This question arises from considering the impedance of a ladder of inductors and capacitors, as shown in the diagram.

Suppose I have two complex numbers $Z_L(\omega) = i \omega L$ and $Z_C (\omega) = 1/i \omega C$. I start a sequence by defining

$$Z_1(\omega) = Z_L(\omega) + Z_C(\omega) \, .$$

I then define subsequent terms in the sequence according to

$$Z_n(\omega) = Z_L(\omega) + \frac{Z_{n-1}(\omega) Z_C(\omega)}{Z_{n-1}(\omega) + Z_C(\omega)} \, .$$

I wish to analyze the position of the poles of $Z_n(\omega)$ as a function of $\omega$ for large finite $n$. How can this be done? Note that a related (but different) question would be: what is $\lim_{n\rightarrow \infty}Z_n(\omega)$?

Surely this could be done via induction, but one needs a guess for the recurrence relation and I have not found a way to make that guess.

enter image description here

  • $\begingroup$ What do you mean by finding the poles, exactly? Do you just want the limit of the $Z_n$? $\endgroup$ – Ian Feb 7 '15 at 22:42
  • $\begingroup$ @Ian: I had left the $\omega$ factors out because I was trying to tidy up the math. I realize now from your comment that this was foolish because asking for poles without a variable is meaningless. $\endgroup$ – DanielSank Feb 7 '15 at 23:20

I assume that you want to calculate $\lim_{n \to \infty} Z_n$.

By taking limits on both sides, you find that any limit $Z$ would have to satisfy

$$Z = Z_L + \frac{Z Z_C}{Z + Z_C}.$$

Doing some algebra:

$$Z = \frac{Z_L Z + Z_L Z_C + Z Z_C}{Z + Z_C} \\ Z^2 + Z Z_C = Z (Z_L + Z_C) + Z_L Z_C \\ Z^2 - Z_L Z - Z_L Z_C = 0.$$

This is a quadratic equation whose roots are

$$Z = \frac{Z_L \pm \sqrt{Z_L^2 + 4 Z_L Z_C}}{2}.$$

So if the limit exists, it is one or the other of these two numbers. Whether the limit exists will depend on the initial condition, and when it does exist, which limit is achieved will depend on the initial condition. Usually you will converge to the possible limit which is closer to the initial condition, and there is always a domain of attraction for each limit (although it could be just the limit itself, if the limit is unstable).

To continue the calculation, it's convenient to make the variables real, so let's do that:

$$Z = \frac{iL \pm \sqrt{-L^2 + 4 L/C}}{2.}$$

Your initial condition is $i(L-1/C)$, so the difference between the roots and the initial condition is

$$Z-Z_1=\frac{iL \pm \sqrt{-L^2 + 4L/C}}{2} - i(L-1/C) = \frac{iL - 2 i L + 2 i/C \pm \sqrt{-L^2 + 4L/C}}{2} \\ = \frac{-i L + 2 i/C \pm \sqrt{-L^2 + 4L/C}}{2}.$$

This means that at least when $L/C$ is small enough, you will get the $+$ root, since

$$\frac{-iL + 2i/C + \sqrt{-L^2 + 4L/C}}{2} = i \frac{-L + 2/C + \sqrt{L^2-4L/C}}{2} \\ \approx i \frac{-L + 2/C + L - \frac{4L/C}{2L}}{2} = 0.$$

This follows by Taylor expanding the square root. I emphasize that it requires that $L/C$ is small enough, in particular it requires that $4L/C \ll L^2$.

  • $\begingroup$ Thanks. In the case that $L$ and $C$ both go to zero at the same rate this answer shows that $\lim_{n\rightarrow \infty}Z_n = \sqrt{L/C}$, which is the well known result (I think the $2i/C$ term in your final answer might be wrong). $\endgroup$ – DanielSank Feb 7 '15 at 23:24
  • $\begingroup$ @DanielSank The $2i/C$ is in $Z-Z_1$, not $Z$. I was using $Z-Z_1$ to determine which limit we should expect given the initial condition $Z_1$ that you gave. I think I read your notation correctly in concluding that $Z_1=i(L-1/C)$. (The slight ambiguity is that $1/iC = 1/(iC)$, but I think that's correct. It's been a few years since I dealt with complex impedances. $\endgroup$ – Ian Feb 7 '15 at 23:34
  • $\begingroup$ Oh, woops, you're right of course. Still, this computation shows that $Z = \sqrt{L/C}$ if $L$ and $C$ go to zero, which is a well known result. I'm really wondering what the poles of $Z(\omega)$ look like for large finite $n$. $\endgroup$ – DanielSank Feb 8 '15 at 1:18
  • $\begingroup$ @DanielSank I think the situation with $\omega$ included is somewhat complicated, because the limit in question will probably depend on $\omega$, and because each of the limits themselves actually have no poles. $\endgroup$ – Ian Feb 8 '15 at 1:41
  • $\begingroup$ @DanielSank Perhaps for just an asymptotic analysis, you could use that $\frac{Z_{n-1} Z_C}{Z_{n-1} + Z_C} = \frac{Z_C}{1 + \frac{Z_C}{Z_{n-1}}}$. So if $\left | \frac{Z_C}{Z_{n-1}} \right | < 1$ then $\frac{Z_{n-1} Z_C}{Z_{n-1} + Z_C} = Z_C \sum_{n=0}^\infty (-1)^n \left ( \frac{Z_C}{Z_{n-1}} \right )^n$. You can get a similar form if $\left | \frac{Z_{n-1}}{Z_C} \right | < 1$. For small $Z_C$ these estimates are pretty tight, which would make your life simpler. Going up to $n=0$ you get the very simple $Z_n = Z_L + Z_C$. Going up to $n=1$ you get $Z_n = Z_L + Z_C - \frac{Z_C^2}{Z_{n-1}}$. $\endgroup$ – Ian Feb 8 '15 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.