Zero shifting for two-port synthesis

Zero shifting by partial removal of a pole. When a driving-point function is expanded into a ladder, removing a pole completely (its full residue) produces transmission zeros only at $s=0$ or $s=\infty$. To place a zero at a chosen finite frequency $\omega_z$, we remove only part of a pole (at $\infty$ or at $0$) so that the remainder becomes zero at $s = \pm j\omega_z$. The reciprocal of the remainder then has a pole at $\pm j\omega_z$, which is removed completely as a shunt series-LC branch (or a series tank), giving a transmission zero exactly at $\omega_z$.

Key facts:

• Partial removal of a pole shifts the zeros of the remaining function towards the pole removed (removing part of the pole at $\infty$ moves zeros up; part of the pole at 0 moves zeros down).
• Poles of the remainder do not move; only zeros shift.
• The part removed is still a realizable element (e.g. a smaller series inductor), and the remainder stays LC because $k < k_\infty$.

Use in two-port synthesis

To synthesize a lossless two-port from $z_{11}$ (or $y_{11}$) with prescribed zeros of $z_{21}$:

1. List the required transmission zeros (at $0$, $\infty$ and finite $\pm j\omega_z$).
2. Zeros at $\infty$ and $0$: remove poles of $Z$ or $Y$ at $\infty$ or $0$ completely (series $L$/shunt $C$ or series $C$/shunt $L$).
3. For each finite zero $\omega_z$: partially remove a pole at $\infty$ or $0$ so that the remainder becomes zero at $s=\pm j\omega_z$.
4. Invert the remainder and remove the new pole at $\pm j\omega_z$ completely as a shunt series-LC branch (or a series tank). This branch blocks transmission at $\omega_z$.
5. Repeat with the remainder until it is fully realized.

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 $Z(s)$.)