Network Topology

ECT2036 Circuits & Signals

Course Structure 1. 2. 3. 4. 5. 6. 7. 8.

Network Topology Intro to Signals App. of Fourier Series Frequency Domain Analysis Two-port Networks State Variable Analysis on RLC Networks Elements of Network Synthesis Filters

Assessment Final Midterm Asign. (group) Lab TOTAL:

60% 15% 15% 10% 100%

ECT2036: CIRCUITS & SIGNALS CHAPTER 1: NETWORK TOPOLOGY

CONTENTS:     

Introduction to Network Graph Network Graph Terminology Node and Mesh Analysis Tree Branch Selection Fundamental Loop & Cutset methods of analysis

NETWORK TOPOLOGY/GRAPH  Network Topology/Graph: The properties that relate to the geometry of the network (circuit).  These properties remain unchanged even if the circuit is bent into any other shape provided that no parts are cut and no new connections are made.  Knowledge about network topology can help one select good analysis procedures (mesh or node analysis) for a given electrical network.

NETWORK GRAPH TERMINOLOGY  GRAPH: a representation of a circuit where each branch is denoted by a line segment.

 ELEMENT: A fundamental unit; eg., a resistor, a capacitor, an inductor.  NODE: A point at which two or more elements are connected.  PATH: A set of elements that may be traversed in order without passing through the same node twice.

 BRANCH: A single path which connects one node to another node (via a single element).  LOOP: A path where the ending node is the same as the beginning node. This is also called a closed path.  MESH: a loop which does not contain any loops within it.  TREE (of a graph): a set of branches (each denoted by a line segment) that connects every node to every other node via some path without forming a loop.  COTREE: Those branches of a graph which are not part of a particular tree. Also known as the complement of the tree.

 LINK/CHORD: Branch that is not considered in a selected tree.  CUTSET: A minimum set of branches that when cut, will divide a graph into two separate parts.  FUNDAMENTAL CUTSET: A cutset containing only a single tree branch & minimum number of links.  FUNDAMENTAL LOOP: A loop that results when a link is put into the tree.

Example 1: Circuit & Graph There is only ONE graph for a circuit – though there may be many ways to draw it.

Example 2: Tree & CoTree TREE: The sub-graph 2. Is connected 3. Connect all nodes 4. DO NOT form loops

COTREE: Complement of a tree

There are SEVERAL trees for a graph, and each tree has its corresponding cotree.

TREE & CoTREE  Let B = number of branches in the graph N = number of nodes in the graph L = number of links  The min. number of branches in a tree Bt = N – 1  The number of links: L = B – (N – 1) = B – N + 1

FUNDAMENTAL CUTSET  Fundamental cutset: A set of branches containing one tree branch and min. no. of links, such that its removal makes the connected graph unconnected.  Total fundamental cutset: = total no. of tree branches 1 2 =N-1 Fundamental Cutset 1: {4, 1}

4

3

5

Fundamental Cutset 2: {3, 1, 2} Fundamental Cutset 3: {5, 2} Tree: Solid lines CoTree: Dotted lines

FUNDAMENTAL LOOP  Fundamental loop: Loop formed with the tree branches and link when one link is added to the selected tree at a time.  Total no. of fundamental loops = total no. of links/chords =B–N+1 Fundamental Loop I: {2, 3, 1} Fundamental Loop II: {4, 3, 5}

FUNDAMENTAL CUTSET & LOOP  Orientation of fundamental cutset: Follow the orientation of tree branch present in the set.

 Orientation of fundamental loop: Follow the orientation of link present in the set.

Example: In figure below, the branches of a tree T marked as continuous lines and its chords as doted lines. (i) List all f-loops of that tree, giving the branches present in each f-loop. (ii) List all the f-cutsets of that tree, giving the branches present in each cutset. 3 b

a

1

6

2 d

c

4 5

7 8

f

e

Solution: i. Loop – a path where the ending node is the same as the beginning node. Also called a closed path. F. Loop – loop formed with tree branches & link when 1 link (cotree) is added to the selected tree at a time. 3 a

b

b

1 2 d

c

1

6 4

8 f

b

7

4

5

a

c 7

4 e

f

fL2 = (2, 5, 4, 1) fL6 = ( 6, 7, 8, 4)

8 f f L3 = (3, 7, 8, 4, 1)

e

ii.

F. Cutset – a set of branches containing one tree branch & min. no. of links, such that its removal makes the connected graph unconnected. fC4

fC8

3

fC1

b

a

1

fC5

c

6

2 d

fC7

4 5 f

fC4 = (4, 2, 3, 6) fC5 = (5, 2)

7 8

fC1 = (1, 2, 3)

e

fC7 = (7, 3, 6) fC8 = (8, 3, 6)

AUGMENTED NODAL INCIDENCE MATRIX, A  For a connected graph of n nodes and b branches, the nodal matrix A will have dimensions n x b  Elements of A depend on the nature of the oriented graph.  The element aij of the matrix A can be defined as:

+ 1 if branch j comes out from node i  aij =  0 if branch j not at all touching node i − 1 if branch j is towards node i 

Example 3: Augmented Nodal Incidence Matrix  Graph: 4 nodes, 6 branches  (4 × 6) matrix A 1  Augmented Nodal Incidence Matrix, A

2

3

4

5

6

1  − 1 0 0 1 0 − 1   2  0 0 −1 0 −1 1  3  0 −1 0 −1 1 0     4  1 1 1 0 0 0  Node

Branc h

 If we select node 4 as reference, − 1 0 A = the matrix A change  3 × 6   0

0 0 1 0 − 1 0 − 1 0 − 1 1  −1 0 −1 1 0 

 Apply KCL for the graph − i1 + i 4 − i6 = 0 branch  current (i1, i2, i3) i3 − i5 + i6 = 0 − i 2 − i 4 + i5 = 0  i1   KCL equation in matrix form: i  0 − 1  2  − 1 0 0 1 i3  A i=0 0  0 − 1 0 − 1 1   = 0  i4    0 − 1 0 − 1 1  0 i 5    i6 

MESH INCIDENCE MATRIX  With the b no. of branches and possible l no. of loops in the graph, the mesh incidence matrix B will have dimensions l × b  The element bij of the matrix B can be defined as: + 1 if branch j is oriented in the direction of the loop i  bij =  0 if branch j does not appear in loop i − 1 if branch j is oriented opposite to that of loop i 

Example 4: Mesh Incidence Matrix  Graph: 3 loops, 6 branches  (3 × 6) matrix B

1

2

3

4 5

6

I  1 −1 0 1 0 0   Mesh Incidence   B = II  0 1 − 1 1 0 0  Matrix, B III  0 0 0 1 1 1  Loop

Branc h

 Matrix B and KVL equation for Loops I, II, III 1 − 1 0 1 0 0 B = 0 1 − 1 1 0 0 0 0 0 1 1 1  In matrix form

B e=0

e1 − e2 + e4 = 0 e2 − e3 + e5 = 0 e4 + e5 + e6 = 0  e1  e  2 1 − 1 0 1 0 0   0 1 − 1 1 0 0 e3  = 0   e  0 0 0 1 1 1  4  e5    e6 

Nodal Matrix A  Branch to node  KCL Mesh Matrix B  Branch to loop  KVL

NODE & MESH ANALYSIS  NODE & MESH analysis: The easiest way to obtain the system variables.  In node analysis: No. of independent voltages = No. of nodes – reference node  Suitable for network with no voltage source.

 In mesh analysis: No. of mesh currents = No. of independent currents  Suitable for network with no current source.

 What if a network has voltage + current sources?  Can use either method.

 Example 5: Write down the node equations for the circuit below. Then, find the branch voltage, e and branch current, i

Graph

   

No. of independent voltages = 3 The current source matrix, I The voltage source matrix, E Branch admittance,1 / 2Y 0 0 0 0   Y =    

0 0 0 0 0

0  1/ 2 0 0 0 0  0 1/ 2 0 0 0   0 0 1/ 2 0 0  0 0 0 1/ 2 0   0 0 0 0 1 / 2

V1  V  V = 0   2 0  V3    3 0  0    I =  0  0  E=  0  0     3 0    0

 The nodal incidence matrix, A reference

− 1 0 0 1 0 1  A =  0 0 − 1 − 1 1 0   0 − 1 0 0 − 1 − 1

 Important equations for nodal analysis:

e = AT V i = Ye − ( I − YE ) Yn = AYAT −1

V = Yn A( I − YE )

e = branch voltage matrix i = branch current matrix V = node voltage matrix Y = branch admittance Yn = node admittance matrix

 First, compute value of V  V = Yn −1 A( I − YE ) where Yn = node admittance matrix = A Y AT Yn =

A

1 0  0 1 − 1 0 0 1 1 0   Yn =  0 0 − 1 − 1 1 0 •  2 0  0 − 1 0 0 − 1 − 1 0  0  3 − 1 − 1 1 = − 1 3 − 1 2 − 1 − 1 3 

Y

AT

0 0 0 0 0  − 1 0 0 1 0 0 0 0  0 0 − 1 0 1 0 0 0  0 − 1 0  •  0 0 1 0 0  1 − 1 0  0 0 0 1 0  0 1 − 1    0 0 0 0 1  1 0 − 1

 ∴ the value of independent voltages, V will be: −1

V = Yn A( I − YE )  0  1   0 0  0 1    V1  2 1 1  − 1 0 0 1 0 1 0 V  = 1 1 2 1   0 0 − 1 − 1 1  0    −   2 2   0 2 0 V3  1 1 2  0 − 1 0 0 − 1 − 1    0  0    3 0 V1   3.0  ∴ V2  =  0.75  V3  − 0.75

Note: Yn-1 = inverse of matrix Yn

0   3   0 0  0 0    0  0   0 0 0 1 0  0     0 0 0 0 1 0 

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

 Finally, the corresponding branch voltage and current are: e = AT V  e1   − 3.0  e   0.75   2  V1    e3  − 0.75  T    = A V2  =   e 2 . 25  4  V3    e5   1.50       3.75  e6 

i = Ye − ( I − YE )  i1   0  i   0.375   2   i3  − 0.375  =  i4   1.125  i5   0.75      i6   − 1.125 

 Example 6: Write down the mesh equations for the circuit below. Then, find the branch voltage, e and branch current, i

Graph

   

No. of mesh currents = 3 The current source matrix, I The voltage source matrix, E Branch impedance, Z 2 0 0 0 0 0 Z = 0 0  0

2 0 0 0 0

0 2 0 0 0

0 0 2 0 0

0 0 0 0 2 0

0 0 0  0 0  2

3 0    0  E=  0  0    0

0  0    0  I =  0  0    3

 im1  im = im 2  im3 

− 1 0 1 − 1 0 0  B =  0 1 − 1 0 − 1 0   0 0 0 1 1 − 1

 The mesh incidence matrix, B

e = Zi − ( E − ZI )

 Important equations for mesh analysis: i = B T i m −1

im = Z m B( E − ZI ) −1

 First, compute value of im  im = Z m B( E − ZI ) where Zm = mesh impedance matrix = B Z BT − 1 Z m =  0  0 3 = 2 − 1 − 1

1 0 0 1 −1 0 0  0 1 − 1 0 − 1 0  • 2 0 0 0 1 1 − 1  0  0 − 1 − 1 3 − 1 − 1 3 

0 0 0 0 0  − 1 0 0 1 0 0 0 0  0 1 0  0 1 0 0 0  1 − 1 0  •  0 0 1 0 0  − 1 0 1  0 0 0 1 0  0 − 1 1     0 0 0 0 1  0 0 − 1

 ∴ the value of mesh currents, im will be: −1

im = Z m B ( E − ZI ) 3 1    0  0   i 2 1 1 − 1 0 + 1 − 1 0 0  m1     1 i  = 1 2 1   0 1 − 1 0 − 1 0  0 − 2 0  m2  8     0  0 im 3  1 1 2  0 0 0 + 1 + 1 − 1     0  0    0 0  im1   0.0  i  = 0.375  m2    im 3  1.125 

Note: Zm-1 = the inverse of matrix Zm

0  0    0 0  0 0    0  0   0 0 0 1 0  0     0 0 0 0 1 3 

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

 Finally, the corresponding branch voltage and current are: i = B T im  i1   0  i   0.375   2   i3  − 0.375  =  i 1 . 125  4   i5   0.75      i6   − 1.125 

e = Zi − ( E − ZI )  e1   − 3.0  e   0.75   2   e3  − 0.75  =  e4   2.25  e5   1.50      e6   3.75 

Compare the results obtained by nodal and mesh equations method!

TREE BRANCH SELECTION 1. Select minimum number of branches to connect all nodes in the graph without forming any loop Voltage source Tree Current source CoTree Capacitance

Tree

Inductance Resistance

CoTree Tree/CoTree (whichever is more suitable)

NUMBERING SEQUENCES    

First number  all voltage sources. Then, follows to other tree branches. Next, cotree branches except current sources. Finally, numbered all current sources. TREE BRANCH SELECTION & NUMBERING SEQUENCES are essential to reduce the complexity and burden in the analysis of the electric network

 Example 7: Illustrate tree branch selection & numbering sequences for the following electrical network.

 No. of tree branch → 4 – 1 = 3 Langarangian tree: All tree branches meet at  One tree branch  voltage source, e1one node The other tree branches  resistances R1 and R2  CoTree  branch 4, 5, 6  Remember the Numbering Sequences!!! voltage s.  other tree branches  cotree/chords  current s.

FUNDAMENTAL CUTSET MATRIX  Fundamental Cutset Matrix, C if branch j does not belong to cutset i 0 Cij = + 1 if the orientation of the branch j is equal to the cutset i  − 1 if the direction of branch j and cutset i does not match 1 2

3

4

5

6

1  1 0 0 − 1 0 − 1   2  0 0 1 1 −1 0  3  0 1 0 0 1 1  Cutset

Branc h

FUNDAMENTAL TIE-SET (LOOP) MATRIX  Fundamental Tie-set (Loop) Matrix, D

if branch j does not appear in fundamental loop i 0 Dij = + 1 if the orientation of the branch j is equal to the loop i  − 1 if the orientation of j and loop i differs

1

2

3

4 5

6

 1 0 −1 1 0 0     0 −1 1 0 1 0 III  1 − 1 0 0 0 1  I II

Cutset

Branc h

 Example 8: Solve the circuit of example 5 by using Cutset analysis method.

Graph

   

No. of tree branch voltages = 3 The current source matrix, I The voltage source matrix, E Branch admittance, Y 1 0 0 0 0 0 1 0 Y=  2 0 0  0

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0  0 0  1

 3 0    0  E=  0  0    0

0  0    0  I =  0  0    3

 et1  et = et 2  et 3 

 The cutset matrix, C

1 0 0 − 1 0 − 1 C = 0 1 0 0 1 1  0 0 1 1 − 1 0 

 Important equations for nodal analysis:

e = C T et i = Ye − ( I − YE ) −1

et = YC C ( I − YE )  First, compute value of et  et = YC −1C ( I − YE ) where YC = cutset admittance matrix = A Y AT 1 0  1 0 0 − 1 0 − 1 1 0 YC = 0 1 0 0 1 1  •  2 0 0 0 1 1 − 1 0  0  0  3 − 1 − 1 1 = − 1 3 − 1 2 − 1 − 1 3 

0 0 0 0 0  1 1 0 0 0 0  0 0 1 0 0 0  0 • 0 0 1 0 0 − 1 0 0 0 1 0  0   0 0 0 0 1 − 1

0 1 0 0 1 0

0 0  1  1 − 1  0 

 ∴ the value of tree branch voltages, et will be: −1

et = YC C ( I − YE )  0  1   0 0   et1  2 1 1  1 0 0 − 1 0 − 1   e  = 1 1 2 1  0 1 0 0 1 1  0 − 1 0  t2  2     0  2 0  et 3  1 1 2 0 0 1 1 − 1 0     0  0    3   0  et1   − 3.0  e  =  0.75   t2    et 3  − 0.75

Note: YC-1 = the inverse of matrix YC

0  3   0 0  0 0    0  0   0 0 0 1 0  0     0 0 0 0 1 0 

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

 Finally, the corresponding branch voltage and current are: e = C T et  e1   − 3.0  e   0.75   2   et1   e3  − 0.75  T   = C et 2  =   2.25  e4   et 3  e5   1.50       3.75  e6 

i = Ye − ( I − YE )  i1   0  i   0.375   2   i3  − 0.375  =  i 1 . 125  4   i5   0.75      i6   − 1.125 

 Example 9: Solve the circuit of example 5 by using Loop analysis method.

Graph

   

No. of chord currents = 3 The current source matrix, I The voltage source matrix, E Branch impedance, Z 2 0 0 0 0 Z = 0 0  0

2 0 0 0 0

0 2 0 0 0

0 0 0 2 0 0

0 0 0 0 2 0

0 0 0  0 0  2

3 0    0  E=  0  0    0

0  0    0  I =  0  0    3

 ic1  ic = ic 2  ic 3 

 The loop matrix, D

1 0 − 1 1 0 0 D = 0 − 1 1 0 1 0 1 − 1 0 0 0 1

 Important equations for mesh analysis:

e = Zi − ( E − ZI ) i = D T ic −1

ic = Z L D( E − ZI ) −1

 First, compute value of ic  iC = Z L D( E − ZI ) where ZL = loop impedance matrix = D Z DT 1 0 1 0 − 1 1 0 0  0 Z L = 0 − 1 1 0 1 0 • 2 0 1 − 1 0 0 0 1  0  0  3 − 1 1 = 2− 1 3 1  1 1 3

0 0 0 0 0  1 0 1  1 0 0 0 0  0 − 1 − 1 0 1 0 0 0  − 1 1 0  •  0 0 1 0 0  1 0 0  0 0 0 1 0  0 1 0     0 0 0 0 1  0 0 1 

 ∴ the value of chord currents, ic will be: −1

iC = Z L D( E − ZI )   3  1    0  0   i 2 1 − 1 1 0 + 1 1 0 0  C1     1 i  =  1 2 − 1 0 − 1 − 1 0 1 0 0 − 2 0  C2  8     0  0 iC 3  − 1 − 1 2  1 − 1 0 0 0 1     0  0    0 0  iC1   1.125  i  =  0.75   C2    iC 3  − 1.125

Note: ZL-1 = the inverse of matrix ZL

0  0    0 0  0 0    0  0   0 0 0 1 0  0     0 0 0 0 1 3 

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

 Finally, the corresponding branch voltage and current are: i = D T iC  i1   0  i   0.375   2   i3  − 0.375  =  i 1 . 125  4   i5   0.75      i6   − 1.125 

e = Zi − ( E − ZI )  e1   − 3.0  e   0.75   2   e3  − 0.75  =  e 2 . 25  4   e5   1.50      e6   3.75 

Compare the results obtained by fundamental cut-set and loop methods! And also by node and mesh analysis… All the answers obtained by the

SUMMARY  There are four types of analysis methods: 1) Node/Nodal Analysis 2) Mesh/Loop Analysis 3) Fundamental Cut-set Analysis 4) Fundamental Loop Analysis  These four-analysis methods will give the same results (branch voltage and current)

IMPORTANT EQUATIONS 1. Node Analysis −1 V = Yn A( I − YE )

2. Mesh Analysis −1 im = Z m B( E − ZI )

e = ATV i = Ye − ( I − YE )

i = B T im

3. Cut-set Analysis −1 et = YC C ( I − YE )

4. Loop Analysis −1 ic = Z L D( E − ZI )

e = Zi − ( E − ZI )

e = C T et

e = Zi − ( E − ZI )

i = Ye − ( I − YE )

i = D T ic