Partial Fraction Expansion of the Transfer Function
2.4
2002/02/04
2002/07/19
Steven
Cox
cox@rice.edu
Charlet
Reedstrom
charlet@rice.edu
Jacob
Grabczewski
jgrab@owlnet.rice.edu
Steven
Cox
cox@rice.edu
Gauss-Jordan method
partial fraction expansion
transfer function
(Blank Abstract)
Partial Fraction Expansion of the Transfer Function
The Gauss-Jordan method informs us that R will be a matrix of rational
functions with a common denominator. In keeping with the
notation of the previous chapters, we assume the denominator
to have the h distinct roots,
λj
j
1
…
h
with associated multiplicities
mj
j
1
…
h
.
Now, assembling the partial fraction expansions of each
element of R we
arrive at
Rs
j
1
h
k
1
mj
R
j
,
k
s
λj
k
where, recalling the equation from Cauchy's Theorem, the matrix
R
j
,
k
equals the following:
R
j
,
k
1
2
j
z
C
j
R
z
z
λ
j
k
1
Concrete Example
As we look at this example with respect to the eigenvalue
problem eqn1 and eqn2, we find
R
1
,
1
100
010
000
R
1
,
2
010
000
000
and
R
2
,
1
000
000
001
One notes immediately that these matrices enjoy some amazing
properties. For example
R
1
,
1
2
R
1
,
1
,
R
2
,
1
2
R
2
,
1
,
R
1
,
1
R
2
,
1
0
, and
R
2
,
1
2
0
Below we will now show that this is no accident. As a
consequence of and
the first resolvent identity, we shall find that these results
are true in general.
R
j
,
1
2
R
j
,
1
as seen above in .
Recall that the
Cj
appearing in is any
circle about
λj
that neither touches nor encircles any other root. Suppose
that
Cj
and
Cj
are two such circles and
Cj
encloses
Cj
. Now
R
j
,
1
1
2
j
z
Cj
R
z
1
2
j
z
Cj
R
z
and so
R
j
,
1
2
1
2
2
z
C
j
R
z
w
Cj
R
w
R
j
,
1
2
1
2
2
z
C
j
w
Cj
R
z
R
w
R
j
,
1
2
1
2
2
z
C
j
w
Cj
R
z
R
w
w
z
R
j
,
1
2
1
2
2
z
C
j
R
z
w
Cj
1
w
z
w
Cj
R
w
z
C
j
1
w
z
R
j
,
1
2
1
2
z
C
j
R
z
R
j
,
1
We used the first resolvent identity, This Transfer
Function eqn, in moving from the second to the third
line. In moving from the fourth to the fifth we used only
w
Cj
1
w
z
2
and
z
C
j
1
w
z
0
The latter integrates to zero because
Cj
does not encircle w.
From the definition of orthogonal projections, which states that matrices
that equal their squares are projections, we adopt the
abbreviation
Pj
R
j
,
1
With respect to the product
Pj
Pk
, for
j
k
, the calculation runs along the same lines. The
difference comes in where, as
Cj
lies completely outside of
Ck
, both integrals are
zero. Hence,
If
j
k
then
Pj
Pk
0
.
Along the same lines we define
Dj
R
j
,
2
and prove
If
1
k
m
j
1
then
D
j
k
R
j
,
k
+
1
.
D
j
m
j
0
.
For k and
l greater than or equal to
one,
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
C
j
R
z
z
λj
k
w
Cj
R
w
w
λj
l
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
C
j
w
Cj
R
z
R
w
z
λj
k
w
λj
l
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
Cj
w
C
j
R
z
R
w
w
z
z
λj
k
w
λj
l
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
C
j
R
z
z
λj
k
w
Cj
w
λj
l
w
z
1
2
2
w
Cj
R
w
w
λj
l
z
C
j
z
λj
k
w
z
R
j
,
k
+
1
R
j
,
l
+
1
1
2
z
C
j
R
z
z
λj
k
l
R
j
,
k
+
l
+
1
because
w
Cj
w
λj
l
w
z
2
z
λj
l
and
z
C
j
z
λj
k
w
z
0
With
k
l
1
we have shown
R
j
,
2
2
R
j
,
3
, i.e.,
Dj
2
R
j
,
3
. Similarly, with
k
1
and
l
2
we find
R
j
,
2
R
j
,
3
R
j
,
4
, i.e.,
Dj
3
R
j
,
4
. Continuing in this fashion we find
R
j
,
k
R
j
,
k
+
1
R
j
,
k
+
2
j
, or
Dj
k
1
R
j
,
k
+
2
. Finally, at
k
m
j
1
this becomes
Dj
m
j
R
j
,
m
j
+
1
1
2
z
C
j
R
z
z
λj
m
j
0
by Cauchy's Theorem.
With this we now have the sought after expansion
R
z
j
1
h
1
z
λj
Pj
k
1
mj
1
1
z
λj
k
1
Dj
k
along with the verification of a number of the properties laid out
in Complex
Integration eqns 1-3.