ee2092!4!2011 matrix analysis

8/13/2019 EE2092!4!2011 Matrix Analysis

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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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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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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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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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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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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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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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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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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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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

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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(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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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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same port.


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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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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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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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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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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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(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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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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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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(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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E l 4


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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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(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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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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(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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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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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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(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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(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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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

ee2092!4!2011 matrix analysis

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