We thus obtain
fort 0
Comments. For complicated functions with denominators involving higher-order
polynomials, partial-fraction expansion may be quite time consuming. In such a case,
use of MATLAB is recommended.
Partial-Fraction Expansion with MATLAB. MATLAB has a command to
obtain the partial-fraction expansion of B(s)/A(s). Consider the following function
B(s)/A(s):
where some of aiandbjmay be zero. In MATLAB row vectors num and den specify the
coefficients of the numerator and denominator of the transfer function. That is,
num = [b 0 b 1 ... bn]
den = [1 a 1 ... an]
The command
[r,p,k] = residue(num,den)
finds the residues (r), poles (p), and direct terms (k) of a partial-fraction expansion of
the ratio of two polynomials B(s)andA(s).
The partial-fraction expansion of B(s)/A(s)is given by
(B–4)
Comparing Equations (B–1) and (B–4), we note that p(1)=–p 1 , p(2)=–p 2 ,p,
p(n)=–pn;r(1)=a 1 , r(2)=a 2 ,p,r(n)=an.[k(s)is a direct term.]
EXAMPLE B–4 Consider the following transfer function,
B(s)
A(s)=
2s^3 +5s^2 +3s+ 6
s^3 +6s^2 +11s+ 6B(s)
A(s)
=
r(1)
s-p(1)
+
r(2)
s-p(2)
+p+
r(n)
s-p(n)
+k(s)
B(s)
A(s)
=
num
den
=
b 0 sn+b 1 sn-^1 +p+bn
sn+a 1 sn-^1 +p+an
=A 1 +t^2 Be-t,
=e-t+ 0 +t^2 e-t
=l-^1 c
1
s+ 1
d +l-^1 c
0
(s+1)^2
d +l-^1 c
2
(s+1)^3
d
f(t)=l-^1 CF(s)D
Appendix B / Partial-Fraction Expansion 871