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Does anyone know if a finite number of parallel-series impedance has a general solution?

I am trying to model a network flow problem that can be described in electrical terms and represented as a finite discrete transmission line with arbitrary impedance.

While it is not difficult to write the equations down for a small number of terms, I'm curious if there are some general solutions or insights into the problem.

It looks like

----Z1---+----Z3---+----Z5---+---...---+
         |         |         |         |
         Z2        Z4        Z6        ZN
         |         |         |         |
---------+---------+---------+---...---+

All the Z's are, in general, different.

I'll need to know the "current" and/or "voltage" at every branch. In this case the bottom is grounded.

What would be cool is to be able to know the voltage and/or current in terms of i, the ith position (the top +).

Any ideas?

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You can use the Modified Nodal Analysis method for solving the nodal voltages, and then back-compute the branch currents.

The beauty of this technique is that it generalizes very well to any well defined netlist and is quite easy to program. This technique can give either an analytical or a numerical solution, depending on your solver, and can be used to solve DC, steady state AC, or full transient problems.

The first link gives good resources on how to solve linear DC and steady-state AC circuits, and if you want to solve linear transient problems I have a blog post about how to approximate the inductors and capacitors to solve a transient problem.

edit:

In response to the OP's comments,

Matrix inverses can be solved analytically, though the degree of difficulty increases drastically for larger matrices (note: difficulty = time, just follow Gaussian elimination).

Some methods for finding matrix inverses analytically:

  1. Gaussian Elimination
  2. LU Decomposition
  3. QR Factorization

... take your pick. The reason I suggested using a computer (even if the computer is solving your system analytically for you) is because for any appreciable size matrix it will take you a very long time to solve by hand.

This is the exact solution if you have a discrete/finite number of nodes, and there's no way around this linear algebra problem.

Maxwell's equations can have a "simple" looking solution because they don't apply to a discrete/finite domain: they rely on continuous functions and thus all the tools of calculus are available.

If you can make the approximation that the number of nodes tends towards infinity and don't care for the discrete nature of your network, you can simulate your transmission line using the Telegrapher's Equations. These will have solutions of a similar form to Maxwell's equations. For example, the solution to a lossless transmission line is: $$ V(x, t) = f_1(x-u t) + f_2(x+u t) $$ $$ I(x, t) = \frac{f_1(x-u t)}{Z_0} + \frac{f_2(x+u t)}{Z_0} $$ A.k.a. the homogeneous wave equation solution.

A lossy transmission line can be described as a non-homogeneous wave equation, who's solution can be found by applying Green's Function to find the particular solution and adding on the homogeneous solution, or by only using Green's Function (details omitted).

Edit 2:

One of the limitations with using Nodal Analysis is you need to fully specify the problem, i.e. specify the entire network and boundary conditions. You have not specified what's connected to the input/output of your "transmission" line, which is why I don't know what your full network will look like and consequently can't offer any advice other than the "general" linear algebra techniques.

In my blog post, I noticed that even though my entire network results in a non-tri-diagonal matrix, I could easily reduce it into one by not solving for IS at the same time as all the nodal voltages, and re-arranging the matrix form. Thomas's Algorithm provides a very efficient method for expressing the solution to a tri-diagonal system.

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  • \$\begingroup\$ While this is on the right track I was hoping that a direct closed form solution existed. Essentially the inversion of your matrix. I'm not 100% sure the results are directly applicable(just redefining terms) but it's a start. ATM I'm not looking for an algorithmic solution. (I'm not even sure if the math is the exact same between the two different problems). \$\endgroup\$ – user3606799 May 21 '14 at 17:52
  • \$\begingroup\$ If you look further into the first link, the author presents SCAM, which uses Matlab to analytically solve a given network by framing it as a linear algebra problem. Linear systems are very well studied and there are many algorithms for solving them. If your question is on methods for solving the system Ax=b using a computer, you might get better results asking on stackoverflow or the comp.sci stack exchange. \$\endgroup\$ – helloworld922 May 21 '14 at 19:39
  • \$\begingroup\$ No, I am looking for a mathematical expression of the solution. For example, one can "solve" maxwell's equations because of en.wikipedia.org/wiki/Jefimenko%27s_equations. It is a "closed form" analytical solution of the pde's. Just "plug and play". Ax=b has the solution x = A^(-1)*b. I am looking for a similar thing. Basically I would like to know what the inverted matrices that you have are. There is probably a closed form expression for it. (Basically it probably involves inverting it by hand and seeing the pattern) \$\endgroup\$ – user3606799 May 23 '14 at 18:44
  • \$\begingroup\$ I don't want just an expression of the problem that requires me to use a computer to solve the problem indirectly. E.g., I could use numerical integration to solve maxwell's equations or I could use jefimenko's equations which require numerical integration. The difference is one is a direct solution and one is not. (Jefimenk's does not involve circular expressions and is directly expressed in known quantities... of course it is actually harder to solve numerically but that's besides the point) \$\endgroup\$ – user3606799 May 23 '14 at 18:47
  • \$\begingroup\$ I've updated my answer: basically a discrete network requires you solve a system Ax=b. Maxwell's equations have a nice form because they rely on a continuum: the non-discrete nature allows the tools of calculus to be applied, and a concise solution can be found. \$\endgroup\$ – helloworld922 May 23 '14 at 19:40
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Yes - it is possible based on "partial fraction" developments (introduced by FOSTER): Please note that for convenience you should replace all even elements by the corresponding conductances (Z2>>>Y2, Z4>>>Y4, ...).

Then, we have

Z= Z1 + 1/D1

D1=Y2 + 1/D2

D2=Z3 + 1/D3

D3=Y4 + 1/D4

....

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  • \$\begingroup\$ Yeah, these equations are basically what comes out of the solution when you write down the total impedance. Is Dn used in any simple way to calculate the voltages and currents at the upper nodes? \$\endgroup\$ – user3606799 May 21 '14 at 17:53

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