Zeros of transmission and their realization

Zeros of transmission of a two-port network are the values of complex frequency $s$ at which the transfer function ($z_{21}$, $y_{21}$, $V_2/V_1$ or $I_2/I_1$) is zero, i.e. a finite input produces no output at that frequency. For an LC ladder they lie on the $j\omega$ axis (including $s=0$ and $s=\infty$).

Realization of transmission zeros

In a ladder, a signal is blocked when a series arm is an open circuit (its impedance has a pole) or a shunt arm is a short circuit (its impedance has a zero, i.e. its admittance has a pole). So the transmission zeros of a ladder are the poles of the series-arm impedances and the poles of the shunt-arm admittances.

 zero at w1=1/sqrt(L1C1)   zero at w2=1/sqrt(L2C2)
o--[L1||C1]--+---------o
             |
            [L2]
            [C2]
             |
o------------+---------o

Zeros at $0$ and $\infty$ are obtained by complete removal of poles of $Z$ or $Y$ at $0$ or $\infty$ (Cauer-type steps). A zero at a finite frequency $\omega_z$ is obtained by zero shifting (partial pole removal) followed by complete removal of the pole at $\pm j\omega_z$.

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 $Z(s)$.)