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A recent question asked how to find the impedance of an infinite chain of series-plus-parallel circuits. The standard trick is to split the chain after the first link, and treat the tail of the circuit as a copy of the original, bigger circuit. That is, the circuit contains a copy of itself, which is fine as it is infinitely long.

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The corresponding equation for the circuit's impedance $Z$ is then

$$ Z=2Z_1+\frac{1}{\frac{1}{Z_2}+\frac{1}{Z}}, $$ which gives an easy quadratic equation in $Z$ with solutions $$ Z=Z_1\left[1\pm\sqrt{1+2Z_2/Z_1}\right]. $$

Now, for purely resistive $Z_1$ and $Z_2$ it is clear enough which sign the square root should have, since then $\sqrt{1+2Z_2/Z_1}>1$ and the minus sign is ruled out by the condition that $\mathrm{Re}(Z)\geq 0$ (unless one is OK with free energy coming out of nowhere).

However, this need not be the case. An easy example is where $Z_1=1/i\omega C$ is capacitive and $Z_2=i\omega L$ is inductive, in which case $$ 1+2Z_2/Z_1=1-2\omega^2 LC $$ can, depending on $\omega$, be smaller than one, giving two distinct possible values for $Z$. This result will hold if both impedances have small but nonzero resistive components, which is a more physical situation. An infinite chain of coupled oscillators has all the makings of a wave line, but that does not mean that the impedance of the line can suddenly have two different values.

How does one resolve this? What are the conditions on $Z_1$ and $Z_2$ for this to be a problem? How does this relate to the continuum limit on the transmission line?

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    $\begingroup$ Add some resistance and take it to zero at the end. That's actually the same thing as pushing the poles of resonances off the real line and letting them go back at the end of the calculation (think retarded Green function). By the way, the real issue here is that impedance only makes sense in an infinite time limit (because it's defined in terms of a Fourier transform) but you have another infinity here in the length of the ladder. What I'm suggesting allows you to switch the order of the limits and get a unique well defined result. I'll write it up later, maybe. $\endgroup$ – DanielSank Jun 17 '15 at 9:34
  • $\begingroup$ For the hapless future reader, I explained this idea in a previous answer. $\endgroup$ – DanielSank Jun 17 '15 at 9:35
  • $\begingroup$ @DanielSank Switching the limits is an interesting thought. On the other hand, I don't think adding resistance helps: if $Z_1=R+1/i\omega C$ and $Z_2=R+i\omega L$, then as long as $R$ is small enough ($R<\sqrt{\tfrac23 \tfrac LC}$) there always exists an open interval in $\omega$ around $\omega=0$ for which $Z$ has two roots of positive real part. $\endgroup$ – Emilio Pisanty Jun 17 '15 at 11:08
  • $\begingroup$ Well, how about loading a short chain into Spice and seeing which of the two solutions is physically sensible? I'm betting one of them isn't -- or you've designed a bistable oscillator, which is quite possible! $\endgroup$ – Carl Witthoft Jun 17 '15 at 14:43
  • $\begingroup$ @EmilioPisanty That's because $+\omega$ and $-\omega$ are the same thing ;) $\endgroup$ – DanielSank Jun 17 '15 at 15:55
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OK. Let us start with the initial equation

$$Z=Z_1\left[1\pm\sqrt{1+2Z_2/Z_1}\right] \tag 1$$
and consider lossless circuit. It is a good idea to determine the value of $Z_1$ as $Z_1=ix_1$ and similarly, $Z_2=ix_2$ where $x$ denotes the reactance.

For the the capacitive reactance $x<0$ and for the inductive reactance $x>0$. We have:

$$Z=ix_1\pm\sqrt{-(x_1^2+2x_1x_2)} \tag 2 $$

Now, we see that the following condition must apply:
$$x_1^2+2x_1x_2>0 \tag 3$$ In the opposite case the impedance $Z$ will contain a real part, and hence the circuit will consume or give energy, depending on the sign of the real part.

The question boils down to the choice of the sign of square root.

For this we assume that the impedance $Z$ contains a small ohmic resistance $r_1$. That means we take $Z_1=r_1+ix_1$ in $(1)$.

The next step is to expand $Z$ into Taylor series with respect to $r_1$. I jump over the math part and write only result

$$Z=r_1+ix_1+ \sqrt{x_1^2+2x_1x_2}\left ( i+\frac{r_1}{x_1}\right ) \tag4 $$

The sign of the square root in $(4)$ should be selected so that the real part of this expression was positive.

Let $x_1>0$. Because $r_1>0$, then in order that the real part of $(4)$ would be positive we must take the sign $+$ in front of the square root.

Let $x_1<0$. Because $r_1>0$, then in order that the real part of $(4)$ would be positive we must take the sign $-$ in front of the square root.

Finally, looking at the process $r_1\rightarrow 0$, we come to the end result:

$$ Z=\left\{ \begin{matrix} ix_1+i\sqrt{(x_1^2+2x_1x_2)} \;\text{ if}\; x_1>0, \\ ix_1-i\sqrt{(x_1^2+2x_1x_2)} \;\text{ if}\; x_1<0. \end{matrix}\right. $$ Remember that $(3)$ must hold.

But what happens when $(3)$ does not apply?

It is impossible in this case to develop a stationary sinusoidal regime in the circuit and the concept of impedance is meaningless.

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  • $\begingroup$ I don't see why it's wrong for $Z$ to have a resistive part as long as it is positive. Sure, the circuit will consume energy, but it would also do that if the $Z_1$ or $Z_2$ were resistive. Even if they are both purely reactive, the chain is infinite so there's plenty of room to put energy by having a continuous flow to the right. $\endgroup$ – Emilio Pisanty Nov 12 '15 at 21:32
  • $\begingroup$ @EmilioPisanty I introduced a resistive part into $Z$ as shown in (4). Note that i considered a limiting process $r_1\rightarrow 0$ not an assumption $r_1=0$. $\endgroup$ – Martin Gales Nov 21 '15 at 20:57
  • $\begingroup$ That's not quite what I mean. If your condition (3) doesn't hold, the only thing that happens is that $Z$ acquires a resistive component, but as long as this doesn't actively create energy (i.e. as long as $\mathrm{Re}(Z)\geq0$) I don't see why that is a problem. In particular, I don't see your claim that it's impossible to sustain a stationary sinusoidal regime - you just need to input energy, but the same holds for a single plain resistor. $\endgroup$ – Emilio Pisanty Nov 21 '15 at 21:05
  • $\begingroup$ @EmilioPisanty You had a probleem with $Z_1$ and $Z_2$ (and consequently $Z$) to be pure imaginary, i.e.,with no ohmic loss in the circuit. I analyzed namely that case. Thus the condition $(3)$ must hold. Otherwise $Z$ gets a real part and there will be a conflict. Further, when one or both of $Z_1$ and $Z_2$ (and consequently $Z$) have a positive real part then $Z$ is determined uniquely because one of its two complex values has always negative real part and must be eliminated as it does not have a physical meaning. $\endgroup$ – Martin Gales Nov 22 '15 at 10:14
  • $\begingroup$ Sorry, I still don't see why it's inconceivable that Z will be resistive even if it's made with purely reactive ingredients. I simply don't see the conflict that you mention. $\endgroup$ – Emilio Pisanty Nov 22 '15 at 21:31

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