Using spectral radius ratio for node degreeIJCNCJournal
In this paper, we show that the spectral radius ratio for node degree could be used to analyze the variation of node degree during the evolution of complex networks. We focus on three commonly studied models of complex networks: random networks, scale-free networks and small-world networks. The spectral radius ratio for node degree is defined as the ratio of the principal (largest) eigenvalue of the adjacency matrix of a network graph to that of the average node degree. During the evolution of each of the above three categories of networks (using the appropriate evolution model for each category), we observe the spectral radius ratio for node degree to exhibit high-very high positive correlation (0.75 or above) to that of the
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Essential concepts in this technique are introduced
Top-down approach and bottom-up approach are used to present a holistic picture of FEM
Using spectral radius ratio for node degreeIJCNCJournal
In this paper, we show that the spectral radius ratio for node degree could be used to analyze the variation of node degree during the evolution of complex networks. We focus on three commonly studied models of complex networks: random networks, scale-free networks and small-world networks. The spectral radius ratio for node degree is defined as the ratio of the principal (largest) eigenvalue of the adjacency matrix of a network graph to that of the average node degree. During the evolution of each of the above three categories of networks (using the appropriate evolution model for each category), we observe the spectral radius ratio for node degree to exhibit high-very high positive correlation (0.75 or above) to that of the
coefficient of variation of node degree (ratio of the standard deviation of node degree and average node degree). We show that the spectral radius ratio for node degree could be used as the basis to tune the operating parameters of the evolution models for each of the three categories of complex networks as well as analyze the impact of specific operating parameters for each model.
Finite Element Method is explained taking a simple example
Essential concepts in this technique are introduced
Top-down approach and bottom-up approach are used to present a holistic picture of FEM
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offering a wide range of dental certified courses in different formats.for more details please visit
www.indiandentalacademy.com
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240164036 ee2092-4-2011-matrix-analysis
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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
Matrix Analysis of Networks – Professor J R Lucas 1 May 2011
2. • 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.
Matrix Analysis of Networks – Professor J R Lucas 2 May 2011
3. Topology
Deals with structure of an interconnected system
Formulates the problem based on non-measurable
properties of network.
Geometric structure of the interconnection of network
elements completely characterises
− number of independent loop currents
− number of independent node-pair voltages
that are necessary to study the network.
Matrix Analysis of Networks – Professor J R Lucas 3 May 2011
4. Figure 1 – Structure of the network
1(a) and (b) have the same structure (or topology).
However elements are quite different.
Matrix Analysis of Networks – Professor J R Lucas 4 May 2011
R1
L1
C
(a) (b)
5. Graph of Network
Figure 2 – Circuit
Matrix Analysis of Networks – Professor J R Lucas 5 May 2011
(a) Network (b) Graph of Network
6. Figure 2
Circuit of figure 2(a) also has same topology.
Figure 2(b) shows structure corresponding to all 3
circuits.
Does not indicate any of the elements in the networks.
Known as the graph of the network
– Has all the nodes of the original network
In obtaining the graph,
− each element of the network is represented by a line
− each voltage source by a short-circuit
Matrix Analysis of Networks – Professor J R Lucas 6 May 2011
7. − and each current source by an open circuit.
Matrix Analysis of Networks – Professor J R Lucas 7 May 2011
8. 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 itself
Third diagram has branches in mid air not joined to main tree.
Matrix Analysis of Networks – Professor J R Lucas 8 May 2011
9. Properties associated with trees.
1. All branches must be part of the tree
2. There cannot be closed loops formed from branches
3. There cannot be branches isolated from the tree
Same 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 links
removed so as to leave all the nodes connected
together by graph, but not to have any loop left.
Matrix Analysis of Networks – Professor J R Lucas 9 May 2011
10. Possible trees for the Graph
When a tree of the network is removed from graph, what remains
is 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.
Matrix Analysis of Networks – Professor J R Lucas 10 May 2011
Graph of
Network Some of the possible trees
11. Analysis structure of network
A single branch is required to join two nodes.
Joining each additional node would require an
additional branch.
Let b = number of branches in the network
n = number of nodes in the network
l = number of independent loops
Thus number of branches in tree = n – 1
number of links removed = b – (n – 1) = b – n + 1
Matrix Analysis of Networks – Professor J R Lucas 11 May 2011
Node 1 Node 2
Node 1 Node 2
New Node
12. 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 in
direction opposite to flow of current
Matrix Analysis of Networks – Professor J R Lucas 12 May 2011
Oriented Graph
1
32
4
6
5
13. Matrix Analysis of Networks
To solve circuit problems,
need to write the equations corresponding to
• Ohm’s Law, and
• Kirchoff’s Current Law
• Kirchoff’s Voltage Law
Same is true even when there are a large number of
branches.
− use matrix analysis
Matrix Analysis of Networks – Professor J R Lucas 13 May 2011
14. k
-1
-1
+1
+1
+1
0
0
i1
i2
i3
i4
i6
i7
i5
Kirchoff’s current Law in matrix form
For any node k
− i1 − i2 − i4 + i6 + i7 = 0
or
i1 + i2 + i4 = i6 + i7
or
i1 + i2 + i4 – i6 – i7 = 0
or
+1. i1 + 1. i2 + 0 .i3 + 1.i4 + 0. i5 – 1.i6 – 1.i7 = 0
Last form is preferred for matrix implementation
Matrix Analysis of Networks – Professor J R Lucas 14 May 2011
15. − all currents in network are included in equation
with different coefficients.
For computer implementation, there must be a unique
method (convention) of obtaining the coefficients ajk.
Ij – current in jth
branch
jth
branch
− directed away from kth
node: ajk = +1
− directed towards kth
node: ajk = –1
− not incident on the kth
node: ajk = 0
Kirchoff’s current law may be written, for the kth
node
Matrix Analysis of Networks – Professor J R Lucas 15 May 2011
16. a1k .i1 + a2k . i2 + a3k .i3 + a4k .i4 + ...... ....... ..... a7k .i7 = 0
or ∑
=
=⋅
b
j
jjk ia
1
0
at kth
node, for all k
Collection of equations, for each node k, would give
[ ] )1()1()(
0
×××
=⋅
nb
b
bn
It
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)
ajk = +1 if jth
current is directed away from the kth
node
ajk = −1 if jth
current is directed towards the kth
node
Matrix Analysis of Networks – Professor J R Lucas 16 May 2011
17. ajk = 0 if jth
current is not incident on the kth
node
Kirchoff’s voltage Law in matrix form
∑
=
=⋅
b
r
rrs vb
1
0
for sth
loop, for all s;
where brs = −1, 0, or +1
[ ] )1()1()(
0
×××
=⋅
lb
b
bl
Vt
B
[B]t
– mesh-branch incidence matrix, (l×b)
[B] – branch-mesh incidence matrix, (b×l)
brs = +1if rth
current is in same direction as sth
loop
brs = −1if rth
current is in opposite direction to sth
loop
brs = 0 if rth
current is in the not part of the sth
loop
Ohm’s Law in matrix form
Matrix Analysis of Networks – Professor J R Lucas 17 May 2011
0
0
0+1
-1
-1
0
-1
+1
s
0
0
0
18. for all branches k = 1, 2, .... ... b
vk = – egk + Zk igk + Zk ik
Either voltage source or current source would normally be used.
Conversion with either Thevenin’s Theorem or Norton’s Theorem.
Matrix Analysis of Networks – Professor J R Lucas 18 May 2011
egk
igk
Zkik
+ igk
ik
vk
Figure - General branch
19. With a voltage source only
vk = – egk + Zk ik
for all branches k = 1, 2, ....... b
and in matrix form as[ ] bbgbb IZEV +−=
With a current source only
ik = Yk vk – igk
for all branches k = 1, 2, .... ... b
and in matrix form as
[ ] bbgbb VYII +−=
, where [Yb] = [Zb]-1
Matrix Analysis of Networks – Professor J R Lucas 19 May 2011
egk
Zk
vk
ik
igk
Ykik
vk
20. In Summary
From Kirchoff’s Laws
[ ] )1()1()(
0
×××
=⋅
nb
b
bn
It
A
(1) (n-1) independent equations
[ ] )1()1()(
0
×××
=⋅
lb
b
bl
Vt
B
(2) l independent equations
and from Ohm’s Law
[ ] bbgbb IZEV +−=
(3) b independent equations
or [ ] bbgbb VYII +−=
(3)* b independent equations
Thus total number of independent equations is
n – 1 + l + b = b + b = 2 b
2b independent equations
2b unknowns (b branch currents and b branch voltages)
Matrix Analysis of Networks – Professor J R Lucas 20 May 2011
21. Can be solved.
Not usual to solve for both current and voltage
simultaneously.
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.
Matrix Analysis of Networks – Professor J R Lucas 21 May 2011
22. Mesh Analysis
− Eliminate the branch voltages from the equations.
− Reduce remaining currents to a minimum using
Kirchoff’s current law.
Apply Kirchoff’s voltage law for solution.
Define a set of mesh currents, mI
.
Branch currents bI
related to mesh currents mI
by an
algebraic summation.
[ ] mb IBI =
(4)
Eliminate Vb from the equations,
Pre-multiply equation (3) by [B]t
.
Matrix Analysis of Networks – Professor J R Lucas 22 May 2011
23. [ ] [ ] [ ] [ ] bb
t
gb
t
b
t
IZBEBVB +−=
from equation (2), [B]t
Vb = 0.
Also [ ] mb IBI =
∴ [ ] [ ] [ ][ ] mb
t
gb
t
IBZBEB =
[B]t
Vb = 0 → sum of voltages around a loop is zero.
i.e. [B]t
Vb → sum of voltages around a loop.
∴ [B]t
Egb → sum of source voltages around a loop.
Defined as mesh source voltage vector Egm .
i.e. Egm = [B]t
Egb
∴ Egm = [ ] [ ][ ] [ ] mmmb
t
IZIBZB =
Matrix Analysis of Networks – Professor J R Lucas 23 May 2011
24. where [Zm] = [ ] [ ][ ]BZB b
t
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 Im
Original 2b equations and 2b unknowns reduced to
l equations and l unknowns.
Elements of [Zm] can be obtained either from above
mathematics, or by inspection as follows.
Simple evaluation of [Zm] and Egm
zjj = self impedance of mesh j
= sum of all branch impedances in mesh j
Matrix Analysis of Networks – Professor J R Lucas 24 May 2011
25. zjk = 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
ej = algebraic sum of the branch voltage sources in
mesh j in mesh direction.
Matrix Analysis of Networks – Professor J R Lucas 25 May 2011
26. Example 1
Solve the circuit using Mesh matrix analysis.
Work from first principles.
Solution
Number the branches and the loops.
Matrix Analysis of Networks – Professor J R Lucas 26 May 2011
j6 Ω
E1
100∠00
V
j20 Ω
-j120 Ω
E2
100∠300
V
10 Ω
20 Ω
10 Ω
27. Write the loop currents in terms of the branch currents.
i1 = I1
i2 = – I3
i3 = I1 – I2
i4 = I2
i5 = I2 – I3
i6 = I3
Matrix Analysis of Networks – Professor J R Lucas 27 May 2011
or in matrix form
−
−
−
=
3
2
1
6
5
4
3
2
1
100
110
010
011
100
001
I
I
I
i
i
i
i
i
i
I1
I2
I3
i1
i4
i3
i5
i6
i2
E1
100∠00
V
j20 Ω j6 Ω
–j120 Ω
E2
100∠36.870
10 Ω
20 Ω
10 Ω
28. This gives the Branch-Mesh incidence matrix [B].
Mesh–Branch incidence matrix [B]t
can also
independently by writing the relation between the
mesh direction and the branch direction.
[ ]
−−
−=
110010
011100
000101
t
B
Notice that this corresponds to the transpose of the
earlier written matrix.
Matrix Analysis of Networks – Professor J R Lucas 28 May 2011
32. Nodal Analysis
− eliminate branch currents from the equations.
− Reduce number of remaining voltages to a minimum
using Kirchoff’s voltage law.
Apply Kirchoff’s current law for solution.
Define a set of nodal voltages, NV
which are node pair
voltages (i.e. voltage across a pair of nodes)
Branch voltages bV
are related to nodal voltages NV
by an
algebraic summation.
[ ] Nb VAV =
(5)
[A] too does not have the reference node.
Matrix Analysis of Networks – Professor J R Lucas 32 May 2011
33. Pre-multiply equation (3)* by [A]t
.
[ ] [ ] [ ] [ ] bb
t
gb
t
b
t
VYAIAIA +−=
from equation (1), [A]t
Ib = 0 .
Substituting from (5)
[ ] [ ] [ ][ ] Nb
t
gb
t
VAYAIA +−=0
[ ] [ ] [ ][ ] Nb
t
gb
t
VAYAIA =
IgN = [YN]VN
where [ ] gb
t
gN IAI =
, and
[ ] [ ] [ ][ ]AYAY b
t
N =
Source nodal current vector IgN and the nodal
admittance matrix [YN] could be written by inspection.
yii = sum of all branch admittances incident at node i
Matrix Analysis of Networks – Professor J R Lucas 33 May 2011
34. yij = negative of the sum of all branch admittances
connecting node i and node j .
Reason for negative sign can be understood as follows:
ik = yk vk = yk (Vi – Vj)
At any node i,
injected current Igi = Σ ik = Σ yk (Vi – Vj)
∑∑
≠
=
≠
=
−=
N
ij
j
jk
N
ij
j
ikgi VyVyI
11
for all j
Since Vi is a constant for a given i,
∑∑∑∑
≠
=
≠
=
≠
=
≠
=
−+
=−=
N
ij
j
jki
N
ij
j
k
N
ij
j
jk
N
ij
j
kigi VyVyVyyVI
1
)
111
(
∑∑
=
≠
=
=+=
N
j
jii
N
ij
j
jiiiiigi VyVyVyI
11 corresponds to nodal equation
As in the case of mesh analysis,
Matrix Analysis of Networks – Professor J R Lucas 34 May 2011
vk
ji
ykik
35. IgN = [YN]VN
is first solved to give VN and the branch voltages
and branch currents then obtained using the matrix equations.
Matrix Analysis of Networks – Professor J R Lucas 35 May 2011
36. Example 2
Example 1 has been reformulated as a problem with
current sources rather than with voltage sources.
[If voltage sources are present, they would first have to
be converted to current sources].
Matrix Analysis of Networks – Professor J R Lucas 36 May 2011
5∠-900
A
j20 Ω
j6 Ω
-j120 Ω
8.575∠5.910
A10 Ω
20 Ω
10 Ω
i1
i4
i3
i5
i2
V1
V2
37. Network may also be drawn in terms of admittances.
The branch-node incidence matrix [A],
branch injected current Igb, and
branch admittance matrix may be written,
with reference selected as earthed node as follows.
Matrix Analysis of Networks – Professor J R Lucas 37 May 2011
∠
∠
i1
i4
i3
i5
i2
V1
V2
38. [ ]
−
−
−
=
10
11
01
10
01
A
, Igb =
∠
−∠
0
0
0
91.5575.8
905
o
o
,
[ ]
−
−
=
1.00000
005.0000
00008333.000
0000441.00735.00
000005.0
j
j
j
Yb
As in mesh analysis, nodal current injection vector and nodal
admittance matrix may be written from first principles.
Left as an exercise for you to work out.
This is worked by inspection.
[ ] [ ]
−++−
−++−
=
∠
−∠
=
0441.00735.01.005.005.0
05.005.000833.005.0
,
91.5575.8
905
j
jj
YI No
o
gN
−++−
−++−
=
∠
−∠
∴
2
1
0441.00735.01.005.005.0
05.005.000833.005.0
91.5575.8
905
V
V
j
jj
o
o
∠
−∠
++−
−++
∆
=
∴ o
o
jj
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) – 0.052
Matrix Analysis of Networks – Professor J R Lucas 38 May 2011
39. = (0.05 – j 0.04167)(0.2235 – j 0.0441) – 0.0025
= 0.06509∠-39.81o
×0.2278∠-11.16o
– 0.0025
= 0.01483∠-50.97 – 0.0025
= 0.00934 – 0.0025 – j 0.01152 = 0.00684 – j 0.01152
= 0.0134∠-59.30o
V1 = (0.2278∠-11.16o
×5∠-90o
+0.05×8.575∠5.91o
)/0.0134∠-59.3o
= (– 0.2205 – j 1.1175 + 0.4265 + j 0.04415) /0.0134∠-59.30o
= (0.2060 – j 1.0733)/0.0134∠-59.30o
= 1.093∠-79.14o
/0.0134∠-59.30o
V1 = 81.6∠-19.84o
V
∴branch current i1 = 20
68.273.23
20
84.196.81100
j
j
j
o
+
=
−∠−
i1
Aj o
09.40809.1165.1384.1 −∠=−=
Matrix Analysis of Networks – Professor J R Lucas 39 May 2011
40. which is the same answer (to calculation accuracy) that
was obtained in example 1.
Matrix Analysis of Networks – Professor J R Lucas 40 May 2011
41. Conversion of Ideal sources
(a) Ideal Voltage sources
No impedance directly in series with voltage source
Ideal voltage sources are distributed to branches
connected to one of the nodes of original ideal source.
Matrix Analysis of Networks – Professor J R Lucas 41 May 2011
E
≡
42. (b) Ideal Current sources
No admittance appears directly in parallel with current source
Ideal current source has been distributed around a loop
connecting the two points of original source.
Matrix Analysis of Networks – Professor J R Lucas 42 May 2011
≡
44. Port
Pair 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 current
coming 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 a country.
Planes that enter at a given port
are the ones that take off from
same port.
Matrix Analysis of Networks – Professor J R Lucas 44 May 2011
45. Two-Port Theory
Convenient to develop special methods for systematic
treatment of networks.
Single-port linear active networks
− Thevenin’s or Norton’s equivalent circuit.
Linear passive networks
− Convenient to study behaviour relative to a pair of
designated ports.
Matrix Analysis of Networks – Professor J R Lucas 45 May 2011
46. Definitions
Driving point impedance is defined as ratio of applied
voltage (driving point voltage) across a node-pair to
the current entering at the same port.
[input impedance of network seen from particular port]
Driving point impedance at Port 1 = V1/I1
Driving point impedance at Port 2 = V2/I2
Driving point admittance is similarly defined as the
ratio of the current entering at a port to the applied
voltage across the same node-pair.
Driving point admittance at Port 1 = I1/V1
Driving point admittance at Port 2 = I2/V2
Matrix Analysis of Networks – Professor J R Lucas 46 May 2011
47. Immittance is sometimes used to represent either an
impedance or an admittance
Transfer impedance is defined as the ratio of the
applied voltage across a node-pair to the current
entering at the other port.
Transfer impedance = V1/I2 , V2/I1
Transfer admittance is similarly defined as the ratio of
the current entering at a port to the voltage appearing
across the other node-pair.
Transfer admittance = I1/V2 , I2/V1
Matrix Analysis of Networks – Professor J R Lucas 47 May 2011
48. Transfer Voltage gain (or ratio) is defined as the ratio
of the voltage at a node pair to the voltage appearing at
the other node-pair.
Transfer voltage gain = V1/V2 , V2/V1
Transfer Current gain (or ratio) is similarly defined
as the ratio of the current at a port to the current at the
other port.
Transfer current gain = I1/I2 , I2/I1
Matrix Analysis of Networks – Professor J R Lucas 48 May 2011
49. Common Two-port parameters
External conditions of a two-port network can be
completely 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.
Matrix Analysis of Networks – Professor J R Lucas 49 May 2011
50. (a) Impedance Parameters (z-parameters)
or Open-circuit parameters
V1 =
z11 I1 + z12 I2
If I2 = 0, then z11 = V1/I1
If I1 = 0, then z12 = V1/I2
It follows that,
Matrix Analysis of Networks – Professor J R Lucas 50 May 2011
52. Example 3
Find impedance parameters of the two port T – network.
With port 2 on open circuit
ba ZZ
II
V
z +=
=
=
021
1
11
bZ
II
V
z =
=
=
021
2
21
similarly with port 1 open,
z12 = Zb
z22 = Zb + Zc
(b) Admittance Parameters (y-parameters)
or Short-circuit parameters
Matrix Analysis of Networks – Professor J R Lucas 52 May 2011
[ ]
+
+
=→
cbb
bba
ZZZ
ZZZ
Z
53. y11, y12, y21,
y22 defined with either V1 or V2 zero.
y-parameters correspond to driving point and transfer
admittances at each port with the other port having zero
voltage (i.e. short circuit) → short circuit parameters.
Matrix Analysis of Networks – Professor J R Lucas 53 May 2011
54. Example 4
Find admittance parameters of the 2 port π–network.
y11 = 021
1
=VV
I
= Ya + Yb
y21 = 021
2
=VV
I
= – Yb
Matrix Analysis of Networks – Professor J R Lucas 54 May 2011
→[Y] =
+−
−+
cbb
bba
YYY
YYY
55. (c) Transmission Line Parameters (ABCD-parameters)
Parameters can be defined using either port 2 on short
circuit or port 2 on open circuit.
In case of symmetrical system, parameter A = D.
For a reciprocal system, A.D – B.C = 1
Matrix Analysis of Networks – Professor J R Lucas 55 May 2011
=
2
2
1
1
I
V
DC
BA
I
V
56. Example 5
Find ABCD parameters.
A = c
cb
Y
YY +
, B = cY
1
C = c
accbba
Y
YYYYYY ++
and D =
c
ca
Y
YY +
[For symmetrical
network, Ya = Yb , A = D].
A.D – B.C = c
accbba
cc
ca
c
cb
Y
YYYYYY
YY
YY
Y
YY ++
⋅−
+
⋅
+ 1
= 2
2
)(
c
accbbaccbacab
Y
YYYYYYYYYYYYY ++−+++
=1
(d) Hybrid Parameters (h-parameters)
Matrix Analysis of Networks – Professor J R Lucas 56 May 2011
57. 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, and
are commonly used in some electronic circuit analysis.
Matrix Analysis of Networks – Professor J R Lucas 57 May 2011
58. Interconnection of two-port networks
(a) Series connection of two-port networks
Series properties are applied to each port
at port 1, Ir1 = Is1 = I1, and Vr1 + Vs1 = V1
at port 2 Ir2 = Is2 = I2, and Vr2 + Vs2 = V2
Matrix Analysis of Networks – Professor J R Lucas 58 May 2011
59. [Z] = [Zr] + [Zs]
Matrix Analysis of Networks – Professor J R Lucas 59 May 2011
60. (b) Parallel connection of two-port networks
at port 1, Ir1 + Is1 = I1, and Vr1 = Vs1 = V1
Matrix Analysis of Networks – Professor J R Lucas 60 May 2011
61. at port 2, Ir2 + Is2 = I2, and Vr2 = Vs2 = V2
[Y] = [Yr] + [Ys]
(c) Cascade connection of networks
Output of one network becomes input to next.
Ir2 = Is1
Vr2 = Vs1
=
2
2
1
1
r
r
rr
rr
r
r
I
V
DC
BA
I
V
,
=
2
2
1
1
s
s
ss
ss
s
s
I
V
DC
BA
I
V
=
2
2
1
1
I
V
DC
BA
DC
BA
I
V
ss
ss
rr
rr
ABCD matrix of component networks
Matrix Analysis of Networks – Professor J R Lucas 61 May 2011
62. A = 022
1
=IV
V
= 1, = 1
B = 022
1
=VI
V
= Z, = 0
C = 022
1
=IV
I
= 0, =Y
D = 022
1
=VI
I
= 1, = 1
In matrix form
DC
BA
=
10
1 Z
, =
1
01
Y
Consider example 5 again
Matrix Analysis of Networks – Professor J R Lucas 62 May 2011
≡