Further Results on Bose s D Stability Test Minoru Yamada, Li Xu, Osami Saito Gifu National College of Technology, Shinsei-cho, Gifu, 50-0495 Japan, Faculty of System Science and Technology, Akita Prefecture University, 84-4 Ebinokuchi, Tsuchiya, Honjo, Akita 05-0055, Japan, Chiba University, Inageku, Chiba, 63-85 Japan myamada@gifu-nct.ac.jp, xuli@akita-pu.ac.jp saito@cute.te.chiba-u.ac.jp Keywords: discrete system, D systems, stability test, Sturm s method. Abstract This paper proposes some further improvements on Bose s D stability test by revealing symmetric properties of the polynomials and resultant occurring in the test and utilizing Sturm s method. Consequently, the improved test can be fulfilled by a totally algebraic algorithm and the complexcity of computation is largely reduced as it involves only certain real variable polynomials with degrees not exceeding half of their previous complex variable counterparts. Nontrivial examples are also illustrated. Introduction Owing to its great importance, the problem of algebraic test for D (two-dimensional) filter stability has been continuously attracting considerable attention,, 4. A D discrete filter given by the transfer function G(z, z ) = N(z, z ) D(z, z ) is (structurally) stable by definition iff () D(z, z ) 0, z, z () where N and D are assumed to be D factor coprime polynomials. It is well known that the condition of () is equivalent to the conditions 5 D(z, 0) 0, z (3) D(z, z ) 0, z =, z. (4) The condition of (3) can be checked by the well-known D methods, and the main difficulty is how to test the condition (4). Based on Schussler s D stability criterion, Bose has developed a resultant-based algebraic test for (4) and further a simplified version of it as follows,. Let D(z, z ) be a D polynomial which may have complex coefficients, n the degree of D(z, z ) in z, and write D (z, z ) = z n D(z, z ) in the form n D (z, z ) = d k (z )z k (5) where d k (z ) are D polynomials in z with perhaps complex coefficients. Furthermore, define D s (z, z ) = n n d k (z )z k + z n d k (z )z k D a (z, z ) = n n d k (z )z k z n d k (z )z k where d k (z ) denotes the complex conjugates of both the coefficients and the indeterminate z. Then, the (simplified) Bose s test for (4) is given by the following theorem. Theorem The polynomial D (z, z ) 0 for z =, z, or equivalently, D(z, z ) 0 for z =, z iff (a) d 0 ()/d n () <, (b) all the zeros of D s (, z ) and D a (, z ) are simple, are located and also alternate on z =, and (c) the resultant R(z ) of D s (z, z ) and D a (z, z ) satisfies that R(z ) 0 for z = e jθ, 0 θ < π. Since D s (, z ) and D a (, z ) are both D polynomials, it is possible to test the conditions of (b) by using e.g. the D continued fraction expansion method. On the other hand, the resultant R(e jθ ) is a trigonometric function and it may not be so straightforward in general to simplify systematically this trigonometric function and carry out the test of condition (c) though a possibility utilizing Chebyshev polynomials has been suggested in. The purpose of this paper is to propose some further improvements on Bose s test so that the test can be fulfilled more efficiently by a totally algebraic algorithm and can be implemented more easily and systematically into a D CAD
system 6. As main results, it is first shown that D s (, z ), D a (, z ) can be expressed in the form D s (, z ) = L (z )F (z ) and D a (, z ) = L (z )F (z ) where F (z ), F (z ) are certain self-inversive polynomials, thus can be easily converted into polynomials F (x ), F (x ) with degrees in the real variable x = (z + z )/ not exceeding n /. This fact leads to a result that the condition (b) holds true iff the zeros of F (x ) and F (x ) are simple, are located and also alternate in certain order within < x <, and this can be tested by using some extended results of Sturm s method. In fact, this is a significant new result even for D stability test as it clarified that by utilizing the symmetric properties of the related polynomials we only need to carry out the test for certain real variable polynomials with degrees not exceeding half of the degrees of the original polynomials. Further, it is shown that the resultant R(z ) is in fact a self-inversive polynomial itself and thus can be converted into a polynomial R(x) with the degree in x = (z + z )/ being only half of the degree of R(z ) in z. Therefore, the test of (c) can be readily accomplished by direct application of Sturm s method to R(x ) on x without any other computational effort which is generally required for testing R(e jθ ) = 0 on 0 θ < π as mentioned above. For simplicity, we will only consider the case for stability test of polynomials with real coefficients in this paper, and will report its extensions to the case with complex coefficients on some other occasion. The paper is organized as follows. In Section, new results for D stability test will be given, while in Section 3 the self-inversive property of the resultant R(z ) will be shown and possible improvements for Bose s D stability test will be summarized. Nontrivial examples for polynomials having both numerical and literal coefficients will be also shown in Section 4 to illustrate the computational advantages of the improved test. Finally, Section 5 concludes the paper with some comments for related further topics. New Results for D Test Let F (z) be a D polynomial of degree n described by F (z) = n f k z k (6) where f k, k = 0,,..., n are real coefficients and define F, F as F (z) = (F (z) + zn F (z )) = F (z) = (F (z) zn F (z )) = n f,k z k (7) n f,k z k (8) For the stability of F (z), the following result has been well known 7. Lemma The polynomial F (z) is stable, i.e., F (z) 0, z > iff f 0 /f n <, and the zeros of F (z) and F (z) are simple, are located and further alternate with each other on the unit circle z =. This stability criterion can be tested by the well-known CFE (continued fraction expansion) approach. However, we want to show here that by revealing and utilizing the symmetrical properties of F (z), F (z), two real variable polynomials F (x), F (x) corresponding respectively to F (z) and F (z) can be constructed which have degrees not exceeding half of the ones for F (z), F (z), and the stability of F (z) can now be tested by just checking if F (x), F (x) satisfy the conditions of Lemma within the real interval < x < which can be easily fulfilled by using some extended results of Sturm s method. Lemma F (z), F (z) can be expressed as follows. F (z) = F (z) = z n/ F (z) z (n )/ (z + ) F (z) n: evne n: odd (9) z n/ (z z ) F (z) n: even z (n )/ (z ) F (z) n: odd (0) where F i (z), i =, are self-inversive polynomials given by F i (z) = m f i,0 + f i,k (z k + z k ) () and when n is even, f,k = f,n/+k, k = 0,,, n/(= m) () f,k = f,n/+k+ + f,n/+k+3 +... + f,n/+l, l = when n is odd, k = 0,,..., n/ (= m) (3) n/ n/ + k : even, (4) n/ n/ + k : odd; f,k = f,(n+)/+k f,(n+)/+k+ +f,(n+)/+k+ + ( ) (n )/+k f,n, k = 0,,..., (n )/(= m) (5) f,k = f,(n+)/+k + f,(n+)/+k+ + + f,n k = 0,,..., (n )/(= m). (6) Proof. F (z), F (z) can be always written as F (z) = n n f k z k + z n f k z k = n n f k z k + f k z n k = n n f k z k + f n k z k
= n (f k + f n k )z k (7) F (z) = n n f k z k z n f k z k = n (f k f n k )z k (8) Comparing (7), (8) with (7), (8), we see f,k = f,n k = f k + f n k (9) f,k = f,n k = f k f n k (0) If n is even, multiplying (7) by z n/ yields z n/ F (z) = n f,k z k n/ = n/ n f,k z k n/ + f,n/ + f,k z k n/ k=n/+ = n/ n/ f,n/ k z k + f,n/ + f,n/+k z k = n/ n/ f,n/+k z k + f,n/ + f,n/+k z k = n/ f,n/ + f,n/+k (z k + z k ) () where f,k = f,n k was used. Setting the right-hand side of () as F (z), we get the results of (9), () and (3). Further, multiplying (8) by z n/ gives z n/ F (z) = n f,k z k n/ = n/ n f,k z k n/ + f,n/ + f,k z k n/ k=n/+ = n/ n/ f,n/ k z k + f,n/ + f,n/+k z k = n/ n/ f,n/+k z k + f,n/+k z k n/ = f,n/+k (z k z k ) () where the relations f,k = f,n k, f,n/ = f n/ f n/ = 0 were applied. Since z k z k = (z z )(z k + z k 3 + + z k+3 + z k+ ) for k, () can be expanded as z n/ F (z) = (z z ) f,n/+ ( ) +f,n/+ ( z +z ) +f,n/+3 ( z + + z ) +f,n/+4 ( z 3 + z +z + z 3 ) +f,n/+5 (z 4 +z + + z + z 4 )... f,n (z + + + + +z n/ n )}, : odd f,n (z n/ + +z+z + +z n/ )}, n : even By collecting all the coefficients of the terms of and (z k + z k ), k =,..., n/ in the parentheses and denoting the calculated coefficients as f,k, the relation z n/ F (z) = (z z ) f,0 + n/ f,k (z k + z k ) (3) can be obtained which implies (0), (). It is ready to verify that f,k satisfy (3). The results corresponding to the case when n is odd can be shown in similar way, thus is omitted here for brevity. Corollary The coefficients f i,k, i =, for F i (z) defined by () can be calculated recursively as follows. f, n +k, k = 0,,..., n n : even f,k = f, n+ +k f,k+, k =0,,..., n n : odd (with f, n = f,n ) f,k = f, n +k+ + f,k+, k = 0,,..., n 3, n : even (with f, n = f,n, f, n = f,n ) +k+ f,k+, k = 0,,..., n, n : odd (with f, n = f,n ). f, n+ Proof. It follows directly from (3) (6). As F i (z), i =,, are self-inversive polynomials and z = z on the unit circle z =, Fi (z), i =, can be converted into polynomials in real variable x = (z + z )/ by the method given in 3. Theorem The polynomial F (z) is stable iff f 0 /f n < and all the zeros of ˆF (x) and ˆF (x) are simple, are located and also alternate within the interval < x <. Further, when n is odd, ˆF (x) has a zero on the most right side, or equivalently, ˆF (x) has a zero on the most left side. Proof. Omitted. We show next the conditions of Theorem can be easily tested by using the extended Sturm s theorem given by Lemma 3. Let g(x), g (x) be two given real variable polynomials, and without loss of generality denote g(x) the one whose degree is not less than the other one. Further, for simplicity, suppose that g(x) and g (x) possess no common zeros. Define the following function series starting from g(x), g (x). g(x), g (x), g (x),, g l (x) (4)
where g i (x) = h i (x)g i (x) g i+ (x), i =,,..., l (5) Lemma 3 8 Let V (x) denote the number of variation of the sign in the polynomial series (4) constructed from g(x) and g (x), and suppose that for the interval a < x < b, g(x)/g (x) 0 at the points x = a, x = b. Then, V (a) V (b) = χ(x 0 ) (6) where x 0 x g(x)/g (x)=0} X (x 0 ) 0, the sign of g(x)/g (x) does not change =, the sign of g(x)/g (x) changes from to +, the sign of g(x)/g (x) changes from + to Based on the above lemmas, the following theorem can now be given. Lemma 4 Let g(x) = ˆF (x), g (x) = ˆF (x), and define V (x) as above for the polynomial series (4) corresponding to ˆF (x), ˆF (x). Then, polynomial F (z) satisfies the conditions of Theorem, i.e., F (z) is stable, iff V ( ) V () = sign(f,n f,n ) n, (7) when n is even, while when n is odd. Proof. Omitted. V ( ) V () = sign(f,n f,n ) n 3 New Results for D Test (8) Another important step in Bose s D stability test is testing whether the resultant R(z ) of D s (z, z ) and D a (z, z ) is devoid of zeros on z =. In this section, we first show that R(z ) is in fact a self-inversive polynomial itself on z =, so it can also be converted into a real variable polynomial with a reduction in the degree to half of the degree for R(z ). Then, an improved version of Bose s D stability test will be given based on the results obtained in this and the previous sections. Lemma 5 The resultant R(z ) is self-inversive on z =. Proof. Since z = z on z =, D s (z, z ), D a (z, z ) can be respectively written as D s (z, z ) = n dk (z ) + d n k(z ) } z k D a (z, z ) = n dk (z ) d n k(z ) } z k Therefore, defining h s,k (z ) = (d k (z ) + d n k(z ))/, h a,k (z ) = (d k (z ) d n k(z ))/, the resultant R(z ) of D s (z, z ) and D a (z, z ) with respect to z is given by R(z ) = h s,n h s,n h s, h s,0 0 h s,n h s, h s, h s,0............. 0 h s,n h s,n h s,n h s,0 h a,n h a,n h a, h a,0 0 h a,n h a, h a, h a,0........... 0 h a,n h a,n h a,n h a,0 where the argument z was omitted for brevity. Then, reversing the order of all the colunms, and reversing the order of the upper n rows, and the lower n rows respectively, we can get the following result no matter n is even or odd. R(z ) = ( ) n h s,0 h s, h s,n h s,n 0 h s,0 h s,n h s,n h s,n........... 0 h s,0 h s, h s, h s,n h a,0 h a, h a,n h a,n 0 h a,0 h a,n h a,n h a,n............ 0 h a,0 h a, h a, h a,n It follows from h s,k (z ) = h s,n k(z a,k(z ) = h a,n k(z R(z ) = ( )n h s,n h s,n h s,0 0 h s,n h s, h s,0............ 0 h s,n h s,n h s,0 h a,n h a,n h a,0 0 h a,n h a, h a,0............ 0 h a,n h a,n h a,0 Multiplying the lower n rows by, we see that R(z ) = R(z ) which means that R(z ) is self-inversive. Now, we can transform R(z ) into a polynomial ˆR(x ) in the real variable x = (z + z )/. It should be clear that to test the condition (c) of Theorem we only need to see whether or not ˆR(x ) 0 on x which can be easily carried out by Sturm s method. Based on the above results and those given in the previous section, we can now give an improved version of Bose s D stability test as follows. Define D s (z ), Da (z ), and ˆD s (x ), ˆDa (x ) for D s (, z ) and D a (, z ) in the same way as the definition
of F (z ), F (z ) and ˆF (x ), ˆF (x ) for F (z ), F (z ) in Section, respectively. Theorem 3 The polynomial D (z, z ) 0 for z =, z, or equivalently, D(z, z ) = 0 for z =, z iff (a) d 0 ()/d n () <, (b) all the zeros of ˆD s (x ) and ˆD a (x ) are simple, are located and also alternate within < x < where x = (z + z )/, particularly when n is odd, ˆD s (x ) has a zero on the most right side, or equivalently, ˆDa (x ) has a zero on the most left side, and further (c) ˆR(x ) constructed from the resulatant R(z ) of D s (z, z ) and D a (z, z ) satisfies that ˆR(x ) 0 for x where x = (z + z )/. Proof. It follows immediately from the results of Theorem and Lemma 5. It is obvious that the condition (b) of Theorem 3 can be tested by the criteria of Theorem and the condition (c) can be checked directly by Sturm s method. i.e., d 0 ()/d n () <. If not, then unstable. 4 Illustrating Examples Several nontrivial examples are given to show the testing procedure and the effectiveness of the proposed method. Example. Consider the stability of the following D polynomial used in,. D(z, z ) = ( + 0z + z ) + (6 + 5z + z )z (9) Step. Obviously, D(z, 0) = + 0z + z = (z + )(z + 3) has zeros of, 3, thus is D stable. Step. Since d 0 () d () = 4 = <, the condition (a) of Theorem 3 is satisfied. Step 3. It is easy to see that D s (, z ) = 8( + z ) D a (, z ) = 6(z ) and D s (z ) = 8, Da (z ) = 6. Clearly, the condition (b) of Theorem 3 is satisfied. Step 4. The resultant R(z ) is calculated as R(z ) = 93 05 (z + z ) 9(z + z ) Let x = (z + z = 93 05 9 z + z z + z )/, then it follows that 3 0 0 = 0 0 0 4 z + z z + z x x and ˆR(x ) = 93 05 9 0 0 0 0 0 4 = 75 05x 36x Then, the Sturm s functions can be constructed as x x f 0 (x ) = ˆR(x ) = 36x 05x 75 (30) f (x ) = f 0(x ) = 7x 05 (3) f (x ) = 5 (3) 44 and it is ready to verify that V ( ) = 0, V () = 0. Then, due to Sturm s Theorem we conclude that ˆR(x ) 0, for x, and therefore the D polynomial given in (9) is stable. Noting that the above ˆR(x ) is only a second degree polynomial, while the one obtained in is a 8th degree polynomial and the one in is a trigonometric function. Example. Show the conditions for the following D polynomial to be stable, where n = n =, and a, b, c are real numbers. D(z, z ) = + az + bz + cz z (33) Step. To ensure that D(z, 0) = + az 0 for z, it is necessary to have that a >. Step. Since d 0 (z ) = b + cz and d (z ) = + az, the following relation must hold. d 0 () d () = b + c + a < (34) Step 3. Since it follows that D (z, z ) = (b + cz ) + ( + az )z (35) D (, z ) = (b + c) + ( + a)z, (36) D s (, z ) = (z + )( + a + b + c) (37) D a (, z ) = (z )( + a + b + c) (38) which have already satisfied the condition (b) of Theorem 3. Step 4. we have D s (z, z ) = (cz + b + + az ) + (az + + b + cz )z } D a (z, z ) = (cz + b az ) + (az + b cz )z }. Therefore, the resultant R(z ) is given by R(z ) = h s (z ) h s0 (z ) h a (z ) h a0 (z ) = (a bc)z + ( + a b c ) + (a bc)z } 4
where h s (z ) = (az + + b + cz ) h s0 (z ) = (cz + b + + az ) h a (z ) = (az + b cz ) h a0 (z ) = (cz + b az ). Letting x = (z + z )/ and using the transformation shown in 3, we have the result ˆR(x ) = 4 ( + a b c ) + (a bc)x } (39) Following Sturm s method, set f 0 (x ) = ˆR(x ), f (x ) = f 0(x ) = (a bc)/. Since f (x ) has already been a constant, to see if ˆR(x ) = 0 for x we only need to see if V ( ) V () = 0 where V (x ) denots the number of the sign variation in the polynomial sequence f 0 (x ), f (x ). It is easy to see that f 0 ( ) = ( + a b c ) a(a bc) } 4 f ( ) = (a bc) f 0 () = ( + a b c ) + a(a bc) } 4 f () = (a bc). As f ( ) = f (), to satisfy that V ( ) V () = 0, f 0 ( ) and f 0 () must have the same sign which is equivalent to requiring that f 0 () f 0 ( ) = + a b c (a bc) + a b c + (a bc) = ( a) (b c) ( + a) (b + c) > 0 (40) Due to (34), we have ( + a) (b + c) > 0. Therefore, for (40) to be true, it must be which is equivalent to ( a) (b c) > 0 (4) b c a <. (4) Now, we see that the necessary and sufficient condition for the polynomial of (33) to be stable is a >, b + c + a <, b c a <. (43) 5 Conclusions Some improvements have been proposed for Bose s D stability test by revealing symmetric properties of the polynomials and resultant occurring in the test and utilizing Sturm s method. Consequently, the improved test can be fulfilled by a totally algebraic algorithm and the complexcity of computation is largely reduced as it involves only certain real variable polynomials with degrees not exceeding half of their previous complex variable counterparts. Kurosawa et al. 9 have recently proposed an efficient algorithm for calculating the determinant of a matrix with D polynomial entrices by making use of DFT (or FFT). It should be clear that this algorithm can be directly applied to compute the resultant R(z ) so that the complexity of Bose s D stability test can be improved even more. Nontrivial examples have also been shown. References Bose N. K. Implementation of a new stability test for two-dimensional filters. IEEE Trans. Acoustics, Speech and Signal Processing, Vol.5, pp.7 0, 977. Bose N. K. Simplification of a Multidimensional Digital Filter Stability Test. Journal of the Franklin Institute, Vol.330, No.5, pp.905 9, 993. 3 Hu X. H. D filter stability test using polynomial array for F (z, z ) on z =. IEEE Trans. Circuits and Systems, Vol.38, No.9, pp.09 095, 99. 4 Hu X. H. On Two-Dimensional Filter Stability Test. IEEE Trans. Circuits and Systems II: Analog and Digital Signal Processing, Vol.4, No.7, pp.457 46, 994. 5 Huang T. S. Stability of two-dimensional recursive filters. IEEE Trans. Audio Electroacoust, Vol.0, pp.58 63, 97. 6 Xu L., Yamada M. and Saito O. Development of nd Control System Toolbox for Use with MATLAB. Proc. of CCA99, pp. 543-548, August -7, Hawaii, 999. 7 Schüssler H. W. Stability Theorem for Discrete Systems IEEE Trans. Acoustics, Speech, and Signal Processing, Vol.4, No., pp.87 89, 976. 8 T. Takagi. Algebra Lecture, Kyoritu Shuppan, Tokyo, 965 (in Japanese). 9 Kurosawa K., Yamada I. and Yokokawa T. A Fast -D stability Test Procedure Based on FFT and its Computational Complexity. IEEE Trans. Circuits and Systems II: Analog and Digital Signal Processing, Vol.4, No.0, pp.676 678, 995. This result is the same as the one given in but the derivation shown here is much simpler.