1 Calculation of the general impedance between adjacent nodes of infinite uniform N-dimensional resistive, inductive, or capacitive lattices Peter M. Osterberg

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Calculation of the general impedance between adjacent nodes of infinite uniform N-dimensional resistive,

inductive, or capacitive lattices

Peter M. Osterberg & Aziz S. Inan

School of Engineering

University of Portland

ASEE 2009 (Austin, Texas)

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Outline• Introduction and previous work

• Case 1: Infinite resistive lattice

• Case 2: Infinite inductive lattice

• Case 3: Infinite capacitive lattice

• Conclusion


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Introduction and previous work

• Special “adjacent node” circuit configurations of uniform, infinite extent are

excellent pedagogical vehicles for teaching and motivating EE students (e.g., superposition, symmetry, etc).

• Infinite 2D square resistive lattice: Reff=R/2 (Aitchison)

• Infinite 2D honeycomb resistive lattice: Reff=2R/3 (Bartis)

• General N-dimensional resistive lattices: Reff=2R/M (Osterberg & Inan)

• This Presentation: Extend previous work to the most general problem of finding the total effective impedance (Reff, Leff, or Ceff) between two adjacent nodes of any infinite uniform N-dimensional resistive, inductive, or capacitive lattice, where N=1, 2, or 3.


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Infinite 2D square lattice


a

b

(M=4)

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Infinite 2D honeycomb lattice


a

b

(M=3)

6ASEE 2009 (Austin, Texas)

Va-b

+

-

a

b

Ia = I

Infinite resistive lattice

Ib = I

Case 1: Infinite N-D resistive lattice

Test circuit

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a

b

(M=3)


Case 1: Infinite N-D resistive lattice

IRRM

I

RM

IR

M

IV

eff

bab-a

2

M

R

I

VR

2b-aeff

For an infinite 2D honeycomb resistive lattice where M=3, the effective resistance between any two adjacent nodes is Reff=2R/3 which agrees with Bartis.

Therefore:


a-b

+

-

a

b

Ia = I

Infinite inductive lattice

Ib = I

Case 2: Infinite N-D inductive lattice

Test circuit

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a

b

(M=3)


Case 2: Infinite N-D inductive lattice

For an infinite 2D honeycomb inductive lattice where M=3, the effective inductance between any two adjacent nodes is Leff=2L/3.

Therefore:

ILLM

I

LM

IL

M

I

eff

bab-a

2

M

L

IL

2b-aeff


Va-b

+

-

a

b

Ia = Q(t)

Infinite capacitive lattice

Ib = Q(t)

Case 3: Infinite N-D capacitive lattice

Test circuit

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a

b

(M=3)


Case 3: Infinite N-D capacitive lattice

For an infinite 2D honeycomb capacitance lattice where M=3, the effective capacitance between any two adjacent nodes is Ceff=3C/2.

Therefore:

eff

bab-a

2C

Q

C

MQC

MQ

C

MQV

2b-aeff

MC

V

QC

• Extended previous work to the most general problem of finding the total effective impedance (Reff, Leff, or Ceff) between any two adjacent nodes of any infinite uniform N-dimensional resistive, inductive, or capacitive lattice, where N=1, 2, or 3.

• Using Kirchhoff’s laws, superposition and symmetry, along with Ohm’s law for a resistive lattice, or the magnetic flux/current relationship for an inductive lattice, or the electrical charge/voltage relationship for a capacitive lattice, a general and easy-to-remember solution is determined for each of these cases, as follows, where M is the total number of elements connected to any node of the lattice:

• These general solutions are simple and of significant pedagogical interest to undergraduate EE education.

• They can serve as illustrations of the simplicity and elegance of superposition and symmetry used with appropriate fundamental physical principles to facilitate more intuitive understanding and simpler solution of such general problems.

• The authors have recently added electrical circuit problems of this type to their first-semester undergraduate electrical circuits course and received good feedback from the students.

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Conclusions


Reff=2R/M Leff=2L/M Ceff=MC/2

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Any questions?


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1 Calculation of the general impedance between adjacent nodes of infinite uniform N-dimensional resistive, inductive, or capacitive lattices Peter M. Osterberg