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Theory of impedance networks: A new formulation F. Y. Wu FYW, J. Phys. A 37 (2004) 6653-6673
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R Resistor network
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Ohm’s law R V I Combination of resistors
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= -Y transformation: (1899) Star-triangle relation: (1944) Ising model = =
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-Y relation (Star-triangle, Yang-Baxter relation) A.E. Kenelly, Elec. World & Eng. 34, 413 (1899)
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1 2 3 4 r1r1 r1r1 r1r1 r1r1 r2r2
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1 2 3 4 r1r1 r1r1 r1r1 r1r1 r2r2 3 1 23 1 3 1
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1 2 3 4 r1r1 r1r1 r1r1 r1r1 r2r2 3 1 23 1 3 1 I I/2 I 1 2 3 4 r1r1 r1r1 r1r1 r1r1 r2r2
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1 2 r r r r r r r r r r r r
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1 2 r r r r r r r r r r r r I I/3 I 1 2 r r r r r r r r r r r r I/6
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Infinite square network I/4 I
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V 01 =(I/4+I/4)r I/4 I I
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Infinite square network
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Problems: Finite networks Tedious to use Y- relation 1 2 (a) (b) Resistance between (0,0,0) & (3,3,3) on a 5×5×4 network is
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I0I0 1 4 3 2 Kirchhoff’s law r 01 r 04 r 02 r 03 Generally, in a network of N nodes, Then set Solve for V i
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2D grid, all r=1, I(0,0)=I 0, all I(m,n)=0 otherwise I0I0 (0,0) (0,1) (1,1) (1,0) Define Then Laplacian
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Harmonic functions Random walks Lattice Green’s function First passage time Related to: Solution to Laplace equation is unique For infinite square net one finds For finite networks, the solution is not straightforward.
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General I1I1 I2I2 I3I3 N nodes
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Properties of the Laplacian matrix All cofactors are equal and equal to the spanning tree generating function G of the lattice (Kirchhoff). Example 1 2 3 c3c3 c1c1 c2c2 G=c 1 c 2 +c 2 c 3 +c 3 c 1
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I2I2 I1I1 ININ network Problem: L is singular so it can not be inverted. Day is saved: Kirchhoff’s law says Hence only N-1 equations are independent → no need to invert L
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Solve V i for a given I Kirchhoff solution Since only N-1 equations are independent, we can set V N =0 & consider the first N-1 equations! The reduced (N-1)×(N-1) matrix, the tree matrix, now has an inverse and the equation can be solved.
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II We find Writing 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
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Example 1 2 3 c3c3 c1c1 c2c2 or The evaluation of L & L in general is not straightforward!
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Spanning Trees: x x x x xx yy y y y y y y x G(1,1) = # of spanning trees Solved by Kirchhoff (1847) Brooks/Smith/Stone/Tutte (1940)
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1 4 2 3 x x yy G(x,y)= + x x x xx x ++ yyyyyy =2xy 2 +2x 2 y 1234 1 2 3 4 N=4
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Consider instead Solve V i ( ) for given I i 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.
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Eigenvectors and eigenvalues of L 0 is an eigenvalue with eigenvector L is Hermitian L has real eigenvalues Eigenvectors are orthonormal
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Consider where Let This gives
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Let = orthonormal Theorem:
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Example 1 2 3 4 r1r1 r1r1 r1r1 r1r1 r2r2
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Example: complete graphs N=3 N=2 N=4
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1 23 N-1 r rrr N
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If nodes 1 & N are connected with r (periodic boundary condition)
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New summation identities New product identity
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M×N network N=6 M=5 r s s I N unit matrix s r rr
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M, N →∞
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Finite lattices Free boundary condition Cylindrical boundary condition Moebius strip boundary condition Klein bottle boundary condition
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Klein bottle Moebius strip
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Orientable surface Non-orientable surface: Moebius strip
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Orientable surface Non-orientable surface: Moebius strip
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Free Cylinder
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Klein bottle Moebius strip
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Klein bottle Moebius strip Free Cylinder Torus on a 5×4 network embedded as shown
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Resistance between (0,0,0) and (3,3,3) in a 5×5 × 4 network with free boundary
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