Download Reduced Incidence Matrix

Reduced Incidence Matrix
Let
G
be
a
connected
digraph with “n” nodes and “b”
branches. Let Aa be the Incidence
Matrix of G. The (n-1) x b matrix
A obtained by deleting any one
row of Aa is called a ReducedIncidence Matrix of G.
1
e1
e2
KCL Equations:
2
1
e3
3
3
2
6
1
2
4
5
3
i1 + i2 − i6 = 0
−i1 − i3 + i4 = 0
−i2 + i3 + i5 = 0
4
Ai = 0 ⇒
node
no. 1
1  1

2 −1

3  0
Branch no.
2 3 4 5 6
1 0 0 0 −1
0 −1 1 0 0
−1 1 0 1 0
A
i1
i 
 2
i3
0
 

i4  = 0
i5 
0
 
i6 
0
i
A is called the reduced Incidence Matrix
of the diagraph G relative to datum node
4
.
Reduced Incidence Matrix
A
Let G be a connected digraph with “n”
nodes and “b” branches. Pick any node as the
datum node and label the remaining nodes
arbitrarily from 1 to n-1 . Label the branches
arbitrarily from 1 to b.
The reduced incidence matrix
A
of G
is an (n-1) x b matrix where each row j
corresponds to node
j
, and each column k,
corresponds to branch k, and where the jkth
element ajk of
1 ,

ajk = −1 ,
0 ,

A is constructed as follow:
if branch k leaves node j
if branch k enters node j
if branch k in not connected to node j
Note:
Each branch k connected to the
datum node has only one
non-zero entry in the kth column
of
A
∴
The (n-1) KCL equations derived
from
Ai = 0
are linearly independent.
Theorem
Ai = 0
gives
the
maximum
possible
number of linearly-independent
KCL equations for a connected
circuit.
Relationship between A and Aa
Let Aa be the n x b Incidence
matrix of a connected digraph G with
“n” nodes and “b” branches.
By
deleting
corresponding to node
any
m
row
from Aa, we
obtain the reduced incidence matrix
A of G relative to the datum node
m
.
e1
−
D6
−
v6
+
1
i1
v1 +
D1
D2
− v3 +
e2
D3
i4
2
+
v4
v2
−
3
i3
e3
i5
D5
D4
−
−
1
+
v5
4
2
1
3
3
KVL around closed node sequence:
2
6
+
i2
4
5
4
1
3
2
1
2
3
4
2
1
3
4
2
: v2 + v3 − v1 = 0
: − v3 + v5 − v4 = 0
1
: v2 + v5 − v4 − v1 = 0
These 3 KVL equations are not linearly-independent because the
3rd equation can be obtained by adding the first 2 equations:
(v2 + v3 − v1 ) + (−v3 + v5 − v4 )
= v2 − v1 + v5 − v4 = 0
1
Choose 4 as datum node for digraph G
e1
e2
2
1
e3
3
3
2
6
KCL Equations:
4
5
1
2
3
i1 + i2 − i6 = 0
−i1 − i3 + i4 = 0
−i2 + i3 + i5 = 0
4
Independent
KCL Equations
Ai = 0 ⇒
Independent
KVL Equations
node
no. 1
1  1

2 −1

3  0
Branch no.
2 3 4 5 6
1 0 0 0 −1
0 −1 1 0 0
−1 1 0 1 0
A
i1
i 
 2
i3
0
 
0
i
=
4
 

i5 
0
 
i6 
0
i
v1 = e1 − e2
 v1   1 −1 0
v   1 0 −1
v2 = e1 − e3
 2 
 e 
 v3   0 −1 1  1 
v3 = −e2 + e3
 =
 e2  ⇒
v4 = e2
v4   0 1 0  e 
 v5   0 0 1  3 
v5 = e3
  
 e
v6 = −e1
 v6   −1 0 0
v
A
T
Note:
Since each equation k from the
system of KVL equations generated
from
T
v =A e
contains the variable vk, which
appears in no other equations,
this system of “b” KVL equations
are guaranteed to be
linearly independent.
KVL
each branch k connected between
node m and the datum node 0
has a KVL equation involving
only one node-to-datum voltage em:
vk = em
How to write An Independent
System of KCL and KVL
Equations
Let N be any connected circuit and let the digraph
G associated with N contain “n” nodes and “b”
branches. Choose an arbitrary datum node and
define the associated node-to-datum voltage
e , the branch voltage vector v , and the
branch current vector i . Then we have the
vector
following system of independent KCL and
KVL equations.
(n-1) Independent KCL Equations :
Ai = 0
b Independent KVL Equations :
T
v = A e
ik
+
For each 2-terminal resistor Rk,
Rk
vk
we have 2 unknown
variables {vk, ik}
_
ik
ij
+
+
For each 3-terminal resistor,
we have 4 unknown
vj
vk
_
ij
+
vj
_
_
ik
variables {vj, vk, ij, ik}
For each 2-port resistor,
+
vk we have 4 unknown
_
variables {vj, vk, ij, ik}
∴
There are more unknown variables
than equations derived from
constitutive relations of the resistors.