Chapter 5 THE APPLICATION OF THE Z TRANSFORM 5.3 Stability Copyright c 2005- Andreas Antoniou Victoria, BC, Canada Email: aantoniou@ieee.org February 13, 2008 Frame # 1 Slide # 1 A. Antoniou Digital Signal Processing Sec. 5.3
Stability A discrete-time system is stable if and only if its impulse response is absolutely summable, i.e., h(nt ) < n=0 Frame # 2 Slide # 2 A. Antoniou Digital Signal Processing Sec. 5.3
Stability A discrete-time system is stable if and only if its impulse response is absolutely summable, i.e., h(nt ) < n=0 Since the transfer function is the z transform of the impulse response, we expect the stability of the filter to depend critically on the transfer function. It actually depends exclusively on the positions of the poles. Frame # 2 Slide # 3 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Consider a causal recursive system characterized by the transfer function H(z) = N(z) D(z) = H Mi=1 0 (z z i ) m i Ni=1 where N M (z p i ) n i Frame # 3 Slide # 4 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Consider a causal recursive system characterized by the transfer function H(z) = N(z) D(z) = H Mi=1 0 (z z i ) m i Ni=1 where N M (z p i ) n i The impulse response is given by the general inversion formula as h(nt ) = Z 1 H(z) = 1 H(z)z n 1 dz 2πj Ɣ Frame # 3 Slide # 5 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Consider a causal recursive system characterized by the transfer function H(z) = N(z) D(z) = H Mi=1 0 (z z i ) m i Ni=1 where N M (z p i ) n i The impulse response is given by the general inversion formula as h(nt ) = Z 1 H(z) = 1 H(z)z n 1 dz 2πj By using the residue theorem, we have { R 0 + N [ h(nt ) = i=1 Res z=pi H(z)z 1 ] for n = 0 Ni=1 Res z=pi [H(z)z n 1 ] for n > 0 where [ ] H(z) R 0 = Res z=0 z if H(z)/z has a pole at the origin and R 0 = 0 otherwise. Ɣ Frame # 3 Slide # 6 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d If we assume that H(z) has simple poles, i.e., n i = 1for i = 1, 2,..., N, then the impulse response can be expressed as { R 0 + N i=1 h(nt ) = p 1 i Res z=pi H(z) for n = 0 N i=1 pn 1 i Res z=pi H(z) for n > 0 where the ith term in the summations is the contribution to the impulse response due to pole p i. Frame # 4 Slide # 7 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d { R 0 + N h(nt ) = i=1 p 1 i Res z=pi H(z) for n = 0 Ni=1 p n 1 i Res z=pi H(z) for n > 0 If we let p i = r i e jψ i then the impulse response can be expressed as { h(0) h(nt ) = Ni=1 r n 1 i e j(n 1)ψ i Res z=pi H(z) for n > 0 where is finite. h(0) = R 0 + N i=1 r 1 i e jψ i Res z=pi H(z) for n = 0 Frame # 5 Slide # 8 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d h(nt ) = n=0 { h(0) N i=1 r n 1 i e j(n 1)ψ i Res z=pi H(z) for n > 0 We can now write N h(nt ) = h(0) + n=1 i=1 r n 1 i e j(n 1)ψ i Res z=pi H(z) Frame # 6 Slide # 9 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d We note that N N ith term ith term i=1 i=1 Frame # 7 Slide # 10 A. Antoniou Digital Signal Processing Sec. 5.3
Example The sum of the magnitudes of the complex numbers is 3 c i = ( 1+j1) + (2+j2) + (2 j1) = 2+ 8+ 5 = 6.479 i=1 Complex plane c 2 =2+j +j2 c 1 =-1+j +j11 c 3 =2-j -j1 Frame # 8 Slide # 11 A. Antoniou Digital Signal Processing Sec. 5.3
Example Cont d On the other hand, the magnitude of the sum of the complex numbers is given by 3 c i = ( 1+j1)+(2+j2)+(2 j1) = 3+j2 = 13 = 3.606 i=1 Complex plane c 2 =2+j +j2 c 1 =-1+j +j11 c 3 =2-j -j1 Frame # 9 Slide # 12 A. Antoniou Digital Signal Processing Sec. 5.3
Example Cont d Therefore, 3 3 c i c i i=1 i=1 Complex plane c 2 =2+j +j2 c 1 =-1+j +j11 c 3 =2-j -j1 Frame # 10 Slide # 13 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d N h(nt ) = h(0) + n=0 Thus we can write n=0 n=1 h(nt ) h(0) + h(0) + h(0) + i=1 n=1 i=1 n=1 i=1 r n 1 i e j(n 1)ψ i Res z=pi H(z) N r n 1 i e j(n 1)ψ i Res z=pi H(z) N N n=1 i=1 r n 1 i e j(n 1)ψ i Res z=pi H(z) r n 1 i Res z=pi H(z) Frame # 11 Slide # 14 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Let us assume that all the poles are inside the unit circle z =1, i.e., r i r max < 1 for i = 1, 2,..., N jim z z plane 1 r max Re z Frame # 12 Slide # 15 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Now if p k is a simple pole of some function F (z), then function (z p k )F (z) is analytic and, therefore, the residue of F (z) at z = p k is finite. Frame # 13 Slide # 16 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Now if p k is a simple pole of some function F (z), then function (z p k )F (z) is analytic and, therefore, the residue of F (z) at z = p k is finite. Consequently, all the residues of H(z) are finite and so Res z=pi H(z) R max for i = 1, 2,..., N where R max is a positive constant. Frame # 13 Slide # 17 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d From the previous two slides r i r max < 1 for i = 1, 2,..., N and Res z=pi H(z) R max for i = 1, 2,..., N Frame # 14 Slide # 18 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d From the previous two slides r i r max < 1 for i = 1, 2,..., N and Res z=pi H(z) R max for i = 1, 2,..., N Therefore, we can write N h(nt ) h(0) + r n 1 i Res z=pi H(z) n=0 h(0) +NR max n=1 i=1 rmax n 1 n=1 Frame # 14 Slide # 19 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d n=0 h(nt ) h(0) +NR max n=1 r n 1 max The sum at the right-hand side is a geometric series with common ratio r max and since we have assumed that r max < 1, the series converges. Frame # 15 Slide # 20 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d n=0 h(nt ) h(0) +NR max n=1 r n 1 max The sum at the right-hand side is a geometric series with common ratio r max and since we have assumed that r max < 1, the series converges. We, therefore, conclude that h(nt ) < n=0 Frame # 15 Slide # 21 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Summarizing, we have assumed that all the poles are inside the unit circle, i.e., r i r max < 1 for i = 1, 2,..., N and demonstrated that in such a case the impulse response is absolutely summable, i.e., h(nt ) < n=0 Frame # 16 Slide # 22 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Summarizing, we have assumed that all the poles are inside the unit circle, i.e., r i r max < 1 for i = 1, 2,..., N and demonstrated that in such a case the impulse response is absolutely summable, i.e., h(nt ) < n=0 Therefore, we conclude that if all the poles are inside the unit circle, the system is stable. Frame # 16 Slide # 23 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d One more thing needs to be done in order to fully establish the role of the pole positions on the stability of the system. Frame # 17 Slide # 24 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d One more thing needs to be done in order to fully establish the role of the pole positions on the stability of the system. The condition established so far is a sufficient condition and one may, therefore, ask: Is it possible for a system to be stable if one or more poles are located on or outside the unit circle? Frame # 17 Slide # 25 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Let us assume that a single pole of H(z), say pole p k,is located on or outside the unit circle, i.e., r k 1. Frame # 18 Slide # 26 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Let us assume that a single pole of H(z), say pole p k,is located on or outside the unit circle, i.e., r k 1. In such a case, as n we have h(nt ) = N i=1 r n 1 i e j(n 1)ψ i Res z=pi H(z) r n 1 k e j(n 1)ψ k Res z=pk H(z) since for a large value of n, r n 1 i 0 for all i k for which r i < 1 whereas r n 1 k is unity or becomes very large since r k 1. Frame # 18 Slide # 27 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Thus h(nt ) n=0 n=0 r n 1 k Res z=pk H(z) e j(n 1)ψ i Res z=pk H(z) n=0 r n 1 k Frame # 19 Slide # 28 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Thus h(nt ) n=0 n=0 r n 1 k Res z=pk H(z) e j(n 1)ψ i Res z=pk H(z) n=0 r n 1 k Since r k 1, the sum at the right-hand side does not converge i.e., the impulse response is not absolutely summable, i.e., h(nt ) n=0 and the system is unstable. Frame # 19 Slide # 29 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d Therefore, we conclude that a discrete-time system is stable if and only if all its poles are inside the unit circle of the z plane. jim z z plane 1 Regions of instability Re z Region of stability Note: Nonrecursive discrete-time systems are always stable since their poles are always located at the origin of the z plane. Frame # 20 Slide # 30 A. Antoniou Digital Signal Processing Sec. 5.3
Example Check the following system for stability: X(z) Y(z) 1 1 2 Frame # 21 Slide # 31 A. Antoniou Digital Signal Processing Sec. 5.3
Example Solution The transfer function of the system can be easily obtained as H(z) = z2 z + 1 z 2 z + 0.5 z 2 z + 1 = (z p 1 )(z p 2 ) where p 1, p 2 = 1 2 ± j 1 2 = 1 2 e ±jπ/4 Since the system is stable. p 1, p 2 < 1 Frame # 22 Slide # 32 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Stability criteria are simple techniques that can be used to determine whether a system is stable or unstable with minimal computational effort. Frame # 23 Slide # 33 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Stability criteria are simple techniques that can be used to determine whether a system is stable or unstable with minimal computational effort. Consider a system characterized by the transfer function where H(z) = N(z) D(z) N(z) = M a i z M i and D(z) = i=0 N b i z N i i=0 Frame # 23 Slide # 34 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Cont d As was demonstrated in previous slides, the stability of a discrete-time system can be determined by finding the poles of the transfer function, namely, the roots of the denominator polynomial D(z). Frame # 24 Slide # 35 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Cont d As was demonstrated in previous slides, the stability of a discrete-time system can be determined by finding the poles of the transfer function, namely, the roots of the denominator polynomial D(z). For a second- or a third-order system this is easily done. Frame # 24 Slide # 36 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Cont d As was demonstrated in previous slides, the stability of a discrete-time system can be determined by finding the poles of the transfer function, namely, the roots of the denominator polynomial D(z). For a second- or a third-order system this is easily done. For higher-order systems, we need to use a computer program that would evaluate the roots of a polynomial, for example, MATLAB. Frame # 24 Slide # 37 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Cont d As was demonstrated in previous slides, the stability of a discrete-time system can be determined by finding the poles of the transfer function, namely, the roots of the denominator polynomial D(z). For a second- or a third-order system this is easily done. For higher-order systems, we need to use a computer program that would evaluate the roots of a polynomial, for example, MATLAB. Alternatively, we can use one of several stability criteria. Frame # 24 Slide # 38 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Cont d Common factors in N(z) and D(z) do not have anything to do with stability because they can be canceled out. Frame # 25 Slide # 39 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Criteria Cont d Common factors in N(z) and D(z) do not have anything to do with stability because they can be canceled out. For example, if N(z) and D(z) have a common factor (z + w), then H(z) = N(z) D(z) = (z + w)n (z) (z + w)d (z) = N (z) D (z) In effect, the poles of H(z) are the roots of D (z) and parameter w will not appear in the impulse response. Frame # 25 Slide # 40 A. Antoniou Digital Signal Processing Sec. 5.3
Test for Common Factors If there are common factors, they must be removed. Frame # 26 Slide # 41 A. Antoniou Digital Signal Processing Sec. 5.3
Test for Common Factors If there are common factors, they must be removed. To test a transfer function H(z) = N(z) D(z) = Mi=0 a i z M i Ni=0 b i z N i for common factors, an (M + N) (M + N) matrix is constructed and its determinant is evaluated where M and N are the numerator and denominator degrees, respectively. Frame # 26 Slide # 42 A. Antoniou Digital Signal Processing Sec. 5.3
Test for Common Factors If there are common factors, they must be removed. To test a transfer function H(z) = N(z) D(z) = Mi=0 a i z M i Ni=0 b i z N i for common factors, an (M + N) (M + N) matrix is constructed and its determinant is evaluated where M and N are the numerator and denominator degrees, respectively. If the determinant of this matrix is zero, then there are common factors. (See Sec. 5.3.4 of textbook for details.) Frame # 26 Slide # 43 A. Antoniou Digital Signal Processing Sec. 5.3
Test for Common Factors If there are common factors, they must be removed. To test a transfer function H(z) = N(z) D(z) = Mi=0 a i z M i Ni=0 b i z N i for common factors, an (M + N) (M + N) matrix is constructed and its determinant is evaluated where M and N are the numerator and denominator degrees, respectively. If the determinant of this matrix is zero, then there are common factors. (See Sec. 5.3.4 of textbook for details.) Hereafter, we assume that N(z) and D(z) do not have any common factors. Frame # 26 Slide # 44 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d The classical stability criteria for continuous-time systems are the Nyquist and Routh-Hurwitz criteria. Frame # 27 Slide # 45 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d The classical stability criteria for continuous-time systems are the Nyquist and Routh-Hurwitz criteria. Corresponding criteria that can be used to check the stability of discrete-time systems and digital filters are the following: Frame # 27 Slide # 46 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d The classical stability criteria for continuous-time systems are the Nyquist and Routh-Hurwitz criteria. Corresponding criteria that can be used to check the stability of discrete-time systems and digital filters are the following: Schur-Cohn criterion (1922) Frame # 27 Slide # 47 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d The classical stability criteria for continuous-time systems are the Nyquist and Routh-Hurwitz criteria. Corresponding criteria that can be used to check the stability of discrete-time systems and digital filters are the following: Schur-Cohn criterion (1922) Schur-Cohn-Fujisawa criterion (1925) Frame # 27 Slide # 48 A. Antoniou Digital Signal Processing Sec. 5.3
Stability Cont d The classical stability criteria for continuous-time systems are the Nyquist and Routh-Hurwitz criteria. Corresponding criteria that can be used to check the stability of discrete-time systems and digital filters are the following: Schur-Cohn criterion (1922) Schur-Cohn-Fujisawa criterion (1925) Jury-Marden criterion (1962) Frame # 27 Slide # 49 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn Stability Criterion For an Nth-order system, N matrices of dimensions ranging from 2 2to2N 2N are constructed using the coefficients of D(z). Frame # 28 Slide # 50 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn Stability Criterion For an Nth-order system, N matrices of dimensions ranging from 2 2to2N 2N are constructed using the coefficients of D(z). The determinants of these matrices, say, D 1, D 2,..., D N, are computed and their signs are determined. Frame # 28 Slide # 51 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn Stability Criterion For an Nth-order system, N matrices of dimensions ranging from 2 2to2N 2N are constructed using the coefficients of D(z). The determinants of these matrices, say, D 1, D 2,..., D N, are computed and their signs are determined. The system is stable if and only if D k < 0 for odd k and D k > 0 for even k Frame # 28 Slide # 52 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn-Fujisawa Stability Criterion The Schur-Cohn-Fujisawa criterion is much more efficient. Frame # 29 Slide # 53 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn-Fujisawa Stability Criterion The Schur-Cohn-Fujisawa criterion is much more efficient. It involves only one matrix of dimension N N, whichis again constructed using the coefficients of D(z). Frame # 29 Slide # 54 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn-Fujisawa Stability Criterion The Schur-Cohn-Fujisawa criterion is much more efficient. It involves only one matrix of dimension N N, whichis again constructed using the coefficients of D(z). The test amounts to checking whether the matrix is positive definite. Frame # 29 Slide # 55 A. Antoniou Digital Signal Processing Sec. 5.3
Schur-Cohn-Fujisawa Stability Criterion The Schur-Cohn-Fujisawa criterion is much more efficient. It involves only one matrix of dimension N N, whichis again constructed using the coefficients of D(z). The test amounts to checking whether the matrix is positive definite. This is based on the Schur-Cohn criterion. Frame # 29 Slide # 56 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion The Jury-Marden criterion is the most efficient of the three criteria. Frame # 30 Slide # 57 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion The Jury-Marden criterion is the most efficient of the three criteria. It involves the computation of a number of 2 2 determinants. Frame # 30 Slide # 58 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion The Jury-Marden criterion is the most efficient of the three criteria. It involves the computation of a number of 2 2 determinants. This is also based on the Schur-Cohn criterion. Frame # 30 Slide # 59 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d Assumptions: Frame # 31 Slide # 60 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d Assumptions: The denominator of the transfer function is given by where b 0 > 0. D(z) = N b i z N i i=0 Frame # 31 Slide # 61 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d Assumptions: The denominator of the transfer function is given by D(z) = N b i z N i where b 0 > 0. The numerator and denominator polynomials of the transfer function, N(z) and D(z), do not have any common factors. i=0 Frame # 31 Slide # 62 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d Assumptions: The denominator of the transfer function is given by D(z) = N b i z N i where b 0 > 0. The numerator and denominator polynomials of the transfer function, N(z) and D(z), do not have any common factors. The first assumption that b 0 > 0simplifies the Jury-Marden stability criterion but it is not a limitation. i=0 Frame # 31 Slide # 63 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d If b 0 < 0 then we can multiply the numerator and denominator polynomials by 1 to get a positive b 0. Frame # 32 Slide # 64 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d If b 0 < 0 then we can multiply the numerator and denominator polynomials by 1 to get a positive b 0. This does not change the pole positions. Frame # 32 Slide # 65 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d If b 0 < 0 then we can multiply the numerator and denominator polynomials by 1 to get a positive b 0. This does not change the pole positions. For example, if we can write H(z) = H(z) = N(z) D(z) = z 2 + 2z + 1 2z 2 + 0.8z 0.4 (z 2 + 2z + 1)( 1) ( 2z 2 + 0.8z 0.4)( 1) = z2 2z 1 2z 2 0.8z + 0.4 = N (z) D (z) where D (z) has a positive b 0. Frame # 32 Slide # 66 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d Row Coefficients 1 b 0 b 1 b 2 b 3 b N 2 b N 1 b N 2 b N b N 1 b N 2 b N 3 b 2 b 1 b 0 3 c 0 c 1 c 2 c N 3 c N 2 c N 1 4 c N 1 c N 2 c N 3 c 2 c 1 c 0 5 d 0 d 1 d 2 d N 3 d N 2 6 d N 2 d N 3 d N 4 d 1 d 0... 2N - 3 r 0 r 1 r 2.. where c i = b i b N b N i b 0 = b 0 b N i b N b i for 0, 1,..., N 1 d i = c i c N 1 c N 1 i c 0 = c 0 c N 1 i for 0, 1,..., N 2 c N 1 c i Frame # 33 Slide # 67 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d The Jury-Marden stability criterion states that polynomial D(z) has roots inside the unit circle of the z plane (i.e., the filter is stable) if and only if the following conditions are satisfied: Frame # 34 Slide # 68 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d The Jury-Marden stability criterion states that polynomial D(z) has roots inside the unit circle of the z plane (i.e., the filter is stable) if and only if the following conditions are satisfied: (i) D(1) >0 Frame # 34 Slide # 69 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d The Jury-Marden stability criterion states that polynomial D(z) has roots inside the unit circle of the z plane (i.e., the filter is stable) if and only if the following conditions are satisfied: (i) D(1) >0 (ii) ( 1) N D( 1) >0 Frame # 34 Slide # 70 A. Antoniou Digital Signal Processing Sec. 5.3
Jury-Marden Stability Criterion Cont d The Jury-Marden stability criterion states that polynomial D(z) has roots inside the unit circle of the z plane (i.e., the filter is stable) if and only if the following conditions are satisfied: (i) D(1) >0 (ii) ( 1) N D( 1) >0 (iii) b 0 > b N c 0 > c N 1 d 0 > d N 2.. r 0 > r 2 Frame # 34 Slide # 71 A. Antoniou Digital Signal Processing Sec. 5.3
Example A discrete-time system is characterized by the transfer function H(z) = z 4 4z 4 + 3z 3 + 2z 2 + z + 1 Check the filter for stability. Solution The denominator polynomial of the transfer function is given by D(z) = 4z 4 + 3z 3 + 2z 2 + z + 1 Since D(1) = 11 > 0 and ( 1) 4 D( 1) = 3 > 0 conditions (i) and (ii) of the test are satisfied. Frame # 35 Slide # 72 A. Antoniou Digital Signal Processing Sec. 5.3
Example Cont d Jury-Marden array: Row Coefficients 1 4 3 2 1 1 2 1 1 2 3 4 3 15 11 6 1 4 1 6 11 15 5 224 159 79 Since b 0 > b 4, c 0 > c 3, d 0 > d 2 condition (iii) is also satisfied and the filter is stable. Frame # 36 Slide # 73 A. Antoniou Digital Signal Processing Sec. 5.3
Example A discrete-time system is characterized by the transfer function H(z) = z 2 + 2z + 1 z 4 + 6z 3 + 3z 2 + 4z + 5 Check the filter for stability. Solution The denominator polynomial of the transfer function is given D(z) = z 4 + 6z 3 + 3z 2 + 4z + 5 In this example, ( 1) 4 D( 1) = 1 Therefore, condition (ii) of the test is violated and the filter is unstable. Note: Note that there is no need to construct the Jury-Marden array! Violating only one of the conditions is enough to demonstrate that filter is unstable. Frame # 37 Slide # 74 A. Antoniou Digital Signal Processing Sec. 5.3
This slide concludes the presentation. Thank you for your attention. Frame # 38 Slide # 75 A. Antoniou Digital Signal Processing Sec. 5.3