Re: generalized Thevenin?


"Jens Tingleff" <jensting@xxxxxxxxxxxx> wrote in message
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In article <fZCni.39974$Um6.1086@xxxxxxxxxxxxxxxxxxxxxxxxxx>, Jon
Slaughter
says...


"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote in message
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"Robert" <Robert@xxxxxxxxx> wrote in message
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"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote in message
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Is there a generalized Thevenin's theorem for arbitrary "black boxes"?


i.e., Suppose I have something like


---> I
V +---[ ]--- 0


where [] is a black box.

I should be able to write something like

V = Z(t, V, I)*I

which sorta resembles ohms law. V and I generally depend on t.


if [ ] is a resistor then Z(t,V,I) = R and in general Z also depends
on
a set of parameters.


But what about more complex black boxes?


If its a resistor and a capacitor then what?


V ---||---/\/\/\/\---- 0


Then Z(t,V,I) = ?

For passive components is Z a linear differential equation?


Any other ways to simplify such expressions?

The reason I ask is I have a circuit that has a lot of these "paths"
that are connected in some way but each path is the same configuration
with only the "constants" of the components that are different.

Thanks,
Jon

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

http://en.wikipedia.org/wiki/Extra_element_theorem

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.



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?


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).

http://users.ecs.soton.ac.uk/mz/CctSim/chap1_2.htm (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.


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 ;/

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

[..]



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.

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


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.



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.

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

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




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.

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


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...



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


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


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


.

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