1 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)

2 Outline Introduction and previous work Case 1: Infinite resistive lattice Case 2: Infinite inductive lattice Case 3: Infinite capacitive lattice Conclusion ASEE 2009 (Austin, Texas)

3 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: R eff =R/2 (Aitchison) Infinite 2D honeycomb resistive lattice: R eff =2R/3 (Bartis) General N-dimensional resistive lattices: R eff =2R/M (Osterberg & Inan) This Presentation: Extend previous work to the most general problem of finding the total effective impedance (R eff, L eff, or C eff ) between two adjacent nodes of any infinite uniform N-dimensional resistive, inductive, or capacitive lattice, where N=1, 2, or 3. ASEE 2009 (Austin, Texas)

4 Infinite 2D square lattice ASEE 2009 (Austin, Texas) a b (M=4)

5 Infinite 2D honeycomb lattice ASEE 2009 (Austin, Texas) a b (M=3)

6 ASEE 2009 (Austin, Texas) V a-b + - a b I a = I Infinite resistive lattice I b = I Case 1: Infinite N-D resistive lattice Test circuit

7 Infinite 2D honeycomb lattice ASEE 2009 (Austin, Texas) a b (M=3)

8 ASEE 2009 (Austin, Texas) Case 1: Infinite N-D resistive lattice For an infinite 2D honeycomb resistive lattice where M=3, the effective resistance between any two adjacent nodes is R eff =2R/3 which agrees with Bartis. Therefore:

9 ASEE 2009 (Austin, Texas) a-b + - a b I a = I Infinite inductive lattice I b = I Case 2: Infinite N-D inductive lattice Test circuit

10 Infinite 2D honeycomb lattice ASEE 2009 (Austin, Texas) a b (M=3)

11 ASEE 2009 (Austin, Texas) 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 L eff =2L/3. Therefore:

12 ASEE 2009 (Austin, Texas) V a-b + - a b I a = Q  (t) Infinite capacitive lattice I b = Q  (t) Case 3: Infinite N-D capacitive lattice Test circuit

13 Infinite 2D honeycomb lattice ASEE 2009 (Austin, Texas) a b (M=3)

14 ASEE 2009 (Austin, Texas) 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 C eff =3C/2. Therefore:

Extended previous work to the most general problem of finding the total effective impedance (R eff, L eff, or C eff ) 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. 15 Conclusions ASEE 2009 (Austin, Texas) R eff =2R/M L eff =2L/M C eff =MC/2

16 Any questions? ASEE 2009 (Austin, Texas)