Group Members
Ashique Ali
Arsalan Ansari
Mukesh kumar
Saddam Hussain
Santosh Kumar
07EL110
07EL126
07EL128
07EL131
07,06EL74
MEHRAN UNIVERSITY OF ENGINEERING
AND
TECHNOLOGY JAMSHORO
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TOPICS
INTRODUCTION
1. NETWORK TOPOLOGY
CUT-SET MATRIX
LOOP MATRIX
E AND I SHIFT
2.
CONCEPT OF STATES
STATE EQNS OF DYNAMIC SYSTEM WITH CONTINOUS
SIGNAL AND CONTINUOUS DATA
STATE EQNS OF HIGHER ORDER SYSTEM
DISCRETE STATE EQNS
STATE EQNS OF ELECTRICAL NETWORK
3.
4.
ANALYSIS OF NETWORK BY LAPLACE
TRANSFORM
CHARACTERISTIC OF LINEAR TIMEINVARIENT TWO-PORTS BY SIX SET OF
PARAETERS
RELATIONS AMONG PARAMETER SETS
INTERCONNECTION OF TWO PORTS
5. CONCEPT OF COMPLEX FREQUENCY
TRANSFORM IMPEDANCE & TRANSFORM CKTS
6. NETWORK FUNCTION OF ONE & TWO PORTS
PARTS OF NETWORKS FUNTIONS
MAGNITUDE & PHASE PLOTS
COMPLEX LOCI PLOTS
7. FOURIER SERIES
EVALUATION OF FOURIER CO-EFFICIENT
WAVEFORM SYMMETRIES
EXPONENTIAL FORM OF FOURIER SERIES
8. STEADY STATE RESPONSE TO PERIODIC SIGNALS
9 INTRODUCTION TO FOURIER TRANSFORMS
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Objective of Subject

The main objective of network analysis is the
determination of the current and voltages at various
points of a network.
Network Analysis

a)
b)
There are two types of methods to analyze a network.
Conventional Network analysis
State variable method
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Basic Terminology
 Graph: “In network analysis, a graph is a circuit
form by replacing each element by a line.”

In constructing the graph of a network special attention
should be placed on the active elements.

For a network with ideal source, the ideal voltage source
should be short circuit & ideal current source should e
replaced by an open circuit.
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Example 1: Draw the graph of given circuit.
d
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ORIENTED GRAPH:

If we indicate a reference direction by an arrow for each
of the lines of the graph, then it is known is “ oriented
graph”.
Example 2: draw the oriented graph for the circuit
shown in fig.
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TOPOLOGICAL PROPERTIES OF
GRAPH:

It deals with the properties of network which are
unaffected, when we stretch twist or otherwise distort
size and shape of the network.

Example 3:

Notice that the graphs of the given circuit are identical.
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These graphs are identical because the relationship between
nodes & are identical.
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Topological properties:
1. PLANAR AND NON PLANAR
GRAPHS
The graphs which may be drawn on a sheet of paper
without crossing lines, called “ planar graphs”
The graphs having crossing lines, called non planar graphs.

2.NODE PAIRS:

The two nodes which we identify for specifying a
voltage variable.
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3.LOOP:
A Loop (or mesh) is a closed path in a graph
(or network ) formed y a number of connected branches.
4.SUB GRAPH:
A Sub graph of a given graph is
formed by removing branches from the original graph.
d
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d
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5.Branch:
“Each line in a graph is identified as a
branch”.
6.Node:
A point in a graph at which two or more lines
(branches) meet is called a node.
7. Node Pair: “A node pair is simply two nodes which
are identified y specifying a voltage variable.
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Tree Of Connected Graph
Tree




The tree is circuit less sub graph of n nodes and n-1
branches.
The tree of connected graph of n nodes has following
properties.
It contains all the nodes of the graph, nodes are not left
and isolated portion.
It contains n-1 branches.
There is no closed path.
Where
n = number of the nodes
&
(n-1)= number of tree branches.
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Example 4:
Draw all possible trees of the given
circuit?
SOLUTION:
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Here
n=4
Tree Branches= n-1
=3
then, all possible trees are given below

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Chord, Link OR Co- Tree

Branches remove from the graph in-forming or preparing
the tree are called links or chords.
 No:
of links
No of links = b-(n-1)
Where
No: of links = L
No: of branches in graph= B
No: of tree branches= (n-1)
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Example 5: Draw at least 5 trees of the following
graph?
Solution:
branches = b = 7
nodes = n =6
Tree Branches =5
No of Links=b-n+1
=2
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The 5 possible trees are given below:
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Example6:Draw the graph of the following circuits.
(A)
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(B)
(C)
R1
b
a
c
R2
d
Solution:
(A)
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Sol:
(B)
a
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Proper or Normal Representation Of
Tree
There are two types of sources.
1.
Current source
2.
Voltage source
Inductor:
1.
2.
3.
4.
Current dependent source.
Stores current.
Represented by short circuit in a circuit.
Represented by link or chord in the tree of connected
graph.
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Capacitor
1.
2.
3.
4.
Voltage dependant source.
Stores voltage
Represented by open circuit in network analysis.
Represented by Tree branch in a tree of connected
graph.
Resistor
Represented by dotted lines or bold lines as a tree
branch.
Current dependent source
Represented by chord or link.
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Voltage dependent source

Represented by tree branch
Excessive capacitor

Excessive capacitor is represented by link.
Excessive inductor

Excessive inductor is represented by tree branch.
Representation of Nodes

Nodes are represented by numbers in circle.
Representation of Branch

Branches are represented by numbers only.
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Example: Draw the graph containing tree branches,
links.
R2
R1
RR33
R1
C1
C2
R4
Sol
branches= b = 1
nodes = n = 5
Tree Branches = n-1
=4
links = b-n+1
=3
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2
1
3
1
4
2
7
3
5
6
4
5
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Cut Set






Cut set of the graph, is the collection of branches and
nodes, such that if these are removed, then the graph
will be divided into
two parts.
Cut set is basically a counter which cuts one tree branch
and maximum number of links at a time.
A tree includes the branches that connects all the nodes
of a graph without forming the closed paths.
The number of cut set in a given network is always
equal to number of tree branches.
Total number of cut sets = number of tree branches
= (n-1)
Different cut sets may e obtained with different tree
branches.
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Example7:
Draw the cut set of the given graph?
Solution:
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branches= b= 8
nodes n = 5
Tree Branches=n-1
=4
Links= 4
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Cut Set Equations
• The set of current equations can be written for each
cut set by applying K.C.L, these equations are called cut
set equations.
• Current is positive when it flows away from cut set &
is negative when it flows towards cut set.
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Cut set Matrix
 “The matrix formed by co efficients of
independent
branch currents in cut set equations is called cut set
matrix.
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ib1
ib6
ib5
ib4
ib7
ib8
ib2
ib3
Cut set equations for above graph
Applying KCL at cut set 1
ib1 –ib4 +ib5 =0
Applying K.C.L at cut set 2
-ib1 +ib2 +ib6 =0
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Applying K.C.L at cut set 3
-ib2 +ib3 +ib7 =0
Applying K.C.L at cut set 4
-ib3 +ib4 +ib8 =0

Cut set matrix for above equations
=0
A ib =0
Where
GROUP 007A is a cut set matrix of order (n-1)xb.
35
Example8: Draw the graph, tree and cut set of the
given circuit?
S0lution:
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nodes =n=4
Tree Branches=n-1
=3
branches= b=6
Links=b-(n+1)
=3
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Graph, Tree & cut set:
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Cut set Equations
Applying K.C.L at cut set 1
ib1 –ib3 +ib4 =0
Applying K.C.L at cut set 2
-ib1 +ib2 +ib5 =0
Applying K.C.L at cut set 3
-ib2 –ib4 +ib6 =0
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Cut set Matrix
=0
Node transformation equations
Node pair voltage
It is the voltage difference between two parts of a graph
that are separated by a cut set.
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The can also be obtained through the node pair voltages &
branch voltages of a graph.
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e1 =V1 – V2
e2 =V2 – V3
e3 =V3 – V4
=
e4 =V4 – V1
e5 = V1
e6 = V2
e7 = V3
e8 = V4
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At Vn = eb
The above equation is known as node
transformation equation.
41
Node transformation equation by
using cut set technique
Vg
Node
Rb
eb
ib
Lb
Cb
ib
Node
ig

Where
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Rb = total resistance in branch
Lb =total inductance in branch
Cb = total capacitance in branch
42
Vg = total series voltage source in branch
ig = current source in branch
eb = branch voltage
ib = branch current
Vrb= Rb (ib - ig)
Lb
b
Lb
b
g
g
Then current for branch is given as
b
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b
b
b
b
g
g
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ib =ig +Yb (eb - Vg)
Multiplying by cut set matrix (A) on oth sides
Aib =Aig +Ayb (eb - Vg)
0 = Aig +Ayb eb -Vg
Ayb Vg – Aig = Ayb eb
Ayb Vg - Aig = Ayb At Vn
A ib = 0
eb = At Vn
…...... Eq (A)
now let
Ayb Vg - Aig = in
&
Ayb At = yn
Eq (A) becomes
in = yn Vn
Where yn is node admittance matrix.
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Example9: write the node equations for the network
shown in fig:
R1
Vg
R2
L3 Il (t)
Sol:
n=3
B=4
T.B = n – 1
=2
Links = b – n + 1
=2
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C4
ig
Vc (t)
At =
Yb
AYb =
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AYb =
AYbAt =
AYbAt =
AYbVg =
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AYbVg =
Aig =
Aig =
ֶin = AYbVg - Aig
in =
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in =
ֶ in = YnVn
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Node equation by using Tie-set technique
Vg
Rb
Lb
Cb
ib
eb
Node
ib
Node
Rb = total resistance in branch
Lb =total inductance in branch
Cb = total capacitance in branch
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Vg = total series voltage source in branch
ig = current source in branch
eb = branch voltage
ib= branch current
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eb=vg+(ib-ig)(Rb+Lb
+
)
eb=vg+(ib-ig)Zb
multiplying tie set matrix on both sides
Beb = BVg + B(ib - ig)Zb
Beb=Bvg+BZbib-BZbig
ֶ Beb=0
0 = BVg + BZbBtiL - BZbig
t
BZ Bti = BZ i - BV ......eq(A) ib =B iL
b
L
b g
g
Assume
BZbBt=ZL
BZbig- BVg=VL
now eq(A) becomes
zL iL=VL
which is loop equations in matrix form, where ZL is loop
impedance matrix
51
Example10: Write the loop equation shown in figure by
means of Tie-set method Assuming R2&L3 Branches as tree
branches
R1
Sol
Vg
b=4
n=3
T.B=2
T.S=2
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R2
ig
C4
L3
52
Since
vL =zL I
ZL=BZbBt
VL=BZbig-BVg
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B=
Bt=
1 1 1 0
0  1  1 1 


1
1

1

0
0 
 1

 1

1 
0
0 
R1 0
 0 R2 0

0
Zb= 

 0 0 DL3
0 


0
0
0
1/DC
4


BZb=
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1 1 1
0  1  1

0

1
0
0 
R1 0
 0 R2 0

0 

 0 0 DL3
0 


0 1/DC4 
0 0
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DL3
0 
R1 R2
 0  R2  DL3 1/DC4 


BZb=
BZb
Bt=
DL3
R1 R2
 0  R2  DL3

0
 1
 1

1
 R2  DL3 
R1  R2  DL3
  R2  DL3 R2  DL3  1/DC4 


BZbBt=ZL=
ig=
1
0  1
1/DC4  1

0
 0 
  ig 


 0 


 0 
BZbig=
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  R2ig 
 R2ig 


55
1 1 1
BVg= 0  1  1

0
1
 vg 
 0 


 0 



v
c


 vg 
BVg= 

  vc 
vL=BZbig-BVg
  R 2i g 
VL=  R 2-ig 


 vg 
  vc 
 
 R 2ig vg 

vL= 
 R 2ig Vc 
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Now
VL = ZL iL
 R 2ig vg  = R 1  R 2  DL3
 R 2ig Vc  

   R 2  DL3
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 R 2  DL3

R 2  DL3  1 / DC 4 
i 1 
i2
 
57
E-SHIFT
It is possible to shift the voltage source generated (vg) to
the branch that are connected in series with it, with out
changing the characteristics of the branch.
Example:
R1
Vg
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R2
R3
R1
Vg
R2
R3
Vg
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There are two possibilities of E-Shifting depending upon
the network
1)Forward E-Shifting
2)Backward E-Shifting
R1
R2
C1
Vg
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C2
R3
R4
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Vg
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I-SHIFT
Current source Ig can be placed in parallel which
each of the branch that formed a close loop with Ig
ig
C1
R1
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C2
R2
R3
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ig
ig
C1
R1
C2
C1
C2
R1
ig
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R3
R2
R2
R3
ig
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Example11:
write the loop & node equations of the
network shown in fig by applying E shift & I shift.
Sol
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n =4
b =7
T.B=n-1
=3
cut sets=3
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Links=b-n+1
=4
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A=
At =
Yb=
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AYb=
AYb=
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Yn=AYbAt
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Yn =
Yn =
Vg=
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AYbVg=
=
Ig=
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Aig=
=
=
In=AYbVg - Aig
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-
=
In=
In=YnVn
=
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