Theory of electric networks: The two-point resistance and impedance

F. Y. Wu

Northeastern UniversityBoston, Massachusetts USA

Z

Impedance network

?Z

Ohm’s law

Z

V

I

I

VZ

Combination of impedances

1z 2z

1z

2z

21 zzz

21

111

zzz

In the phasor notation, impedance for inductance L is

Ljz

Impedance for capacitance C is

Cjz /1

where 1j .

Impedances

=

-Y transformation: (1899)

z

zzZ Z

Z Zz

ZZ

ZZz

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

z1 z1

z1

z1

z2

?13 Z

1

2 3

4

z1 z1

z1

z1

z2

3

1

2 3

1

3

1

113 zZ ?13 Z

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

z01z04

z02

z03

04

40

03

30

02

20

01

10

040302010

z

VV

z

VV

z

VV

z

VV

IIIII

Generally, in a network of N nodes,

N

ijjji

iji VV

zI

,1

1

Then set )( iii II I

VVZ

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

,1

,1,1

N

ijjjijii

N

ijjji

iji VYVYVV

zI

ijij

N

ijj iji z

Yz

Y1

,1

,1

NNNNN

N

N

I

I

I

V

V

V

YYY

YYY

YYY

2

1

2

1

21

2221

1121

The sum of each row or column is zero !

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

y3

y1

y2 G=y1y2+y2y3+y3y1

2112

1313

2332

yyyy

yyyy

yyyy

L

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

General I1 I2

I3

N nodes

,1

,1,1

N

ijjjijii

N

ijjji

iji VYVYVV

zI

ijij

N

ijj iji z

Yz

Y1

,1

,1

NNNNN

N

N

I

I

I

V

V

V

YYY

YYY

YYY

2

1

2

1

21

2221

1121

The sum of each row or column is zero !

I2I1

IN

network

Problem: L is singular so it cannot 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

YYY

YYY

YYY

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

1,2221

1,1121

NNNNN

N

N

I

I

I

V

V

V

YYY

YYY

YYY

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

Example1

2 3

y3

y1

y2

2112

1313

2332

yyyy

yyyy

yyyy

L

133221211

1311 yyyyyy

yyy

yyyL

2112 yyL

133221

21

1

1212 yyyyyy

yy

L

Lz

32112

111

zzzz

or

The evaluation of L & L in general is not straightforward!

)( iii II

I I Kirchhoff result:

Writing

L

LZ

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.

But the evaluation of Lfor general network is involved.

trees)spanning(

) and rootsh forest wit (spanning

G

G

NNNNN

N

N

I

I

I

V

V

V

YYY

YYY

YYY

2

1

2

1

21

2221

1121

)(

)(

)(

For resistors, z and y are real so L is Hermitian, we can then consider instead the eigenvalue equation

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

YYY

YYY

YYY

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

)(

Z and

0 sinceout drops 1

Term i

iIN

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

0 and

,,2,1

1

Ni

L iii

For resistors let

iN

i

i

i

2

1

= orthonormal

Theorem for resistor networks:

2

2

1

ii

N

i i

R

This is the main result of FYW, J. Phys. A37 (2004) 6653-6679 whichmakes use of the fact that L is hermitian and is orthonormal

Corollary:

)1

)((1 2

22

ii

N

i i

N

iiN

) and rootsh forest wit spanning( G

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 →∞

Resistance between two corners of an N x N

square net with unit resistance on each edge

2ln

2

141ln

4NRNxN

......082069878.0ln4

N

where ...5772156649.0 Euler constant

N=30 (Essam, 1997)

Finite lattices

Free boundary condition

Cylindrical boundary condition

Moebius strip boundary condition

Klein bottle boundary condition

Klein bottleMoebius 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

In the phasor notation, impedance for inductance L is

Ljz

Impedance for capacitance C is

Cjz /1

where 1j .

Impedances

For impedances, Y are generally complex and the matrixL is not hermitian and its eigenvectors are not orthonormal; the resistor result does not apply.

But L^*L is hermitian and has real eigenvelues.We have

N1,2,..., , 0 , ^* α LL

with

.

1

:

:

1

1

N

1 0, 11

N1,2,..., , 0 , ^* α LL

Theorem

Let L be an N x N symmetric matrix with complexMatrix elements and

Then, there exist N orthnormal vectors u

satisfying the relation

NuuL ,...,2,1 *,

where * denotes complex conjugate and

real. ,

ie

Remarks:

For nondegenerate one has simply

u

For degenerate

,

, one can construct

as linear combinations of u

ieLv *)(

0 and

,,2,1

*

1

Ni

uLu iii

For impedances let

iN

i

i

i

u

u

u

u

2

1

= orthonormal

Theorem for impedance networks:

2,0 if ,

2 ,0 if ,)(1 2

2

i

iuuZ

i

iii

N

i i

This is the result of WJT and FYW, J. Phys. A39 (2006) 8579-8591

The physical interpretation of Z

is the occurrence of a resonance such asin a parallel combination of inductance Land capacitance C the impedance is

Z =

Lj

Cj

Cj

Lj

))((

LC/1at , =

Generally in an LC circuit there can exist multiple resonances at frequencies where . 0i

In the circuit shown, 15 resonance frequencies at

M=6N=4

.3,..,1;5,..,1 ,1

)2/sin(

)2/sin( nm

LCMm

Nnmn

Summary

• An elegant formulation of computing two-point impedances in a network, a problem lingering since the Kirchhoff time.

• Prediction of the occurrence of multi-resonances in a network consisting of reactances L and C, a prediction which may have practical relevance.

FYW, J. Phys. A 37 (2004) 6653-6673

W-J Tzeng and FYW, J. Phys. A 39 (2006), 8579-8591