--- title: "Steps in realizing transmission zeros" subject: "Filter Design" chapter: 4 key: "semester_5/filter_design:transmission-zeros-realization-steps" source: https://www.bctengineer.com/q/semester_5/filter_design/transmission-zeros-realization-steps site: BCT Engineering --- # Steps in realizing transmission zeros Filter Design · Chapter 4: Properties and Synthesis of Passive Networks - Asked in: 2076 Chaitra, 1+4 marks; 2070 Chaitra, 5 marks (asked 2 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/transmission-zeros-realization-steps - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-4 ## Question What is transmission zeros in two port network? What are the steps involved in realizing transmission zeros in (a lossless) two port network? Explain with suitable example. ## Short answer Transmission zeros are frequencies where transfer function is zero. Steps: get Z11/Y11, check LC, remove poles at 0/∞ completely, for finite zero partially remove a pole to shift a zero to ωz, then remove the resulting pole as a shunt series-LC or series parallel-LC tank. Example: Z(s)=(s²+1)(s²+3)/(s(s²+2)) with ωz=2 yields L1=3/8 H, shunt L2=7/8 H, C2=2/7 F, L3=35/16 H, C3=8/21 F. ## 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, the transmission zeros are the poles of the series-arm impedances and the zeros of the shunt-arm impedances. Each element type gives its own transmission zero: - series $L$ or shunt $C$: transmission zero at $s=\infty$; - series $C$ or shunt $L$: transmission zero at $s=0$; - series parallel-LC tank or shunt series-LC branch: transmission zero at $\omega = 1/\sqrt{LC}$. #### Steps to realize transmission zeros in a lossless two-port 1. From the specification, find the driving-point function ($z_{11}$ or $y_{11}$) and the required zeros of $z_{21}$ (or $y_{21}$). 2. Check that the driving-point function is a valid LC function. 3. For zeros at $s=\infty$: remove the pole at $\infty$ completely (series $L$ from $Z$ or shunt $C$ from $Y$). 4. For zeros at $s=0$: remove the pole at the origin completely (series $C$ or shunt $L$). 5. For a zero at a finite $\omega_z$: **partial removal** — subtract only $ks$ (or $k/s$) with $k$ chosen so that the remainder is zero at $s=\pm j\omega_z$ (zero shifting). 6. Invert the remainder; it now has a pole at $\pm j\omega_z$. Remove it **completely** as a shunt series-LC branch (or series parallel-LC tank) with $1/\sqrt{LC} = \omega_z$. 7. Continue with the remainder until nothing is left; check that all element values are positive. **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. How are transmission zeros at s = infinity and s = 0 realized? A. By partial removal of poles B. By complete removal of the pole at infinity (series L / shunt C) and at the origin (series C / shunt L) C. By adding resistors D. By removing poles at +-jw Answer: B. Complete removal of poles at infinity and 0 gives zeros at those points. (Key point: Zero at infinity or origin) 2. For a finite zero wz, what is the first step on the driving-point function? A. Remove the pole at wz completely from Z B. Subtract only ks (or k/s) with k chosen so the remainder is zero at +-jwz C. Divide Z by s D. Set the load to zero Answer: B. Partial removal (zero shifting) makes the remainder vanish at wz, so that the reciprocal can have a pole at wz. (Key point: Finite-frequency zero) 3. After inverting the remainder, the pole at +-jwz is removed completely as: A. a series resistor B. a shunt series-LC branch (or series parallel-LC tank) with 1/sqrt(LC) = wz C. a shunt capacitor D. a series inductor Answer: B. Such a branch has zero shunt impedance (or infinite series impedance) at wz, blocking transmission. (Key point: Element for the zero) 4. What should be checked at the end of the realization? A. That the elements are all negative B. That the remainder is a resistor C. That nothing is left and all element values are positive D. That the number of elements is even Answer: C. Continue until the remainder is exhausted and verify every element value is positive. 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