Finds the partial fraction expansion of a

RESIDUEZ(b, a)

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

b |
- |
A series. The numerator (i.e. zero) coefficients in ascending powers of |

a |
- |
A series. The denominator (i.e. pole) coefficients in ascending powers of |

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. |

(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

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

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[1], a/a[1]}}.

(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.

(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.

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:

Given the rational polynomial H(z) = b(z) / a(z) where:

z |
= |
e |

N |
= |
number of numerator terms |

M |
= |
number of denominator terms |

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

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