Transcript Lecture4.ppt

EE 529 Circuit and
Systems Analysis
Lecture 4
EASTERN MEDITERRANEAN UNIVERSITY
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
(r )
Bf =  B

and
()

U  ( )

r    
Af =  U A   r 


If the column orderings are identical then
A = B T
Matrices of Oriented Graphs
Consider the following graph
e3
e6
 e2 e4 e5 e1

1
0
0

1
1
0

 e2
A f   0 1 0 1 0 1 e4


0
0
1
0

1
1

 e5


v1
e2
e3
e1
v0
e5
e4
v2
e6
e5
e3
e6
e1
 e2 e4

1

1
0
1
0
0

 e1
Bf   1 0 1 0 1 0  e3


0
1

1
0
0
1

 e6


v3
FUNDAMENTAL POSTULATES
Now, Let G be a connected graph having e
edges and let
xT   x1  t  , x2  t  ,
and
xe  t  
y T   y1  t  , y2  t  ,
ye  t  
be two vectors where xi and yi, i=1,...,e,
correspond to the across and through
variables associated with the edge i
respectively.
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
Bx = 0
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
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
xT   xTb
y T   y Tb
xTc 
y Tc 
 where xb (yb) and xc (yc) correspond to the across
(through) variables associated with the branches and
chords of the tree T, respectively.
 Then
yb 
 xb 
B U  x   0
 c
xc  Bxb
fundamental circuit
equation
and
U
A   = 0
 yc 
y b = -Ay c
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.
v0
ei
ek
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.
vi
ek
ei
vk
(n+1) edges connected in series
(x1,y1)
(x2,y2)
(x0,y0)
(xn,yn)
 1
 x0 
x 
 1
x 
1  2   0
 x3 
 
 
 xn 
1 1 1
n
x0   xi
i 1
1
1

1

1


1
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
0   y0 
 y 
  1
0   y2 
  0
0   y3 
 
 
1  yn 
 y0  y1  y2 
 yn
(n+1) edges connected in parallel
(x0,y0)
1
(x1,y1)
 y0 
y 
 1
y 
1  2   0
 y3 
 
 
 yn 
1 1 1
n
 y0   yi
i 1
(x2,y2)
 1
 1

 1

 1


 1
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
x0  x1  x2 
(xn,yn)
0   x0 
x 
  1
0   x2 
  0
0   x3 
 
 
1  xn 
 xn
Mathematical Model of a Resistor
A
a
v(t)
i(t)
B
b
v(t )  Ri (t )
Mathematical Model of an
Independent Voltage Source
v(t)
A
a
v(t)
Vs
Vs
i(t)
i(t)
B
b
Mathematical Model of an
Independent Voltage Source
v(t)
a
A
v(t)
Is
i(t)
Is
i(t)
B
b
Circuit Analysis
A-Branch Voltages Method:
Consider the following circuit.
2 k
15 V
30 V
4 k
3 k
1 k
20 V
10 mA
Circuit Analysis
A-Branch Voltages Method:
1. Draw the circuit graph
2 k
15 V
30 V
4 k
10 mA
3 k
1 k
20 V
2
a
3
b
There are:
•5 nodes (n)
4
c
1
6
5
d
7
•3 voltage sources (nv)
e
8
•8 edges (e)
•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
2
a
3
b
4
c
1
6
5
d
7
e
8
Circuit Analysis
A-Branch Voltages Method:
2. Write the fundamental cut-set equations for the tree branches which do
not correspond to voltage sources.
i2 
i 
 3
1 1 1 1 1 i5   0
 
i6 
i7 
2
a
3
b
4
c
1
6
5
d
7
e
8
Circuit Analysis
A-Branch Voltages Method:
2. Write the currents in terms of voltages using terminal equations.
i2
i3
i5
i6
i7
v2

2k
v3

4k
v5

1k
v6

3k
 10mA
2
a
3
b
4
c
1
6
5
d
7
e
8
Circuit Analysis
A-Branch Voltages Method:
2. Substitute the currents into fundamental cut-set equation.
v3 v5 v6
v2

   10m
2k 4k 1k 3k
2
a
3
b
4
c
1
6
5
d
7
e
8
3. v3, v5, and v6 must be expressed in terms of branch voltages using
fundamental circuit equations.
Circuit Analysis
A-Branch Voltages Method:
v3  v2  v1  v2  30
v5  v4  v2  1  15  v2  30  v2  15
v6  v2  v1  v8  v2  30  20  v2  50
v
v
v
v

12k  2  3  5  6  10m 
 2k 4k 1k 3k

6v2  3v3  12v5  4v6  120
6v2  3(v2  30)  12(v2  15)  4(v2  50)  120
 6  3  12  4  v2  90  180  200  120
25v2  350
v2 
2
a
3
b
4
c
1
6
5
d
7
e
8
350
 14 V
25
Find how much power the 10 mA current source delivers to the circuit
Circuit Analysis
A-Branch Voltages Method:
Find how much power the 10 mA current source delivers to the circuit
v7  v8  v1  v2  20  30  14  36V
P10 mA  v7i7  36 10m  360 mW
2
a
3
b
4
c
1
6
5
d
7
e
8
Circuit Analysis
 Example: Consider the following circuit. Find
ix in the circuit.
20 V
2
i1
4
4
ix
15 V
5
i2
i3
3
10 V
Circuit Analysis
 Circuit graph and a proper tree
20 V
2
1
2
3
i1
6
4
4
4
ix
15 V
5
7
8
5
i3
i2
3
ix  i5
10 V
Circuit Analysis
 Fundamental cut-set equations
1
2
3
i2  i6  i8  i7
i3  i6  i8
6
4
5
7
8
v2
i2 
2
v6
i6 
4
v8
i8 
3
v3
i3 
4
v7
i7 
5
Circuit Analysis
 Fundamental cut-set equations
1
2
3
6
4
5
7
8
v6 v8 v7
v2
  
2
4 3 5
v3 v6 v8
 
4 4 3
Circuit Analysis
 Fundamental circuit equations
1
2
3
6
4
5
7
8
v6  v3  v1  v2  v3  v2  20
v8  v4  v3  v1  v2  v5  15  v3  20  v2  10  v3  v2  15
v7  v1  v2  v5  20  v2  10  v2  30
Circuit Analysis
v3  v2  20 v3  v2  15 v2  30 
 v2
60   



4
3
5 
2
 v3 v3  v2  20 v3  v2  15 
12  


4
4
3


30v2  15v3  15v2  300  20v3  20v2  300  12v2  360
35v3  77v2  960...................(1)
3v3  3v3  3v2  60  4v3  4v2  60
10v3  7v2  120......................(2)
v3= 9.5639V v2=-8.1203 V
Circuit Analysis
v7 v8
v2  30 v3  v2  15
i5  i7  i8     

5 3
5
3
i5  3.48A
ix  3.48 A