I am interested in finding the impedance as a function of frequency of the \$n^{th}\$ circuit in this series:


simulate this circuit – Schematic created using CircuitLab

(By request: this is intended to be a model of the internal impedance of a battery.)

I've tried simulating this circuit in LTSpice, which worked well. However, I can't easily change \$n\$ that way: I have to manually edit the schematic to increase the number of branches. I want a more general solution, where I can plug in values for \$n, R_1, R_2, R\$, and \$C\$.

  • How can I attempt to find an analytic solution for the impedance of this circuit? Are there any tricks that might help here?
  • If an analytic solution is impossible (or too ugly to be useful), what kind of techniques can I use for a numerical solution? I believe there will be a way to calculate this with some sort of loop and a bunch of complex-valued math.
  • \$\begingroup\$ I don't know much about LTspice, but some simulators allow you to add a multiplicity (M) parameter to a subcircuit. Multiplicity can be used just like your 'n' parameter. \$\endgroup\$
    – curtis
    Apr 22, 2015 at 19:09

3 Answers 3


Here's an attempt to finding an analytic solution (and how it becomes ugly):

Without any RC (easy)

$$R_0 = R_1 + R_2$$

With one RC (using Z for impedence from now on)

$$Z_1 = R_1 + X + R_2$$


$$X = R||Z_c=\frac{R \times Z_c}{R + Z_c}$$


$$Z_c = \frac{1}{j\omega C}$$

With two RCs + additional \$R_1\$ and \$R_2\$ (getting ugly now)

$$Z_2 = R_1 + (X + R_2)~||~(R_1 + X) + R_2$$

The problem with such circuits is that things are added both in parallel and in row, which is usually not easily simplified.

This becomes a problem in the next step adding a third RC (getting very ugly now). \$R_2\$ is in row, but \$R_1 + X\$ is in parallel (well, partially, at least). At the core of the problem, you have two triangles sharing one edge.

The cool thing (cool as in "makes very ugly things possible, not more beautiful") is that you can convert triangles to stars

Take a look at this image and your schematic: enter image description here

[image from here: http://en.wikipedia.org/wiki/File:Wye-delta-2.svg ]

The goal is to convert the right side (triangle) of the circuit to a star. The image corresponds as follows to your schematic:

$$R_b = X ~(the~ one~ in~ the~ middle)$$

$$R_a = R_1 + X ~(the~ right~ most~ R_1~ and ~the~ rightmost~ X)$$

$$R_c = R_2 ~(the~ middle~one~ )$$

After the conversion, the Z is simpler to calculate. To prevent confusion with the indices, I add the term "triangle" to each indice 1,2,3 that is corresponding to thos in the image above.

$$Z_3 = R_1 + (R_1 + R_{3,triangle})~||~(X + R_2 + R_{1,triangle}) + R_{2,triangle} + R_2$$

With those formulas (according to the wikipedia article above) $$R_{1,triangle} = \frac{X R_2}{R_{sum}}$$ $$R_{2,triangle} = \frac{(R_1+X) R_2}{R_{sum}}$$ $$R_{3,triangle} = \frac{(R_1+X) X}{R_{sum}}$$ $$R_{sum} = X + R_1 + X + R_2 = R_1 + R_2 + 2X$$

You get $$Z_3 = R_1 + \left(R_1 + \frac{(R_1+X) X}{R_{sum}}\right)~||~\left(X + R_2 + \frac{X R_2}{R_{sum}}\right) + \frac{(R_1+X) R_2}{R_{sum}} + R_2$$

It should not come as a surprise at this point that \$Z_4\$ won't be any prettier. This is not the way to go if your goal is a formula for \$Z_n\$

In order to get a formula \$Z_n\$, you want to employ more systematic strategies of network analysis. Take a look at mesh analysis (n+1 means adding another mesh) and/or two port theory (n+1 means adding another two port to the chain).

I'm a bit rusty on those to say the least, but I hope they are a starting point for you.

  • \$\begingroup\$ Yeah, this is exactly what I have written on my paper for Z_3 - +1 for the effort. I think I'm hoping to turn this into a sum or product of some kind, but I'm not seeing it yet. \$\endgroup\$
    – Greg d'Eon
    Apr 22, 2015 at 19:01
  • 1
    \$\begingroup\$ @Gregd'Eon that sounds good then. This approach is probably not the way you want to go though. I added two more approaches of network analysis that are more systematic, which you should probably employ instead of finding formulas by hand then trying to see patterns emerge from them. Please also add the topic of your paper to the question and what this circuit actually represents (as much as you can/want to at least). Chances are somebody did something similar before. \$\endgroup\$ Apr 22, 2015 at 19:17
  • \$\begingroup\$ Added. I think this might work for a numerical method - at each step, the wye-delta transform strips another impedance off the side, so it should be possible to follow this all the way to the end. \$\endgroup\$
    – Greg d'Eon
    Apr 22, 2015 at 19:33

It's been way too long since I have solved problems of this kind, but it brings back fond memories. Hence:

For an analytic solution, try drawing this circuit in more of a star configuration and see whether you see anything (handle the R-C combination simply as a complex impedance Z and think of everything as a complex resistance as to not get confused).

See something? If not, try to apply a Y-Δ transform, to see whether you end up with something more palatable.

Still nothing? Kirchhoff is your friend. Kirchhoff's Current Law and Voltage Law will let you state some of the properties of the circuit in equation form. The voltages along any closed loop add up to zero, as do the currents into/out of any node.

The solution is somewhere along this path.

To simulate it, I would hack up a script that takes n as a parameter in addition to R1, R2, R and C and generates the circuit. Then feed it into your simulator. Try a few values of n and look whether anything converges.


I wrote a program to solve this after one of the previous answers convinced me it would work.

(In these schematics, I'm drawing resistors to represent complex impedances. I'm also combining the resistor and capacitor into one block as $$ Z_{RC} = \frac{1}{j\omega C} \big|\big| R $$ .)

If we keep a running total of the impedance, we can start by stripping \$R_1\$ and \$R_2\$ off the ends. Then, the circuit is


simulate this circuit – Schematic created using CircuitLab

The left-most block looks like


simulate this circuit

where $$ Z_{top} = R_1 $$$$ Z_{bottom} = Z_{RC} + R_2 $$ From the Y-\$\Delta\$ transform, this can be changed into


simulate this circuit

so we can add \$Z_A\$ onto the total impedance and set $$ Z_{top} = Z_B + R_1 $$ $$ Z_{bottom} = Z_C + R_2 $$ which brings us back to the original circuit with one less \$Z_{RC}\$ branch. This process can keep going until we get to the last branch:


simulate this circuit

where we can add \$(Z_{top} + Z_{RC}) || Z_{bottom}\$ to the impedance. This procedure seems to be giving me correct numerical answers, but I can't imagine getting a nice symbolic formula out of it.


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