Synthesize two-port LC ladder from z21 and z22

Idea. The poles of $z_{21}$ and $z_{22}$ are the same; the zeros of $z_{21}$ are the transmission zeros, which must be produced by the ladder elements. We synthesize $z_{22}$ (looking in from port 2, port 1 open) so that the required transmission zeros appear.

Step 1: Transmission zeros

$$z_{21}(s) = \frac{k(s^2+9)}{s(s^2+4)}, \qquad z_{22}(s) = \frac{s^2+1}{s(s^2+4)}$$

Zeros of $z_{21}$: at $s = \pm j3$ (finite) and at $s = \infty$ (numerator degree 2, denominator degree 3, so one zero at infinity).

• Zero at $s = \infty$: produced by a shunt capacitor.
• Zero at $\omega = 3$: produced by a series parallel-LC tank resonating at $\omega = 3$ (it becomes an open circuit there).

Step 2: Zero shifting by partial removal of shunt capacitor

$$y_{22} = \frac{1}{z_{22}} = \frac{s(s^2+4)}{s^2+1}$$

$y_{22}$ has a pole at infinity. Removing it fully would give zeros at the wrong places, so remove only part of it, a shunt $C_a$, such that the remainder becomes zero at $s = j3$:

$$C_a = \left.\frac{y_{22}(s)}{s}\right|_{s^2=-9} = \frac{-9+4}{-9+1} = \frac{5}{8} = 0.625\ \text{F}$$ $$y'(s) = y_{22} - \tfrac58 s = \frac{s^3+4s - \tfrac58 s^3 - \tfrac58 s}{s^2+1} = \frac{3s(s^2+9)}{8(s^2+1)}$$

The zero of $y'$ at $s = \pm j3$ now matches the required transmission zero.

Step 3: Remove the pole at $\pm j3$ as a series tank

$$z'(s) = \frac{1}{y'} = \frac{8(s^2+1)}{3s(s^2+9)} = \frac{8/27}{s} + \frac{(64/27)\,s}{s^2+9}$$

The term $\dfrac{(64/27)s}{s^2+9}$ is a parallel LC tank in the series arm:

$$C_b = \frac{27}{64} = 0.4219\ \text{F}, \qquad L_b = \frac{1}{9C_b} = \frac{64}{243} = 0.2634\ \text{H}$$

It blocks transmission at $\omega = 1/\sqrt{L_bC_b} = 3$ rad/s.

Step 4: Remainder

$$z''(s) = \frac{8/27}{s} \;\Rightarrow\; \text{shunt capacitor } C_c = \frac{27}{8} = 3.375\ \text{F across port 1}$$

Final network

 port 1        series tank        port 2
 o----+----+---UUU---+-----+----o
      |    |   Lb    |     |
      |    +---||----+     |
     ===       Cb         ===
     Cc                    Ca
      |                    |
 o----+--------------------+----o
 Cc = 3.375 F, Lb = 0.2634 H,
 Cb = 0.4219 F, Ca = 0.625 F

Check

Analysing this ladder gives $z_{22} = \dfrac{s^2+1}{s(s^2+4)}$ and

$$z_{21} = \frac{(1/9)(s^2+9)}{s(s^2+4)}$$

So the required $z_{21}$ is realized with the constant $k = 1/9 \approx 0.111$ (the constant of $z_{21}$ is fixed by the network; another $k$ needs an ideal transformer at port 1).

Answer: shunt $C_a = 0.625$ F at port 2, series tank $L_b = 0.2634$ H $\parallel$ $C_b = 0.4219$ F, shunt $C_c = 3.375$ F at port 1, with $k = 1/9$.