eastern mediterranean university ee 529 circuit and systems analysis lecture 4
TRANSCRIPT
Matrices of Oriented Graphs
THEOREM: In a graph G let the fundamental circuit and cut-set matrices with respect to a tree to be written as
( ) ( )
( )
and
If the column orderings are identical then
r
r
r
f
f
T
B = B U
A = U A
A = B
Matrices of Oriented Graphs
Consider the following graph
e3
v1
v2
v3
v0
e1
e2
e4e5
e3
v1
v2
v3
v0
e1
e2
e4e5
e6
5 3 62 4 1
2
4
5
1 0 0 1 1 0
0 1 0 1 0 1
0 0 1 0 1 1
e e ee e e
e
e
e
fA
5 3 62 4 1
1
3
6
1 1 0 1 0 0
1 0 1 0 1 0
0 1 1 0 0 1
e e ee e e
e
e
e
fB
FUNDAMENTAL POSTULATES
Now, Let G be a connected graph having e edges and let
be two vectors where xi and yi, i=1,...,e, correspond to the across and through variables associated with the edge i respectively.
1 2
1 2
, ,
and
, ,
Te
Te
x t x t x t
y t y t y t
x
y
FUNDAMENTAL POSTULATES
2. POSTULATE Let B be the circuit matrix of the graph G having e edges then we can write the following algebraic equation for the across variables of G
3. POSTULATE Let A be the cut-set matrix of the graph G having e edges then we can write the following algebraic equation for the through variables of G
Bx = 0
Ay = 0
FUNDAMENTAL POSTULATES
2. POSTULATE is called the circuit equations of electrical system. (is also referred to as Kirchoff’s Voltage Law)
3. POSTULATE is called the cut-set equations of electrical system. (is also referred to as Kirchoff’s Current Law)
Fundamental Circuit & Cut-set Equations
Consider a graph G and a tree T in G. Let the vectors x and y partitioned as
where xb (yb) and xc (yc) correspond to the across (through) variables associated with the branches and chords of the tree T, respectively.
Then
T T Tb c
T T Tb c
x x x
y y y
0
b
c
c b
xB U
x
x Bx
b
c
b c
yU A = 0
y
y = -Ay
and
fundamental circuit equation
fundamental cut-set equation
Series & Parallel Edges Definition: Two edges ei and ek are said to be
connected in series if they have exactly one common vertex of degree two.
ei
ek
v0
Series & Parallel Edges Definition: Two edges ei and ek are said to be
connected in parallel if they are incident at the same pair of vertices vi and vk.
ei
ek
vi
vk
(n+1) edges connected in series
(x1,y1)
(x2,y2)
(xn,yn)
(x0,y0)
0
1
2
3
1 1 1 1 1 0
n
x
x
x
x
x
0
1
2
3
1 1 0 0 0
1 0 1 0
1 0 0 1 0
1 0 0 0 0
1 0 0 0 1 n
y
y
y
y
y
0
01
n
ii
x x
0 1 2 ny y y y
(n+1) edges connected in parallel
0
1
2
3
1 1 1 1 1 0
n
y
y
y
y
y
0
1
2
3
1 1 0 0 0
1 0 1 0
1 0 0 1 0
1 0 0 0 0
1 0 0 0 1 n
x
x
x
x
x
0
01
n
ii
y y
0 1 2 nx x x x
(x0,y0) (x1,y1) (x2,y2) (xn,yn)
Circuit Analysis
A-Branch Voltages Method:
Consider the following circuit.
30 V
2 k
4 k
1 k
15 V
20 V
3 k 10 mA
Circuit Analysis
A-Branch Voltages Method:
1. Draw the circuit graph
30 V
2 k
4 k
1 k
15 V
20 V
3 k 10 mA
1
2
3 4
56
7
8
a b
c
d e
There are:
•5 nodes (n)
•8 edges (e)
•3 voltage sources (nv)
•1 current source (ni)
Circuit Analysis
A-Branch Voltages Method:
1. Select a proper tree: (n-1=4 branches)
Place voltage sources in tree
Place current sources in co-tree
Complete the tree from the resistors
1
2
3 4
56
7
8
a b
c
d e
Circuit Analysis
A-Branch Voltages Method:
2. Write the fundamental cut-set equations for the tree branches which do not correspond to voltage sources.
1
2
3 4
56
7
8
a b
c
d e
2
3
5
6
7
1 1 1 1 1 0
i
i
i
i
i
Circuit Analysis
A-Branch Voltages Method:
2. Write the currents in terms of voltages using terminal equations.
1
2
3 4
56
7
8
a b
c
d e
22
33
55
66
7
2
4
1
310
vi
kv
ikv
ikv
ik
i mA
Circuit Analysis
A-Branch Voltages Method:
2. Substitute the currents into fundamental cut-set equation.
1
2
3 4
56
7
8
a b
c
d e
3 5 62 102 4 1 3
v v vvm
k k k k
3. v3, v5, and v6 must be expressed in terms of branch voltages using fundamental circuit equations.
Circuit Analysis
A-Branch Voltages Method:
1
2
3 4
56
7
8
a b
c
d e
3 2 1 2
5 4 2 2 2
6 2 1 8 2 2
30
1 15 30 15
30 20 50
v v v v
v v v v v
v v v v v v
3 5 62
2 3 5 6
2 2 2 2
2
2
2
12 102 4 1 3
6 3 12 4 120
6 3( 30) 12( 15) 4( 50) 120
6 3 12 4 90 180 200 120
25 350
35014
25
v v vvk m
k k k k
v v v v
v v v v
v
v
v V
Find how much power the 10 mA current source delivers to the circuit
Circuit Analysis
A-Branch Voltages Method:
1
2
3 4
56
7
8
a b
c
d e
7 8 1 2 20 30 14 36v v v v V
Find how much power the 10 mA current source delivers to the circuit
10 7 7 36 10 360mAP v i m mW
Circuit Analysis
Example: Consider the following circuit. Find ix in the circuit.
4 4
ix
2 20 V
3
15 V
5 10 V
i1
i2 i3
Circuit Analysis
Circuit graph and a proper tree
4 4
ix
2 20 V
3
15 V
5 10 V
i1
i2 i3
5xi i
12
3
4
5
6
7 8
Circuit Analysis
Fundamental cut-set equations
2 6 8 7
3 6 8
i i i i
i i i
322 3
6 76 7
88
2 4
4 5
3
vvi i
v vi i
vi
12
3
4
5
6
7 8
Circuit Analysis
Fundamental cut-set equations
6 8 72
3 6 8
2 4 3 5
4 4 3
v v vv
v v v
12
3
4
5
6
7 8
Circuit Analysis
Fundamental circuit equations1
2
3
4
5
6
7 8
6 3 1 2 3 2
8 4 3 1 2 5 3 2 3 2
7 1 2 5 2 2
20
15 20 10 15
20 10 30
v v v v v v
v v v v v v v v v v
v v v v v v
Circuit Analysis
3 2 3 22 2
3 3 2 3 2
2 3 2 3 2 2
3 2
3 3 2 3 2
3 2
20 15 3060
2 4 3 5
20 1512
4 4 3
30 15 15 300 20 20 300 12 360
35 77 960...................(1)
3 3 3 60 4 4 60
10 7 120..........
v v v vv v
v v v v v
v v v v v v
v v
v v v v v
v v
............(2)
v3= 9.5639V v2=-8.1203 V