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^{ jω} complex frequency |
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