residuez

z-transform partial-fraction expansion

Syntax

[r,p,k] = residuez(b,a)
[b,a] = residuez(r,p,k)

Description

residuez converts a discrete time system, expressed as the ratio of two polynomials, to partial fraction expansion, or residue, form. It also converts the partial fraction expansion back to the original polynomial coefficients.

    Note:   Numerically, the partial fraction expansion of a ratio of polynomials is an ill-posed problem. If the denominator polynomial is near a polynomial with multiple roots, then small changes in the data, including roundoff errors, can cause arbitrarily large changes in the resulting poles and residues. You should use state-space (or pole-zero representations instead.

[r,p,k] = residuez(b,a) finds the residues, poles, and direct terms of a partial fraction expansion of the ratio of two polynomials, b(z) and a(z). Vectors b and a specify the coefficients of the polynomials of the discrete-time system b(z)/a(z) in descending powers of z.

B(z)=b0+b1z1+b2z2++bmzmA(z)=a0+a1z1+a2z2++anzn

If there are no multiple roots and a > n-1,

B(z)A(z)=r(1)1p(1)z1++r(n)1p(n)z1+k(1)+k(2)z1++k(mn+1)z(mn)

The returned column vector r contains the residues, column vector p contains the pole locations, and row vector k contains the direct terms. The number of poles is

n = length(a)-1 = length(r) = length(p)

The direct term coefficient vector k is empty if length(b) is less than length(a); otherwise:

length(k) = length(b) - length(a) + 1

If p(j) = ... = p(j+s-1) is a pole of multiplicity s, then the expansion includes terms of the form

r(j)1p(j)z1+r(j+1)(1p(j)z1)2++r(j+sr1)(1p(j)z1)s

[b,a] = residuez(r,p,k) with three input arguments and two output arguments, converts the partial fraction expansion back to polynomials with coefficients in row vectors b and a.

The residue function in the standard MATLAB® language is very similar to residuez. It computes the partial fraction expansion of continuous-time systems in the Laplace domain (see reference [1]), rather than discrete-time systems in the z-domain as does residuez.

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Algorithms

residuez applies standard MATLAB functions and partial fraction techniques to find r, p, and k from b and a. It finds

  • The direct terms a using deconv (polynomial long division) when length(b) > length(a)-1.

  • The poles using p = roots(a).

  • Any repeated poles, reordering the poles according to their multiplicities.

  • The residue for each nonrepeating pole pi by multiplying b(z)/a(z) by 1/(1 - piz−1) and evaluating the resulting rational function at z = pi.

  • The residues for the repeated poles by solving

    S2*r2 = h - S1*r1
    

    for r2 using \. h is the impulse response of the reduced b(z)/a(z), S1 is a matrix whose columns are impulse responses of the first-order systems made up of the nonrepeating roots, and r1 is a column containing the residues for the nonrepeating roots. Each column of matrix S2 is an impulse response. For each root pj of multiplicity sj, S2 contains sj columns representing the impulse responses of each of the following systems.

    11pjz1,1(1pjz1)2,,1(1pjz1)sj

    The vector h and matrices S1 and S2 have n + xtra rows, where n is the total number of roots and the internal parameter xtra, set to 1 by default, determines the degree of over-determination of the system of equations.

References

[1] Oppenheim, Alan V., Ronald W. Schafer, and John R. Buck. Discrete-Time Signal Processing. 2nd Ed. Upper Saddle River, NJ: Prentice Hall, 1999.

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