General Methods of Network Analysis Node Analysis Mesh Analysis Loop Analysis Cutset Analysis State variable Analysis Non linear and time varying network

General Methods of Network Analysis Node Analysis Mesh Analysis Loop Analysis Cutset Analysis State variable Analysis

Non linear and time varying network

Linear and time invariant network

Node and Mesh analyses

Source transformation Basic facts of node analysis Implication of KCL Implication of KVL Node analysis of linear time invariant networks Duality Basic facts of mesh analysis Implication of KVL Implication of KCL Mesh analysis of linear time invariant networks

Node and Mesh analyses

KCL

KVL

SNode Equation

Mesh Equation

Branch Equation

Fig. 1

Source transformation

+1

-2

3

4

+

1'-

3

4

+-

se

se

se

Ideal voltage source branch can be eliminatedFig. 2

Source transformation

Ideal current source branch can be eliminatedFig. 3

1 2

1 2

3

si1 2

1 2

3

sisi

Source transformation summary By source transformation we can modify any given

network in such a way that each voltage source is connected in series with an element which is not a source and each current source is connected in parallel with an element which is not a source.

If a current source in connected in series with a voltage source or an element the voltage source or that element can be ignored in analyzing the circuit.

If a voltage source in connected in parallel with a current source or an element the current source or that element can be omitted in analyzing the circuit.

Source transformation summary

+

-

kR

skv

+

-

kvskj

kj

k skj j

k sk k sk k kv v R j R j

kR

kv

kj

+

-sk k skv R j

+

-

Fig. 4

Source transformation summary

Fig. 5

+

-

kG

skv

+

-

kvskj

kj

k skv v

k sk k sk k kj j G v G v

kGkv

kj

sk k skj G v

+

-

+

-

Basic facts of node analysisFor any network with nodes and branches pick an arbitrary nodecalled the datum node. Assign to all node as

tn b1 2 n…..

Where 1tn n

Implications of KCL

Apply KCL to nodes 1 2 n….. a system of linear algebraic equationn

of unknowns b bjjj ,.....,, 21 is obtained

“The n linear homogenous algebraic equations in bjjj ,.....,, 21 obtained by

applying KCL to each node except the datum node constitute a set of linearly independent equation.”

Basic facts of node analysis

k4j

6j7j

At node k

0764 jjj

For all node KCL is written in the form

Aj 0

1

2

.

.

n

j

j

j

J

A is the reduced incident matrix.

(Aa with datum node deleted)

Fig. 6

(KCL)

Basic facts of node analysisExample 1

Consider the graph in Fig. 8. The graph has 4 nodes and 5 branches. WriteThe node incident matrix Aa and the KCL in matrix form.

2 31

4Datum node

1

2

3

4

5

Fig. 8

1 1 0 0 0

0 1 1 1 0

0 0 0 1 1

1 0 1 0 1

a

A

Incident matrix Aa

Basic facts of node analysis1

2

3

4

5

j

j

j

j

j

j

Aj 0KCL

1

2

3

4

5

1 1 0 0 0

0 1 1 1 0

0 0 0 1 1

j

j

j

j

j

0

0

0

0

54

432

21

jj

jjj

jjor


Implications of KVL

Let be the node voltage at nodes neee ,......., 21 1 2 n…..

respect to the datum node. The kth branch voltage is always the differenceBetween the nodes connected to it. Therefore all branch voltage can be written in the matrix form

Tv A e(KVL)

i

ik e

ev

if branch k leaves node i

if branch k enters node i


nbnbb

n

n

b e

e

e

ccc

ccc

ccc

v

v

v

.

.

...

......

......

...

...

.

.2

1

21

22221

11211

2

1

If branch leaves node and enter node then k i j jik eev

In matrix form

0

1

1

kic

If branch leaves node k i

If branch enters node k i

If branch is not incident with node k i

Basic facts of node analysisExample 2 For the graph of example 1

1

2

3

4

5

v

v

v

v

v

v

3

2

1

e

e

e

e

T

1 0 0

1 1 0

0 1 0

0 1 1

0 0 1

A Tv A e

35

324

23

212

11

ev

eev

ev

eev

ev

or

KVL


Proof of Tellegen theorem

Tv A eFrom KVL and from KCL

T

1

( )

( )

b

k kk

T T

T T T

T

v j

v j

A e j

e A j

e Aj

Aj 0

Node analysis of linear time invariant networks

In linear time invariant network all element except the independent sourceare linear and time invariant. The combination of branch equations to KCL and KVL forms a general linear simultaneous equation for neee ,......., 21

Resistive network

In a resistive network the branch equation takes the formthk

+

-

kR

skv

+

-

kvskj

kj

k skj j

, 1, 2....k k k sk k skv R j v R j k b

, 1, 2....k k k sk k skj G v j G v k b or

In matrix form

s s j Gv j Gv (1)Fig. 9

Node analysis of linear time invariant networksG is called the branch conductance matrix and it is a diagonal matrix

1

2

1

0 0 . 0

0 0 0

. . . .

. . . 0.

0 0 . 0.b

b

G

G

G

G

G

The and are source vectorssj sv

1

2

1

.

s

ss

sb

j

j

j

j

1

2

1

.

s

ss

sb

v

v

v

v

Substitute and pre-multiply by A in (1) yieldsTv A e

T 0s sA AGA e + Aj Gvor T

s sAAGA e = Gv - Aj

(2)

(3)

Node analysis of linear time invariant networksT

nY AGALet

and s s si AGv - Aj

be the node admittance matrix

be the node current source vector

Then the node equation becomes

snY e = i (4)

Once the node voltages are known the branch currents can be found from

s s j Gv j Gv

Tv A e

and (5)

Node analysis of linear time invariant networksExample 3

Consider the circuit in example 1 with elements shown in Fig.10,

solve the circuit for the node voltages and branch currents by nodeanalysis.

2 3

4

1

G1

G1= 2SG2

G2=1S G3G3=3S

G4

G4=1S G55 1sv V

G5=1S2sj A

Fig.10

Node analysis of linear time invariant networks1) Write KCL

1

2

3

4

5

1 1 0 0 0

0 1 1 1 0

0 0 0 1 1

j

j

j

j

j

Aj 0

2) Write KVL1

T2

3

1 0 0

1 1 0

0 1 0

0 1 1

0 0 1

e

e

e

v A e

Node analysis of linear time invariant networks3) Write branch equation in the form

s s j Gv j Gv

Branch 1 )0(222 11 vj

Branch 55 51 0 1(1)j v

Thus

1 1

2 2

3 3

4 4

5 5

2 0 0 0 0 2 2 0 0 0 0 0

0 1 0 0 0 0 0 1 0 0 0 0

0 0 3 0 0 0 0 0 3 0 0 0

0 0 0 1 0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0 0 0 1 1

j v

j v

j v

j v

j v

Node analysis of linear time invariant networks4) Make the form snY e = i

2 0 0 0 0 1 0 0

1 1 0 0 0 0 1 0 0 0 1 1 0

0 1 1 1 0 0 0 3 0 0 0 1 0

0 0 0 1 1 0 0 0 1 0 0 1 1

0 0 0 0 1 0 1

210

151

013

TnY AGA


s s si AGv - Aj5) and

2 0 0 0 0 0 2

1 1 0 0 0 1 1 0 0 00 1 0 0 0 0 0

0 1 1 1 0 0 1 1 1 00 0 3 0 0 0 0

0 0 0 1 1 0 0 0 1 10 0 0 1 0 0 0

0 0 0 0 1 1 0

1

0

2

6) The node equation is1

2

3

3 1 0 2

1 5 1 0

0 1 2 1

e

e

e

Node analysis of linear time invariant networks7) Solve for e

1

9 2 1 2 171 1

2 6 3 0 125 25

1 3 14 1 12n s

e Y i

8) Solve for v

1T

2

3

1 0 0 17

1 1 0 161

0 1 0 125

0 1 1 13

0 0 1 12

e

e

e

v A e

Node analysis of linear time invariant networks9) Solve for j s s j Gv j Gv

1

2

3

4

5

2 0 0 0 0 17 2 2 0 0 0 0 0

0 1 0 0 0 16 0 0 1 0 0 0 01

0 0 3 0 0 1 0 0 0 3 0 0 025

0 0 0 1 0 13 0 0 0 0 1 0 0

0 0 0 0 1 12 0 0 0 0 0 1 1

j

j

j

j

j

13

13

3

16

16

25

1

Node analysis of linear time invariant networksNode equation by inspection

The node equation snY e = i can be written in scalar form

11 12 1 1 1

12 22 23 2 2 2

1 2

. .

.

. . .

.

. .

n s

n s

ij in i si

n n nn n sn

y y y e i

y y y y e i

y y e i

y y y e i

(6)


iiy = sum of admittance at node

jky = negative sum of admittance between the node and the node

where thi

thj thk

and ski = equivalent current source injected at node k

2 3

4

1

2S

1S

3S

1S

1S

5 1si A2sj A

Redraw the circuit in example 3 By inspection

1

0

2

210

151

013

3

2

1

e

e

e

Node analysis of linear time invariant networksSinusoidal steady state analysis

In RLC circuit with sinusoid excitations, branch voltage and branch currentare in the form of phasors and branch admittances are thefunction of frequency

kV kJ

The branch equation take the form

, 1, 2......k k k ks k skj Y V j Y V k b

In the matrix form b bs s J Y V J Y V

and the node equation becomes snY E = I

Tn bY AY A s s s bI AY V - AJ

(7)

(8)

(9)


Fig. 11

+

-

1V

1 G2

1J

C1I

2J+ -

2V 3J

3LJ

3L

2mg V

4J 4L

5L1'mg V

5J

2 3

5LJ +

-

5V

+ -4V

Consider the circuit shown in Fig.11. The sinusoid current source ofphasor is applied at node 1. The inductors are coupled as shownby its inductance matrix

IL

5 43 3

543 3

1 1 1

1

1

L

Write the node equation of the circuit.

Node analysis of linear time invariant networks1 1 0 0 0

0 1 1 1 0

0 0 0 1 1

A

Branch equations

IVCjj 111

2 2 2j G VInductor branch equations

' LjV LJ

5

4

3

35

34

34

35

5

4

3

1

1

111

V

V

V

L

L

J

J

J

j


11'L j

J L V

3 3

4 4

5 5

3 1 1 V1

1 2 1 V

1 1 2 V

L

L

J

Jj

J

5432

233

113V

jV

jV

jVg

VgJJ

m

mL

5 5 1

1 3 4 5

'

1 1 2'

L m

m

J J g V

g V V V Vj j j

Node analysis of linear time invariant networksBranch equation in matrix form

b s J Y V J

0

0

0

0

0'

00

0

0000

0000

5

4

3

2

1

211

121

1132

1

5

4

3

2

1 I

V

V

V

V

V

g

g

G

Cj

J

J

J

J

J

jjjm

jjj

jjjm


1

23 1 1

1 2 1

1 1 2

0 0 0 0 1 0 00 0 0 01 1 0 0 0 1 1 000 1 1 1 0 0 1 00 00 0 0 1 1 0 1 1

0 0 1' 0

m j j j

j j j

m j j j

j C

G

g

g

The node admittance matrix

1 2 2

2 2

0

7 3

3 2'

n m m

m

j C G G

G g G gj j

gj j

Y

Tn bY AY A

Node analysis of linear time invariant networksThe node equation is

snY E = I

1 2 21

2 2 2

3

0E I

7 3E 0

E 03 2

'

m m

m

j C G G

G g G gj j

gj j

Solve for E and substitute in

and b s J Y V JTV A E

Node analysis of linear time invariant networksIntegrodifferential form equations

In general node analysis of a linear network lead to a set of integro-differential equation. The equation involves unknown functions , theirsderivatives and integrals. e.g.

,..')'(,')'(...,..,0

20

12121 tt

dttedtteeeee

Example 5

The linear time-invariant network shown in Fig.12 has the reciprocalInductance matrix

44 45

45 54

Γ


Fig. 12

+

-

1v

1 C3

1j

G1

3j+ -

3v

2j

4Lj

44

3mg v4j 5j

2 3

+

-

5v

+ -2v

1sj55

45

G2

KCL: 1

2

3

4

5

1 0 1 0 0

0 1 1 1 0

0 1 0 0 1

j

j

j

j

j

0Aj 0

Node analysis of linear time invariant networksKVL:

3

2

1

5

4

3

2

1

100

010

011

110

001

e

e

e

v

v

v

v

v

Branch equation

1111 sjvGj

2 2 2j G v

3 3 3j C v

4 3 44 4 45 5 4

0 0

( ') ' ( ') ' (0)t t

m Lj g v v t dt v t dt j 5 45 4 55 5 5

0 0

( ') ' ( ') ' (0)t t

j v t dt v t dt j

Tv A e


It is convenient to write

dt

dvvDv 3

33

and 14 4 4

0

1( ') '

t

v v t dt D vD

Therefore the branch equations are

11 1 1

22 2

33 31 1

4 44 45 4 4

1 15 5 545 55

0 0 0 0

0 0 0 0 00 0 0 0 0

0 0 (0)

(0)0 0

s

m L

Gj v jGj v

C Dj v

j g D D v j

j v jD D

1 1DD D D

Note

1

0( ) ( )

tdDD f f d f t

dt

1

00( ) ( ) ( ) (0)

t tD Df f d f f t f

Node analysis of linear time invariant networksIn the matrix form

( )b sD j Y v j

Multiple by A and from KVL

Tb( ) sD AY A e Aj

sDnY ( )e = ior

T( ) ( )n bD DY AY A


1

2

3

1 144 45

1 145 55

0 0 0 0 1 0 00 0 0 01 0 1 0 0 0 1 10 0 0 00 1 1 1 0 1 1 0

0 1 0 0 1 0 0 0 1 0

0 0 10 0

m

G

G

C D

g D D

D D

1552

1452

1452

144323

331

0

0

DGDG

DGDgDCGgDC

DCDCG

mm

T( ) ( )n bD DY AY A

Node analysis of linear time invariant networksThe node equation

s(D)nY e = i

1 3 3 1 11 1

3 2 3 44 2 45 2 4

1 13 52 45 2 55

0

(0)

(0)0

s

m m L

G C D C D e j

C D g G C D g D G D e j

e jG D G D

The cut set for branches1, 4, 5 gives initial conditions

1 1 4 3 51

1(0) [ (0) (0) (0) (0)]s L me j j g v j

G

)0()0()0( 312 vee

3 2 5 2(0) (0) (0) /e e j G

(a)

Node analysis of linear time invariant networksNotes

If we define new variables

2 2 3 3

0 0

( ) ( ') ' , ( ) ( ') 't t

t e t dt t e t dt

Then 2 12 2 2 2 2 2, , andDe D e D D e

The node equation becomes

1 3 3 1 12

3 3 2 44 2 45 2 4

2 45 2 55 3 5

0

( (0)

0 (0)

s

m m L

G C D C D e j

C D g C D G g D G D j

G D G D j

0)0()0( 22 e 3 3(0) (0) 0e )0(1e as in (a)


The short cut method

If the circuit involves only few dependent sources the node equation canalso be written by inspection.

Example 6

Fig.13

Write the node equation for Fig. 13 in sinusoid steady state.

+

-

1V

1 G2

1J

C1I

2J+ -

2V 3J

32mg V

4J

1'mg V

5J

2 3

+

-

5V

+ -4V

4

5


2 1 21

3 4 42 2 2 3

3 54 54

0

0

s

s

G j C GE I

G G E Jj j

E J

j j

By inspection

)( 2123 EEgVgJ mms

5 1 1' 's m mJ g V g E

consider as independent sources


2 1 21

3 4 42 2 2

34 54

'

0

0

0m m

m

G j C GE I

g G G g Ej j

E

gj j

Rearrange the equation


Fig.14

Write the integro-differential equation by inspection for the circuit in Fig.14

+

-

1V

1 C3

1j

G11sj

3j

+ -3v 4j

43mg v

2j

5j

2 3

+

-

5V

+ -2v

5

4Lj

G2

Node analysis of linear time invariant networksSince )0(')'()(

0

L

t

LL idttvtj Then by inspection

1 3 3 1 11

3 2 4 3 2 2 4 4

13 52 2 5

0

(0)

(0)0

s

s L

G C D C D e j

C D G D C D G e j j

e jG G D

substitute )( 2134 eegvgj mms

1 3 3 1 11

3 2 4 3 2 2 4

13 52 2 5

0

(0)

(0)0

s

m m L

G C D C D e j

g C D g G D C D G e j

e jG G D

Then

Node analysis of linear time invariant networks Duality

A graph of a circuit can be drawn in many ways but it has the same results.

Planar graph, Meshes, outer mesh

A planar graph can be drawn on the plane without branch intersection.

A mesh is the smallest closed path (loop) in a graph and a outer meshIs a loop formed outside the graph.

1

2 3

4

5

a

b cd

ef

g(a)

1

2 3

4

5ab cd

ef

g(b)

1

2

3

4

5

a

b cd

e

f g

(c)

Fig.15

Outer Mesh

Node analysis of linear time invariant networksFig.15 (a),(b),and (c) have the same incident matric and of the same graph.

But they have different topology. In Fig.15 the loop is not a meshbut this loop is a mesh in (b) while this loop is an outer mesh in (c).

bcef

Hinged and unhinged graph

Hinged graph can be partitioned into two subgraph by one node.

Fig.16 Hinged graph Unhinged graph

G1 G2

G1G2

G1 G2


The number of meshes is equal to 1tl b n Fundamental property of an unhinged planar graph

Each branch belongs to exactly two meshes including the outer mesh

5

6

7

8

1 2

3

4

1 2

34

5Outer mesh

Fig.17

Node analysis of linear time invariant networksAssigned reference direction for meshes

Each mesh has clockwise direction but the outer mesh has counter clockwise direction.

A mesh matrix Ma can be written its element is defined by

0

1

1

ikm

If branch is in mesh and their direction coincidek i

If branch is in mesh and their direction oppositek i

If branch is not in meshk i

Node analysis of linear time invariant networksFor the graph in Fig.17, the mesh matrix Ma there are four mesh

and 8 branches. The elements of the matrix Ma are

1 0 0 0 1 0 0 1

0 1 0 0 1 1 0 0

0 0 1 0 0 1 1 0

0 0 0 1 0 0 1 1

1 1 1 1 0 0 0 0

a

M

It can be observed that the Mesh matrix Ma has the same properties as

The node incident matrix Aa . It element is 1 or -1 or 0 .

DualityDual graph

Dual graph has some properties being demonstrated in example 8.

Example 8

Consider the linear time invariant circuit in Fig. 18. In sinusoid steadyState, write the node equation of the circuit and find its dual circuit.

1 L

C1sI C2

1E 2E

network

Fig. 18

DualityThe node equation written by inspection is

0)1

(1

1)

1(

221

211

ELj

GCjELj

IELj

ELj

Cj s

By changing

1 1 2 2ˆ ˆ, , ,C L L C C L G R

The node equation becomes

ss EIandIEIE ˆˆ,ˆ2211

Duality1 1 2

1 2 2

1 1ˆ ˆ ˆ ˆ( )ˆ ˆ

1 1ˆ ˆ ˆ ˆ( ) 0ˆ ˆ

sj L I I Ej C j C

I j L R Ij C j C

These equation are the loop equation for the circuit in Fig.19

ˆsE

network

R

2L1L

C+- 2I1I

Fig.19Fig.19 is the dual graph of Fig 18.

A graph have node branch and the is the dual graph of if

Duality

There is a one-to-one correspondence between the meshes of and the node of

There is a one-to-one correspondence between the meshes of and the node of

Branch between mesh of correspond to branch between node of

G

G1nnt b G1

GG1

GG1

G1G

algorithm

• write node in each mesh of and node for the outer mesh• for each branch of common to mesh and mesh there is a branch connected between node and node of• if the graph is oriented the direction of the branch of is rotated 90o clockwise

1

1 2... G L+1

i ji j G1

G G1

G

DualityExample 9

From the planar graph G in Fig.20 construct the dual graph G

1 2

3

b=5n=2l=3

G

Fig.20

Duality

Step 1 assign node of

1 2

3

12

3

41 2

3

12

3

4

G

G Fig.21

Duality

Step 2 draw branches of G

1 2

3

12

3

41 2

3

12

3

4

1 2 3

4 ˆ 5b ˆ 2l ˆ 2n ˆ 3tn

G

Fig.22

DualityNotes

In general a given topological graph G has many duals. But if thedatum node and elements belong to the outer mesh is specified, TheDual graph is always unique

The correspondence between the graph G and involvesGBranches versus Branches

Meshes versus Nodes

Datum node versus Outer mesh

The incident matrix Aa of equal the mesh matrix Ma of G G

DualityDual network

A network is the dual of the network if the topological graph of is dual of the topological graph of

G G

And the branch equation of obtained form the correspondingequation of by performing the following substitution

qvj

qjv

ˆˆ

ˆˆ

Where andqjv ,, are voltage, current, charge and flux linkage

DualityExample 10

Fig.23

Consider the nonlinear time varying network shown in Fig 23 draw the dualnetwork.

1 1tanh j

+

-( ) ( )se t f t 2R

1j

2j

2

3 31 tq e v

3j

4 ( ) 2 cosR t t

4j1i 2i

24321211 ijjiijij

DualityThe mesh equations are

2 2

12 1 22

1

32 2 1 2 2

0

1( ) ( ) ( )

cosh

(0) 10 ( ) ( ') ' (2 cos ) ( )

1 1

s

t

t t

die t f t R i i

dti

qR i i i t dt t i t

e e

2 2

12 1 22

1

32 2 1 2 2

0

ˆ1 ˆˆ ˆ ˆ( ) ( ) ( )ˆcosh

ˆ (0) 1ˆ ˆ ˆ ˆ ˆ0 ( ) ( ') ' (2 cos ) ( )1 1

s

t

t t

dei t f t G e e

dte

G e e e t dt t e te e

Dual equation

Recall that

11 1

1

sinhtanh tanh

cosh

ij i

i

11

1

sinhtanh

cosh

d d ii

dt dt i

1 1 112

1

cosh cosh sinh sinh

cosh

i i i i

i

2 21 1

2 21 1

cosh sinh 1

cosh cosh

i i

i i

u

v

vdu udv

v2

DualityThe dual network

1 1ˆ ˆtanhq eˆ ( ) ( )si t f t

2 2G R

23 3ˆ ˆ1 te j

4ˆ ( ) 2 cosG t t

4j1e+

-

2v+ -

3v+

-3j

4v+

-

Fig.24

Basic facts of mesh analysisIf we apply KVL to meshes 1,2,3,…l (omit the outer mesh), a system ofl linear homogeneous equations in b unknowns is obtained. bvvv ,..., 11

The KVL can be written in the matrix form as

Mv 0 or bivmb

kkik ..2,1,0

1

0

1

1

ikm

If branch is in mesh and their direction coincidek i

If branch is in mesh and their direction oppositek i

If branch is not in meshk i

Note that the mesh matrix M is obtained from the matrix Ma with the outerMesh deleted.

Basic facts of mesh analysisExample 11

Fig.25

Obtain the KVL for the graph shown in Fig.25

2j 4j

3j

1j

5j1i 3i

2i

5

4

3

2

1

v

v

v

v

v

v

Basic facts of mesh analysis1 1 0 0 0

0 1 1 1 0

0 0 0 1 1

M

KVL

1

2

3

4

5

1 1 0 0 0

0 1 1 1 0

0 0 0 1 1

v

v

v

v

v

Mv 0

or

0

0

0

54

432

21

vv

vvv

vv

Basic facts of mesh analysisImplication of KCL

Let be the mesh currents in clockwise direction. These currentsare linearly independent . Thus KCL can not be written in terms of meshCurrents. Since each mesh current runs around a loop if it crosses a cutset in a positive direction it will also cut that cutset in a negative direction too.

1 2, ,... li i i

However, the branch current can be calculated by the equation

Tj M iThis is some what similar for the node analysis in which

Tv A e

Basic facts of mesh analysisExample 12

Write the KCL for the graph in the example 11

35

324

23

212

11

ij

iij

ij

iij

ij

Or

1

2

3

1 0 0

1 1 0

0 1 0

0 1 1

0 0 1

T

i

i

i

j M i

Mesh analysis of linear time invariant networks

Mesh analysis of a linear time-invariant network is the dual of the node analysis.

Sinusoidal steady state analysis

A linear time invariant network with branch and node whosegraph is unhinged and planar is in sinusoid steady state at frequency .

b tn

The phasors of voltage and current vector can be used. If thephasors of mesh current vector is chosen, the Kirchhoff’s laws give

JV,I

(KVL) Mv 0(KCL) TJ M IBranch eqn.

s( ω) ( ω)b b sj j V Z J Z J V


The matrix bb )(b jZ is called the branch impedance matrix.

Substitution of equation yields

s( ( ) ) ( )Tb b sj j MZ M I MZ J V

or s( )m j Z I E

( ) ( ) Tm bj j Z MZ M

s s( )b sjω E MZ J MV

Mesh analysis of linear time invariant networksExample 1

Consider the circuit of Fig 26. the phasor represent the sinusoid voltage write the mesh equation of the circuit.

Fig 26

1sV

1 1 1( ) | | cos( )s s sv t V t V 2F

1J

1sV

2J

+ -2V 3J

3L

4J 4L

5L 24V

5J

5LJ +

-

5V+-

3W

1I 2I

Mesh analysis of linear time invariant networksLet the inductance matrix of the branch 3,4,5 is

3 1 1

1 4 2

1 2 5

L

The mesh matrix1 1 1 0 0

0 0 1 1 1

M

The branch 3,4,5 voltages

5

4

3

5

4

3

52

24

3

LJ

J

J

jjj

jjj

jjj

V

V

V


252552

4 Jj

JVJJ L and

Then the branch equation becomes

0

0

0

0

52100

2440

320

0002

10

000031

5

4

3

2

1

5

4

3

2

1 sV

J

J

J

J

J

jjj

jjj

jjjj

V

V

V

V

V

Mesh analysis of linear time invariant networks3 0 0 0 0

1 01

0 0 0 0 1 021 1 1 0 0

( ) 1 10 2 30 0 1 1 1

0 10 4 4 2

0 10 10 2 5

m

jj

j j j

j j j

j j j

Z

15 3 3

2

16 3 16

j jj

j j


The mesh equation becomes

s( )m j Z I E

016316

32

135 1

2

1 sV

I

I

jj

jj

j

1s s s( )

0s

b s

Vjω

E MZ J MV MV

Mesh analysis of linear time invariant networksThe properties of mesh impedance matrix

If the network has no coupling elements is diagonal and is symmetric if there is no coupling element, the mesh impedance matrix can be written by inspection:

( )b jZ( )m jZ

iiZ is the sum of all impedance in mesh i

ikZ is the negative sum of all impedance common between mesh iand mesh k

Current source is converted to Thevenin source and is the algebraic sum of all voltage sources whose reference direction push the current flows in the mesh k

ske

Mesh analysis of linear time invariant networksIntegrodifferential Equations

Fig. 27

Consider the linear time-invariant circuit shown in Fig. 27 where the Inductance matrix is

1

2

L M

M L

L

3R

3sv

3j1j 1L 2j2L

3j

5J

+

-

4v+-

1I 2I4C

4j

5R

Mesh analysis of linear time invariant networksStep 1 Write the KVL Mv 0

011010

01101

5

4

3

2

1

v

v

v

v

v

Step 2 Write the KCLTj M i

2

1

5

4

3

2

1

10

11

01

10

01

i

i

j

j

j

j

j

Mesh analysis of linear time invariant networksStep 3 Write the Branch equations

0

)0(

0

0

000

01

000

0000

000

000

4

3

5

4

3

2

1

55

4

3

2

1

5

4

3

2

1

v

v

j

j

j

j

j

RRDC

R

DLMD

MDDL

v

v

v

v

v

s

Or the forms( )b D v Z j v

Combine the equation in the form

s s( ) ( )Tb b sD D MZ M i MZ j Mv Mv


Or in scalar form

s( )m D Z i e

( ) ( ) Tm bD DZ MZ M ss e Mv

)0(

)0(11

11

4

43

2

1

452

45

4431

v

vv

i

i

DCRDL

DCRMD

DCMD

DCRDL

s

)0()(')'(1

')'(1

43

0

24

2

0

14

131

1 vtvdttiCdt

diMdtti

CiR

dt

diL s

tt

1 2

5 1 1 2 5 2 2 44 40 0

1 1( ') ' ( ') ' (0)

t tdi di

M R i i t dt L R i i t dt vdt C dt C

Mesh analysis of linear time invariant networksIf we define new variables

1 1 2 2

0 0

( ) ( ') ' , ( ) ( ') 't t

q t i t dt g t i t dt

Then 2 11 1 1 1 1 1

2 12 2 2 2 2 2

, ,

, ,

Dq i Di D q and D i q

Dq i Di D q and D i q

The mesh equation becomes

1(0) 0q 1 1(0) (0)q j

1 5 1 1 2 2 5 2 2 44 4

1 1(0)Mq R q q L q R q q v

C C

1 1 3 1 1 2 2 3 44 4

1 1( ) (0)sL q R q q Mq q v t v

C C

2 (0) 0q 2 2(0) (0)q j

Documents

General Methods of Network Analysis Node Analysis Mesh Analysis Loop Analysis Cutset Analysis State variable Analysis Non linear and time varying network