generalized Thevenin?

There's Middlebrook's Theorems but I don't think that's what you're asking about.

Robert

Actually it looks very similar and it seems close to my problem. Essentially I have a circuit where each "branch" looks identical(Actually its not but uses identical topology... its almost fractal like) and I am trying to use that symmetry to make it easier to solve. I'll have to read up on it to see what exactly it doing though.

Thanks, Jon

hmm... actually it doesn't seem to be what I want. My problem is similar to his but each branch in the graph has the same "structure".

For example, take any graph and treat an edge as a black box that contains passive elements. Then each edge is described by a linear differential equation L_E where E is the edge indicator which really only depends on the passive elements characteristics. Then is there a way to simplify the graph/solve the system for the voltages and currents?

for example, you all know about the delta wye transformations... lets take osmething similar

/\\ Z1 Z2 / \\

--Z3--

Now Z1, Z2, and Z3 are a series of passive components. They all are structurally equivilent. So if Z1 is a resistor than Z2 and Z3 are resistors(although not necessarily of the same resistance). If Z3 is a capacitor and inductor then so are Z2 and Z1.

All we know are the "voltage inputs" and I want to be able to calulate the other variables in the circuit(node voltages and currents).

So essentially between any two nodes we have Vb - Va = Z*I which is just ohms law in some sense but Z is somewhat arbitrary. The problem is that in general Z depends on the nodes, current, and time along with all the other components that exist on that branch.

You can think of it very similar to solving the graphs of resistors that most people do in basic electronics or physics courses except I want to replace them with more general components. In the case of a resistor its very simple. In the case of a capacitor it is not. Since I want to take into account the transients I get system of integeral equations(because of the non-constant/non-sinusodial input voltages and unknown node voltages). Of course these integral equations are equivilent to a system of differential equations.

Essentially you can think of each branch representing a linear differential equation. But each linear DE "looks" the same as every other one except for the constant coefficients and it might depend on different currents and voltages. It seems though I should be able to recursively simplify the circuit until I find all the unknowns. well, this should be obvious but the issue is, is the size. Just hoping for a way to reduce the complexity because of the "symmetry" that exists.

Anyways, probably no way to do what I want but I can wish...

Thanks, Jon

In a passive circuit, the complexity of finding the solution grows with the number of nodes - sorry :-( Think of finding the voltages along a string of resistors. For two resistors, the voltage at the middle point depends on the value of two resistors. For three resistors (same structure in each branch...) both of the two internal voltages depend on the value of three resistors, etc etc. (However, if you know in advance that all the resistors have the same value, you can easily calculate the solutions for all the internal nodes.)

Solving the network to find all node voltages / branch currents can certainly be done the "hard way." This uses admittance matrices, as in i = Y v (with "v" the vector of internal voltages and "i" the vector of external currents and "Y" the admittance matrix).

(for instance)

I think there's actually hope for solving your problem if you have a very simple and regular structure. In that case, you might be able to find the inverse of the Y matrix (this is what you need in order to solve the circuit equations) in a way simpler than the general way. An advantage of this approach is that you don't need to use the actual elements in each branch, you can use a symbol for each distinct branch admittance and enter the actual branch admittances in the solution.

Symbolic simplification of matrix inversions is pretty difficult to do, though.

So, the short answer is that going from simple circuits like a resistive divider to more general circuits is not simple.

[..]

Worse - the voltages at the ends of this branch depend on all other voltages.

Actually, for a circuit with only capacitors, it's not different from a purely resistive circuit (just a lot more unknown voltages at DC...).

For some very special circuits, yes (e.g. resistor strings, delta or star shaped circuits, etc). In general, I think not.

It's worth thinking about this. I studied in a group where the professor made a Big Discovery as a young student when hearing about a deceptively simple result at a conference (for all passive circuits, the sum of voltages in the solution is always constant - or something[1]) and asking himself if that could really be true and then wondering what the consequences were. The answer was the companion circuit and its use for sensitivity analysis :-). He thought it was so trivial that he didn't publish it right away. Tut tut. Two lessons in that anecdote ;-)

Good luck

Best Regards

Jens

[1] You look it up. Annoyingly, I cannot remember right now.

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For a linear 2-terminal circuit, what you're looking for is the Laplace transform of the impedance. This can express any impedance as an algebraic equation using "s"

John

Well, luckily I can use a CAS to do most of the work. I'm actually not interested in extremly complex circuits so it won't actually be to much work. But of course its still not that easy ;/

Yes, at all nodes there will be unknown voltages except at end nodes were the voltages will be known. But because all paths/branches/edges "look the same" all the equations will be the same and there is probably someway to exploit this.

Well, for a capacitor if you want to know the transient voltages it is much more complicated because your dealing with integrals over unknowns.

No, I do not mean that the topology is simplifiable but that because of the symmetry of the equations it reduces the overall complexity. Its like comparing a network of resistors to something else. Because they are all resistors it makes it "easy" to solve(sure it could still be a bitch though). In this case we don't have a resistor but its like a resistor in that each component is similar to all the others. (its not like I have some transistors on one branch, and inductor on another, etc...

Well, I doubt that will happen here ;/ Chances are I won't be able to make any progress but its always fun to try ;)

Thanks, Jon

Yes, I think that might work. If I laplace transform all the equations then I'll end up with a system of linear equations which can be solved pretty easily.

i.e., say for a capacitor in series with a resistor then we have

(V2 - V1) - Vc = R*C*dVc/dt

taking the laplace of each one results in a similar equation

v2 - v1 - vc = R*C*(s*vc - vc(0-))

or

v1 + (1 + RCs)vc - v2 = R*C*Vc(0-)

which can treat as a system(assuming one of the v's are unknown besides vc)

of course the issue is going to be actually transforming the functions but maybe I don't have to worry about that to much.

I'll think about it some more and see what I come up with.

Thanks, Jon

You could try Tellegen's theorem, unfortunately I can't find an explanation on the web. I had a link but can't seem to find it now. I will not try explain it here because it looks ridiculous at first reading (at least it did to me).

Ray

here :