---
title: "Transmission zeros and zero shifting by partial pole removal"
subject: "Filter Design"
chapter: 4
key: "semester_5/filter_design:transmission-zeros-zero-shifting"
source: https://www.bctengineer.com/q/semester_5/filter_design/transmission-zeros-zero-shifting
site: BCT Engineering
---

# Transmission zeros and zero shifting by partial pole removal

Filter Design · Chapter 4: Properties and Synthesis of Passive Networks

- Asked in: 2080 Bhadra, 1+3 marks; 2075 Asoj, 1+3 marks; 2071 Chaitra, 1+4 marks (asked 3 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/transmission-zeros-zero-shifting
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-4

## Question

Define zeros of transmission in two port network. What is zero shifting by partial removal of pole? Explain with suitable example.

## Short answer

Zeros of transmission are s-values where transfer function is zero. Zero shifting: partially remove a pole at ∞ or 0 to create a desired zero in remainder, then invert and fully remove that pole as a shunt series-LC branch. Example: removing 3/8 H series inductor shifts zero to ω=2 rad/s; extracting shunt L=7/8 H, C=2/7 F provides transmission zero.

## Answer

**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 $s=\infty$).

In a ladder, a zero of transmission appears when a series arm has a pole of impedance (open circuit) or a shunt arm has a zero of impedance (short circuit), e.g. a series $L$ gives a zero at $\infty$, a shunt series-LC branch gives a zero at $1/\sqrt{LC}$.

#### Zero shifting by partial removal of pole

**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 $k < k_\infty$.

**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:

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

```text
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 $Z(s)$.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A transmission zero of a two-port is a value of s where:
   A. the transfer function is zero, so there is no output
   B. the driving point impedance is infinite
   C. the network becomes unstable
   D. the input current is zero
   Answer: A. Transmission zeros are zeros of the transfer function (z21, y21 etc.), where a finite input produces no output. (Key point: Definition of transmission zero)

2. Complete removal of a pole of a driving-point function can only give transmission zeros at:
   A. any finite jw
   B. s = 0 or s = infinity
   C. the negative real axis
   D. the poles of the remainder
   Answer: B. Complete removal gives zeros only at 0 or infinity. To place a zero at finite frequency, only part of a pole is removed (zero shifting). (Key point: Partial vs complete pole removal)

3. When part of the pole at infinity is removed, what happens to the zeros of the remaining function?
   A. They move down toward the origin
   B. The poles move up
   C. The zeros shift up toward the pole removed, while poles stay fixed
   D. Nothing changes
   Answer: C. Partial removal shifts the zeros toward the pole removed; the poles of the remainder do not move. (Key point: Effect of partial removal)

4. In the example with transmission zero at 2 rad/s, which shunt branch creates the zero?
   A. L2 = 7/8 H in series with C2 = 2/7 F, since 1/sqrt(L2C2) = 2
   B. L2 = 3/8 H in series with C2 = 2/7 F
   C. A shunt capacitor of 2/7 F alone
   D. L3 = 35/16 H with C3 = 8/21 F
   Answer: A. The shunt series L-C branch resonates at 2 rad/s (sqrt(7/8 x 2/7) = 1/2), shorting the signal there. (Key point: Example result)

## Related in Filter Design

- [Zero shifting by partial removal of pole](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Azero-shifting-partial-pole-removal) — Filter Design, chapter 4, asked 3×
- [Zero shifting for two-port synthesis](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Azero-shifting-two-port-synthesis) — Filter Design, chapter 4, asked 3×
- [Partial and complete removal of pole](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Apartial-complete-pole-removal) — Filter Design, chapter 4
- [Zeros of transmission and their realization](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Azeros-of-transmission-realization) — Filter Design, chapter 4, asked 10×
- [Steps in realizing transmission zeros](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atransmission-zeros-realization-steps) — Filter Design, chapter 4, asked 2×

## Related in other subjects

- [Two-port network model and parameters](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Atwo-port-network-model) — RF and Microwave Engineering, chapter 3