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Matrix Analysis of Networks Professor J R Lucas 1 May 2011
Matrix Analysis of Networks J. R. Lucas
Used to have a compact and neat form of solution. necessary to know the structure of a network, and formulate the problem based on the structure
Because Large networks are tedious to analyse using normal equations easier/more convenient to formulate in matrix form.
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Matrix Analysis of Networks Professor J R Lucas 2 May 2011
Topology
Deals with structure of an interconnected systemFormulates the problem based on non-measurable
properties of network.
Geometric structure of the interconnection of networkelements completely characterises
number of independent loop currents number of independent node-pair voltages
that are necessary to study the network.
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Matrix Analysis of Networks Professor J R Lucas 3 May 2011
Figure 1 Structure of the network
1(a) and (b) have the same structure (or topology).However elements are quite different.
R 1
L1
C
(a) (b)
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Matrix Analysis of Networks Professor J R Lucas 4 May 2011
Graph of Network
Figure 2 Circuit
Figure 2Circuit of figure 2(a) also has same topology.
Figure 2(b) shows structure corresponding to all 3circuits.
(a) Network (b) Graph of Network
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Matrix Analysis of Networks Professor J R Lucas 5 May 2011
Does not indicate any of the elements in the networks.Known as the graph of the network
Has all the nodes of the original networkIn obtaining the graph,
each element of the network is represented by a line
each voltage source by a short-circuit and each current source by an open circuit.
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Matrix Analysis of Networks Professor J R Lucas 6 May 2011
Tree of a Network
Which of the diagrams would also represent a normal
tree (without leaves) ? and why ?
Only first diagram would fully satisfy the requirements.Second diagram has branches closing on itself
only a tree like Nuga might appear to close on itselfThird diagram has branches in mid air not joined to main tree.
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Matrix Analysis of Networks Professor J R Lucas 7 May 2011
Properties associated with trees.1. All branches must be part of the tree
2. There cannot be closed loops formed from branches3. There cannot be branches isolated from the treeSame properties apply in defining a tree of a network
can be many trees associated with a given network. need not have a trunk coming from the ground and branches coming from the trunk. a reduced graph of network with some of the linksremoved so as to leave all the nodes connectedtogether by graph, but not to have any loop left.
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Matrix Analysis of Networks Professor J R Lucas 8 May 2011
Possible trees for the Graph
When a tree of the network is removed from graph, what remainsis called the co-tree of the network.
Co-tree is graph of removed links compliment of the tree.A co-tree may contain closed loops, and disconnected branches.
Graph of Network
Some of the possible trees
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Matrix Analysis of Networks Professor J R Lucas 9 May 2011
Analysis structure of network A single branch is required to join two nodes.
Joining each additional node would require anadditional branch.
Let b = number of branches in the networkn = number of nodes in the networkl = number of independent loops
Thus number of branches in tree = n 1 number of links removed = b (n 1) = b n + 1
Node 1 Node 2
Node 1 Node 2
New Node
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Matrix Analysis of Networks Professor J R Lucas 10 May 2011
Formation of Independent Loops
If any one of removed links are added to the tree, then
a new loop is formed. number of links removed from graph to form the
tree is equal to the number of independent loops.l = b n + 1
Oriented Graph Numbered branches with assigned
directions to currents.
Voltage considered to increase indirection opposite to flow of current
Oriented Graph
1
32
4 65
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Matrix Analysis of Networks Professor J R Lucas 11 May 2011
Matrix Analysis of Networks
To solve circuit problems,
need to write the equations corresponding to Ohms Law , and Kirchoffs Current Law
Kirchoffs Voltage LawSame is true even when there are a large number of
branches.
use matrix analysis
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Matrix Analysis of Networks Professor J R Lucas 12 May 2011
k -1 -1
+1 +1
+1
0
0 i 1
i 2
i 3
i 4
i 6
i 7
i 5
Kirchoffs current Law in matrix form For any node k
i1 i2 i4 + i 6 + i 7 = 0or
i1 i2 i4 = i 6 + i 7 or
i1 + i 2 + i 4 i6 i7 = 0or+1 . i1 + 1 . i2 + 0 .i 3 + 1 .i4 + 0 . i5 1 .i6 1 .i7 = 0
Last form is preferred for matrix implementation all currents in network are included in equation withdifferent coefficients.
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Matrix Analysis of Networks Professor J R Lucas 13 May 2011
For computer implementation, there must be a uniquemethod (convention) of obtaining the coefficients a jk .I j current in j th branch
jth branch directed away from k th node: a jk = +1 directed towards k th node: a jk = 1 not incident on the k th node: a jk = 0
Kirchoffs current law may be written , for the k th nodea1k . i1 + a 2k . i 2 + a 3k . i3 + a 4k . i4 + ...... ....... ..... a 7k . i7 = 0
or
b
j j jk ia
10
at k th node, for all k
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Matrix Analysis of Networks Professor J R Lucas 14 May 2011
Collection of equations, for each node k, would give
)1()1()(
0nb
b
bn
I t A
In [A] t, row vectors are dependant, since sum is zero.[A] t written with one row less, giving only (n-1) rows.
[A] t node-branch incidence matrix, (n-1) b.[A] branch-node incidence matrix, b (n-1)
a jk = +1 if j th current is directed away from the k th node
a jk = 1 if j th current is directed towards the k th nodea jk = 0 if j
th current is not incident on the k th node
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Matrix Analysis of Networks Professor J R Lucas 15 May 2011
Kirchoffs voltage Law in matrix form b
r r rs vb
1
0
for s th loop, for all s;where b rs = 1, 0, or +1
)1()1()(0l b
b
bl
V t
B [B] t mesh-branch incidence matrix, ( l b) [B] branch-mesh incidence matrix, ( b l)
b rs = +1 if r th
current is in same direction as sth
loop b rs = 1 if r
th current is in opposite direction to s th loop b rs = 0 if r
th current is in the not part of the s th loop
0
0
0+1
-1
-1
0
-1
+1
s
0
0
0
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Matrix Analysis of Networks Professor J R Lucas 16 May 2011
Ohms Law in matrix form
for all branches k = 1, 2, .... ... bvk = egk + Z k igk + Z k ik
Either voltage source or current source would normally be used.
egk
igk
Zkik + i gk ik
vk
Figure - General branch
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Matrix Analysis of Networks Professor J R Lucas 17 May 2011
Conversion with either Thevenins Theorem or Nortons Theorem.
With a voltage source onlyv
k = e
gk + Z
k i
k
for all branches k = 1, 2, ....... band in matrix form as
bb gbb I Z E V With a current source only
ik = Y k vk igk for all branches k = 1, 2, .... ... band in matrix form as
bb gbb V Y I I , where [Y b] = [Z b]-1
egk Zk
vk
ik
igk
Ykik
vk
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Matrix Analysis of Networks Professor J R Lucas 18 May 2011
I n SummaryFrom Kirchoffs Laws
)1()1()(0
nb bbn
I t
A (1) (n-1) independent equations
)1()1()(
0l b
b
bl
V t B (2) l independent equations
and from Ohms Law bb gbb I Z E V (3) b independent equationsor bb gbb V Y I I (3)* b independent equationsThus total number of independent equations is
n 1 + l + b = b + b = 2 b
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Matrix Analysis of Networks Professor J R Lucas 19 May 2011
2b independent equations2b unknowns ( b branch currents and b branch voltages )
Can be solved. Not usual to solve for both current and voltagesimultaneously.Reductions can be done in two ways.
1) Eliminate voltages and solve for currents mesh analysis
2) Eliminate currents and solve for voltages. nodal analysis.
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Matrix Analysis of Networks Professor J R Lucas 20 May 2011
Mesh Analysis Eliminate the branch voltages from the equations.
Reduce remaining currents to a minimum usingKirchoffs current law .A pply Kirchoffs voltage law for solution.
Define a set of mesh currents, m I .Branch currents b I related to mesh currents m I by analgebraic summation.
mb I B I (4) Eliminate V b from the equations,
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Matrix Analysis of Networks Professor J R Lucas 21 May 2011
Pre-multiply equation (3) by [B] t.
bbt gbt bt I Z B E BV B from equation (2), [B] t V b = 0.
Also mb I B I mb
t gb
t
I B Z B E B [B] t V b = 0 sum of voltages around a loop is zero.i.e. [B] t V b sum of voltages around a loop.
[B] t Egb sum of source voltages around a loop.Defined as mesh source voltage vector E gm .
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Matrix Analysis of Networks Professor J R Lucas 22 May 2011
i.e. E gm [B]t Egb
E gm = mmmbt I Z I B Z B
where [Z m] = B Z B bt corresponds to l equations
[B] also known as the tie-set matrix(as its elements tie the loop together)
Unknowns are l values of current I m Original 2b equations and 2b unknowns reduced tol equations and l unknowns.
Elements of [Z m] can be obtained either from abovemathematics, or by inspection as follows.
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Matrix Analysis of Networks Professor J R Lucas 23 May 2011
Simple evaluation of [Z m] and E gm z jj = self impedance of mesh j
= sum of all branch impedances in mesh jz jk = mutual impedance between mesh j and mesh k
= sum of all branch impedances common to mesh j
and mesh k and traversed in mesh direction sum of all branch impedances common to mesh j and mesh k, and traversed in opposite direction
e j = algebraic sum of the branch voltage sources inmesh j in mesh direction.
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Matrix Analysis of Networks Professor J R Lucas 24 May 2011
Example 1
Solve the circuit using Mesh matrix analysis.Work from first principles.Solution
Number the branches and the loops.
j6
E1 100 00
j20
-j120 E2 100 300 V
10
20
10
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Matrix Analysis of Networks Professor J R Lucas 25 May 2011
Write the loop currents in terms of the branch currents.i1 = I 1 i2 = I3i3 = I 1 I2
i4 = I 2 i5 = I 2 I3 i6 = I 3
or in matrix form
3
2
1
6
5
4
3
2
1
100
110010
011
100
001
I
I
I
i
ii
i
i
i
I1 I2 I3
i1 i4
i3 i5i6
i2
E1 100 00
20 6
j120 E2
100 36.87 0
10
20
10
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Matrix Analysis of Networks Professor J R Lucas 26 May 2011
This gives the Branch-Mesh incidence matrix [B].Mesh Branch incidence matrix [B] t can also
independently by writing the relation between themesh direction and the branch direction.
110010 011100
000101t
B
Notice that this corresponds to the transpose of the
earlier written matrix.
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Matrix Analysis of Networks Professor J R Lucas 27 May 2011
Vector of branch source voltages is
Branch impedance matrix is
600000
0100000
0020000
00012000
00001000000020
j
j
j
Z b
0
00
0
87.36100
01000
0
gb E
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Matrix Analysis of Networks Professor J R Lucas 28 May 2011
Egm = [B]t Egb , and [Z m] = [B]
t [Z b] [B]
0
0
0
0
87.36100
0100
110010
011100000101
0
0
gm E
, Egm = 0
0
87.36100
0
0100
100
110
010
011
100
001
600000
0100000
0020000
00012000
0000100
0000020
110010
011100
000101
j
j
j
Z m
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Matrix Analysis of Networks Professor J R Lucas 29 May 2011
60010100
0200
0120120
1000
0020
110010
011100
000101
j
j j
j
Z m
= 6201001012030120
0120100
j
j j
j j
Both Egm and Zm could have been written by inspection.Thus
0
0
87.36100
0
0100
=3
2
1
620100
1012030120
0120100
I
I
I
j
j j
j j
Equations may be solved by inversion or otherwise.
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Matrix Analysis of Networks Professor J R Lucas 30 May 2011
0
0
2
2
3
2
1
87.36100
0
0100
)120()12030(1001010010120
10100)620(100)620(120
10120)620(12010)620)(12030(1
j j j j j
j j j j j
j j j j j
I
I
I
6080
0
100
1440012000300010001200
100060020007202400
12007202400222012201
3
2
1
j j j j
j j j
j j j j
I
I
I
= (1220 j2220) ( j100) + (720 j2400) (j120) + ( j1200) 0= j122000 222000 + j 86400 + 288000= 66000 j 35600 = 74989 -28.34 o
I1 = (122000 j 222000 + 0 + j 96000 72000)/74989 -28.34o
= (50000 j 126000)/ 74989 -28.34 o = 135558 -68.36 o/74989 -28.34 o = 1.808 -40.02 o A
[ Note: Inversion has not been checked so answers may be in error.]Currents I 2 and I 3 can be similarly determined. The branch currents i 1, i2, ..... may then be determined from the matrix equation.[Normally branch 6 would have been marked as part of branch 2]
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Matrix Analysis of Networks Professor J R Lucas 31 May 2011
Nodal Analysis eliminate branch currents from the equations.
Reduce number of remaining voltages to a minimumusing Kirchoffs voltage law .A pply Kirchoffs current law for solution. Define a set of nodal voltages, N V which are node pairvoltages (i.e. voltage across a pair of nodes)Branch voltages bV are related to nodal voltages N V byan algebraic summation.
N b V AV (5) [A] too does not have the reference node.
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Matrix Analysis of Networks Professor J R Lucas 32 May 2011
Pre-multiply equation (3)* by [A] t.
bbt gbt bt V Y A I A I A from equation (1), [A] t I b = 0 .Substituting from (5)
N bt gbt V AY A I A 0 N bt gbt V AY A I A
IgN = [Y N]V N
where gbt
gN I A I , and AY AY bt N
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Matrix Analysis of Networks Professor J R Lucas 33 May 2011
Source nodal current vector I gN and the nodaladmittance matrix [Y N] could be written by inspection.
yii = sum of all branch admittances incident at node i
yij = negative of the sum of all branch admittancesconnecting node i and node j .
Reason for negative sign can be understood as follows:
ik = y k vk = y k (V i V j)At any node i,injected current Igi ik = yk (V i V j)
N
i j j
jk
N
i j j
ik gi V yV y I 11 for all j
vki
ykik
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Matrix Analysis of Networks Professor J R Lucas 34 May 2011
Since V i is a constant for a given i,
N
i j j jk i
N
i j j k
N
i j j jk
N
i j j k i gi
V yV yV y yV I 1 )111(
N
j
jii
N
i j j
jiiiii gi V yV yV y I
11 corresponds to nodal equation
As in the case of mesh analysis,IgN = [Y N]V N
is first solved to give V N and the branch voltagesand branch currents then obtained using the matrix equations.
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Matrix Analysis of Networks Professor J R Lucas 35 May 2011
Example 2
Example 1 has been reformulated as a problem withcurrent sources rather than with voltage sources.
[If voltage sources are present, they would first have to be converted to current sources].
5 -90 0 A
j20
6
-j120 8.575 5.910 A10
20
10
i1 i4
i3 i5
i2 V1 V2
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Matrix Analysis of Networks Professor J R Lucas 36 May 2011
Network may also be drawn in terms of admittances.
The branch-node incidence matrix [A], branch injected current I gb, and branch admittance matrix may be written,with reference selected as earthed node as follows.
5 -90 0 A
-j0.05S
0.0735 j0.0441 S
j0.00833S
8.575 5.91 0 A0.1 S
0.05S i1 i4
i3 i5
i2V1 V2
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Matrix Analysis of Networks Professor J R Lucas 37 May 2011
1011
01
10
01
A
,Igb =00
0
91.5575.8
905o
o
,
1.00000
005.0000
00008333.000
0000441.00735.00
000005.0
j
j
j
Y b
As in mesh analysis, nodal current injection vector and nodaladmittance matrix may be written from first principles.Left as an exercise for you to work out.
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Matrix Analysis of Networks Professor J R Lucas 38 May 2011
This is worked by inspection.
0441.00735.01.005.005.0
05.005.000833.005.0,
91.5575.8
905 j
j jY I N o
o
gN
2
1
0441.00735.01.005.005.0
05.005.000833.005.0
91.5575.8
905V
V
j
j jo
o
o
o
j j
j
V
V
91.5575.8
905
05.000833.005.005.0
05.00441.00735.01.005.01
2
1
= ( j0.05+j0.00833+0.05)(0.05+0.1+0.0735 j0.0441) 0.05 2 = (0.05 j 0.04167)(0.2235 j 0.0441) 0.0025= 0.06509 -39.81 o 0.2278 -11.16 o 0.0025
= 0.01483 -50.97 0.0025= 0.00934 0.0025 j 0.01152 = 0.00684 j 0.01152= 0.0134 -59.30 o
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Matrix Analysis of Networks Professor J R Lucas 39 May 2011
V1 = (0.2278 -11.16o 5 -90 o+0.05 8.575 5.91 o)/0.0134 -59.3 o
= ( 0.2205 j 1.1175 + 0.4265 + j 0.04415) /0.0134 -59.30 o = (0.2060 j 1.0733)/0.0134 -59.30 o
= 1.093 -79.14 o/0.0134 -59.30 o
V1 = 81.6 -19.84o V
branch current i 1 = 20
68.273.23
20
84.196.81100
j
j
j
o
i1 A j
o09.40809.1165.1384.1 which is the same answer (to calculation accuracy) that
was obtained in example 1.
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Matrix Analysis of Networks Professor J R Lucas 40 May 2011
Conversion of Ideal sources (a) Ideal Voltage sources
No impedance directly in series with voltage source
Ideal voltage sources are distributed to branchesconnected to one of the nodes of original ideal source.
E
E E E E E
or
Z1Z3
Z2
Z5Z4
Z1
Z3
Z2
Z5Z4
Z1Z3
Z2
Z5Z4
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Matrix Analysis of Networks Professor J R Lucas 41 May 2011
(b) Ideal Current sources No admittance appears directly in parallel with current source
Ideal current source has been distributed around a loopconnecting the two points of original source.
Is
Is
Is
Is
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Matrix Analysis of Networks Professor J R Lucas 42 May 2011
PortPair of nodes across which a device can be connected.
Voltage is measured across the pair of nodes.Current going into one node is the same as the currentcoming out of the other node in the pair.These pairs are entry (or exit) points of the network.Compare with an Airport or a Sea Port.Entry and exit points to acountry.
Planes that enter at a given portare the ones that take off from
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Matrix Analysis of Networks Professor J R Lucas 43 May 2011
same port.
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Matrix Analysis of Networks Professor J R Lucas 44 May 2011
Two-Port Theory Convenient to develop special methods for systematictreatment of networks.Single-port linear active networks
Thevenins or Nortons equivalent circuit.Linear passive networks
Convenient to study behaviour relative to a pair ofdesignated ports.
LinearPassive Network
I1 I2
V1 V2ort 1 Port 2
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Matrix Analysis of Networks Professor J R Lucas 45 May 2011
Definitions
Driving point impedance is defined as ratio of applied
voltage ( driving point voltage) across a node-pair tothe current entering at the same port.[input impedance of network seen from particular port]
Driving point impedance at Port 1 = V 1/I1
Driving point impedance at Port 2 = V 2/I2 Driving point admittance is similarly defined as theratio of the current entering at a port to the applied
voltage across the same node-pair.Driving point admittance at Port 1 = I 1/V1 Driving point admittance at Port 2 = I 2/V2
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Matrix Analysis of Networks Professor J R Lucas 46 May 2011
Immittance is sometimes used to represent either animpedance or an admittance
Transfer impedance is defined as the ratio of theapplied voltage across a node-pair to the currententering at the other port.
Transfer impedance = V 1/I2 , V 2/I1 Transfer admittance is similarly defined as the ratio ofthe current entering at a port to the voltage appearingacross the other node-pair.
Transfer admittance = I 1/V2 , I 2/V1
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Matrix Analysis of Networks Professor J R Lucas 47 May 2011
Transfer Voltage gain (or ratio) is defined as the ratioof the voltage at a node pair to the voltage appearing atthe other node-pair.
Transfer voltage gain = V 1/V2 , V 2/V1
Tr ansfer Current gain (or ratio) is similarly definedas the ratio of the current at a port to the current at theother port.
Transfer current gain = I 1/I2 , I 2/I1
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Matrix Analysis of Networks Professor J R Lucas 48 May 2011
Common Two-port parametersExternal conditions of a two-port network can becompletely defined by currents and voltages at the 2 ports.A general two port network can be characterised by four
parameters, derived from the network elements.With symmetry, number of parameters will be reduced.
(a) Impedance parameters(b) Admittance parameters(c) Transmission Line parameters
(d) Hybrid parameters.
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Matrix Analysis of Networks Professor J R Lucas 49 May 2011
(a) Impedance Parameters (z-parameters)or Open-circuit parameters
2
1
2221
1211
2
1
I I
z z z z
V V
V1 = z 11 I1 + z 12 I2
If I 2 = 0, then z 11 = V 1/I1If I 1 = 0, then z 12 = V 1/I2
LinearPassive
Network
I1
I2
V1 V2 Port 1 Port 2
f ll h
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Matrix Analysis of Networks Professor J R Lucas 50 May 2011
It follows that,
0211
11 I I
V z
, 0121
12 I I
V z
,
021
221
I I
V z
, 012
222
I I
V z
.z-parameters correspond to the driving point andtransfer impedances at each port with the other port
having zero current (i.e. open circuit). open circuit parameters.
l
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Matrix Analysis of Networks Professor J R Lucas 51 May 2011
Example 3Find impedance parameters of the two port T network .With port 2 on open circuit
ba Z Z I I V
z 021
111
b Z I I V
z 021221
similarly with port 1 open, z 12 = Z b z 22 = Z b + Z c
I1 I2
V1 V2 Port 1 Port 2
Za Zc
Z b
cbbbba
Z Z Z
Z Z Z Z
(b) Ad i P ( )
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Matrix Analysis of Networks Professor J R Lucas 52 May 2011
(b) Admittance Parameters (y-parameters)or Short-circuit parameters
2
1
2221
1211
2
1
V V
y y y y
I I
y11 , y12, y21, y22 defined with either V 1 or V 2 zero.
y-parameters correspond to driving point and transfer admittances at each port with the other port having zerovoltage (i.e. short circuit) short circuit parameters.
LinearPassive
Network
I1 I2
V1 V2ort 1 Port 2
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Matrix Analysis of Networks Professor J R Lucas 53 May 2011
E l 4
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Matrix Analysis of Networks Professor J R Lucas 54 May 2011
Example 4Find admittance parameters of the 2 port network.
y11 = 0211
V V
I = Y a + Y b
y21 = 0212
V V I = Y b
I1 I2
V1 V2
Y b
Ya Yc
I1 I2
V1 V2=0
Y b
Ya Yc
with ort 2 on short circuit
[Y] = cbbbba
Y Y Y
Y Y Y
( ) T i i Li P (ABCD )
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Matrix Analysis of Networks Professor J R Lucas 55 May 2011
(c) Transmission Line Parameters (ABCD-parameters)
Parameters can be defined using either port 2 on shortcircuit or port 2 on open circuit.In case of symmetrical system, parameter A = D.For a reciprocal system, A.D B.C = 1
Linear
Passive Network
I1 I2
V1 V2 Port 1 Port 2
2
2
1
1
I
V
DC
B A I
V
E l 5
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Matrix Analysis of Networks Professor J R Lucas 56 May 2011
Example 5Find ABCD parameters.
A = ccb
Y Y Y
, B = cY 1
C = caccbba
Y Y Y Y Y Y Y
and D = cca
Y Y Y
[For symmetrical network , Y a = Y b , A = D ].
A.D B.C = caccbba
cc
ca
c
cb
Y Y Y Y Y Y Y
Y Y Y Y
Y Y Y 1
= 22 )(
c
accbbaccbacab
Y Y Y Y Y Y Y Y Y Y Y Y Y Y
=1
I1 I2
V1 V2
Yc
Ya Y b
(d)
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Matrix Analysis of Networks Professor J R Lucas 57 May 2011
(d)
Hybrid Parameters (h-parameters)
Linear
Passive Network
I1 I2
V1 V2 Port 1 ort 2
Th h b id t t i b itt
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Matrix Analysis of Networks Professor J R Lucas 58 May 2011
The hybrid parameter matrix may be written as
2
1
2221
1211
2
1
V
I
hh
hh
I
V
h-parameters can be defined as in other examples, andare commonly used in some electronic circuit analysis.
I t ti f t t t k
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Matrix Analysis of Networks Professor J R Lucas 59 May 2011
Interconnection of two-port networks (a) Series connection of two-port networks
Series properties are applied to each port
at port 1, I r1 = I s1 = I 1, and V r1 + V s1 = V 1 at port 2 I r2 = I s2 = I 2, and V r2 + V s2 = V 2 [Z] = [Z r ] + [Z s]
LinearPassive
Networkr
Ir1 Ir2V r1 V r2ort r 1 Port r 2
LinearPassive
Networks
Is1 Is2Vs1 Vs2ort s 1 Port s 2
V2V1
(b) Parallel connection of two port networks
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Matrix Analysis of Networks Professor J R Lucas 60 May 2011
(b) Parallel connection of two-port networks
at port 1, I r1 + I s1 = I 1, and V r1 = V s1 = V 1 at port 2, I r2 + I s2 = I 2, and V r2 = V s2 = V 2 [Y] = [Y r ] + [Y s]
Linear
Passive Network
Ir1 Ir2
V r1 V r2
LinearPassive Network
s
Is1 Is2
Vs1 Vs2
V2V1
I1 I2
(c) Cascade connection of networks
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Matrix Analysis of Networks Professor J R Lucas 61 May 2011
(c) Cascade connection of networksOutput of one network becomes input to next.
Ir2 = I s1 V r2 = V s1
2
2
1
1
r
r
r r
r r
r
r
I
V
DC
B A
I
V
, 2
2
1
1
s
s
s s
s s
s
s
I
V
DC
B A
I
V
2
2
1
1
I V
DC B A
DC B A
I V
s s
s s
r r
r r
LinearPassive
Network
Is1 Is2
Vs1 Vs2 Port s1 Port s2 LinearPassive
Network
Ir1 Ir2
V r1 Vr2 Port r1 Port r2
ABCD matrix of component networks
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Matrix Analysis of Networks Professor J R Lucas 62 May 2011
ABCD matrix of component networks
A = 0221
I V V
= 1, = 1
B = 0221
V I V
= Z, = 0
C = 022
1
I V I
= 0, =Y
D = 0221
V I I
= 1, = 1
V1 V2
I1 I2
Z V1 V2
I1 I2
Y
In matrix form
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In matrix form
DC
B A
=10
1 Z
, =1
01
Y Consider example 5 again
DC
B A
= 1
01
aY 10
11bY 1
01
cY Simplification of matrix product would give the same answer as in example 5.
I1 I2
V1 V2
Y b
Ya Yc
Y b
Ya Yc