Closed form expression for a resistor ladder network

Is it possible to find a closed form expression for the resistance between A and B for the general n-section ladder network, below, where all resistors are $\small 1\Omega$? An iterative expression is quite easy to derive ($\small R_n =2 + 1//R_{n-1})$

• Looks like it should be possible. After all, the series converges in real life too. – Olin Lathrop Feb 9 '16 at 15:17
• converges to 2xR, or in your case 2OHM as 1R=1Ohm. – ursusd8 Feb 9 '16 at 15:20
• @ursusd8, it must be >2 no matter how big (or small) n is. – Chu Feb 9 '16 at 15:21
• yes it is always bigger, it just converges to 2R. Read here. With every new n, you add, the resistance of the decreases. @n=2 you are pretty far off, 2.75; at n=3 you get to 2.6 etc etc.. you will never hit the 2. – ursusd8 Feb 9 '16 at 15:27
• @urusd8, actually, it converges to $\small 1+\sqrt{3}$, next to 2.75 comes 2.733 – Chu Feb 9 '16 at 15:29

By inspection,the maximum possible value of resistance is $\small3\Omega$ (for $\small n=1$), and for $\small n>1$ the resistance must be $\small >2\Omega$, hence $\small 2< R_n\le 3$

Calculating the effective resistance values, $\small R_n$, for $\small n=1, \:2,\: 3,\: 4,\: 5,\: 6\:...$ gives:

$$\small[3,\:2\frac{3}{4},\: 2\frac{11}{15},\: 2\frac{41}{56},\: 2\frac{153}{209},\: 2\frac{571}{780}\:... ]$$

Initially considering just the fractional parts, and noting that $\small 3= 2\frac{1}{1}$, we may write the sequence:$$\small [1,\:1,\:3,\:4,\:11,\:15,\:41,\:56,\:153,\:209,\:571,\:780\:...]$$

Searching a catalogue of z-transform generating functions for this particular sequence (http://www.lacim.uqam.ca/~plouffe/articles/MasterThesis.pdf) gives:

$$\small F(z)=\frac{z^2+z-1}{z^4-4z^2+1}$$

The denominator factorises to a very convenient form (what luck!), and the generating function may be expressed in partial fractions as:

$$\small F(z)=\frac{A}{z-a}+\frac{B}{z+a}+\frac{C}{z-b}+\frac{D}{z+b}$$

where the constants: a, b, A, B, C, D are:

$\small a=\sqrt {2+\sqrt3}$,

$\small b= \sqrt{2-\sqrt3}$,

$\small A=\frac{a^2+a-1}{2a(a^2-b^2)}$,

$\small B=\frac{a^2-a-1}{2a(b^2-a^2)}$,

$\small C=\frac{b^2+b-1}{2b(b^2-a^2)}$,

$\small D=\frac{b^2-b-1}{2b(a^2-b^2)}$

Inverse z-transforming $\small F(z)$ gives the closed form expression for the sequence:

$$\small f(k)=A(a)^k +B(-a)^k+C(b)^k+D(-b)^k$$

The resistance value, $\small R_n$, for n sections can now be obtained by evaluating the last equation with $\small k=2n$ and $\small k=2n-1$, to form the denominator and numerator of the fractional part of $\small R_n$; and then adding $\small 2\Omega$:

$$\small R_n = 2+\frac{A(a)^{2n-1}+B(-a)^{2n-1}+C(b)^{2n-1}+D(-b)^{2n-1}}{A(a)^{2n}+B(-a)^{2n}+C(b)^{2n}+D(-b)^{2n}}$$

This is the required closed form expression.

• A really great answer. It can be written a bit more simply: $$\frac{2 \sqrt{3} \left(2+\sqrt{3}\right)^n}{\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n}-\sqrt{3}+1$$ (But I cheated and used Mathematica.) – Oleksandr R. Feb 10 '16 at 1:39
• @Oleksandr R. that's neat. – Chu Feb 10 '16 at 1:43
• I had wanted to write it like this, but the comment box was resisting at first: $$\frac{ \left(-2+\sqrt{3}\right)^n \left(-1+\sqrt{3}\right) + \left(-2-\sqrt{3}\right)^n \left(1+\sqrt{3}\right)} {\left(-2-\sqrt{3}\right)^n - \left(-2+\sqrt{3}\right)^n}$$ I think this way it has a pleasing symmetry to it. – Oleksandr R. Feb 10 '16 at 1:49
• Just goes to show how important 'proper' fractions are! The analysis would not be possible with decimal fractions. – Chu Feb 10 '16 at 1:57

You can solve this by assuming that the series converges. If this is true, then adding one more section makes an infinitesimal change in the result. Therefore, the value that they are converging to must satisfy:

$$X = 2 + \frac{1}{1 + \frac{1}{X}}$$

Solving for X gives $1 \pm \sqrt{3}$. Since only a positive result makes sense, the value is 2.7320508...

But that isn't the question you actually asked. I'm afraid my math skills aren't up to the task of coming up with a closed form solution for a finite number of sections. Just working out the first few steps gives:

• $R_1 = 3$
• $R_2 = \frac{11}{4}$
• $R_3 = \frac{41}{15}$
• $R_4 = \frac{153}{56}$

The numbers don't seem to form a power series, so there's probably an exponential involved...

• Dave, I've found the generating function for the sequence. I'll post the answer when I get to a proper keyboard later. – Chu Feb 9 '16 at 16:51

Great answers. You might wonder how Mathematica is able to find the answer. We have $$R_{n+1} = 2 + (1\;\|\;R_n),\;\text{with }R_0 = 3$$ or $$R_{n+1} = 2 + \frac{1}{1 + \frac{1}{R_n}} = 2 + \frac{R_n}{R_n+1}=2+\frac{R_n+1-1}{R_n+1}=3-\frac{1}{R_n+1},$$ Let's look at the sequence $x_n = R_n + 1$, to get a proper divisor: $$x_{n+1} = R_{n+1} + 1 = 4 - \frac{1}{R_n+1}=4 - \frac{1}{x_n}$$ I.e.: $$\{x_n\}_0^\infty = \left\{4,\;4-\frac{1}{4},\;4-\frac{4}{15},\;\ldots\right\}$$ The trick now is to define another sequence $$y_{n+1} = y_{n} x_{n},\qquad y_0 = 1,$$ Then $$\frac{y_{n+2}}{y_{n+1}} = x_{n+1} = 4 - \frac{1}{x_n} = 4 - \frac{y_n}{y_{n+1}},$$ Which gives $$y_{n+2} - 4y_{n+1} + y_n = 0$$ This linear recurrence can be solved by standard techniques. Its characteristic equation ($z^2-4z+1=0$), has zeros $z_{1,2}=2\pm\sqrt{3},$ so the recurrence can be written $$y_n = A(2-\sqrt{3})^n+B(2+\sqrt{3})^n,$$ for some $A$ and $B$. Plugging in $y_0 = 1$ gives $B=1-A$. From $y_1 = y_0 x_0 = 4$ we get $$4 = A(2-\sqrt{3})+(1-A)(2+\sqrt{3})\\ A = \frac{1}{2} - \frac{1}{3}\sqrt{3}$$

Finally, we get the closed form solution from $$R_n = x_n - 1 = \frac{y_{n+1}-y_n}{y_n} = \frac{A(1-\sqrt{3})(2-\sqrt{3})^n+B(1+\sqrt{3})(2+\sqrt{3})^n}{A(2-\sqrt{3})^n+B(2+\sqrt{3})^n}$$