Errata

Node2

Next: Bibliography Up: Equivalent resistance for infinite Previous: Infinite ladder of resistors

Subsections


Infinite 2D square grid of $1\Omega$ resistors

The infinite 2-dimensional square grid of resistors, assumed to be $1\Omega$ each, shown in figure 2.



Figure 2: Infinite resistor grid (2D square)
\begin{figure}\psfig{file=ResistorGrid2D.ps,width=7.5in,angle=270}\end{figure}

In this section, the Fourier technique used in section 1 is expanced to the 2-dimensional grid. This method sets up an integral for this resistance between any two nodes, and a solution is found for the resistance across the diagonal of a single square (between nodes (0,0) and (1,1) in Figure 2.

The set-up

Let $m$ be the horizontal position, and $n$ be the vertical position, of a given node (where 4 resistors connect). If current $I_{m,n}$ is injected at that node (and allowed to flow out toward infinity), then this current induces voltages as follows from Kirchoff's and Ohm's laws through the $1\Omega$ resistors:
$\displaystyle I_{m,n}$ $\textstyle =$ $\displaystyle (V_{m,n}-V_{m+1,n})+(V_{m,n}-V_{m-1,n})+(V_{m,n}-V_{m,n+1})+(V_{m,n}-V_{m,n-1})$ (25)
  $\textstyle =$ $\displaystyle 4V_{m,n}-V_{m+1,n}-V_{m-1,n}-V_{m,n+1}-V_{m,n-1}$ (26)

where $V_{m,n}$ is the potential at grid node $(m,n)$ due to the current. The potential difference between nodes $(m,n)$ and $(m',n')$, due to injected current, is $V_{m,n}-V_{m',n'}$.

Defining the 2-dimensional discrete-space Fourier transform pairs

$\displaystyle I_{m,n}$ $\textstyle =$ $\displaystyle \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} du dv I(u,v)e^{\jmath(mu+nv)}$ (27)
$\displaystyle I(u,v)$ $\textstyle =$ $\displaystyle \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} I_{m,n} e^{-\jmath(mu+nv)}$ (28)
$\displaystyle V_{m,n}$ $\textstyle =$ $\displaystyle \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} du dv V(u,v)e^{\jmath(mu+nv)}$ (29)
$\displaystyle V(u,v)$ $\textstyle =$ $\displaystyle \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} V_{m,n} e^{-\jmath(mu+nv)}$ (30)

Then, using equation 29, equation 26 becomes


$\displaystyle \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} du dv I(u,v)e^{\jmath(mu+nv)}$ $\textstyle =$ $\displaystyle \left.\left. \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi}... ...h(mu+nv)} \right[4-e^{\jmath u}-e^{-\jmath u}-e^{\jmath v}-e^{-\jmath v}\right]$ (31)

In order for the above relation to hold for any set of $I_{m,n}$ (or, equivalently, $I(u,v)$), the integrands must be equal.

$\displaystyle I(u,v)$ $\textstyle =$ $\displaystyle 2V(u,v)\left[2-\cos(u)-\cos(v)\right]$ (32)

Current sources of 1A at (0,0) and -1A at $(n,m)$ are written as $I_{m,n} = \delta_{m,n}-\delta_{m-M,n-N}$, which has a Fourier transform (by equation 28) of $I(u,v) = 1-e^{-\jmath(Mu+Nv)}$.

$\displaystyle V(u,v)$ $\textstyle =$ $\displaystyle \frac{1}{2} \frac{1-e^{-\jmath(Mu+Nv)}}{2-\cos(u)-\cos(v)}$ (33)

The equivalent resistance between nodes (0,0) and (M,N) is the voltage between the nodes

$\displaystyle R_{M,N}$ $\textstyle =$ $\displaystyle V_{0,0}-V_{M,N}$ (34)
  $\textstyle =$ $\displaystyle \left.\left. \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} du dv  V(u,v)\right[1-e^{\jmath(Mu+Nv)}\right]$ (35)
  $\textstyle =$ $\displaystyle \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} du dv  \frac{1}{2}\frac{(1-e^{-\jmath(Mu+Nv)})(1-e^{\jmath(Mu+Nv)})}{2-\cos(u)-\cos(v)}$ (36)
  $\textstyle =$ $\displaystyle \frac{1}{4\pi^2}\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} du dv  \frac{1-\cos(Mu+Nv)}{2-\cos(u)-\cos(v)}$ (37)

This has ``simplified'' the resistance to a definite integral. The rest of this effort is just bending the integral to produce the number of interest.

As an aside, superposition can be more directly invoked to find the equivalent resistance between adjacent nodes, say (0,0) and (0,1), in Figure 2. If 1A is injected in node (0,0), by symmetry the current is evenly divided among the four connected resistors. This 0.25A current results in a 0.25V potential between nodes (0,0) and (1,1). Similarly, the -1A sink at node (0,1) draws current equally through each connected resistor, resulting in a 0.25V potential difference between nodes (0,0) and (0,1). Superposition has these acting in concert to produce a potential difference of 0.5V across that resistor, resulting in the equivalent resistance of $0.5\Omega$. This is the same answer as obtained by evaluating equation 37 with either (M,N) = (0,1) or (1,0).

As another aside, it is straightforward to see the extension for N-dimensional case. If the N-dimensional nodes are represented by the coordinates $(n_1, n_2, n_3, \cdots, n_N)$ in lattice space and $(u_1, u_2, u_3, \cdots, u_N)$ in Fourier space, then the resistance separated by $(n_1, n_2, n_3, \cdots, n_N)$ is

$\displaystyle R_{n_1, n_2, \cdots, n_N}$ $\textstyle =$ $\displaystyle \left(\frac{1}{2\pi}\right)^N \int_{-\pi}^{\pi}\cdots\int_{-\pi}^... ...1-\prod_{i=1}^N \cos\left(n_i u_i\right)} {N-\sum_{i=1}^N \cos\left(u_i\right)}$ (38)

The general solution is left for the student. The particular solution for $N=2, n_1=1, n_2=1$ (or, equivalently from equation 37, $R_{1,1}$) is given below.

Finding the equivalent diagonal resistance: integration grinding

In this section, the details for the solution to equation 37 with $M=N=1$ are shown. Since the denominator is even, and the limits are even, the odd parts of the numerator will not contribute to the integral. The odd parts of $\cos(u+v)=\cos(u)\cos(v)+\sin(u)\sin(v)$ are $\sin(u)\sin(v)$, and that term can be discarded. Since the resulting integrand is even in both $u$ and $v$, each integral evaluates to twice the value when the limits for each is changed to $(0,\pi)$.
$\displaystyle R$ $\textstyle =$ $\displaystyle \frac{1}{\pi^2}\int_0^{\pi}\int_0^{\pi} du dv \frac{1-\cos(u)\cos(v)}{2-\cos(u)-\cos(v)}$ (39)
  $\textstyle =$ $\displaystyle \frac{1}{\pi^2} \int_0^{\pi} du \int_0^{\pi} dv \frac{1}{[2-\... ... \int_0^{\pi} du \cos(u)\int_0^{\pi} dv \frac{\cos(v)}{[2-\cos(u)]-\cos(v)}$ (40)

From any table of integrals (for example, the CRC Handbook of Chemistry and Physics, 60th ed. #341 pg. A-69, using case #1 for which $a^2 > b^2$, with $a=2-\cos(u)$ and $b=-1$),
$\displaystyle \int\frac{1}{[2-\cos(u)]+(-1)\cos(v)}$ $\textstyle =$ $\displaystyle \frac{2}{\sqrt{[2-\cos(u)]^2-1}}\tan^{-1}\left[\frac{3-\cos(u)}{\sqrt{[2-\cos(u)]^2-1}}\tan\left(\frac{v}{2}\right)\right]$ (41)

Evaluating this at the endpoints of the integration limits, note that $\tan^{-1}[C(u)\tan(0)]=0$ and $\tan^{-1}[C(u)\tan(\pi/2)] = \pi/2$ when $u < \pi$. Then
$\displaystyle \int_0^{\pi} dv \frac{1}{[2-\cos(u)]-\cos(v)}$ $\textstyle =$ $\displaystyle \frac{\pi}{\sqrt{[2-\cos(u)]^2-1}}$ (42)

Also, back to the integral tables (e.g. CRC #370 pg. A-72, with $a=2-\cos(u)$ and $b=-1$),
$\displaystyle \int dv \frac{\cos(v)}{[2-\cos(u)]-\cos(v)}$ $\textstyle =$ $\displaystyle -x+[2-\cos(u)]\int dx \frac{1}{[2-\cos(u)]-\cos(v)}$ (43)
$\displaystyle \int_0^{\pi} dv \frac{\cos(v)}{[2-\cos(u)]-\cos(v)}$ $\textstyle =$ $\displaystyle -\pi+[2-\cos(u)]\int_0^{\pi} dx \frac{1}{[2-\cos(u)]-\cos(v)}$ (44)
  $\textstyle =$ $\displaystyle -\pi + [2-\cos(u)]\frac{\pi}{\sqrt{[2-\cos(u)]^2-1}}$ (45)
  $\textstyle =$ $\displaystyle \pi\left[\frac{2-\cos(u)}{\sqrt{[2-\cos(u)]^2-1}}-1\right]$ (46)

Combining these integration results in equation 40.
$\displaystyle R$ $\textstyle =$ $\displaystyle \frac{1}{\pi}\int_0^{\pi} du \left[\frac{1}{\sqrt{[2-\cos(u)]^2-1}}-\cos(u)\left[\frac{2-\cos(u)}{\sqrt{[2-\cos(u)]^2-1}}-1\right]\right]$ (47)
  $\textstyle =$ $\displaystyle \frac{1}{\pi}\int_0^{\pi} du  \frac{[1-\cos(u)]^2}{\sqrt{[2-\cos(u)]^2-1}}+\cos(u)$ (48)
  $\textstyle =$ $\displaystyle \frac{1}{\pi}\int_0^{\pi} du \frac{[1-\cos(u)]^2}{\sqrt{[2-\cos(u)]^2-1}}$ (49)

where the $\cos(u)$ term integrated to 0. Changing variables to $x=1-\cos(u)$, then $[2-\cos(u)]^2-1 = [1+x]^2-1 = x(2+x)$ and the limits $(0,\pi)$ become $(0,2)$ and the Jacobian is

\begin{displaymath} dx = \sin(u) du = \sqrt{1-\cos^2(u)} du = \sqrt{1-(1-x)^2} du = \sqrt{x(2-x)} du \end{displaymath}

Equation 49 gets transformed to
$\displaystyle R$ $\textstyle =$ $\displaystyle \frac{1}{\pi} \int_0^2 dx  \frac{1}{\sqrt{x(2-x)}}\frac{x^2}{\sqrt{x(2+x)}}$ (50)
  $\textstyle =$ $\displaystyle \frac{1}{\pi} \int_0^2 dx \frac{x}{\sqrt{4-x^2}}$ (51)
  $\textstyle =$ $\displaystyle \frac{1}{\pi}\left[-\sqrt{4-x^2}\right]_0^2$ (52)
  $\textstyle =$ $\displaystyle \frac{2}{\pi}$ (53)

The resistance across the diagonal of one square in an infinite grid of 1-Ohm resistors is \framebox{$R = \frac{\mbox{\normalsize $2$}}{\mbox{\normalsize $\pi$}}$}.


Next: Bibliography Up: Equivalent resistance for infinite Previous: Infinite ladder of resistors
frooha@yahoo.com 2004-04-30
Comments