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Suppose I have a resistor mesh that is acting as a two-port network. I'm feeding in \$V_\text{IN}(t)\$ and I'm reading out \$V_\text{OUT}(t)\$. Here's an example of a potential circuit:

enter image description here

Is it true that for this particular network and for any two-port resistor network in general, the transfer function can be written as a rational function of \$R_1,R_2,...,R_{12}\$?

My gut tells me this should be true, but I could be wrong. Even if my intuition is correct, I'm not sure how to arrive at it formally. I don't think the circuit such I showed satisfies the conditions for Thevenin's Theorem (which I believe is meant to cover one-port resistor networks), and the resistors aren't quite in parallel or series so I can't just fold them all together. You could solve this using mesh current analysis but at that point it's a system of equations and not a nice rational function.

It's been a while since I've been in a circuits class (and even then two port networks weren't my strong suit). Is there some theorem I'm missing?

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3 Answers 3

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You could show this via nodal analysis. This isn't a particularly rigorous proof by any stretch of the imagination, but I hope it is at least somewhat convincing.

If you write the KCL equation for each node, you will come up with equations of the form

$$ 0 = \frac{V_i-V_a}{R_{\alpha}} + \frac{V_i-V_b}{R_{\beta}} +... $$

where \$V_i\$ is the voltage at a particular node \$i\$ and \$a, b, ...\$ are neighboring nodes connected through resistors \$R_\alpha, R_\beta, ...\$ respectively. For the two input and output ports, we would set the LHS of the equation to the proper voltage defined by the voltage sources connected to those ports.

For \$n\$ nodes, we will have \$n\$ linear equations relating the node voltages to the voltage sources at the input and output ports, in terms of the reciprocal of the resistances. Solving this system could be done (laboriously) through elementary row operations of the system matrix. That is, we can add/subtract rows, multiply rows by scalars, and rearrange rows.

Now we know that adding reciprocals of resistances can get us terms like $$ \frac{1}{R_\alpha} + \frac{1}{R_\beta} + ... $$ which, as one recalls from the parallel resistor formulas, can always be written as a rational function of the resistances. And naturally, even if we were to take the reciprocal of these sums, we'd still have a rational function. So we can conclude (with sufficient hand waving) that each node voltage can be linearly related to another node voltage via a rational function of the resistances.

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  • \$\begingroup\$ I see, so in this case, we're banking on the fact that a system of linear equations can be solved to get a closed form expression? \$\endgroup\$
    – rcplusplus
    Jul 22 at 19:18
  • \$\begingroup\$ @rcplusplus Any n-port circuit consisting entirely of LTI components (purely resistors in this case) can be solved using the usual linear analysis. Think of it this way: if I physically built this circuit on a breadboard, applied a voltage on one port, and measured the voltage on the other port, I WILL get a well defined output voltage for a given input voltage. And that output voltage WILL be linearly related to the input voltage, since the entire network is LTI. Such a network can always be described by a system of linear (perhaps differential) equations, and there is always a solution. \$\endgroup\$ Jul 22 at 21:18
  • \$\begingroup\$ ah I see. That's a really clean explanation but I guess it relies on the claim that there is not way to connect linear components to get a linear circuit. And also that the solution in analytic form is itself a linear function. But that second point I think you could prove by using linear algebra and gaussian elminiation via elementary row operation matrices \$\endgroup\$
    – rcplusplus
    Jul 26 at 15:01
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Yes. A proof might go something like this:

  1. The only available operations are series (\$R_1 + R_2 = R_{eq}\$), parallel (\$R_1 \parallel R_2 = \frac{R_1 R_2}{R_1 + R_2}\$), and higher-order transformations such as delta/wye, etc. (which are also rational polynomials). Or in general, solutions of linear algebraic equations (simultaneous equations).
  2. This forms a[n algebraic?] field, where all inputs and outputs are members of that field.
  3. There are no irrational or transcendental members of that field.

Or maybe reference to fields aren't needed (and, honestly I probably shouldn't be mentioning them anyway given my limited understanding..), but the general idea that everything is in a set, defined by rational operations (addition, multiplication and division), which is closed under those operations (every result is another member of the set). There's no [finite] sequence of operations you can do to produce a result outside that set.

Put more simply: there's no \$\sqrt x\$ or \$e^x\$ or whatever going on, so there's no way to produce such oddities. There's also no infinite series or other shenanigans like that.

The opposite is not true, of course: if you need to solve a rational polynomial for the value of some parameter(s), you'll in general get an algebraic (irrational) result. But as you asked for the transfer function in terms of parameters, the above is fine.

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Yes, it will eventually reduce to a ratio of consituent elements. For your particular circuit, the symbolic analysis will give a transfer function that is very large. I won't bother transcribing it into MathJax, so I'll just post a screenshot:

enter image description here

That is a rational function, but my screen is not that large to allow a proper screenshot.

What is currently not known (as far as I know) is to recreate such circuits (combinations of series-parallel elements) from rational transfer functions. The above expression stands true and can be verified:

verification with SPICE


(edit)

I realize I haven't sad why, while the other two answers do so. However, the reasoning is quite simple(r): the transfer function involves \$V_{\text{OUT}}/V_{\text{IN}}=\text{TF}\$ (transfer function) and the circuit is comprised of unknown elements. Since all are unknowns (you mean to describe a generic way), it follows that the transfer function is also a ratio of unknwons and, since a fraction has one numerator and one denominator, it folows that the transfer function can be reduced to a rational function. Whether this can be simplifed to have no fractions, or not, that would be a particular case.

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  • \$\begingroup\$ I'd argue that, at least for a 2 port, one could come up with a simple implementation for any transfer function. Since you can fully describe a resistive 2 port with a Z matrix (or any of the other matrix descriptions), it comes down to coming up with a network with four degrees of freedom and solving for those four values in terms of the Z matrix elements. Though if you allow for reactive components... then things get more complicated. \$\endgroup\$ Jul 22 at 18:27
  • \$\begingroup\$ @LetterSized Sure, but that Z matrix will be made of each resistor, in particular, so the only thing that you'd achieve by it would be a symbolic representation, it won't change the underlying transfer function. That will be made of as many elements as the network has. Of course, I am presuming you (and me) are talking about a symbolic expression. \$\endgroup\$ Jul 22 at 19:01
  • \$\begingroup\$ What I mean is that given a purely resistive network, the 2x2 Z matrix will contain exactly four elements whose values are constant. So regardless of how complex the resistor arrangement may be, there are exactly four numbers that describe the entire network. The transfer function from input to output can be derived from just those four numbers without any other information. And what I'm saying is that you could come up with a simple "canonical" circuit with four degrees of freedom to implement those four Z matrix elements. \$\endgroup\$ Jul 22 at 21:14
  • \$\begingroup\$ @LetterSized I understood what you meant but, those four degrees of freedom can only come if you can evaluate the Z parameters, and that can only be done numerically. If you want a generalized view then you have to treat all the elements as unknowns. But, if Z can be evaluated, then even those 4 degrees of freedom vanish into only one, also numerical, since all Z are known, now. Therefore you end up with a constant describing your attenuation. So, it's either symbolic, or numeric, and for the former, the number of unknown elements in the system determine the transfer function. \$\endgroup\$ Jul 23 at 6:17
  • \$\begingroup\$ If it's a symbolic approach, even if you mean to solve the system in terms of Z parameters, you'll end up with each Z being represented by equations in more than one unknown, which means that the four degrees of freedom are just an illusion, a mathematical abstraction that hides the greater number of unknowns. So, while I understand that your aim is to simplify, what I'm saying is that you can only go so far. And, when you do so, you have to decide whether it's symbolic, or numeric. \$\endgroup\$ Jul 23 at 6:21

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