Partial and complete removal of pole

In ladder synthesis of a lossless two-port, a driving-point function is reduced step by step by removing its poles as elements. A pole can be removed with its full residue (complete removal) or with only part of it (partial removal).

Complete removal of a pole

The whole term of the pole is subtracted, e.g. $Z_1 = Z - k_\infty s$ with $k_\infty = \lim Z/s$. The remainder no longer has that pole, its degree drops, and the removed element (series $L$, series $C$, tank, etc.) produces a transmission zero at the pole frequency ($0$, $\infty$ or $\omega_i$). Cauer ladders use complete removal at every step.

Example: $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$. Here $k_\infty = 1$, so

$$Z_1 = Z - s = \frac{2s^2+3}{s(s^2+2)}$$

A series 1 H inductor is removed and $Z_1$ now has a zero at $\infty$ (degree 2 over 3) — this is the first step of Cauer-I.

Partial removal of a pole

Only $k s$ with $0 < k < k_\infty$ (or $k/s$ with $k < k_0$) is removed. The pole remains in the remainder, the degree does not drop, but the zeros of the remainder shift towards the removed pole. $k$ is chosen so that a zero lands exactly at a desired frequency $\omega_z$ (zero shifting); the next step then removes the pole of the reciprocal at $\pm j\omega_z$ completely to give a transmission zero there.

Example. Realize $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$ as a ladder with a transmission zero at $\omega = 2$ rad/s.

The zeros of $Z$ are at $\omega = 1$ and $1.732$; we need one at $\omega=2$, so remove part of the pole at $\infty$: $Z_1(s) = Z(s) - k s$ with $Z_1(j2) = 0$.

$$\begin{aligned} Z(j2) &= \frac{(-4+1)(-4+3)}{j2(-4+2)} = \frac{3}{-j4} = j0.75 \\ Z_1(j2) &= j0.75 - j2k = 0 \;\Rightarrow\; k = \tfrac{3}{8} \\ Z_1(s) &= Z(s) - \tfrac{3}{8}s = \frac{(s^2+4)(5s^2+6)}{8s(s^2+2)} \end{aligned}$$

So series $L_1 = 3/8$ H, and $Z_1$ now has a zero at $s=\pm j2$ (the other zero moved from 1 to 1.095 rad/s). Its reciprocal has a pole at $\pm j2$, removed completely as a shunt branch:

$$\begin{aligned} Y_1(s) &= \frac{8s(s^2+2)}{(s^2+4)(5s^2+6)} = \frac{\tfrac{8}{7}s}{s^2+4} + \frac{16s}{7(5s^2+6)} \\ \tfrac{8}{7}\tfrac{s}{s^2+4} &\Rightarrow L_2 = \tfrac{7}{8}\ \text{H},\ C_2 = \tfrac{2}{7}\ \text{F}\quad (1/\sqrt{L_2C_2} = 2) \\ Z_2(s) &= \frac{7(5s^2+6)}{16s} = \tfrac{35}{16}s + \frac{21}{8s} \Rightarrow L_3 = \tfrac{35}{16}\ \text{H},\ C_3 = \tfrac{8}{21}\ \text{F} \end{aligned}$$

o--[L1=3/8H]--+--[L3=35/16H]--[C3=8/21F]--+
              |                           |
          [L2=7/8H]                       |
          [C2=2/7F]                       |
              |                           |
o-------------+---------------------------+

The shunt branch $L_2$–$C_2$ shorts the signal at $\omega=2$ rad/s, which is the required transmission zero. (Check: rebuilding the ladder gives back the original $Z(s)$.)