# RESIDUEZ

## Purpose:

Finds the partial fraction expansion of a Z-transform.

## Syntax:

RESIDUEZ(b, a)

(r, p, k) = RESIDUEZ(b, a)

 b - A series. The numerator (i.e. zero) coefficients in ascending powers of z-1. a - A series. The denominator (i.e. pole) coefficients in ascending powers of z-1.

## Alternate Syntax:

RESIDUEZ(r, p, k)

(b, a) = RESIDUEZ(r, p, k)

 r - A series. The residues representing the numerator terms of the partial fraction expansion. p - A series. The poles of the partial fraction expansion. k - A series. The numerator coefficients for the direct terms of the partial fraction expansion.

## Returns:

(r, p, k) = RESIDUEZ(b, a) returns the partial fraction expansion of the rational polynomial.

R, p and k are series where r represents the residues of the partial fraction expansion, p are the pole locations and k represents the direct terms (if any).

(b, a) = RESIDUEZ(r, p, k) returns the inverse partial fraction expansion, converting the partial fraction expansion back into b(z) / a(z) form. Series b and a are the numerator and denominator terms of the rational polynomial with the partial fraction expansion represented by r, p, and k. Series k represents the remainder terms (if any).

RESIDUEZ(r, p, k) with one or zero output arguments returns b and a in one series of two columns where b == col(1) and a == col(2).

RESIDUEZ(f) or (b, a) = RESIDUEZ(f) assumes f is a three column series with r, p and k as each of the columns. Thus:

residuez(residuez(b, a)) == {{b/a, a/a}}.

## Example: (r, p, k) = residuez({1, -1}, {1, -5, 6})

r == {2, -1}

p == {3, 2}

k == {}

Representing the partial fraction expansion: From this partial fraction expansion, the impulse response can be derived by inspection: Because the poles lie outside the unit circle, the system is unstable.

We can evaluate the expression directly and compare with the IMPZ function:

n := 0..5

y1 = 2 * 3^n –2^n

y2 = impz({1, -1}, {1, -5, 6}, length(n))

y1 == y2 == {1, 4, 14, 46, 146, 454}

Now, performing the inverse transform:

(b, a, c) = residuez(r, p, k)

b == {1, -1}

a == {1, -5, 6}

c == {}

The series b and a represent the numerator and denominator terms of the original rational polynomial.

## Example: (r, p, k) = residuez({2, 3, 4}, {1, 3, 3, 1})

r == {4, -5, 3}

p == {-1, -1, -1}

k == {}

Since H(z) contains 3 repeated poles, the resulting partial fraction expansion becomes: Now, performing the inverse transform:

(b, a, c) = residuez(r, p, k)

b == {2, 3, 4}

a == {1, 3, 3, 1}

c == {}

The series b and a represent the numerator and denominator terms of the original rational polynomial.

## Example: or Since RESIDUEZ expects H(z) to be in terms of z-1 and the first denominator term cannot be zero, we form: Now find the partial fraction expansion for G(z):

(r, p, k) = residuez({1, -10, -4, 4}, {2, -2, -4})

r == {-1.5, 0.5}

p == {2, -1}

k == {1.5, -1}

Representing the partial fraction expansion: Since zG(z) = H(z) we have: ## Remarks:

Given the rational polynomial H(z) = b(z) / a(z) where: z = e jω complex frequency N = number of numerator terms M = number of denominator terms

If a ≠ 1, the numerator and denominator terms are normalized by dividing each coefficient by a.

If there are no repeated roots, the partial fraction expansion of the rational polynomial is of the form: If there are K repeated roots (closer than 1.0e-3), then the partial fraction expansion includes terms such as: See RESIDUE to find the partial fraction expansion of the Laplace rational polynomial H(s) = b(s) / a(s).