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