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\square$$

The next theorem determines the values of the coefficients $k_j$ of (3) in addition to some important relations which will prove useful in generating the required coefficients.

Theorem 3.

$$k_j=\frac{a_{j,j+1}-h_j\left(b_{j,j+1}-b_{j,j+2}\right)}{b_{j,j+1}}for0\le j\le n,$$

where $h_j$as determined in Theorem 2.

In addition, the following two relations hold:

$$a_{j+1,j+l}=b_{j,j+l}\mathit{for}0\le j\le n\mathit{and}1\le l\le n-j,$$

and

$$b_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}\mathit{for}0\le j\le n\mathit{and}2\le l\le n-j.$$

Proof.

Consider the expression of $T_j\left(s\right)$ as in (5) from which we subtract $h_j·\frac{s-1}{s},$

$$\begin{array}{c}{T}_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^l}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l}-h_j·\frac{s-1}{s}\hfill \\ =\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^{l+1}-h_j\left(s-1\right)({{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l)}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}}\hfill \\ =\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^{l+1}+h_j({{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l)-h_j({{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1})}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}}\hfill \end{array}$$

Isolate the first term in the first two summations in the numerator of the last expression,

$$\begin{array}{c}{T}_j\left(s\right)-h_j·\frac{s-1}{s}\hfill \\ =\frac{\left(a_{j,j}+h_j·b_{j,j+1}\right)s+{{\displaystyle \sum }}_{l=1}^{n-j}{a}_{j,j+l}{s}^{l+1}+h_j({{\displaystyle \sum }}_{l=2}^{n-j}{b}_{j,j+l}{s}^l)-h_j({{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1})}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}}\hfill \end{array}$$

$\mathrm{By}\mathrm{Theorem}1,h_j=-\frac{a_{j,j}}{b_{j,j+1}}.\mathrm{That}\mathrm{leads}\mathrm{to}{a}_{j,j}+h_j·b_{j,j+1}=0.\mathrm{Therefore},$

$$T_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{{{\displaystyle \sum }}_{l=1}^{n-j}{a}_{j,j+l}{s}^{l+1}+h_j({{\displaystyle \sum }}_{l=2}^{n-j}{b}_{j,j+l}{s}^l)-h_j({{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1})}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}}$$

By changing indices in the second summation in the numerator, we get

$$T_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{{{\displaystyle \sum }}_{l=1}^{n-j}{a}_{j,j+l}{s}^{l+1}+h_j({{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j,j+l+1}{s}^{l+1})-h_j({{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1})}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}}$$

Combining like terms leads to

$$T_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l+1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j+1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l+1}}$$

Taking $s^2$ as a common factor in both the numerator and denominator leads to:

$$\begin{array}{c}{T}_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{s^2\left\{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l-1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j-1}\right\}}{s^2\left[{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}\right]}\hfill \\ =\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l-1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}.\hfill \end{array}$$

Therefore,

$$T_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l-1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}$$

(6)

On the other hand, returning to the expression of $T_j\left(s\right)$ as in (4),

$$T_j\left(s\right)=h_j\frac{s-1}{s}+k_j+\frac{1}{T_{j+1}\left(s\right)},$$

We get,

$$\frac{1}{T_{j+1}\left(s\right)}=T_j\left(s\right)-h_j\frac{s-1}{s}-k_j$$

(7)

In addition, using the expression of $T_j\left(s\right)\mathrm{in}\left(5\right),\mathrm{we}\mathrm{can}\mathrm{write}{T}_{j+1}\left(s\right)\mathrm{in}\mathrm{the}\mathrm{form}$

$$T_{j+1}\left(s\right)=\frac{{{\displaystyle \sum }}_{l=0}^{n-j-1}{a}_{j+1,j+l+1}{s}^l}{{{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j+1,j+l+1}{s}^l}$$

which is inverted to,

$$\frac{1}{T_{j+1}\left(s\right)}=\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j+1,j+l+1}{s}^l}{{{\displaystyle \sum }}_{l=0}^{n-j-1}{a}_{j+1,j+l+1}{s}^l}$$

(8)

$\mathrm{By}\mathrm{comparing}\mathrm{the}\mathrm{two}\mathrm{forms}\mathrm{of}\frac{1}{T_{j+1}\left(s\right)}\left(7\right)\mathrm{and}\left(8\right),\mathrm{we}\mathrm{get}$

$$T_j\left(s\right)-h_j\frac{s-1}{s}-k_j=\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j+1,j+l+1}{s}^l}{{{\displaystyle \sum }}_{l=0}^{n-j-1}{a}_{j+1,j+l+1}{s}^l}.$$

We already proved relation (6), which is

$$T_j\left(s\right)-h_j·\frac{s-1}{s}=\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l-1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}$$

hence,

$$\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l-1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}-k_j=\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j+1,j+l+1}{s}^l}{{{\displaystyle \sum }}_{l=0}^{n-j-1}{a}_{j+1,j+l+1}{s}^l}.$$

Therefore,

$$\begin{array}{c}\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)]s^{l-1}+\left(a_{j,n-}{h}_j{b}_{j,n}\right)s^{n-j-1}-k_j{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}\hfill \\ =\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j+1,j+l+1}{s}^l}{{{\displaystyle \sum }}_{l=0}^{n-j-1}{a}_{j+1,j+l+1}{s}^l}.\hfill \end{array}$$

We write it as

$$\begin{array}{c}\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}\left[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}]s^{l-1}+\right[\left(a_{j,n-}{h}_j{b}_{j,n}\right)-k_j{b}_{j,n}]s^{n-j-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}.\hfill \\ =\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}{b}_{j+1,j+l+1}{s}^l}{{{\displaystyle \sum }}_{l=0}^{n-j-1}{a}_{j+1,j+l+1}{s}^l}.\hfill \end{array}$$

By changing indices in the numerator and denominator of the right-hand side, we get

$$\begin{array}{c}\frac{{{\displaystyle \sum }}_{l=1}^{n-j-1}\left[a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}]s^{l-1}+\right[\left(a_{j,n-}{h}_j{b}_{j,n}\right)-k_j{b}_{j,n}]s^{n-j-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^{l-1}}\hfill \\ =\frac{{{\displaystyle \sum }}_{l=2}^{n-j}{b}_{j+1,j+l}{s}^{l-1}}{{{\displaystyle \sum }}_{l=1}^{n-j}{a}_{j+1,j+l}{s}^{l-1}}.\hfill \end{array}$$

The above equation leads to the following three conclusions:

1. $a_{j,j+1}-h_j\left(b_{j,j+1}-b_{j,j+2}\right)-k_j{b}_{j,j+1}=0.$

Hence,

$$k_j=\frac{a_{j,j+1}-h_j\left(b_{j,j+1}-b_{j,j+2}\right)}{b_{j,j+1}}\mathrm{for}0\le j\le n,\mathrm{as}\mathrm{desired}.$$

In addition,

$2.a_{j+1,j+l}=b_{j,j+l}\mathrm{for}0\le j\le n\mathrm{and}1\le l\le n-j,$ and

$3.b_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}\mathrm{for}0\le j\le n\mathrm{and}2\le l\le n-j.\square$

4. Feasibility of the Procedure

We shall illustrate the feasibility of the above procedure by applying it to a 3rd degree Schur polynomial.

We reconsider the same example we addressed in Example 2 of [10], but instead of the trial-and-error approach we used there, we shall apply the above systematic procedure to obtain the coefficients of the continued fraction expansion.

Consider the Schur polynomial

$$g\left(z\right)=4z^3-6z^2+4z-1$$

whose zeros are

$$\frac{1}{2},\frac{1}{2}+\frac{1}{2}i,\frac{1}{2}-\frac{1}{2}i$$

all lying inside the unit disc.

The reciprocal of g is

$$g^{\tau }\left(z\right)=z^n\overline{g\left(1/\overline{z}\right)}=-z^3+4z^2-6z+4$$

Therefore, the test function can be written as

$$\Psi \left(z\right)=\frac{g\left(z\right)-g^{\tau }\left(z\right)}{g\left(z\right)+g^{\tau }\left(z\right)}=\frac{5z^3-10z^2+10z-5}{3z^3-2z^2-2z+3}$$

Apply the transformation

$$T\left(s\right)=\Psi \left(2s-1\right).$$

Then,

$$T\left(s\right)=\Psi \left(2s-1\right)=\frac{5{\left(2s-1\right)}^3-10{\left(2s-1\right)}^2+10\left(2s-1\right)-5}{3{\left(2s-1\right)}^3-2{\left(2s-1\right)}^2-2\left(2s-1\right)+3}.$$

$T\left(s\right)\mathrm{can}\mathrm{now}\mathrm{be}\mathrm{written}\mathrm{as}$

$$T\left(s\right)=\frac{20s^3-50s^2+45s-15}{12s^3-22s^2+11s}.$$

We would like to expand $T\left(s\right)$ in the form

$$T\left(s\right)=h_0\frac{s-1}{s}+k_0+\frac{1}{h_1\frac{s-1}{s}+k_1+\frac{1}{h_2\frac{s-1}{s}+k_2}}$$

$\mathrm{We}\mathrm{seek}\mathrm{the}\mathrm{values}\mathrm{of}{h}_0,k_0,h_1,k_1,h_2,k_2\mathrm{using}\mathrm{the}\mathrm{above}\mathrm{procedure}.$

$\mathrm{By}\left(5\right),\mathrm{we}\mathrm{have}{T}_j\left(s\right)=\frac{{{\displaystyle \sum }}_{l=0}^{n-j}{a}_{j,j+l}{s}^l}{{{\displaystyle \sum }}_{l=1}^{n-j}{b}_{j,j+l}{s}^l}.$

$\mathrm{By}\left(4\right),T\left(s\right)=T_0\left(s\right),\mathrm{so}$

$$T\left(s\right)=T_0\left(s\right)=\frac{{{\displaystyle \sum }}_{l=0}^3{a}_{0,l}{s}^l}{{{\displaystyle \sum }}_{l=1}^3{b}_{0,l}{s}^l}=\frac{a_{0,0}+a_{0,1}s+a_{0,2}{s}^2+a_{0,3}{s}^3}{b_{0,1}s+b_{0,2}{s}^2+b_{0,3}{s}^3},$$

$\mathrm{which}\mathrm{compared}\mathrm{with}T\left(s\right)=\frac{20s^3-50s^2+45s-15}{12s^3-22s^2+11s}\mathrm{leads}\mathrm{to}$

$$a_{0,0}=-15,a_{0,1}=45,a_{0,2}=-50,a_{0,3}=20,$$

$$b_{0,1}=11,b_{0,2}=-22,b_{0,3}=12.$$

$\mathrm{Calculation}\mathrm{of}{h}_0\mathrm{and}{k}_0:$

$\mathrm{By}\mathrm{Theorem}1,h_0=-\frac{a_{0,0}}{b_{0,1}}=\frac{15}{11}.$

$\mathrm{By}\mathrm{Theorem}2,k_0=\frac{a_{0,1}-h_0\left(b_{0,1}-b_{0,2}\right)}{b_{0,1}}=\frac{45-\frac{15}{11}\left(11+22\right)}{b_{0,1}}=0.$

$\mathrm{Calculation}\mathrm{of}{h}_1\mathrm{and}{k}_1:$

$\mathrm{Again}\mathrm{by}\mathrm{Theorem}1,h_1=-\frac{a_{1,1}}{b_{1,2}}.$

$\mathrm{By}\mathrm{the}\mathrm{formulas}\mathrm{of}\mathrm{Theorem}3,\mathrm{namely}$

$$a_{j+1,j+l}=b_{j,j+l},$$

$\mathrm{and}$

$$b_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l},$$

$\mathrm{We}\mathrm{get},a_{1,1}=b_{0,1}=11.$

$$b_{1,2}=a_{0,2}-h_0\left(b_{0,2}-b_{0,3}\right)-k_0{b}_{0,2}=-50-\frac{15}{11}\left(-22-12\right)-0=-\frac{40}{11}.$$

Then,

$$h_1=-\frac{a_{1,1}}{b_{1,2}}=-\frac{11}{-\frac{40}{11}}=\frac{121}{40}.$$

In addition, by Theorem 2, $k_1=\frac{a_{1,2}-h_1\left(b_{1,2}-b_{1,3}\right)}{b_{1,2}}.$ $b_{1,2}=-\frac{40}{11},\mathrm{already}\mathrm{calculated}.$

$$a_{j+1,j+l}=b_{j,j+l}\mathrm{leads}\mathrm{to}{a}_{1,2}=b_{0,2}=-22$$

$$\begin{array}{c}{b}_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}\mathrm{leads}\mathrm{to}{b}_{1,3}\\ =a_{0,3}-h_0\left(b_{0,3}-b_{0,4}\right)-k_0{b}_{0,3}\end{array}$$

$$b_{1,3}=20-\frac{15}{11}\left(12-0\right)-0·b_{0,3}=\frac{40}{11}.$$

$\mathrm{Therefore},k_1=\frac{-22-\frac{121}{40}\left(-\frac{40}{11}-\frac{40}{11}\right)}{b_{1,2}}=0.$

$\mathrm{Calculation}\mathrm{of}{h}_2\mathrm{and}{k}_2:$

$\mathrm{Now},h_2=-\frac{a_{2,2}}{b_{2,3}},\mathrm{where}$

$$a_{2,2}=b_{1,2}=-\frac{40}{11}.$$

$$\begin{gathered} b_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}\mathrm{leads}\mathrm{to}{b}_{2,3} \\ =a_{1,3}-h_1\left(b_{1,3}-b_{1,4}\right)-k_1{b}_{1,3} \end{gathered}$$

$$a_{1,3}=b_{0,3}=12.$$

$$b_{1,3}=\frac{40}{11}\mathrm{already}\mathrm{calculated}.$$

$$\begin{gathered} b_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}\mathrm{implies}{b}_{1,4} \\ =a_{0,4}-h_0\left(b_{0,4}-b_{0,5}\right)-0·b_{1,3} \end{gathered}$$

$$b_{1,4}=a_{0,4}-h_0\left(b_{0,4}-b_{0,5}\right)-0·b_{1,3}=0-\frac{15}{11}\left(0-0\right)=0.$$

$\mathrm{Then},b_{2,3}=12-\frac{121}{40}·\frac{40}{11}=1.$

$\mathrm{Therefore},h_2=-\frac{a_{2,2}}{b_{2,3}}=-\frac{-\frac{40}{11}}{1}=\frac{40}{11}.$

$$k_2=\frac{a_{2,3}-h_2\left(b_{2,3}-b_{2,4}\right)}{b_{2,3}}.$$

$\mathrm{We}\mathrm{know}\mathrm{that},b_{2,3}=1.$

$$a_{2,3}=b_{1,3}=\frac{40}{11}\mathrm{was}\mathrm{already}\mathrm{calculated}.$$

$$\begin{gathered} b_{j+1,j+l}=a_{j,j+l}-h_j\left(b_{j,j+l}-b_{j,j+l+1}\right)-k_j{b}_{j,j+l}\mathrm{implies}{b}_{2,4} \\ =a_{1,4}-h_1\left(b_{1,4}-b_{1,5}\right)-k_1{b}_{1,4}=0. \end{gathered}$$

$$k_2=\frac{a_{2,3}-h_2\left(b_{2,3}-b_{2,4}\right)}{b_{2,3}}=\frac{\frac{40}{11}-\frac{40}{11}\left(1-0\right)}{1}=0.$$

Therefore,

$$T\left(s\right)=\frac{15}{11}\frac{s-1}{s}+0+\frac{1}{\frac{121}{40}\frac{s-1}{s}+0+\frac{1}{\frac{40}{11}\frac{s-1}{s}+0}}$$

or

$$\Psi \left(z\right)=\frac{15}{11}\left(\frac{z-1}{z+1}\right)+\frac{1}{\frac{121}{40}\left(\frac{z-1}{z+1}\right)+\frac{1}{\frac{40}{11}\left(\frac{z-1}{z+1}\right)}}$$

It is no surprise that the ks are all zeros since by Theorem 1 they are either 0 or pure imaginary. So, they must be zero because the given polynomial is of real coefficients. We wanted to show via calculations that the k’s are zero to match the application with the theory.

5. Application of the Procedure

In this section, we present an application of the above results to test the bounded input bounded output (BIBO) stability of two-dimensional digital filters. Specifically, we shall prove the relationship between BIBO stability and the procedure introduced in Section 3.

BIBO stability has been characterized in several equivalent ways of which we shall mention the following two.

Theorem 4

[13]. The general two-dimensional complex digital filter $F\left(z_1,z_2\right)={\displaystyle \sum }_{j=0}^m{\displaystyle \sum }_{k=0}^n{a}_{jk}{z}_1^j{z}_2^k$ is BIBO stable if and only if $F\left(z_1,z_2\right)\ne 0forall\left\{\left(z_1,z_2\right):\left|z_1\right|\le 1,\left|z_2\right|\le 1\right\}.$

Based on the work of Ansell [14], Huang [15] proved an equivalent result to that of Justice and Shanks which states the following:

Theorem 5

[15]. $ThedigitalfilterF\left(z_1,z_2\right)isBIBOstableifandonlyif$ the following two conditions hold:

$$F\left(z_1,0\right)\ne 0\mathit{for}\left|z_1\right|\le 1,$$

(9)

and

$$F\left(z_1,z_2\right)\ne 0\mathit{for}\left|z_1\right|=1,\mathit{and}\left|z_2\right|\le 1.$$

(10)

Other equivalent conditions for BIBO stability can also be found in Strintzis [16] and DeCarlo et al. [17].

In this section, we shall follow the conditions of Huang [15] to assess BIBO stability.

Consider the two-dimensional digital filter characterized by

$$F\left(z_1,z_2\right)={\displaystyle \sum }_{j=0}^m{\displaystyle \sum }_{k=0}^n{a}_{jk}{z}_1^j{z}_2^k,\mathrm{where}{b}_{00}=4,b_{10}=-6,b_{20}=4,b_{30}=-1.$$

(11)

Theorem 6.

$ThesystemF\left(z_1,z_2\right)={\displaystyle \sum }_{j=0}^m{\displaystyle \sum }_{k=0}^n{a}_{jk}{z}_1^j{z}_2^{k}where{b}_{00}=4,b_{10}=-6,b_{20}=4,b_{30}=-1satisfiescondition\left(9\right).$

Proof.

$\mathrm{Note}\mathrm{that}F\left(z_1,0\right)=b_{00}+b_{10}{z}_1+b_{20}{z}_1^2+b_{30}{z}_1^3=4-6z_1+4z_1^2-z_1^3.$

It can be verified that the zeros of $F\left(z_1,0\right)\mathrm{are}1-i,1+i\mathrm{and}2$ which all lie outside the closed unit disc.

$\mathrm{Therefore},F\left(z_1,0\right)\ne 0\mathrm{for}\left|z_1\right|\le 1,\mathrm{and}\mathrm{condition}\left(9\right)\mathrm{is}\mathrm{verified}.$ $\square$

The following theorem illustrates the relationship between condition (9) of BIBO stability, and the procedure introduced in Section 3.

Theorem 7.

Condition (9) of BIBO stability of the system

$$F\left(z_1,z_2\right)={\displaystyle \sum }_{j=0}^m{\displaystyle \sum }_{k=0}^n{a}_{jk}{z}_1^j{z}_2^k\mathrm{where}{b}_{00}=4,b_{10}=-6,b_{20}=4,b_{30}=-1,$$

is satisfied if and only if the test function associated with the function$F\left(z_1,0\right)$can be expanded in the continued fraction form

$$\Psi \left(z\right)=\frac{15}{11}\left(\frac{z-1}{z+1}\right)+\frac{1}{\frac{121}{40}\left(\frac{z-1}{z+1}\right)+\frac{1}{\frac{40}{11}\left(\frac{z-1}{z+1}\right)}}.$$

Proof. 

We need to realize that the function $F\left(z_1,0\right)$ is the reciprocal of the polynomial

$$g\left(z_1\right)=4z_1^3-6z_1^2+4z_1-1$$

introduced in Section 4.

Following the steps in the procedure of Section 4, the theorem is established. $\square$

Theorem 8.

$F\left(z_1,z_2\right)asdefinedin\left(11\right)hasnozerosfor\left|z_1\right|=1,and\left|z_2\right|\le 1ifandonlyifthetestfunction\Psi \left(z\right)$defined in (2) is a complex discrete reactance function.

Proof.

We apply the procedure used by Reddy and Rajan [11]. On top of page 1691 in [11], the functions $B_{*o}\left(e^{i{\theta }_1},z_2\right)\mathrm{and}{B}_{*e}\left(e^{i{\theta }_1},z_2\right)$ which Reddy and Rajan call the para-odd and para-even parts are exactly the numerator and denominator, respectively, of our test function (2) defined in Section 2, and therefore their function, $F_o\left(e^{i{\theta }_1},z_2\right)=\frac{B_{*e}\left(e^{i{\theta }_1},z_2\right)}{B_{*o}\left(e^{i{\theta }_1},z_2\right)}\mathrm{is}\mathrm{exactly}\mathrm{our}\mathrm{test}\mathrm{function}\Psi \left(z\right).$

Since $\Psi \left(z\right)$ is already $\mathrm{a}\mathrm{complex}\mathrm{discrete}\mathrm{reactance}\mathrm{function},$ condition (10) is thus satisfied, and that proves the theorem.     □

Now, system (11) satisfies both conditions (9) and (10) and is therefore BIBO stable.

The author thanks the referees for their suggestions, which certainly improved the quality of the paper.