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(the motivation section turned out a little long, the mathematical question is at the end)

I need to work with electrical circuts at the moment, computing effective impedances etc. From electrodynamics, we have Kirchhoffs law and so on, which result in two rules: If you have two impedances $Z_1,Z_2$ in series, then the total impecance is given by $$s(Z_1,Z_2)=Z_1+Z_2.$$

If you have them in parallel you get $$p(Z_1,Z_2)=\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}$$

I was using the more general rule $\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}+\frac{1}{Z_3}}$ until I dediced I should derive it from the above one, and indeed

$$p(Z_1,Z_2,Z_3):=p(Z_1,p(Z_2,Z_3))=\frac{1}{\frac{1}{Z_1}+\frac{1}{\frac{1}{\frac{1}{Z_2}+\frac{1}{Z_3}}}}=\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}+\frac{1}{Z_3}}.$$

Now we have $s(Z_1,Z_1)=2Z_1$ and $p(Z_1,Z_1)=\tfrac{1}{2}Z_1$ and these operations are highly symmetric, in that for example $p(Z_1,p(Z_2,p(Z_3,Z_4))=p((Z_1,Z_2),p(Z_3,Z_4))$. And we can build complicated stuff, e.g. $p(p(Z_1,Z_2),s(Z_1,Z_2))=\frac{Z_1^2+Z_1Z_2}{3Z_1+2Z_2}$. Might be we can fit the coefficients there to any natural we like.

Now we have the passive elements resistor $R$, capacity $C$ and inductivity $L$ with

$$Z_R=R,\ \ Z_C(\omega)=\frac{1}{iC\omega},\ \ Z_L(\omega)=i\omega L.$$

and electrical circuits are used to compose these to various functions, so that the electrical resistence is selective w.r.t $\omega$. E.g. we have filters which effectively are functions which have only restricted support. Or see e.g. the two element LC-circuit for a simple construction. I was thinking if I can construct a frequency independend impedance without using constant elements $Z_R$, and the obvious idea would have been to let $\omega$ cancel out in

$$Z_C\cdot Z_L=\frac{1}{iC\omega}i\omega L=\frac{L}{C}.$$

For a moment I was thinking about which circuit combination would give my $Z_1Z_2$, but I quickly realized that both $s$ and $p$ map a unit Ohm back to Ohm (e.g. $\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}=(Z_1+Z_2)^{-1}Z_1Z_2$, which effectively still has units Ohm to the power of 1), and the unit of $Z_1Z_2$ would be Ohm to the power of 2. But it does for example make it possible to construct $\frac{1}{1+i\omega C}$.

So my question now is: What is the function space I can generate? I assume it's some extension of some restriction of the rational polynomials generated by symmetrical functions. Notice that we got "$+$" but not "$-$", and a weak version of "$/$" but not "$\cdot$". For $x,y\in \mathbb R_{>0}$, we have $p(x,y)=p(y,x)>x$ and $s(x,y)=s(y,x)<x$.

What is the function space in $\omega$ I can generate with the operations $$s(Z_1,Z_2)=Z_1+Z_2\ \text{and }\ p(Z_1,Z_2)=(Z_1+Z_2)^{-1}Z_1Z_2,$$ where $Z$'s are of the form $$a,\ b\ (i\omega),\ c\ (i\omega)^{-1}$$ where $a,b,c$ are e.g. in $\mathbb R_{>0}$ and $i^2=-1$?

This <link> is a more broad SE question on electrical circuits.


Edit (11.7.13):

I worked on it further and here some insights: In terms of $Z_j$'s, the expressions look always of the form $\tfrac{P}{Q}$, where both $P$ and $Q$ are polynmials of the form

$$\sum_i^N\prod_{j=\sigma(i)}^M Z_j,$$

e.g.

$$s(p(Z_1,Z_2),Z_3)=\frac{Z_1\cdot Z_2}{Z_1+Z_2}+Z_3=\frac{Z_1 \cdot Z_2+Z_1 \cdot Z_3+Z_2 \cdot Z_3}{Z_1 + Z_2}=\frac{\sum_{i=1}^3\prod_{j=i}^{i+1} Z_j}{\sum_{i=1}^2\prod_{j=i}^{i} Z_j},$$

An interesting subquestion pops up: Given $P$, can we determine the form of Q? (Up to overall multiplicative factors, which we know we can rescale) If yes, does to what extend does this still hold one we substiture the $Z$'s for $R$, $i\omega L$ or $\tfrac{1}{i\omega C}$?

The structure of every expression $s(p(*,*),*)$ clearly is of the form of a simple tree which we can enumerate. For any fixed number of inputs which all can be of the form $a,\ b\ (i\omega),\ c\ (i\omega)^{-1}$ (e.g. three $Z_1,Z_2,Z_3$ in the above case) and where the real values $a,b,c$ can be adjusted as we like, we can generate all possible expression.

This suggest a more explicit formulation of the question: For a given number of arguments, what is the function mapping the index of the tree to the polynomial $P/Q$?

I post this update now that I've discovered that that function exibits some nontrivial features: For exmaple, if two certain $Z$'s in the expression $s(*,p(*,*))$ have the same dependence in $i\omega$, then it turns out that (by rescaling of the numbers $a,b,c$), this is equivalent to the combination $p(*,s(*,*))$:

en.wikipedia.org/wiki/Equivalent_impedance_transforms

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Can the circuit contain an infinite number of dipoles like a transmission line? –  metacompactness Jul 11 '13 at 22:03
1  
I suspect a good place to start would be to prove that you cannot get anything with a negative real part (when $Z$s themselves have positive real parts). Perhaps, this was "one of the reasons" behind the invention of the transistor (or in general, an amplifier) itself. Once you have an amplifier, you can get a negative resistance, and conversely, once you have negative resistance, you can build an amplifier. –  Lord Soth Jul 16 '13 at 19:35
    
Take a look at Section 4 of front.math.ucdavis.edu/1203.1256 –  john mangual Aug 5 '13 at 3:49
    
Votes on the partial answers. 1. The bounty offers a reward to partial answers. 2. There are two answers with some partial work done. 3. It would be good if they get some votes to help award the bounty. 4. Evaluating partial advance is subjective even if a complete answer was known. A collective evaluation may be more fair than my unilateral evaluation. –  ABC Aug 6 '13 at 15:14

2 Answers 2

You are asking for the closure of the vector space of functions spanned by $1, \omega, \omega^{-1}$ closed under addition $h = f + g$ and harmonic mean $\frac{1}{h} = \frac{1}{f} + \frac{1}{g}$.

This will be some subspace of the rational functions $\mathbb{C}(\omega)$ possibly all of them.

  • you can get any $a + b\omega + c\omega^{-1}$.

  • i'm not even sure you can construct $\omega^2$ this way.


Let me try a little harder. For any function $Z(\omega)$ you can get $$Z + 1 \text{ and } \frac{1}{1 + \frac{1}{Z}} = \frac{Z}{Z+1}$$ These are examples of fractional linear transformations. These correspond to matrices

\[ \left(\begin{array}{cc}1 & 1 \\ 0 & 1 \end{array} \right) \text{ and } \left(\begin{array}{cc}1 & 0 \\ 1 & 1 \end{array} \right)\] These matrices may be easier to fiddle around with than rational functions. I then tried:

\[ \left(\begin{array}{cc}1 & 1 \\ 1 & 0 \end{array} \right) \left(\begin{array}{cc}1 & 0 \\ -1 & 1 \end{array} \right)= \left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \] This would correspond to the function $\frac{1}{Z}$. In terms of rational functions you get the identity

\[ \frac{\frac{z}{-z+1}+1}{\frac{z}{-z}+1} = \frac{z + (-z+1)}{z} = \frac{1}{z} \]

You can imagine the circuit this corresponds to. So your space of rational functions is closed under reciprocals even if you can't multiply.

There are tiny bits of planar algebras and continued fractions in here.


Here's a way to get $\frac{Y^2}{Z}$.

\[ \frac{1}{\frac{1}{y} + \frac{1}{z} } = \frac{zy}{z + y} \]

If we make the replacement $z \mapsto z - y$, the z-transform now reads

\[ \frac{(z-y)y}{z} = \left( 1 - \frac{y}{z} \right) y \text{ and so } -1\left[\left( 1 - \frac{y}{z} \right) y - y\right] = \frac{y^2}{z} \]


If we set $z = 1$, The identity $ \frac{-1}{\frac{1}{y} + \frac{1}{1-y}}+y = y^2 $ means we set $y = \omega$ and build $y = \omega^2$.


You should be able to get $\omega^k$ for any $k \in \mathbb{Z}$ and add to get any Laurent series. Your ring is $\mathbb{C}(\omega)$

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You're still assuming that he can subtract (presumably because of the confusing mention of 'inverse' in the question's title); without the ability to generate $-z$ I don't see how you get, e.g., $1/z$. –  Steven Stadnicki Aug 5 '13 at 15:05
    
@StevenStadnicki Another problem is these circuits will get very complex, definitely not minimal. They will be finite for rational functions and infinite for Laurent series. –  john mangual Aug 5 '13 at 15:12
    
@StevenStadnicki Is it theoretical possible to switched $Z \mapsto -Z$ ? Electrical engineers do study complex-valued voltage and resistance‌​. Wikipedia does have an entry on Negative resistance –  john mangual Aug 5 '13 at 15:15
    
@StevenStadnicki Richard Kenyon mentions cluster algebras arise in the rational functions related to circuits. –  john mangual Aug 5 '13 at 15:20
    
@StevenStadnicki: It's easy to get $1/z$ by checking that the set of multiplicatively invertible functions is closed w.r.t. the operations. –  Alexander Shamov Aug 5 '13 at 23:59

The space of holomorphic functions that preserve the right half-plane $\mathbb{C}_{+} := \mathbb{R}_{+} + i \mathbb{R}$ is obviously closed under your two operations, and contains the functions $1$, $z$ and $z^{-1}$.

As far as I know, there is a theorem that describes all holomorphic functions that preserve this half-plane as $f(z) = \intop_\mathbb{R} f_\alpha(z) \mu(d \alpha) + b + cz$, where $f_\alpha(z) = (z - i \alpha)^{-1}$, $\Re b \ge 0$, $c \ge 0$ and $\mu$ are positive measures (see, e.g., Operator monotone function). The functions $1$, $z$ and $z^{-1}$ also preserve $\mathbb{R}$, thus in your case $\mu$ should be symmetric. Now note that $\frac{1}{2} (f_\alpha + f_{-\alpha}) = (z + \frac{\alpha^2}{z})^{-1}$, and this function can be constructed using our operations (at least, if $\alpha$ is rational). Therefore, the closure of your space of functions (in the topology of uniform convergence on compact subsets) is the whole space of holomorphic functions from $\mathbb{C}_+$ to itself "with real coefficients".

Edit: I should also mention that for my construction you need $\alpha z$ for all $\alpha > 0$, but starting from any function you can get all positive rational multiples of it.

It would be interesting to figure out what happens if we don't allow approximation...

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I don't see how you can make the leap to the closure that you're talking about. To borrow my example from the other answer, the (positive) rationals are closed under addition - but starting with $1$ and using only addition you can't get all the positive rationals; you can't even get 1/2. Just because two semigroups are closed under the same set of operations doesn't mean they're identical or that their closures are identical, and I don't see where your argument generates all the functions you're talking about from his starting point. –  Steven Stadnicki Aug 6 '13 at 0:51
    
@StevenStadnicki: Of course I don't claim that all holomorphic functions are generated by these operations. But the $\frac{1}{2}(f_\alpha + f_{-\alpha})$ are there (for rational $\alpha$), as well as their positive rational multiples. Hence at least we can approximate these integrals by sums, so I only claim that the space that we want is dense there. –  Alexander Shamov Aug 6 '13 at 1:32

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