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
Basic facts of node analysis
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
Basic facts of node analysis
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
Basic facts of node analysis
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
Node analysis of linear time invariant networks
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)
Node analysis of linear time invariant networks
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)
Node analysis of linear time invariant networksExample 4
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
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networks
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)
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networks
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
Node analysis of linear time invariant networksExample 7
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
Node analysis of linear time invariant networks
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
Mesh analysis of linear time invariant networks
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
Mesh analysis of linear time invariant networks
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
Mesh analysis of linear time invariant networks
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
Mesh analysis of linear time invariant networks
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