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Network Analysis
Net work Topology
eNotes
By
Prof.R.V. Srinivasa Murthy
Assistant Professor, Dept. of E & C
A.P.S. College of Engineering
Bangalore
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Net work Topology
Definition:
The term circuit topology refers to the science of placement of elements and is a study of the
geometric configurations.
“Circuit topology is the study of geometric properties of a circuit useful for describing the circuit
behavior“
Terms used in Topology:
The following terms are often used in network topology
Graph:
In the given network if all the branches are represented by line segments then the resulting figure
is called the graph of a network (or linear graph). The internal impedance of an ideal voltage
source is zero and hence it is replaced by a short circuit and that of an ideal current source is
infinity and hence it is represented by an open circuit in the graph.
Example: Network
Graph
Node
It is a point in the network at which two or more circuit elements are joined. In the graph shown 1,
2, 3 and 4 are nodes.
Branch (or Twig):
It is a path directly joining two nodes. There may be several parallel paths between two nodes.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Oriented Graph
If directions of currents are marked in all the branches
of a graph then it is called an oriented (or directed) graph .
Connected graph
A network graph is connected if there is a path between any two nodes .In our further
discussion,let us assume that the graph is connected. Since, if it is not connected each disjoint part
may be analysed separately as a connected graph.
1
2
4
3
Unconnected graph
If there is no path between any two nodes,then the graph is called an unconnected graph.
1
2
4
1
2
4
3
3
5
3
5
5
Planar graph
A planar graph is a graph drawn on a two dimensional plane so that no two branches intersect at
a point which is not a node.
B
A
C
A
B
E
D
D
E
E
C
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Non – planar graph
A graph on a two – dimensional plane such that two or more
branches intersect at a point other than node on a graph.
Tree of a graph
Tree is a set of branches with all nodes not forming any loop or closed path.
(*) Contains all the nodes of the given network or all the nodes of the graph
(*) No closed path
(*) Number of branches in a tree = n-1 , where n=number of nodes
A
2
B
C
A
B
2
5
6
D
C
4
5
D
Two possible trees
Graph
Co- tree
A Co- tree is a set of branches which are removed so as to form a tree or in other words, a co- tree
is a set of branches which when added to the tree gives the complete graph. Each branch so
removed is called a link.
Number of links = l = b – (n-1) where b = Total number of branches
n = Number of nodes
Incidence Matrix
Incidence matrix is a matrix representation to show which branches are connected to which nodes
and what is their orientation in a given graph
(*) The rows of the matrix represent the nodes and the columns represents the branches of the
graph.
(*) The elements of the incidence matrix will be +1, -1 or zero
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
(*) If a branch is connected to a node and its orientation is away from the node the corresponding
element is marked +1
(*) If a branch is connected to a node and its orientation is towards the node then the corresponding
element is marked – 1
(*) If a branch is not connected to a given node then the corresponding element is marked zero.
Incidence Matrix
Complete Incidence matrix
Reduced incidence matrix
( i )Complete incidence matrix:
An incidence matrix in which the summation of elements in any column in zero is called
a complete incidence matrix.
( ii )Reduced incidence matrix:
The reduced incidence matrix is obtained from a complete incidence matrix by eliminating a
row. Hence the summation of elements in any column is not zero.
Example 1: Consider the following network and the oriented graph as shown
Network
Oriented graph
(*) There are four nodes A, B, C and D and six branches 1, 2, 3, 4, 5 and 6. Directions of
currents are arbitrarily chosen.
(*) The incidence matrix is formed by taking nodes as rows and branches as columns
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Nodes
Branches
1
2
3
4
5
6
A
-1
+1
+1
0
0
0
B
0
-1
0
-1
+1
0
C
0
0
-1
+1
0
+1
D
+1
0
0
0
-1
-1
P =
-1 1 1
0
0
0
0 -1 0 -1
1
0
0
0
1
0 -1 1
1 0 0 0
-1 -1
In the above example the fourth row is negative of sum of the first three rows. Hence the fourth
can be eliminated as we know that it can be obtained by negative sum of first three rows. As a
result of this we get the reduced incidence matrix.
PR =
-1 1 1
0
0 0
0 -1 0 -1 1 0
0
0 -1 1
0 1
Properties of a complete incidence matrix
(*) Sum of the entries of each column is zero
(*) Rank of the matrix is (n-1), where n is the no of nodes
(*) Determinant of a loop of complete incidence matrix is always zero
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Example 2 : The incidence matrix of a graph is as shown. Draw the corresponding graph.
Solution :
1
0
0
0
1
-1
-1
1
1
0
0
0
given matrix is zero. Hence it
0 -1 0
-1
0
1
is a compete incidence matrix.
0
-1 1 -1
0
The sum of each column of the
0
Number of nodes = n = 4 ( say A. B. C and D)
Number of branches = b = 6 ( say 1, 2, 3, 4, 5 and 6)
Prepare a tabular column as shown.
Nodes
Branches
1
2
3
4
5
6
A
1
0
0
0
1
-1
B
-1
1
1
0
0
0
C
0
-1
0
-1
0
1
D
0
0
-1
1
-1
0
From the tabular column, the entries have to be interpreted as follows:
From the first column the entries for A and B are one’s . Hence branch 1 is connected between
nodes A and B . Since for node A entry is +1 and for node B it is -1, the current leaves node
A and enters node B and so on.
From these interpretations the required graph is drawn as shown.
6
A
B
1
C
2
5
3
4
D
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Example 3: The incidence matrix of a graph is as shown. Obtain the corresponding graph
1 1 0 0 0
0 0
0 -1 1 1
0
0
0
0 0 0 -1 1 1 0
0 0 0 0 0 -1 1
Solution:- Given incidence matrix is a reduced incidence matrix as the sum of each column is not
zero. Hence it is first converted in to a complete incidence matrix by adding the deleted row. The
elements of each column of the new row is filled using the fact that sum of each column of a
complete incidence matrix is zero.
In the given matrix in first, third, fifth and the seventh column the sum is made zero by adding –1
in the new row and the corresponding node is E. The complete incidence matrix so obtained and
also the graph for the matrix are as shown.
Nodes
Branches
1
2
3
4
5 6
7
A
1
1
0
0
0
0
0
B
0
-1
1
1
0
0
0
C
0
0
0
-1
1
1
0
D
0
0
0
0
0
-1
1
E
-1
0
-1
0
-1
0
-1
Graph:
6
7
E
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Net work Topology
SESSION :2
Tie – set Analysis:
In order to form a tree from a network several branches need to be removed so that the
closed loops open up. All such removed branches are called links and they form a Cotree. Alternatively when a link is replaced in a tree, it forms a closed loop along with
few of the tree branches. A current can flow around this closed loop. The direction of
the loop current is assumed to be the same as that of the current in the link. The tree –
branches and the link that form a loop is said to constitute a tie – set.
Definition
A tie – set is a set of branches contained in a loop such that the loop has at least one link
and the remainder are twigs (tree branches)
6
6
4
B
4
5
A
C
C
x
1
1
3
5
A
2
Graph
z
B
y
2
3
D
tree (In thick lines) Co-tree (In dotted lines)
D
We see that by replacing the links 1, 4 and 5 three loops are formed and hence three loop
currents x, y and z flow as shown. The relationships obtained between loop currents, tree
branches and links can be scheduled as follows
Tie – set schedule
Tie – set
1
2
3
Tree – branches
2, 3
1, 4
4, 5
Link
5
2
6
Loop current
x
y
z
Tie – set matrix (Bf)
The Tie – set schedule shown above can be arranged in the form of a matrix where in the
loop currents constitute the rows and branches of the network constitute the columns
Entries inside the matrix are filled by the following procedure :
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Let an element of the tie – set matrix be denoted by mik
Then
mik
=
1 Branch K is in loop i and their current directions are same
-1 If branch K is in loop i and their current directions are opposite.
0 If the branch K is not in loop i
By following this procedure we get the Tie – set matrix which is shown below:
Loop
Branches
currentss1
2
3
4
5
6
x
0
+1
+1
0
+1
0
y
+1
+1
0
+1
0
0
z
0
0
0
+1
-1
-1
Or
0 1 1 0 1 0
Bf =
1 1 0 1 0 0
0 0 0 1 -1 -1
Analysis of the net work based on Tie – set schedule
From the tie –set schedule we make the following observations.
(i) Column wise addition for each column gives the relation between branch and loop
currents
That is
i1
i2
i3
i4
i5
i6
=
=
=
=
=
=
y
x+y
x
y+ z
x-z
-z
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Putting the above equations in matrix form, we get
i1
i2
i3
i4
i5
i6
=
0
1
1
0
1
0
1
1
0
1
0
0
0
0
0
1
-1
-1
x
y
z
In compact form IB = B fT IL
Where
IB = Branch current matrix
B fT = Transpose of the tie- set matrix
IL = Loop current column matrix
(ii)Row wise addition for each row gives the KVL equations for each fundamental
loop
Row - 1
Row - 2
Row - 3
:
:
:
0 1 1 0 1 0
1 1 0 1 0 0
0 0 0 1 -1 -1
In compact form
V1 + V2 + V3 = 0
V1 + V2 + V4 = 0
V4 - V5 - V6 = 0
V1
V2
V3
V4
V5
V6
= 0
B f VB = 0
- - - (1)
Where VB = Branch voltage column matrix and
B f = Tie - set matrix
Equilibrium equations
Let us consider a network
having b- branches. Each branch
of the network has a representation
as shown in figure
Vk = Ik Zk
+
Ik
Zk
Vk +
Vsk
±
-
VSK- - - (2)
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Since the network has b branches one such equation could be written for every branch of
the network. That is
V1 = Z1 I1 + Vs1
V2 = Z2 I2 + Vs2
‘
,
‘
,
‘
,
,
,
,
, ,
,
Vb = Zb Ib + Vsb
Putting the above set of equations in a matrix form we get
VB = ZB IB+ VS
- - - (3)
Where VB = Branch – Voltage column matrix of order b x 1
Z1 0
0 Z2
0
0
ZB =
0
-
0
-
-
0
0 = Branch Impedance matrix (bX b)
ZB
IB = Branch current matrix of order b x 1
and Vs
=
Vs 1
Vs 2
Vs3
‘
‘
‘
=
Voltage source column matrix of order bx1
Vs b
The elements in the voltage source matrix are positive if the branch currents enter the
positive terminal of the source otherwise negative
From equation (1) BfVB =0
Substituting equation (3) we get, Bf ZB IB + Vs = 0
………(4)
Also the relation between branch and loop currents is given by IB = B fT IL
Substituting for IB in equation (4)
Bf [ ZB B fT IL + Vs ] = 0
Bf ZB B fT IL+ Bf Vs = 0
ZL IL + Bf Vs = 0 …………….(5)
impedance matrix
Where ZL = Bf ZB B
f
T
is defined as Loop
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Action Plan for tie –set analysis
(*) Form the Tie-set Matrix (Bf )
(*) Construct the branch impedance matrix (ZB)
(*) Loop impedance matrix is formed by using the relation ZL = Bf ZB B fT
(*) Form the KVL or Equilibrium equations using the relation ZL IL + Bf Vs = 0
(*) The branch currents are then found using the matrix equation IB = B fT IL
(*) Finally the branch voltages are found using the matrix equation
VB = ZB IB+ VS
A
5
Example: For the network shown in figure, write a
Tie-set schedule and then find all the branch 50 ±
Currents and voltages
V
5
10
B
10
Solution: The graph and one possible tree is shown:
C
5
5
D
A
1
2
1
5
B
4 6
C
5
x
y
6
4
z
D
3
2
3
Loop
current
x
y
z
1
1
0
0
Tie set Matrix:
2
0
1
0
Branch Numbers
3
4
0
1
0
0
1
-1
5
-1
1
0
6
0
-1
1
Bf = 1 0 0 1 -1 0
0 1 0 0 1 -1
0 0 1 -1 0 1
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Branch Impedence matrix
ZL =
ZB =
1 0 0 1 -1 0
0 1 0 0 1 -1
0 0 1 -1 0 1
5
0
0
0
0
0
0 0 0 0
10 0 0 0
0 5 0 0
0 0 10 0
0 0 0 5
0 0 0 0
0
0
0
0
0
5
5 0 0 0 0 0
0 10 0 0 0 0
0 0 5 0 0 0
0 0 0 10 0 0
0 0 0 0 5 0
0 0 0 0 0 5
1 0 0
0 1 0
0 0 1
1 0 -1
-1 1 0
0 -1 1
20 -5 -10
-5 20 -5
-10 -5 20
Loop Equations :
ZL IL = - Bf Vs
ZL
=
20 -5 -10
-5 20 -5
-10 -5 20
x
y
z
=-
1 0 0 1 -1 0
0 1 0 0 1 -1
0 0 1 -1 0 1
-50
0
0
0
0
0
20x-5y-10z =50
-5x+20y-5z = 0
-10x-5y+20z =0
Solving the equations, we get x = 4.17 Amps
y = 1.17 Amps
And z = 2.5 Amps
Branch Currents :
I1
I2
I3
I4
I5
I6
=
IB = B fT IL
1 0 0
0 1 0
0 0 1
1 0 -1
-1 1 0
0 -1 1
x
y
z
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Which gives
I1 = 4.17 A
I2 = 1.17 A
I3 = 2.5 A
Branch Voltages:
V1
V2
V3
V4
V5
V6
=
I4 = 1.67 A
I5 = -2.5A
I6 = 0.833A
VB = ZB IB+ VS
5 0 0 0 0 0
0 10 0 0 0 0
0 0 5 0 0 0
0 0 0 10 0 0
0 0 0 0 5 0
0 0 0 0 0 5
I1
I2
I3
I4
I5
I6
+
-50
0
0
0
0
0
V1 = 29.17 volts
V2 = 16.67 volts
V3 = 12.5 volts
V4 = 16.67 volts
V5 = - 12.5 volts
V6 = 4.17 volts
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Net work Topology
SESSION :3
Cut – set Analysis
A cut – set of a graph is a set of branches whose removal , cuts the connected graph in to two
parts such that the replacement of any one branch of the cut set renders the two parts
connected.
A
A
1
5
Example
4
B
3
D
4
B
D
2
6
3
C
C
Directed graph
Two separate graphs created by the cut set (1, 2, 5, 6)
Fundamental cut – set is a cut – set that contains only one tree branch and the others are
links
Formation of Fundamental cut – set
(*) Select a tree
(*) Select a tree branch
(*) Divide the graph in to two sets of nodes by drawing a dotted line through the selected
tree branch and appropriate links while avoiding interruption with any other tree – branches.
Example 1 : For the given graph write the cut set schedule
A
4
1
B
5
3
A
D
2
1
4
6
C
The fundamental cut –set of the
Selected tree is shown in figure
5
B
2
3
D
6
C
Note that FCS - 1 yields node A and the set of nodes ( B, C, D)
The Orientation of the fundamental cut – set is usually assumed to be the same as the
orientation of the tree branch in it, Which is shown by an arrow. By following the same
procedure the FCS- 2 and FCS -3 are formed as shown below:
A
A
1
5
1
5
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
4
B
3
D
B
3
2
2
4
D
6
6
C
C
FCS -2
FCS -3
It should be noted that for each tree branch there will be a fundamental cut – set. For a graph
having ‘n’ number of nodes the number of twigs is (n-1).Therefore there will be (n-1)
(n-1) fundamental cut-sets.
Once the fundamental cut sets are identified and their orientations are fixed, it is
possible to write a schedule, known as cut – set schedule which gives the relation between
tree – branch voltages and all other branch voltages of the graph.
Let the element of a cut – set schedule be denoted by Qik then,
Qik = 1
If branch K is in cut – set I and the direction
of the current in the branch K is same as cut – set direction .
-1 If branch k is in cut – set I and the direction of the current in
k is opposite to the cut – set direction.
O If branch is k is not in cut – set i
Cut set Schedule
Tree
Branch Voltages
branch 1
2
3
4
5
6
voltage
e1
1
0
0
-1
-1
0
e2
0
1
0
1
0
1
e3
0
0
1
0
1
-1
The elements of the cut set schedule may be written in the form of a matrix known as the cut
set matrix.
Qf =
1 0
0
-1
-1 0
0 1
0
1
0 1
0 0
1
0
1 -1
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Analysis of a network using cut set schedule
(*) Column wise addition of the cut set schedule gives the relation between tree branch
voltage and the branch voltages for the above cut the schedule
V1 = e1
V2 = e2
V3 = e3
V4 = - e1 + e2
V5 = - e1+ e3
V6 = e2 - e3
In matrix form
V1
V2
V3
V4
V5
V6
1
0
0
= -1
-1
0
0
1
0
1
0
1
0
0
1
0
1
-1
e1
e2
e3
In compact form
VB = QTf VT
………. (1)
Where VB =Branch voltage Matrix
QTf = Transpose of cut set matrix
and VT = Tree branch voltage matrix.
(*) Row wise addition given KCL at each node
I1 - I4 - I5 =
0
I2 + I4 + I6 =
0
I3 + I5 + I6 =
0
In matrix form
1 0
0
-1
-1 0
I1
0 1
0
1
0 1
I2
0 0
1
0
1 -1
I3
=
0
I4
I5
I6
In compact form Qf IB = 0 …………………(2) Where Qf = cut set matrix
IB = Branch current matrix
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
IK
VK
(*) Let us consider a network having ‘b’ branches.
Each of the branches has a representation as shown in figure
YK
ISK
Referring to the figure Ik = Yk Vk + Isk
Since the network has b branches, one such equation could be written for every branch
of the network ie
I1 = Y1 V1 + Is1
I2 = Y2 V2 + Is2
-----------------------------…………………………
Ib = Yb Vb + Isb
Putting the above set of equations in a matrix form we get
………………………
IB = YB VB + Is
(3)
IB = Branch current matrix of order bx1
YB =
Y1
0
0
0 0 - - 0
Y2 0 - - 0 =
- - - - 0 - - - Yb
Branch admittance
matrix of order bx b
VB = Branch Voltage column matrix of order bx1 and
IB = Source current matrix of order bx1.
Substituting (3) in (2) we get
Qf
YB VB + IS =
0
Qf YB VB + Qf IS =
0
But from (1) we have VB = Qf T VT
Hence equation (4) becomes
Qf YB QTf VT + Qf IS
YC VT + QfIS
=
=
or
…………………….
YC VT = - Qf IS
0
0
(5)
Where YC = Qf YB Qf T is called cut - set admittance matrix
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Action Plan for cut – set analysis.
(*) Form the cut- set matrix Qf
(*) Construct the branch admittance matrix YB
(*) Obtain the cut – set admittance matrix using the equation YC = Qf YB Qf T
(*) Form the KCL or equilibrium equations using the relation YC VT = - Qf IS
The elements of the source current matrix are positive if the directions of the branch
current and the source connect attached to that branch are same otherwise negative.
(*) The branch voltages are found using the matrix equation VB = Qf T VT
(*) Finally the branch currents are found using the matrix equation IB = YB VB + IS
Example 2 : For the directed graph obtain the cut set matrix
A
5
6
1
D
E
B
4
2
8
3 7
C
Solution : The tree (marked by thick lines)and the link (marked by doffed lines)are as
shown. The fundamental cut sets are formed at nodes A B C and D keeping ‘E’ as
reference node
Fcs - 1 --
( 1, 5, 6)
Fcs - 2 --
(2, 6,
7)
Fcs - 3 -- ( 3, 7,
8)
Fcs - 4 -- (4,
8)
5,
A
FCS-1
1
D
FCS4
4 E 2
B FCS2
FCS3 C
Hence the cut – set schedule is as follows:
Tree
Branch
branch 1
2
3
4
5
6
7
8
0
0
0
-1
1
0
0
voltage
e1
1
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
e2
0
1
0
0
0
-1
1
0
e3
0
0
1
0
0
0
-1
1
e4
0
0
0
1
1
0
0
-1
Hence the required cut –set matrix
Qf = 1
0
0
0
0
1
0
0
0 0 -1 1 0 0
0 0 0 -1 1 0
1 0 0 0 -1 1
0 1 1 0 0 -1
Example 3: Find the branch voltages using the concept of cut-sets
1
1
1
1
1
1V ±
1
Solution : The voltage source is Transformed in to an equivalent current source. It should be
noted that all the circuit Passive elements must be admittances and the net work should
contain only current sources.
The graph for the network is shown. A possible tree (shown with thick lines) and co tree
(shown by dotted lines) are shown
1mho
FCS 1 = 3, 1, 5
1mho
FCS 2 = 4, 2, 5
1mho
1mho
FCS 3 =
1 mho
1mho
6, 1, 2
1A
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Cut set schedule
Tree
Branches
branch 1
2
3
4
FCS2
5
5
6
voltage
e3
-1
0
1
0
-1
0
e4
0
1
0
1
1
0
e6
1
-1
0
0
0
1
3
A
FCS1
4
8
6
1
2
FCS3
-1 0 1 0 -1 0
0 10 1 1 0
1 -1 0 0 0 1
Qf =
1 0
0 1
0 0
0 0
0 0
0 0
YB =
0
0
1
0
0
0
0 0
0 0
0 0
1 0
0 1
0 0
0
0
0
0
0
1
-1 0 1
-1 0 1 0 -1 0
YC=
=
0 1 0 1 1 0
1 -1 0 0 0 1
3
-1
-1
-1
3
-1
-1
-1
3
Equilibrium Equations
1 0 0 0 0 0
0 1 0 0 0 0
0 0 10 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0
1
0
-1
0
1
0
1
1
0
-1
0
0
0
1
; YC VT = - Qf IS
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
3 -1 -1
-1 3 -1
-1 -1 3
e3
e4
e6
=
-1 0 1 0 -1 0
0 1 0 1 1 0
1 -1 0 0 0 1
-
-1
0
0
0
0
0
3 e3 – e4 – e6 = - 1
-e3 + 3 e4 – e6 = 0
- e3 – e4 + 3e6 = 1
Solving we get e3=-0.25 volt
e4=0
e6=0.25 volt
Branch voltages
VB = Qf T VT
V1
-1 0
V2
1
0
1 -1
V3
1
0
0
V4
0
1
0
V5
-1
1
0
V6
0
=
0
-0.25
0
0.25
1
0.5
-0.25
= -0.25
0.0
0.25
0.25
V1 = 0.5 Volts
V2 = -0.25 Volts
V3 = -0.25 Volts
V4 = 0 Volts
V5 = 0.25 Volts
V6 =0.25 Volts
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
SESSION- 4
DUALITY CONCEPT
Two electrical networks are duals if the mesh equations that characterize one have the same
mathematical form as the nodal equations of the other.
Example 1
Consider an R-L-C series network excited by a voltage
so source V as shown in the figure. The equation generating
the circuit behavior is Ri+Ldi +1 ∫idt =V ……..(1)
dt C
Figure 1
Now consider the parallel G-C-L network
fed by a
Current Source i is shown in the figure. The
equation generating the
Circuit behavior is GV+ CdV +1 ∫vdt = i
……..(2)
dt L
Figure 2
Comparing the equations (1) and (2),we get the similarity between the networks of fig(1) and
fig(2).The solution of equation (1) will be identical to the solution of equation (2) when the
following exchanges are made
R → G, L→ C, C→L and V →i
Hence networks of figure (1) and (2) are dual to each other.
Table of dual Quantities
1.Voltage Source
2.Loop currents
3.Iductances
4.Resistances
5.Capacitances
6. On KVL basis
7.Close of switch
Current source
Node voltages
Capacitances
Conductances
Inductances
On KCL basis
Opening of switch
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Note:Only planar networks have duals.
Procedure for drawing dual network
The duals of planar networks could be obtained by a graphical technique known as the dot
method. The dot method has the following procedure.





Put a dot in each independent loop of the network. These dots correspond to
independent nodes in the dual network.
Put a dot outside the network. This dot corresponds to the reference node in the dual
network.
Connect all internal dots in the neighboring loops by dotted lines cutting the common
branches.
These branches that are cut by dashed lines will form the branches connecting the
corresponding independent nodes in the dual network.
Join all internal dots to the external dot by dashed lines cutting all external branches.
Duals of these branches will form the branches connecting the independent nodes and
the reference node.
Example 1:
Draw the exact dual of the electrical
circuit shown in the figure.
3ohm
2sin6t
4F
4ohm
±
1
6H
O
2
Solution: Mark two independent nodes 1 and
2 and a reference node 0as shown in the
figure.
 Join node 1 and 2 by a dotted line
passing through the inductance of
6H.this element will appear as
capacitor of 6 F between node 1 and 2
in the dual.
 Join node 1 and reference node
through a dotted line passing the
voltage source of 2sin6t volts. This
will appear as a current source of
2sin6t amperes between node1 and
reference node.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore

Join node 1 and reference node through a dotted line passing through 3 ohms resistor.
This element appears as 3mho conductance between node1 and reference node in the
dual.

Join node 2 and reference node through a dotted line passing through the capacitor of
4 Farads. This element will appear as 4 Henry inductor between node 2 and
reference node in the dual

Join node 2 and reference node through a dotted line passing through the resistor of 4
ohms. This element will appear as 4 mho conductance between node 2 and reference
node.
The Dual network drawn using these procedural steps is shown.
Loop equations for the original network
Applying Kirchoff’s Voltage Law to the two loops we get 3i1 + 6 di1 -6di2 =2 Sin6t
dt
dt
and -6 di1 + 6 di2 + 1 ∫ i 2 dt =0
dt
dt 4
N0DE EQUATIONS
Applying Kirchoff’s Current Law at nodes 1 and 2 ,we get, 3V1 + 6 dV1 -6dV2 =2 Sin6t
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
dt
dt
-6 dV1 +6dV2 + 1 ∫V2 dt + 4V2 =0
dt
dt
4
By looking at these equations it can be further understood that dual network drawn is the
correct one.
Example 2: Construct the dual electrical network for the network shown
K
L1
i1
i2
C1
vs
R1
±
R3
L2
R2 R2
i3
iS
Solution: Mark three independent nodes 1,2and 3 and a reference node 0 as shown in figure.
The dual of the network is then drawn using the dot-method.
L1
1
VS ±
C1
L2
2
R2
R3
R
R3
3
iS
0
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Finally the dual circuit is redrawn in a more standard form as shown.
C1
L2
L1
1
iS
2
G1
3
C1
G3
±± VS
DUAL GRAPHS
For a given graph a dual graph can be constructed by converting each tie-set of the given
graph as
Cut-set in the dual graph. Two network graphs are said to be dual to each other when the
eqations
Written for one on loop current basis and the other on node-pair voltage basis are identical.
# ) Total number of branches in the graph is equal to the number of branches in the dual
graph
# ) The number of independent loops in the graph is equal to the number of node pairs in the
dual graph.
Procedure for constructing dual graphs
1.
2.
3.
4.
5.
Mark node-pairs equal to number of loops
Mark one reference node
Assign each of the nodes to each of the meshes in the graph
For each mesh note the tie-set and the corresponding cut-set in the dual graph
Common branch between loops appears as a branch between
corresponding nodes
6. Peripheral branch confined to a loop appears as a branch
between that node and the reference node.
7. Directions of branches in the dual are marked as follows:
4
i)
If the loop current and branch current directions
8
are same orientation is away from that node
ii)
On the other hand if they are opposite then the
orientation is into the node
7
Example1: Draw the oriented dual graph of the graph
Q
In figure
1
Solution:
In the given graph
n=4(P,Q,R and
b=8(
3
2
5
R
6
shown
S)
1,2,3,4,5,6,7and8)
d
a
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
b
e
c
c
l=b-(n-1)
=8-(4-1)=5
(*) Hence there are five tie-sets in the given graph. Correspondingly
there will be five node-pair voltages in its dual graph with one ref
node.
(*) Let the loop currents be a,b,c,d and e as shown.
(*) Mark nodes a,b,c,d and e as shown in figure
(*) Mark ‘f ‘ as the reference node
(*) For loop ‘ a ‘ branch 1 is peripheral and 4 is common
to a and b.Hence branch 1 is connected
between nodes a and f . Branch 4 is connected between
a and b.
P
S
(*) Loop current ‘a ‘ and branch 1 current are in same direction.Hence
the orientation of the branch 1 is away from node a,where as for branch 4
the directions are opposite.Hence the orientation of branch 4 is towards
node ‘a ‘(*) Following the same procedure other branches are drawn and their orientation
is marked
to get the required graph as shown.
8
b
4
a
c
5
7
6
d
e
1
2
3
f
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Example 2 :
Determine the dual graph for the graph shown in figure
a
Solution: For the given graph b= 7
n =4
2
l
1
= 7-(4-1) = 4
b
(*) Hence there are 4 tie-set schedules and hence four loop currents
l= b-(n-1)
( i, j , k, l ) as shown
e
k
c
d
j
i
f
4
3
g
(*) Mark four nodes i , j , k and l on the sheet and also the reference
node o as shown in figure
(*) By following the same procedure as explained in the previous
example the dual oriented graph is drawn as shown in figure.
i
j
k
l
d
e
b
c
g
f
a
o
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Resonant Circuits
Resonance is an important phenomenon which may occur in circuits containing both
inductors and capacitors.
In a two terminal electrical network containing at least one inductor and one capacitor, we define resonance as the condition,
which exists when the input impedance of the network is purely resistive. In other words a network is in resonance when the voltage and
current at the network in put terminals are in phase.
Resonance condition is achieved either by keeping inductor and capacitor same and
varying frequency or by keeping the frequency same and varying inductor and capacitor.
Study of resonance is very useful in the area of communication. The ability of a radio
receiver to select the correct frequency transmitted by a broad casting station and to
eliminate frequencies from other stations is based on the principle of resonance.
The resonance circuits can be classified in to two categories
1. Series – Resonance Circuits.
2. Parallel – Resonance Circuits.
R
L
C
1.Series Resonance Circuit
A series resonance circuit is one in which a coil and
a capacitance are connected in series across an alternating
I
voltage of varying frequency as shown in figure.
The response ‘I’ of the circuit is dependent on the impedance of the circuit,



V
Where Z= R +jXL - jXC
and I= V
at any value of frequency
Z
We have XL =2πfL
XL varies as f
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
and
XC
= 1
XC varies inversely as f
2πfC
In other words, by varying the frequency it is possible to reach a point where X L = XC
In that case Z = R
.
and hence circuit will be under resonance. Hence the series A.C.
circuit is to be under resonance, when inductive reactance of the circuit is equal to the
capacitive reactance. The frequency at which the resonance occurs is called as resonant
frequency ( fr)
Expression for Resonant Frequency ( fr )
At resonance XL = XC
i.e. ω r L = 1
ω rC
ω2 r L C = 1
ω2 r =
1
L C
f2r =
1
4π 2LC
fr =
1
2π √LC
Salient Features of Resonant circuit
(*) At resonance XL = XC
(*) At resonance Z = R i.e. impedance is minimum and hence I = V
is maximum
Z
(*) The current at resonance (Ir) is in phase with the voltage
(*) The circuit power factor is unity
(*) Voltage across the capacitor is equal and opposite to the voltage across the
inductor.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Frequency response of a series resonance circuit
For a R-L-C series circuit the current ‘I’ is given by
V
I=
____________
R + j ( XL- XC )
At resonance XL = XC and hence the current at resonance (Ir) is given by Ir = V/R
At off resonance frequencies since the impedance of the circuit increases the current
in the circuit will reduce. At frequencies f Where f> fr , the impedance is going to be more
inductive. Similarly at frequencies f < fr the circuit impedance is going to be more
capacitive. Thus the resonance curve will be as shown in figure.
I
Ir
Xc > Xl
Xl > Xc
f
fr
Qualify – factor (or Q – factor)
Another feature of a resonant circuit is the Q – rise of voltage across the resonating
elements.
If V is the applied voltage across a series resonance circuit at resonance, I r = V
R
The voltage across the inductance ‘L’, =V L = I r XL = V ω r L
R
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
ω rL
VL = V
R
V L = QV where Q = ω r L is the Q factor
R
From the above equation it is seen that the voltage across the inductive coil is Q times the applied voltage (V) The response of the series
resonant circuit is largely dependent on Q of the coil.
Band – width concept
Any circuit response, which is frequency dependent, has certain limitations. The output
response during limited band of frequencies only will be in the useful range. If the out put
power is equal to or more than half of the maximum powerout put that band of frequencies
is considered to be the useful band. If I r is the maximum current at resonance then
Power at resonance = Pmax = I2 r R
But If I= I r Then corresponding power is given by I2 R = Ir
√2
I2 R = 1
2
R
√2
I2 r R = 1 ( Max Power)
2
2
Hence the useful range of frequencies will be frequencies where current will be equal to or
more than Ir
= 0.707 Ir
√2
Consider the frequency response characterstic of a series resonant circuit as shown in figure
Ir
0.707 Ir
freq
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
F1
F2
Fr
In the figure it is seen that there are two frequencies where the out put power is half of the
maximum power. These frequencies are called as half power points f1 and f2
A frequency f1 which is below fr where power is half of maximum power is called
as lower half power frequency (or lower cut – off frequency). Similarly frequency f2
which is above fr is called upper half power frequency (or upper cut-off frequency)
The band of frequencies between f2 and f1 are said to be useful band of frequencies since during these frequencies of operation
the out put power in the circuit is more than half of the maximum power. Thus their band of frequencies is called as Bandwidth.
i.e Band width =B.W = f2 - f1
Selectivity : Selectivity is a useful characteristic of the resonant circuit. Selectivity is
defined as the ratio of band width to resonant frequency
Selectivity
= f 2- f 1
fr
It can be seen that selectivity is the reciprocal of Quality factor. Hence larger the value of Q
Smaller will be the selectivity.
The Selectivity of a resonant circuit depends on how sharp the out put is contained
with in limited band of frequencies. The circuit is said to be highly selective if the
resonance curve falls very sharply at off resonant frequencies.
Relation between Resonant frequency and cut-off frequencies
Let fr be the resonant frequency of a series
Ir
resonant circuit consisting of R,L and C
elements .From the Characteristic it is
0.707Ir
seen that at both half frequencies
f2 and f1 the out put current is 0.707 Io
which means that the magnitude of the
impedance is same at these points.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
At lower cut-off frequency f1
———————
Impedance is given by √ R2 +
XC1 – XL1
f1
fr
f2
2
At upper cut-off frequency f2
___________________
Impedance is given by √R2 + ( XL2 – XC2 ) 2
But these impedances are equal
——————————
Hence
√ R2 + ( XC1 – XL1)
2
—————————
√ R2 + ( XL2 – XC2 )
=
2
XC1 – XL1 = XL2 – XC2
XC1 + XC2
1
1
C
= XL1 + XL2
=
+1
ω1
ω1 ω2
L
ω1 + ω2
ω2
= 1
LC
ωr
We have fr = 1
=
2π √LC
Hence ω r
√LC
= _1__
2
1
ωr
2
= ω1 ω2
or f 2r = f 1 f 2
LC
fr = √ f1 f2
i.e. Resonant frequency is the geometrical mean of half power frequencies.
Relation between resonant frequency (f r) Band width and quality factor (Q)
Let f 1and f 2 be the lower and upper
Ir
half power frequencies and f r the
resonant frequency .At both half
0.707Ir
power frequencies the magnitude
of impedance can be calculated as follows.
At lower cut- off frequency (f 1)
V
I=
——————————
√ R2 +
Xc1 – XL1
2
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
f1
But I1 = Ir / √ 2
Hence V
=
fr
f2
V/ √ 2 R
=
V
√ 2 R
——————————
√ R2 +
√ R2 + Xc1 – XL1
2
Xc1 – XL1
2
√ 2 R
=
i.e. Xc1 – XL1 = R ………(1)
Similarly at upper cut-off frequency,
I2 =
V
——————————
√ R2 +
XL2 – XC2
V
= V/ √ 2 R
2
=
√ 2 R
But I2 = Ir / √ 2
V
——————————
√ R2 + XL2 – XC2
2
——————————
√ R2 + XL2 – XC2
XL2 – XC2
=
2
= √ 2 R
R………(2)
Adding equations (1) and (2),we get Xc1 – XL1 + Xc2 – XL2
1/ω 1 C - 1/ω 2C + ω 2L - ω 1L
=2R
1/C 1/ω 1 - 1/ω 2 + L ω 2 - ω 1
1/C ω 2 - ω 1
+ Lω2-ω1
= 2R
=2R
= 2R
ω2ω1
ω 2 - ω 1 = 2R
1/ ω 2 ω 1C +L
=
2R/L
=
R/L
1/ ω 2 ω 1LC + 1
ω2-ω1 = R
ωr
ωr L
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
ω2-ω1 = ωr
ω r L/R
f2 – f1 = fr / Q
Band width = f r / Q
Resonance by varying Inductance
Resonance in RLC series circuit can also be obtained by varying resonating circuit
elements . Let us consider a circuit where in inductance is varied as shown in figure.
R
L
C
Ir
V, f Hz



0.707Ir
At resonance XL =Xc , ω L r = 1/ ωC
L r = 1/ ω2C,where L r = value of inductance at resonance.
L1
Lr
L2
At lower – half point
Let L1 be the value of inductance.
I= I0/ √ 2 and I0 = V/R
V
√R2 +
1
=
Xc1 – XL1
-ωL1 = R
ωC
2
V/
√2 R
L1 = 1
-R
ω 2C ω
At upper – half point
Let L 2 be the value of inductance. Then ω L 2 - 1 = R
ωC
L2 = 1
+ R
ω 2C
ω
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Resonance by varying capacitance
R
L
V, f Hz
C



C1
Cr
C2
C
Consider an RLC series circuit in which
capacitance ‘C’ is varied.
At resonance Xc = XL
1
= ωL
ω Cr
Cr = 1
Cr =
Capacitance value at resonance.
ω2L
1
At lower – half point
1
= R+ωL .
-ωL = R
ω C1
ω C1
C1 =
1
Farad
ω 2L + ωR
At upper – half point
ωL- 1
-= R
1
= ω L- R.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
ω C2
C2 =
ω C2
1
Farad
ω 2L - ωR
Expressions for maximum capacitor voltage and maximum inductor voltage
From the figure it is clear that voltage across L and voltage
across C are not maximum at resonant frequency (fr).
Rather voltages VL and VC are equal in magnitude and
opposite in phase at fr. The voltage Vc is maximum at
a frequency fc max which is less than fr and the voltage
VL is maximum at a frequency fL max which is greater than fr
Ir
VC
VL
fc
fr
max
f
l
max
Freq
Expression for fcmax
Voltage across capacitor Vc is given by Vc = I Xc = I 1/ ω C
But I =V/Z. So Vc =
V
ω C √R2 + ωL –1/ ωC
=
2
To find the frequency at which Vc is maximum we have to differentiate Vc with respect to ω and equate it to zero
V2 .
VC2 = _________________________
2 C2[ R2 + ( L – 1/ C )2 ]
V2.
=
______________________________
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
2 R2 C2 + ( 2 L C – 1 )2
V2 [ 2  R2 C2 + 2 ( 2 L C – 1 ) ( 2  L C ) ]
d VC2
____
=
_____________________________________
d
=0
[  R C +( LC–1) ]
2
2
2
2
2 2
2R2C2 + 2( 2LC) (2LC- 1) =0
2 2 L2 C2 =2 LC - R2 C2
2 =1/ (LC - R2 / 2L2 )
Cmax= √ 1/LC – R2/2L2
rad/sec
fCmax = 1/2 √ 1/LC – R2 / 2L2 Hz
Expression for fLmax
The voltage across inductor VL is given by
VL = I XL = I (L)
But I = V/ Z
Hence VL =
V(L)
√ R2 + ( L – 1/ C)2
V22 L2
VL2 =
______________
R2 + ( L – 1/ C)2
= V2 4 L2 C2
______________
2R2C2 + (2LC- 1)2
Differentiating with respect to time and equating it to zero we get
dVL2/ d = 0 which simplifies to
22LC -2 R2C2 – 2 = 0
2 ( 2LC - R2C2) = 2
2 =
2
or fLmax =
1/ 2√ LC - R2C2 /2
2LC - R2C2
Example 1 :
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
60
0.5H
40F
A series RLC circuit shown in figure is connected
across an A.C. variable frequency supply of 200
votes. Calculate the resonant frequency and half –



power frequencies.
Solution:
fr=
200V,
1
=
1_____
f
=35.6Hz
2π√0.5x40x10 –6
2π√LC
Q= ωr L = 2πx35.6 x0.5
R
= 1.86
60
B.W = fr =
35.6
Q
= 19
1.86
f2- f1 =
19………………(1)
Also fr 2 = f1f2
(35.6) 2 = f1f2
=
1266…………..(2)
Solving equations (1) and (2) we get,
f1 = 27.33 Hz
f 2 = 46.33 Hz
Example 2:
20
Inductor
Capacitor
A 20 ohm resistor is connected in series with
an inductor, a capacitor and an Ammeter across
25 volts variable frequency a.c supply .When the
A
I
frequency is 400 Hz the current is at its maximum
value of 0.5 A and the potential difference across the
capacitor is 150 volts .Calculate
i)
The capacitances of the capacitor
ii)
The resistance and inductance of the inductor.
25V



Solution : Given Ir = 0.5 A, fr = 400Hz
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
(1)
We have V C =I 0 X C
X C = V C = 150
1
I0
0.5
= 300Ω
or C=
ω0 C
= 300 ohms
1
= 3.33 microfarad
300 ω0
(ii) At resonance circuit impedance is pure resistance given by Z =R +r =V
=25
Io
=50 Ω
0.5
Where r =Internal resistance of the inductor.
R =50 –R =50-20 =30 Ohms
At resonance XL =XC = 300 ohms
L= 300 = 300
ω0
= 0.119 henry
2π x 400
100
0.02H
0.02F
Example 3:
A Series RLC circuit consists of R= 100 ohms
L= 0.02H and C = 0.02 μF. Calculate frequency
of resonance. A variable frequency sinusoidal
voltage
of
rms
value
of
50
volts
is
applied
to



the
5
0V
circuit. Find the frequency at which voltage across L and C is maximum
Solution:
Given R=100 ohms,L =0.02H,C=0.02μF
(i) Resonant frequency fo = 1
________ =
2π √LC
1
——————
2π √0.02x0.02x10-6
fo = 7.957 KHz
(ii) The frequency at which voltage across C is maximum is given by
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
fC = 1 /2π √ 1/ ( LC – R2 / 2 L2 )
= 1 /2π √ (1/ 0.02 X 0.02 X 10-6 ) – ( 1002 ) / 2( 0.002)2
= 7.937 KHz
(iii) The frequency at which voltage across L is maximum is given by
fL =
1
2π √ LC – (R2 C2 / 2)
= 1 /2π √ (0.02 X 0.02 X 10-6 ) – ( 1002 ) ( 0.002 x 10-6)2 /2
= 7.977 KHz
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Parallel Resonance
A parallel resonant circuit is one in which a coil and a capacitance are connected in parallel
across a variable frequency A.C. Supply. The response of a parallel resonant circuit is
somewhat different from that of a series resonant circuit.
I .A coil in parallel with a pure capacitor
R
L
Consider the parallel circuit shown in the figure.
Let ZL be the impedance of the coil given by ZL =R +j ω L
Then YL=1 =
1
= R-j ω L
ZL R +j ω L
R2+ ω2L2
Similarly let ZC = -j /
I
C
ωC
1 = jωC
ZC
Total admittance of the circuit = Y= YL + YC
YC
=
Y = R-j ω L
R2+ ω2L2
V
+ jωC
+ jωC -
R
R2+ ω2L2
ωL
R2+ ω2L2
At resonance the impedance ( or admittance) of the circuit is purely resistive( or conductive)
.For this to be true the j part of equation (1) should be zero.
ω 0C - ω 0L
R2+ ω 02 L
ω 0C
=
=0
2
ω 0L
R + ω 02 L
2
R2 +ω0 2 L
ω0 2 =
2
=
L/C - R2
L2
2
L
C
= 1 LC
R2
L2
f0 = 1/ 2 √ (1/LC – R2 / L2 )
Impedance at resonance
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
We know that at resonance the susceptive part of the admittance is zero.
Hence Y0 =
R
R2 +ω0 2 L 2
But R2 +ω0 2 L
2
= L/C
So Y0 = RC/L or Zo = L/RC
Where Zo is called the dynamic resistance. when coil resistance R is small, dynamic
resistance of the parallel circuit becomes high. Hence the current at resonance is minimum.
Hence this type of circuit is called rejector circuit.
Frequency –response characterisitics
The frequency response curve of a parallel resonant circuit is as
shown in the figure. We find that current is minimum at resonance. The half –power points
are given by the points at which the current is √2 Ir .From the above characteristic it is clear
that the characteristic is exactly opposite to that of series resonant.
I
2Ir
Ir
Frequency
f1
fr
f2
Quality factor ( Q-factor)
The quality factor of a parallel resonant circuit is defined as the current
magnification
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Q = Current through capacitance at resonance
Total Current at resonance
= IC0 / I0
= V / ( 1/ 0C ) V / Z0
= Z00C = (L / RC) 0C
= 0L / R
Hence the expression for the Q- factor for both series and parallel resonant circuit are the
same
Also Band width= f0 / Q
II A coil and a Practical Capacitor in parallel
IL
Consider a parallel resonant circuit in which
the resistance of the capacitance is also considered
Impedance of the coil = ZL = RL + j L
YL = 1 / ZL = 1/ RL+j L
= RL – j L / RL2 +2L2
I
RL
XL
IC
RC
Impedance of the Capacitor = ZC = RC – j / C
= RC + j / C
XC
V
Rc2 +1/2C2
Therefore total admittance = Y=YL+YC
=( RL – j L / RL2 +2L2)+ RC + j / C
Rc2 +1/2C2
At resonance the susceptance part of the total admittance is zero, which gives
1/0C
0L
=
RC2 + 1/20C2
RL2 +20L2
1/LC [ RL2 +20L2] = 20[ RC2 + 1/20C2 ]
20 ( RC2 – L/C ) = RL2/ LC – 1/ C2
20 = 1/LC(RL2 – L/C )
( RC2 – L/C)
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
0 = 1/√ LC
f0 =
RL2 – L/C
√ ( RC2 – L/C)
( RL2 – L/C)
1
2 √ LC
( RC2 – L/C)
At Resonance the admittance is purely conductive given by
Y0 =
RL
RC
+
RL2 +20L2
RC2 + 1/20C2
Example 1 : Determine the value of RC in the
Shown in figure to yield Resonance
Solution: Let ZL be the impedance of the inductive branch
then ZL = 10+ j 10
RC
10
YL = 1/ (10 +j 10)
= 10 – j 10 = 10- j 20
102 +102
200
Let ZC be the impedance of the capacitive branch then
-j2
j10
ZC = RC – j 2
YC =
1
RC – j 2
=
RC – j 2
R C2 + 4
Total admittance of the circuit = Y = YL + YC
For the circuit under Resonance the Susceptance part is zero
( 2/ RC2 + 4) - (10 / 200) = 0
RC2 = 36
25
25
mH
RC = 6 ohms
Answer
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Example 2: An Impedance coil of 25 ohms
Resistance and 25 mH inductance
is connected in parallel with a variable
capacitor. For what value of Capacitor
will the circuit resonate.If 90 volts,400
Hz source is used, what will be the line
Current under these conditions
Volts,400Hz
Solution:
C
90
0 = 2f0 = 2( 400)
ω0 2 = 1 LC
6.316 x 106 = 1 LC
R2
L2
R2
L2
1
= ω0 2 + R2
LC
L2
= 6.316 x 106 + 252 / (25 x 10-3)2
= 7.316 x 106
C= 5.467 F
Z0 = L/RC = (25 x 10-3)/ 25 x 5.467 x 10-6
= 182.89 ohms
I0 = V0/ Z0 = 90/ 182.89 = 0.492 ampere
Example 3: In a parallel resonant circuit
R, L and C are all in parallel
Half – power frequencies are
R
L
C
103 and 118 rad / sec respectively
The magnitude of impedance at
105 rad/ sec is 10 ohms.Find R,L and C
Solution : The given circuit is an ideal parallel resonantCircuit.
Total admittance= Y = YR + YL + YC
YR = 1/R
YL = 1/ j XL
YC = 1/ -j XC
So Y = 1/R + j (1/ XC - 1/ XL)
At Resonance the j part of the equation is zero
(1/ XC - 1/ XL)=0
XC= XL
1/ 2f0C =2f0L
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Or f0= 1/ 2√ LC
Q= IC / I or IL / I is the current magnification
At Resonance Y0 = 1/ R or Z0 =R
Since (2 - 1) = 0/ Q
118 – 103 = 105 / Q
Q= 7
But Q= 0L / R
7= 105 x L / 10
L= 0.67 Henry
0= 1/√ LC
105= 1 / √ 0.67 x C
C=1.35 x 10-4 farard
Effect of Generator resistance on Band width and Selectivity
Rg
L
Ig
V
C
Rg
ZL
R
Consider parallel resonant circuit driven by a generator having internal resistance
Rg as shown in figure.
At anti- resonance the parallel resonant circuit behaves like a resistive network
Zr = L / CR
Q = rL / R = 1 / rRC
And B.W= f2 – f1= fr / Q
But now as the parallel circuit is driven by a generator having internal
resistance(Rg)
Which comes in parallel with Zr at resonance.
Hence the total impedance becomes Zr’ = Rg11 Zr
= Rg11( L / CR)
Rg L
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
=
= L / C( R+ L/ CRg)-----------(1)
RgCR+L
From equation (1) it is clear that driving a parallel resonant circuit with a
Generator having internal Rg is nothing but to increase a resistance in series
With an inductor by value L /RgC
Hence Quality factor of parallel circuit driven by generator with internal
Resistance Rg is given by
Q’ = rL
R+ L / RgC
= ( rL / R)
( 1+ L/CR)
Rg
Q’ =
Q0
( 1 + Zr / Rg)
Bandwidth = B.W= fr / Q’ = fr ( 1+ Zr / Rg )
Q
’
B.W = B.W ( 1+ Zr / Rg)
100 k
Example 4: A coil of 10 henry and Resistance
of 10 ohm is in parallel with 100
picofarad capacitor.The combination
10 H
is connected across a generator of 100
volts, having internal resistance 100 k
100
Volts
100
Determine (i)Voltage across the ckt
picofarad
10
(ii)Band-width
ohm
Solution:
fr = 1/ 2 √ (1/LC – R2 / L2 )
= 1/ 2√( 1 / 10 x 100 x 10-12 )– (102 / 102)
= 5.032 khZ
Dynamic resistance =Zr= L / CR
= 10 / 100 x 10-12 x 10 = 1010 ohms
Q=rL / R
= 2 x 5.032 x 103 x10
10
= 31617
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
100 k
100 V
Zr
(i)Voltage across parallel circuit at Resonance = 100 x 1010
1010 + 100 x103
= 100 volts
(ii) Band width = fr / Q [ 1+ Zr / Rg)
5.032 x 103 { 1 + 1010 / 100 x 103 }
31617
= 15.91 KHz
=
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
TRANSIENT RESPONSE
Electrical circuits are connected to supply by closing the switch and disconnected
from the supply by opening the switch. This switching operation will change the
current and voltage in the device.
A purely resistive device will allow instantaneous change in current and voltage.
An inductive device will not allow sudden change in current or delay the change in
current.
A capacitive device will not allow sudden change in voltage or delay the change in
voltage.
Hence when switching operation is performed in inductive or capacitive device the current
and voltage in the device will take a certain time to change from preswitching value to steady
value after switching. This study of switching condition in network is called transient
analysis. The state (or condition) of the current from the instant of switching to attainment of
steady state is called transient state or transient. The current and voltage of circuit elements
during transient period is called transient response.
The transient may also occur due to variation in circuit elements. Transient analysis is an
useful tool in electrical engineering for analysis of switching conditions in Circuit
breakers, Relays, Generators etc.
It is also useful for the analysis of faulty conditions in electrical devices. Transient analysis is
also useful for analyzing switching Conditions in analog and digital Electronic devices.
R
R-L Series circuit transient:
Consider The R-L series circuit shown in the fig. Switch K
is closed at t=0. Referring to the circuit, balance equation
ldit 
using Kirchoff’s law can be written as V t   Ri t  
dt
Taking Laplace Transform we get
V s
 Is  R  LSIs  i0 
s
Assuming there is no stored energy in the inductor
I(0)=0
V s
 RIs  LSIs
s
K
t=0
L
i(t)
V
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore




V s
V s 
1
Is 



R
SR  SL
L  
S S  
 
L  
A
B
Is  
S SR
L
R
V s

A  S    BS 
L
L

 R  V s 
A  
Put s=0
L
L
V s
A
R
V s 
R
 R
S
B
put
 
L
L
 L 
 V s
B
R


V s  1
1 
Therefore Is 
 

R S S  R 
L

Taking inverse Laplace we get
R
  t 
V 
it  
1 e  L  
R



The equation clearly indicates transient nature of current, which is also shown in
figure.
L
 Tune constant of the circuit, which is denoted by Z given in seconds.
Where
R
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Hence
it  
V
1  e
R 
Z 
t


V
V
Where
= steady state current. Hence Time
R
R
constant for an R-L series current circuit is defined as the time taken by the circuit to
reach 63.2% of its final steady value.
Putting t=z we get
i(z) = 0.632
R-C series circuit Transient
Consider the RC circuit shown. Let the switch be closed at t=0.
Writing the balance equation using Kirchoff’s voltage law ,
1
v t   iR   i dt
c
Taking Laplace transform, we get
V s
1  Is Q 0  
 IsR  


S
c s
s 
Let us assume that there is no stored energy in the circuit.
Q o  =0
Hence
V s
Is
1 

 IsR 
 IsR 

s
CS
CS 

 
 


V s 
1 


Is 
R S 1 


RC 

Taking Laplace inverse we get
 1 
V t   RC  t
it  
e
R
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
The sketch of transient current is shown in figure Where
the circuit. Putting  
1
  the time constant of
RC
1
in the current equation we get
RC
V
R
Hence time constant of RC series current can be defined as the time taken by
current transient to fall to 36.7% of its initial value.
i(z) = 0.367
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Example 1:
In the circuit shown in figure the switch ‘K’ is moved from position 1 to position 2 at
time t = 0. The steady state current having been previously established in R-L circuit.
Find the current i(t) after switching.
Solution:
From the given data the circuit is under steady state when switch K is in position 1
10
under steady state condition inductance is a short and hence i(0) =
= 1 Amp.
10
When the circuit is switched to position 2, this 1 Amp current constituted the stored
energy in the coil.
Writing the balance equation for position 2 we get 20i  4
di
0
dt
Taking Laplace transformation
20Is  4s Is  i0  0
20Is  4s Is  1
4
1
Is 

4s  5 s  5
taking inverse Laplace we get
it   e5 t
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Example 2:
A series R-C circuit is shown in figure. The capacitor has an initial charge of
800µCoulombs on its plates, at the time the switch is closed. Find the resulting
current transient.
Solution: From the data given q(0) = 800  106 C
Writing the balance equation we get
1
100  10it  
 i(t )dt
4  106
Taking Laplace transformation
100
1
I(s)  Q(0)
 10I(s) 
S
4  10 6 S

106  100 800  10 6
I(s)10 


4S 
S
4  10  6 S

100  200
5
 40S  106  30
I(s)

4S

 S
1200
1200
I(s) 

6

40S  10
106 


40 S 

40



Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
30
S  25000
Taking Inverse Laplace we get
i( t )  30e25000 t
I(s) 
Example3:
For the circuit shown in figure the relay coil is adjusted to operate at a current of 5
Amps. Switch K is closed at t = 0 and the relay is found to operate at t = 0.347
seconds. Find the value of inductance ‘L’ of the relay.
Soln: Writing the balance equation for the relay circuit
di
V( t )  Ri( t )  L
dt
Applying Laplace transformation
V( s)
 RI(S)  LSI(S)  i(0)
S
Since there is no mention of initial current in the coil i(0) =0
10
 I(s)  I(s)LS
Hence
S
10
I(S)SL  1 
S
10
10
A
B
L
I(s) 

 
1 S 
1
S1  SL

S S  
S  
L
L


Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
10
1

 A S    BS
L
L

A=10 B= -10
10
I(S)  10 
1
S
L
Taking Inverse Laplace we get

t
L
i( t )  10  10e
The relay operates at t = 0.347 seconds when the current value reaches 5A. Hence
5  10  10e
0.347
10e L
0.347
L
 10  5  5
0.347
5
e L
Solving the equation we get
L=0.5H
Example 4:
In figure the switch ‘K’ is closed. Find the time when the current in the circuitry
reaches to 500 mA
Soln: When the switch is closed Vc (0) = 0
When the switch is closed at t = 0
I1 (t)×50 = 10
I2 ( t )  70 
1
 10
100  10 6  i2dt
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Taking Laplace for both the equations
10
0. 2
I1(S) 

       (1)
50S
5
1 I (s) 10
I2 (S)  70  2

c S
5
I2 (S)
10
70I2 (s) 

6
100  10 s 5
10
1
1
1
I2 (S) 

 '
(2)
4
3
70S  10
7s  10 7 S  142.86
Taking inverse Laplace for equation (1) and (2)
I1 (t) =0.2 A
1
I2 ( t )  e 142.86 t
7
Total current from the battery i(t) =I1 + I2
1
i( t )  0.2  e142.86 t
7
when this current reaches 500 mA
1
500  10 3  0.2  e142.86 t
7
-3
Solving we get t = 5.19 × 10 Seconds.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
R-L-C Series Transient circuit:
Assuming zero initial conditions when switch K is closed the balanced equation is
di 1
given by V  iR  L   idt
dt C
Taking Laplace transformation we get
V ( s)
I(s)
 1(s)R  LSI(s) 
s
CS
1 

 I(s)R  SL 
CS 

V( s)
V ( s)
L
I(s) 

1
R
1
S(R  SL 
) S2  S 
CS
L
LC
The time response of the circuit depends on the poles or roots of the characteristic
equation
R
1
S2  S 
0
L LC
Roots of the characteristic equation are given by
2
R
1
R
     4
L
LC
L
S1, S 2 
2
2
R
1
R
S1, S2  
   
2L
 2L  LC
2
1
R
Case1: If   
the roots S1 and S2 are real positive and unequal.
LC
 2L 
The response of such a case is called is known as over damped.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
2
1
R
Case 2 : If   
the root S1 and S2 are complex conjugate the response of
LC
 2L 
such a case is called as undamped.
2
1
R
Case 3 : If   
then the roots S1 and S2 are complex conjugate. The
LC
 2L 
response of such a case is called as under damped.
Example 5 :
A circuit is made up of an inductance of 10 H, a resistance of 2000 Ω and a capacitance
of 15.62 µF all connected in series. A voltage of 100 Volts is suddenly applied across this
circuit. Calculate the circuit current. What is its maximum Value?
Soln : At t = 0 switch ‘K’ is closed. There is no initial inductor current and capacitor
voltage.
di( t ) 1
it R  L
  it dt  100
dt
c
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Taking Laplace transformation we get
106
100
I(s)2000  10SI(s) 

15.625
s
 31240s  156.252  106  100
I(s)

15.625
s


10
1562
I(s) 
 2
2
156.2 s  200s  6402
S 200s  6402
2
S + 200s +6402=0 is the characteristic equation

S1 
 

 200  2002  4  6402
 40
2
 200  2002  4  6400
 160
2
S2  200S  6402  S  40S  160
10
K1
K2
I(s) 


S  40S  160 S  40 S  160
K1=0.0833 and K2=-0.0833
1 
 1
I(s)  0.0833 


 S  40 S  160 
Taking inverse Laplace
i( t )  0.0833 e40 t  e160 t
S2


To find the maximum value of i(t)



di d

0.0833 e 40 t  e160 t
dt dt
di
0
for i(t) to be maximum
dt
0.0833 [ 40e 40 t  160e 160 t ]  0
e40t  4e160t  0
e40t
 4  e120t  4
160 t
e
Solving we get t = 0.0166 seconds
Substituting this value of ' t ' in the current equation we get
imax  0.0833 [e400.0166  e1600.0166 ]
imax = 0.04A
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
(*) In the R-L-C circuit the capacitor has an initial voltage of 40 Volts when the switch
is closed at t=0 find an expression for the current.
Solution : Writing the balance equation for the circuit
1 
40  100
2I(s)  10sI(s) 
I
(
s
)


4S 
S 
S
1  100 40

I(s)2  10S 


4S 
S
S

60
S
I(s) 

6
S  0.25  0.025
1
4S
Hence the characteristic equation is
S2+0.25+0.025=0
2  10S 
2
 0.2  0.22  4  0.025
 0.1  j0.1225
2
Since the roots are complex conjugate the current i(t) will be damped sinusoid.
6
I(s)  2
S  2  0.15  0.12  0.025  0.12
6
6


2
2
S  0.1  0.015 S  0.1  0.015 2
6
6
0.1225



2
2
S  0.1  0.1225 0.1225 S  0.1  0.12252
S



 48.96 

0.1225
S  0.12  0.12252
Taking inverse Laplace we get
I(t)=48.96 sin 0.1225t e- 0.1t.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Initial conditions
The reason for studying initial and final conditions in a network is to evaluate the arbitrary constants that appear in the general solution of
the differential equations written for the network.
In this chapter we concentrate on finding the change in selected variables in a circuit when a switch is thrown from open to
closed or vice versa position. Please note that t = 0 indicates the time of throwing the switch
t = 0- indicates time immediately before throwing the switch and
t =0+ indicates time immediately after throwing the switch.
We are very much interested in the change in currents and voltages of energy storage elements (inductor and capacitor) after the
switch is thrown since these variables along with the sources will dictate the circuit behavior for t > 0.
Initial conditions in a network depend on the past history of the circuit (before t= 0-)
and structure of the network at t = 0+.Past history will show up in the form of capacitor
voltages and inductor currents.
Initial and final conditions in elements
i)
The resistor:
The cause effect relation for the ideal resistor is given by v = Ri. From this equation
we find that the current through a resistor will change instantaneously, if the voltage
changes instantaneously.Similarly voltage will change instantaneously if current
changes instantaneously.
ii)
The inductor :
K
L
V
Initial condition
The switch is closed at t= 0
t
The expression for current through the inductor is given by i(t) = 1
∫v dt
L -∞
0-
t
= 1/L ∫ vdt + 1/L ∫ vdt
-∞
t
0-
= i(0-) + 1/L ∫ vdt
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
0Putting t = 0+
0+
i (0+) = i (0-) + 1/L ∫ vdz
0i (0+) = i (0-)
The above equation indicates that the current in an inductor can not change
instantaneously. Hence if i (0-) =0, then i(0+) = 0. This means that at t = 0+ inductor
will act as an open circuit, independent of voltage across the terminals.
O.C
L
If i (0-) = I0 (i. e. if a residual current is present) then i (0+) = I0 , meaning that an
inductor at t = 0+ can be thought of as a current source of I0 which is as shown
I0
II
I0
L
Final (or steady state) condition
The final –condition equivalent circuit of an inductor is derived from the basic relation
ship V = L di/ dt
Under steady state condition di = 0 which means v = o and hence L acts as a short
dt
S.C
at t = ∞ ( final or steady state)
I0
I0
L
S.C
At t= ∞
iii)
The capacitor
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
The switch is closed at t = 0 . The expression
At t=0
For voltage across the capacitor is given by
t
v
i(t)
C
v = 1/ C ∫ i dt
-∞
0-
t
v(t) = 1/ C ∫ i dt + 1/ C ∫ i dt
0-
-∞
Putting t= 0+
0+
v(0+ ) = v(0-) + 1/C ∫ i dt
0V(0+)
= V (0-) which means that the voltage across the capacitor can not change
instantaneously. If V(o-) = o then V (o+) = o
indicating that the Capacitor acts as a
short at t=0+
S.C at t=0+
C
-+
V0= Q0 / C
V0
Final (or steady state ) condition
The final-condition equivalent network is determined from the basic relationship
i = C dv/dt
Under steady state condition dv / dt = 0 which means at t= ∞ the Capacitor acts
as a open circuit.
C
O.C
V0
-
+ V0
-
-+
-
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
O.C
Example 1 : In the given circuit ‘ k ‘ is
K
R=8
Closed at t=0 with zero current in the inductor
R=8
±
Find the values of i , di/dt ,d2i / dt2 at t=0+
12V
L= 0.2H
Solution : The symbol for the switch implies that it is open at t = 0- and then closes at
t = 0+.From thedata given it is also clear that i (0-) = 0. Hence from the circuit i(0+) = i(0-) = 0
To find di (0+) and d2i (0+)
dt
dt2
K
R
O.C
Applying KVL to the circuit we get
R i (t) + L di (t) = 12
dt
8 i (t) + 0.2 di (t) = 12
dt
i(0+)
±
……….(1)
K
R
At t = 0+
i(t)
8 i (0+) + 0.2 di (0+) = 12
dt
8 x 0 + 0.2
±
12V
L
di (0+) = 12
dt
di (0+) = 12 = 60 A/sec
dt
0.2
Differentiating equation (1) with respect to time
8 di + 0.2 d2i
dt
dt2
= 0
At t = 0+ the above equation becomes
8 di (0-) + 0.2 d2i (0+) = 0
dt
dt2
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
8x 60 + 0.2 d2 i(0+) = 0
d t2
Hence d2 i (0+) = - 2400 A / Sec2
dt2
Example 2:
K
t =0
In the network shown in figure the switch is
closed at t = 0 determine i, di and d2 i at t =0+
dt
dt2
R =10
L =1H
10V
±
C=1F
At t = 0+ The circuit appears as shown in figure. From the circuit i (0+) = 0 = i (0-)
O.C
10
±
10V
S.C
Writing KVL clock wise for the circuit when the switch is closed, we get
t
Ri + Ldi +1 ∫ i(t) dt =10 …………..(1)
dt C
Ri + Ldi +VC (t) =10 …………..(2)
dt
Putting t =0+ in equation (2), we get Ri (0+)+ Ldi(0+) +VC(0+) =10
dt
0+L di(0+) +0=10
dt
di(0+) = 10 =10 = 10A/sec
dt
L 1
Differentiating equation (1) with respect to time, we get, R di + L d2i +i(t ) =0
dt
dt2 C
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
At t= 0+ the above equation becomes, , Rdi (0+)+ Ld2i(0+) +i(0+) =0
dt
dt2
C
10R +Ld2i (0+) +0 =0
dt2
Hence d2i (0+) = -10 x10
dt2
1
= -100A/Sec2
10
Example :3
1
Refer the circuit shown in figure . The switch K is
20V
2
changed from portion 1 to position 2 at t = 0
1 F
±
steady – state condition having been reached in position 1.
Find the values of i, di and d2i at t = 0+
dt
dt2
K
Solution ; The symbol for switch K implies that it is in
position 1 at t = 0 - and in position 2 at t = 0+ .
Under steady – state conditions, inductor acts as a short – circuit.
1H
K 10
Hence at t = 0 - the circuit diagram appears as shown in figure
-
i(0 ) = 20 = 2A
10
±
i =(0-)
20V
S.C
Since the current through an inductor can not change instantaneously
i (0-) = 20 = 2A
10
Since the current through an inductor can not change instantaneously
i (0+) =
i (0-) = 2A
Since there is no initial change on capacitor
Vc (0-) = 0
Since the voltage across a capacitor can not change instantaneously
Vc (0+) = Vc (0-) = 0
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Hence at t = o+ the circuit appears as shown
1
10
2
2A
S.C
i (0+)
For t > o+ the circuit diagram is as shown
10
1F
i(t)
1H
Applying KVL to the above circuit we get
Ri (t) + L di (t) +1 ∫ i(t) dt =0
dt
C
Ri (t) + L di (t) + VC= 0…………(3)
dt
At t =0+ Ri(0+)+L di(0+) +VC(0+) =0
dt
+
2R +L di(0 ) +0 = 0
dt
di(0+) = -2R/L = -20 A/sec
dt
Differentiating equation(3) with respect to time we get, R di +L d2i + i = 0
At t =0+ Rdi(0+) +L d2i(0+) + i(0+) = 0
dt
dt2
C
2
dt
dt
C
R(-20)+ L d2i(0+) + 2 = 0
dt2
C
20
d2i(0+) = -2 x 106 A/sec
dt2
1
40V
2
Example 4 : In the network the switch is moved
±
From position 1 to position 2 at t =0.The steady state
1H
1F
having reached before switching.
Calculate i , di/dt,d2i/dt2 at t = 0 +
Solution:The symbol for switch K implies it is in
position 1 at t = 0-and position 2 at t = 0 +.Under
steady state condition a capacitor acts as an open circuit.
K
1
20
40V
Hence at t = 0-,the circuit diagram is as shown in figure.
We know that the voltage across a capacitor cannot change
instantaneously. This means that VC(0+) = VC(0-) =40Volts
Also i(0-)=0.Hence i(0+)=0.
±
VC(0-) =40V
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
20
+
The circuit diagram for t= 0 is as shown.
For t≥0+ the circuit diagram appears as shown.
t
Applying KVL we get, Ri(t) +Ldi(t) +1 ∫V(t)dt = 0
dt
C 0+
+
VC(0+) =40V
……(1)
Ri(0+) +Ldi(0+) +VC(0+) = 0
dt
i(0+)
-
R
20x0 + 1x di(0+) +40 = 0
dt
+
di(0 ) =- 40 A/Sec
dt
i(t)
C
L
Differentiating equation (1) with respect to time we get,Rdi +L d2i + i =0
dt
dt2
C
At t =0+ Rdi(0+) +L d2i(0+) + i(0+) =0
dt
dt2
C
2
+
R(-40)+ L d i(0 ) + 0 = 0
dt2
d2i(0+) = 800 A/sec2
dt2
Example 5: The Network shown in the figure has the switch
K opened at t = 0.Solve for v, dv and d2v at t = 0+
dt
dt
dt2
1A
100
K
1H
Solution: Initial condition for Inductor is open circuit and hence the
Circuit at t = 0+ reduces to the form shown in figure.
V(0+) = IR = 1x100 =100Volts
O.C
K
Consider the given circuit. Let ‘V’ be the node voltage.
Writing KCL for the node, V +L dv = 1………(1)
R
dt
V
At t =0+, V(0+ )+L dv( 0+ )= 1
R
dt
I
R
L
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
L dv( 0+ )= 1- V(0+ ) =1- 100
dt
R
100
dv( 0+ )= 0
dt
Differentiating equation(1) we get,1 dv +L d2v = 0
R dt
dt2
+
2
+
At t = 0 , 1 x 0 +L d v(0 ) = 0
R
dt2
Hence d2v(0+) = 0
dt2
Example 6 :In the network shown the switch ‘K’ is I
opened at t =0.At t= 0+,solve for the values of V,
K
dv and d2v if I =10A,R =100 ohms and C= 1μF.
dt
dt2
C
R
V(0+)
Solution:
S.C
At t =0+,switch K is opened.Equivalent circuit at
t = 0+ is as shown.Since capacitor is a short at
t =0+we get V(0+) =0
Consider the given network.Let’ V’ be the node
voltage.
Applying KCL we get, V +C dv = I ……..(1)
R
dt
V
At t = 0+, V(0+) + C dV(0+) =10
R
dt
R C
IK
dV(0+) =10
= 107 V/Sec
dt
1x10-6
Differentiating the equation(1) with respect to time, we get
1 dv + C d2V = 0
R dt
dt2
At t =0+, 1 dv (0+)+C
d2 V(0+) = 0
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
R dt
dt2
1/R x 107 +C d2 V(0+) =0
dt2
d2 V(0+) = -1/CR x 107 = -1010 V/sec2
dt2
****************************************
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
LAPLACE TRANSFORMATION
Laplace transform is a very useful and powerful tool in circuit analysis.Integro-differential equations
Can be transformed in to algebraic equations using the technique of Laplace transformation and
complete solution involviong both natural response and forced response is obtained in one step
Definition of Laplace Transform :
Let f(t) be a function of time.Assuming the value of function to be zero for t<0,the
Laplactransform of f(t) is given as
∞
L [ f(t) ] = F(S) = ∫ f(t) e-St dt
0
f(t) is a function in time-domain and F(S) is a function in complex frequency domain. Complex
frequency ‘S’ is given by S= ¬ +j
From the above it is obvious that Laplace transformation changes a function in time domain into
a function in frequency domain.
Important properties of Laplace transform
1)
Linearity Property:
If L { f1(t) }= F1(S)
And L{ f2(t) }=F2(S)
Then L {a1 f1(t) }+ a2 f2(t)}= a1 F1(S)+ a2 F2(S)
∞
Proof:
L {a1 f1(t) }+ a2 f2(t)}=∫ {a1 f1(t) }+ a2 f2(t) }e-St dt
0 ∞
∞
= a1 ∫ f1(t) e-St dt +a2 ∫ f2(t) e-St dt
0
0
= a1 F1(S) + a2 F2 (S)
2) Time-Shifting Property:
u(t- to)
If L{ x(t) } = X(S) then for any real number t0
L{ x(t- t0) u(t- t0)}= e-to S X(S)
t = to
time
∞
Proof: Let L{ x(t-to) u(t- to) } = ∫ x(t- t0) u(t- t0)}e-St dt
0
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Since u(t- to)= 1 }t>t0
0 }t<t0
∞
We get L [ x(t-t0) u(t-to)] = ∫ x(t- t0)e-St dt
to
Let t=t0 + τ dt = dτ
As t
t0 τ
0
As t
∞
τ
∞
∞
Hence we get L { x(t-t0) u(t-to)}= ∫ x(τ) e-S (τ + t0) dτ
0
∞
= e-Sto ∫ x(τ) e-S τ dτ
0
= X( S) e-Sto
3) Frequency-domain shifting property
If L{x(t)}= X(S) then
L{eSot x(t) }= X(S-So)
∞
Proof: L{eSot x(t) }= ∫ eSot x(t) e-St dt
0
∞
= ∫ x(t) e-( S-So)t
0
= X( S- S0)
4. Time-Scaling Property
If L[ x(t) ] = X(S) then
Proof:
L [ x(at) ] = 1/a X( S/a)
∞
L [ x(at) ] =
∫ x(at) e-St dt
0
Put at= τ
a dt = d τ
∞
Hence L [ x(at) ] = ∫ 1/a x(τ) e- Sτ / a d τ
0
∞
Hence L [ x(at) ] = 1/a ∫ x(τ) e- Sτ / a d τ
0
L [ x(at) ] = 1/ a X( S/ a)
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
5. Time-Differentiation Property
If L[ x(t) ] = X(S)
L[dx/dt] = S X(S) – x(0)
Proof: Let y(t) = dx/dt
∞
Then L[ y(t)] = Y(S) = ∫ y(t) e-St dt
0
∞
= ∫ (dx/dt) e-St dt
0
∞ ∞
= [ e-st x(t) ] - ∫ x(t) { -se-St}dt
0 0
∞
= [ x(∞) – x(0) ] + s ∫ x(t) e-St dt
0
Y(S)= SX(S) – x(0)
i.e.
L[dx/dt] = S X(S) – x(0)
6.Time-integration Property
∞
For a Causal signal x(t) , if y(t) = ∫ x(τ) d τ Where ‘τ ‘is a dummy variable of ‘t '
0
Then L[ y(t)] =Y(S) = X(S)
S
∞
Proof
L{ x(t) }= X(S) = ∫ x(t) e-St dt
0
Dividing both sides by S yields
∞
X(S)= ∫ x(t) e-St dt
S
0 S∞
∞ ∞
= e-St ∫ x(t)dt - ∫ [∫ x(t) e-St(-s) dt]
S
0
0 S
∞
= e-St y(t) - ∫ y(t) e-St (-s) dt
S
0 S
= e-St y(t) - Y(S)
S
X(s)= Y(S)
S
as e-St y(t) =0 at t=∞
S
Therefore Y(S) = X(S)
S
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
7. Time-Periodicity Property
x(t)
time
T
2T
3T
4T
x1(t)
0
T
x2(t)
T
2T
x3(t)
2T
3T
Let us consider a Function x(t) that is periodic as shown in figure.The function
x(t) can be represented as the sum of time-shifted functions as shown in figure.
Hence x(t) = x1(t) + x2(t) + x3(t) + …………………
Where x2(t) = x1(t- T) u(t-T)
x3(t) = x1(t- 2T) u(t-2T)
……………………………
…………………………… and so on
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
Hence x(t) = x1(t) + x1(t- T) u(t-T)+ x1(t- 2T) u(t-2T) + ………………..
Where x1(t) is the waveform described over the first period of x(t).
Taking Laplace transformation on both sides of the above equation we get
X(s) = X1(s) + X1(s) e-TS + X1(s) e-2TS + X1(s) e-3TS + ………………
= X1(s)[1+ e-TS + e-2TS +e-3TS + ………………]
But 1+ a2 +a3 + a4 + ……………………..= 1
1-a
Hence we get X(S) = X1(S)
1
1- e-TS
For a< 1
Initial –value Theorem:
The Initial -value theorem allows us to find the initial value x(0) directly from the Laplace
Transform X(S).If x (t) is a casual signal, then x(0) = Lim S X(S)
S
∞
Proof: WE have L dx(t) = S X(S) –x(0)
dt
∞
∫ (dx/dt) e-St dt
0
= S X(S) –x(0)
Taking Limit on both sides as S
∞ ,we get
∞
Lim
∫ (dx/dt) e-St dt = Lim S X(S) –x(0)
S
∞ 0
S
∞
Hence 0 = Lim S X(S) -x(0) or Lim S X(S) = x(0)
S
∞
S
∞
Final Value theorem
The final value theorem allows us to find the value of x(∞) directly from its Laplace
Transformation X(S)
If x(t) is a casual signal, Lim x(t) =Lim S X(S)
t
∞
S 0
Proof:
We have L dx(t) = S X(S) –x(0)
dt
∞
∫ (dx/dt) e-St dt = S X(S) –x(0)
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
0
Taking the limits S
0 on both sides we get,
∞
Lim [ S X(S) -x(t)] = Lim ∫ (dx/dt) e-St dt
S
0
S 0
∞
∫ (dx/dt) Lim
0
S
e-St dt
0
= S X(S) –x(0)
∞
∞
= ∫ (dx/dt) dt = x(t)
0
0
Lim S X(S) –x(0) = x(∞ ) –x(0)
S
0
x(∞) = Lim S X(S)
S
0
Inverse Laplace Transformation
The inverse Laplace Transform of X(S) is defined by an integral operation with respect to
Variable ‘S’ as follows
¬+∞
x (t) = 1/ 2 ∫ X(S) eSt ds ……………(1)
¬-∞
Since ‘S’ is a complex quantity the solution requires a knowledge of complex variables.In
Other words the evaluation of integral in equation (1) requires the use of contour integration
In the complex plane, which is very difficult.Hence we will avoid using equation(1) to compute
Inverse Laplace transform.We go for indirect methods to get the inverse Laplace transform of
The given function,which are
i)
Partial – Fraction method
ii)
Convolution integral method
Partial – Fraction method:
In many situations,the Laplace transform can be expressed in the form
X(S) = P(S)
…………………(2)
Q(S)
Where P(S)= bmSm + bm-1Sm-1+ bm-2Sm-2 ………………..+ b0
Q(S)= an Sn + an-1 Sn-1 + an-2 Sn-2 +………………..+ a0
The function X(S)as defined by equation(2) is said to be rational function of ‘S’,since
It is a ratio of two polynomials.The denominator Q(S) can be factored in to linear factors.
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
A partial fraction expansion allows a strictly proper rational function P(S) to be expressed
Q(S)
As a factor of terms whose numerators are constants and whose denominator corresponds to
Linear or a combination of linear and repeated factors.This in turn allows us to relate such terms
To their corresponding inverse Laplace transform.
For performing partial fraction technique on X(S) the function X(S) has to meet
the following conditions.
i) X(S) must be a proper fraction i.e. m< n . When X(S) is improper we can use long division
to reduce it to proper fraction.
ii) Q(S) should be in the factored form.
Convolution-integral method:
If L { x(t) } = X(S)
L { h(t) } = H(S)
Then L { x(t) * h(t) }= X(s) H(S)
Where ‘* ‘ is the convolution of two functions given by
∞
x(t) * h(t) = ∫ x ( τ) h( t-τ ) d τ
-∞
∞
Proof: { x(t) * h(t) }= ∫ x ( τ) h( t-τ ) dτ
-∞
Taking Laplace transformation we get
∞ ∞
Proof: L { x(t) * h(t) }= ∫ [ ∫ x ( τ) h( t-τ ) dτ ] e-St dt
0 0
Interchanging the order of integration we get
∞
∞
L { x ( τ) h( t-τ ) } = ∫ x ( τ)[ ∫ h( t-τ ) e-St dt ] d τ
0
0
Let t- τ =  dt = d
∞
∞
L { { x(t) * h(t)} = ∫ x(τ ) [ ∫ h( ) e-S e-Sτ d ] dτ
0
0
∞
∞
= ∫ x(τ ) e-Sτ d τ ∫ h( ) e-S d
0
0
L { { x(t) * h(t)} = X(S) H(S)
Three important Singularity functios
The three important singularity functions employed in circuit analysis are
(i)
the unit step function u(t)
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
(ii)
the delta function (t)
(iii)
the ramp function r(t)
They are called Singularity functions because they are either not finite or they do not
posess finite derivative everywhere.
u (t)
Unit step function:
The Unit-step function is defined as u(t) = 0 t< 0
=1 t>0
The step function can have a discontinuiy
For example in sequential switching .The unit
Step function that occurs at t=a is expressed as
u(t-a) which is expressed as
t
u(t-a)
u( t-a ) = 0 (t-a )< 0
= 1 (t-a)> 0
t
We use step function to represent an abrupt change in voltage or current , like the changes that
Occur in the circuit of control engineering and digital systems.
∞
Laplace transformation of unit step function is given by L { u( t) } = ∫ e-St dt
0
= - 1/ S [ e-∞ - e0 ]
=
1
S
∞
Similarly L{ u (t-a) } = ∫ u ( t-a) e-St dt
a
∞
-St
=-1 e
= 1 e- a S
S
a
S
Impulse function:
The derivative of the unit step function is the unit impulse function (t)
i.e. (t) = d/dt { u(t) } = 0 t< 0
(t)
=1 t=0
=0 t>0
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
The unit impulse may be visualized as very short duration pulse of unit area
This may be expressed mathematically as 0+
∫ (t) dt = 1
0Where t= 0- indicates the time just before t=0 and t=0+ denotes the time
Just after t=0. Since the area under the unit impulse is unity, it is practice to
write
‘1’ beside the arrow. When the impulse has a strength other than unity the area
of the impulse function is equal to its strength.
Since (t) = d/dt { u(t) }
L { (t) } = L [d/dt { u(t) }] = S X 1 / S = 1
Ramp function:
Integrating the unit step function results in the unit ramp function r(t)
t
r (t) = ∫ u (τ ) dτ = t u ( t)
r(t)
-∞
= 0 t<0
= t t>0
In general a ramp is a function that changes at a
Constant rate.
A delayed ramp function is shown in figure
Mathematically it is described as follows
r(t-t0 ) = 0
0
t
t< 0
r (t- t0 )
= t – t0 t > 0
Laplace transformation of a ramp function is given by
t
L { r ( t ) } = L [ ∫ u ( t ) dt ]
0
t0
= 1 / S X 1 / S = 1 / S2
L { r ( t – t 0 ) } = 1 / S X 1 / S e-t0S
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore
t
= 1 e- toS
S2
Prof.R.V. Srinivasa Murthy, Assistant Professor, Dept. of E & C, A.P.S. College of Engineering
Bangalore