--- title: "Partial and complete removal of pole" subject: "Filter Design" chapter: 4 key: "semester_5/filter_design:partial-complete-pole-removal" source: https://www.bctengineer.com/q/semester_5/filter_design/partial-complete-pole-removal site: BCT Engineering --- # Partial and complete removal of pole Filter Design · Chapter 4: Properties and Synthesis of Passive Networks - Asked in: 2079 Bhadra, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/partial-complete-pole-removal - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-4 ## Question What do you mean by partial removal and complete removal of pole in the synthesis of 2-port lossless ladder network? Explain with suitable examples. ## Short answer Complete removal subtracts full residue, degree drops, creates transmission zero at pole frequency (e.g., series 1 H for Z at ∞). Partial removal subtracts fraction of residue, degree unchanged, shifts zeros. Example: remove 3/8 of ∞ pole to place zero at ω=2, giving series L=3/8 H, then shunt LC (L=7/8 H, C=2/7 F) and remaining L=35/16 H, C=8/21 F. ## Answer In ladder synthesis of a lossless two-port, a driving-point function is reduced step by step by removing its poles as elements. A pole can be removed with its **full residue** (complete removal) or with only **part** of it (partial removal). #### Complete removal of a pole The whole term of the pole is subtracted, e.g. $Z_1 = Z - k_\infty s$ with $k_\infty = \lim Z/s$. The remainder no longer has that pole, its degree drops, and the removed element (series $L$, series $C$, tank, etc.) produces a transmission zero at the pole frequency ($0$, $\infty$ or $\omega_i$). Cauer ladders use complete removal at every step. **Example:** $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$. Here $k_\infty = 1$, so $$ Z_1 = Z - s = \frac{2s^2+3}{s(s^2+2)} $$ A series 1 H inductor is removed and $Z_1$ now has a zero at $\infty$ (degree 2 over 3) — this is the first step of Cauer-I. #### Partial removal of a pole Only $k s$ with $0 < k < k_\infty$ (or $k/s$ with $k < k_0$) is removed. The pole remains in the remainder, the degree does not drop, but the **zeros of the remainder shift** towards the removed pole. $k$ is chosen so that a zero lands exactly at a desired frequency $\omega_z$ (zero shifting); the next step then removes the pole of the reciprocal at $\pm j\omega_z$ completely to give a transmission zero there. **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 $Z(s)$.) | Point | Complete removal | Partial removal | |---|---|---| | Part removed | full residue $k$ | fraction of residue | | Pole in remainder | disappears | stays | | Degree of remainder | reduced | unchanged | | Effect on zeros | — | shifted towards removed pole | | Transmission zeros | only at $0$, $\infty$, $\omega_i$ of poles removed | at any chosen $\omega_z$ | | Use | Foster/Cauer forms | zero shifting, elliptic filters | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In complete removal of a pole: A. Only a fraction of the residue is removed and the pole stays B. The zeros of the remainder shift toward the pole C. The whole residue is removed, the pole disappears and the degree drops D. Only poles on the jw axis are removed Answer: C. Complete removal subtracts the full term, eliminating the pole; Cauer ladders use it at every step. (Key point: Complete removal) 2. Partial removal of a pole at infinity (removing k*s with k less than k_inf) leaves: A. No pole in the remainder B. The pole in the remainder, the degree unchanged, and the zeros shifted C. A pole at the origin D. The same zeros as before Answer: B. It is used for zero shifting: the zeros of the remainder move toward the removed pole. (Key point: Effect of partial removal) 3. Why is partial removal used in ladder synthesis? A. To make the ladder have fewer elements B. To get a transmission zero at a chosen finite frequency C. To convert an LC function to an RC one D. To remove the need for a termination Answer: B. k is chosen so that a zero lands at the desired frequency, then the next step removes that pole completely to give a transmission zero there. (Key point: Purpose of zero shifting) 4. Z = (s^2+1)(s^2+3)/(s(s^2+2)), with a transmission zero wanted at w = 2. Z(j2) = j0.75, so the series inductor removed first is: A. L1 = 3/8 H B. L1 = 3/4 H C. L1 = 1 H D. L1 = 8/3 H Answer: A. Z1(j2) = j0.75 - j2k = 0 gives k = 3/8. 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