theory of impedance networks: a new formulation f. y. wu fyw, j. phys. a 37 (2004) 6653-6673
TRANSCRIPT
Theory of impedance networks:A new formulation
F. Y. Wu
FYW, J. Phys. A 37 (2004) 6653-6673
R
Resistor network
?R
Ohm’s law
R
V
I
I
VR
Combination of resistors
1r 2r
1r
2r
21 rrr
21
111
rrr
=
-Y transformation: (1899)
r
rrR R
R Rr
RR
RRr
3
12
)2(2
Star-triangle relation: (1944)
1
32
Ising model
J
JJR
R
R
1
2 3
=
)()( 133221321
RJ Fee =
)(cosh2 321 J
1r
2r3r2R
3R
1R
-Y relation (Star-triangle, Yang-Baxter relation)
A.E. Kenelly, Elec. World & Eng. 34, 413 (1899)
321
321 RRR
RRr
321
132 RRR
RRr
321
213 RRR
RRr
133221
321321
11 111
111)(
1
rrrrrr
rrrrrr
rR
133221
321321
22 111
111)(
1
rrrrrr
rrrrrr
rR
133221
321321
33 111
111)(
1
rrrrrr
rrrrrr
rR
1
2 3
4
r1 r1
r1
r1
r2
?13 R
1
2 3
4
r1 r1
r1
r1
r2
3
1
2 3
1
3
1113 rR
?13 R
1
2 3
4
r1 r1
r1
r1
r2
3
1
2 3
1
3
1113 rR
?13 R
I
I/2I/2
I/2
I/2I
1
2 3
4
r1 r1
r1
r1
r2
112
112
1
13 rI
IrIrR
1
2
r
r
r
r
r
r
r
r
r
r
r
r
?12 R
1
2
r
r
r
r
r
r
r
r
r
r
r
r
rI
VR
IrrI
rI
rI
V
6
56
5
363
1212
12
I
I/3
I/3
I/3
I/3
I
1
2
r
r
r
r
r
r
r
r
r
r
r
r
?12 R
I/3
I/6I/6
Infinite square network
I/4I/4
I/4I/4
I
V01=(I/4+I/4)r
I/4I/4
I/4I/4
I
I
I/4
201
01
r
I
VR
Infinite square network
2
1
38
2
1
2
24
2
17
4
2
4
2 2
14
2
14
0
24
2
17 4
3
46
43
46
823
2
1
3
4
2
1
3
4
Problems:
• Finite networks• Tedious to use Y- relation
1
2
r
rR
)7078.1(
027,380,1
898,356,212
(a)
(b) Resistance between (0,0,0) & (3,3,3) on a 5×5×4 network is
r
rR
)929693.0(
225,489,567,468,352
872,482,658,687,327
I0
1
4
3
2 Kirchhoff’s law
r01r04
r02
r03
04
40
03
30
02
20
01
10
040302010
r
VV
r
VV
r
VV
r
VV
IIIII
Generally, in a network of N nodes,
N
ijjji
iji VV
rI
,1
1
Then set )( iii II I
VVR
Solve for Vi
2D grid, all r=1, I(0,0)=I0, all I(m,n)=0 otherwise
I0
(0,0)
(0,1) (1,1)
(1,0)
00,0,),(4)1,()1,(),1(),1( InmVnmVnmVnmVnmV nm
Define
)1()(2)1()(
)1()()(2
nfnfnfnf
nfnfnf
n
n
Then 00,0,22 ),()( InmV nmnm
Laplacian
Harmonic functionsRandom walksLattice Green’s functionFirst passage time
• Related to:
• Solution to Laplace equation is unique
• For infinite square net one finds
2
0
2
02 2coscos
)(exp
)2(2
1),(
nmiddnmV
• For finite networks, the solution is not straightforward.
General I1 I2
I3
N nodes
N
ijjjijii
N
ijjji
iji VCVCVV
rI
,1,1
1
ijij
N
ijj iji r
Cr
C1
,1
,1
NNNNN
N
N
I
I
I
V
V
V
CCC
CCC
CCC
2
1
2
1
21
2221
1121
Properties of the Laplacian matrix
All cofactors are equal and equal to the spanning tree generating function G of the lattice (Kirchhoff).
Example1
2 3
c3
c1
c2 G=c1c2+c2c3+c3c1
2112
1313
2332
cccc
cccc
cccc
L
I2I1
IN
network
Problem: L is singular so it can not be inverted.
Day is saved:
Kirchhoff’s law says 01
N
jjI
Hence only N-1 equations are independent → no need to invert L
NNNNN
N
N
I
I
I
V
V
V
CCC
CCC
CCC
2
1
2
1
21
2221
1121
Solve Vi for a given I
Kirchhoff solutionSince only N-1 equations are independent, we can set VN=0 & consider the first N-1 equations!
1
2
1
1
2
1
12,11,1
2221
1121
NNNNN
N
N
I
I
I
V
V
V
CCC
CCC
CCC
The reduced (N-1)×(N-1) matrix, the tree matrix, now has an inverse and the equation can be solved.
0
0
131211
131211
131211
zcycxc
zbybxb
Izayaxa
0
0
333231
232221
131211 I
z
y
x
aaa
aaa
aaa
333231
232221
131211
aaa
aaa
aaa
3332
2322
1312
0
0
aa
aa
aaI
x
3331
2321
1311
0
0
aa
aa
aIa
y
0
0
3231
2221
1211
aa
aa
Iaa
z
)( iii II
I I
We find
Writing
L
LR
Where L is the determinant of the Laplacian with the -th row & column removed
L= the determinant of the Laplacian with the -th and -th rows & columns removed
Example1
2 3
c3
c1
c2
2112
1313
2332
cccc
cccc
cccc
L
133221211
1311 cccccc
ccc
cccL
2112 ccL
133221
21
1
1212 cccccc
cc
L
Lr
32112
111
rrrr
or
The evaluation of L & L in general is not straightforward!
Spanning Trees:
x
x
x
x
xx
y y
y
y
y
y
y
y
xS.T all
21),( nn yxyxG
G(1,1) = # of spanning trees
Solved by Kirchhoff (1847) Brooks/Smith/Stone/Tutte (1940)
1
4
2
3x
x
y y G(x,y)= +
x
x
x
xx
x
+ +yyyy y y
=2xy2+2x2y
yxxy
xyxy
yyxx
yxyx
yxL
0
0
0
0
),(
1 2 3 4
1
2
3
4
LN
yxG of seigenvalue nonzero ofproduct 1
),(
N=4
1
2
1
1
2
1
12,11,1
2221
1121
)(
)(
)(
NNNNN
N
N
I
I
I
V
V
V
CCC
CCC
CCC
Consider instead
Solve Vi () for given Ii and set =0 at the end.
This can be done by applying the arsenal of linear algebra and deriving at a very simple result for 2-point resistance.
Eigenvectors and eigenvalues of L
1
1
1
0
1
1
1
21
2221
1121
NNN
N
N
CCC
CCC
CCC
L
0 is an eigenvalue with eigenvector
1
1
1
1
N
L is HermitianL has real eigenvaluesEigenvectors are orthonormal
IGV
IVL
)()(
)()(
Consider
where1)]([)( LG
i
i
2i
1 :)( of sEigenvalue
:)( of sEigenvalue
, ,0 :)0( of sEigenvalue
G
L
L
LLet
This gives
N
i i
ii
NG
2
*1
)(
R and
0 sinceout drops 1
Term i
iIN
0 and
,,2,1
1
Ni
L iii
Let
iN
i
i
i
2
1
= orthonormal
Theorem:
2
2
1
ii
N
i i
R
Example
1
2 3
4
r1 r1
r1
r1
r2
21121
111
21211
111
2
20
2
02
ccccc
ccc
ccccc
ccc
L
)1,0,1,0(2
1 ),(2
)0101(2
1 ,2
)1111(2
1 ,4
3214
313
212
cc
,,,c
,,,c
)(4
)32()(
1)(
1)(
1
)(1
)(1
)(1
21
21124441
4
23431
3
22421
214
12
43414
23331
3
22321
213
rr
rrrr
rr
Example: complete graphs
111
111
111
1
N
N
N
rL
N=3
N=2
N=4
110
121 ),/2exp(1
121 ,
,00
,N-,,α
,,N-,,nNniN
,N-,,nr
N
n
n
rN
R2
1 2 3 N-1r rr r N
100
021
011
1
rL
r
Nn
Nn
Nn
N
rR
N
i
1
1
2
cos1
)21
cos()21
cos(
N
n
N
N
N
n
n
n
)2
1cos(
2
1
cos12
0
If nodes 1 & N are connected with r (periodic boundary condition)
][ /1
2cos1
/2exp/2exp
2
1
1
2Per
Nr
Nn
NniNni
N
rR
N
iαβ
NniN
N
n
n
n
/2exp1
2cos12
201
021
112
1
rL
New summation identities
1
0 coscos
cos1
)(N
n
Nn
Nn
l
NlI
NlNN
lNlI
l
20 ,2/cosh4
)1(1
sinh
11
sinhsinh
)cosh()(
221
NlN
lNlI
0 ,
2/sinhsinh
)2/cosh()(2
New product identity
2sinh
2coscosh
1
0
2 N
N
nN
n
M×N network
N=6
M=5
r
s
sNMNMNM TIs
ITr
L 11
1000
0210
0121
0011
NT IN unit matrix
1,,2,1 ,2
1cos
2
0 ,1
2cos1
22cos1
2
)(
)()(),(
),(
NN
n
N
N
N
n
rM
m
s
Nn
Nn
Mmnm
nm
s
r
rr
M, N →∞
Finite lattices
Free boundary condition
Cylindrical boundary condition
Moebius strip boundary condition
Klein bottle boundary condition
Klein bottleMoebius strip
Orientable surface
Non-orientable surface: Moebius strip
Orientable surface
Non-orientable surface: Moebius strip
Free
Cylinder
Klein bottle
Moebius strip
Klein bottle
Moebius strip
Free
Cylinder
Torus
)3,3)(0,0(R on a 5×4 network embedded as shown
Resistance between (0,0,0) and (3,3,3) in a 5×5×4 network with free boundary