--- title: "Terminated ladder with finite transmission zeros" subject: "Filter Design" chapter: 5 key: "semester_5/filter_design:finite-transmission-zeros-ladder" source: https://www.bctengineer.com/q/semester_5/filter_design/finite-transmission-zeros-ladder site: BCT Engineering --- # Terminated ladder with finite transmission zeros Filter Design · Chapter 5: Design of Resistively-Terminated Lossless Filters - Asked in: 2073 Shrawan, 2+4 marks; 2071 Shrawan, 2+4 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/finite-transmission-zeros-ladder - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-5 ## Question Define transmission and reflection coefficient. Explain how resistively terminated ladder network can be realized with finite transmission zeros. ## Short answer The transmission coefficient t(s)=2√(R1/R2) V2/V1 gives fraction of available power delivered; reflection coefficient ρ(s)=(R1−Z1)/(R1+Z1) gives reflected power, with |ρ|²+|t|²=1. Finite transmission zeros are realized by series parallel-LC tanks (open at ωz) or shunt series-LC branches (short at ωz), inserted by partial pole removal and zero shifting in the impedance expansion. ## Answer #### Definitions The **transmission coefficient** $t(s)$ is the ratio (in power terms) of the power actually delivered to the load $R_2$ to the maximum available power of the source $P_{\max} = |V_1|^2/4R_1$: $$ |t(j\omega)|^2 = \frac{P_2}{P_{\max}} = \frac{4R_1}{R_2}\left|\frac{V_2(j\omega)}{V_1(j\omega)}\right|^2 , \qquad t(s) = 2\sqrt{\frac{R_1}{R_2}}\,\frac{V_2(s)}{V_1(s)} $$ The **reflection coefficient** $\rho(s)$ of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance $Z_1$) from a source of resistance $R_1$: $$ \rho(s) = \frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}} $$ Since the ladder is lossless, all power entering it reaches the load, so the two coefficients are linked by **Feldtkeller's equation**: $$ |\rho(j\omega)|^2 + |t(j\omega)|^2 = 1 $$ #### Realizing finite transmission zeros in a terminated ladder A **transmission zero** is a frequency where no signal reaches the load ($t = 0$, $|\rho| = 1$). All-pole filters (Butterworth, Chebyshev) have all zeros at $s = \infty$, made by series inductors and shunt capacitors. Filters such as inverse Chebyshev and elliptic (Cauer) also need **finite zeros on the $j\omega$ axis**. In a ladder a transmission zero is produced when either: - a **series arm becomes an open circuit**: a parallel LC tank in the series arm, resonant at $\omega_z = 1/\sqrt{LC}$; or - a **shunt arm becomes a short circuit**: a series LC branch from line to ground, resonant at $\omega_z$. ```text series tank (open at wz) shunt series-LC (short at wz) o--+--UUU--+--o o-----+-----o | | | +--||---+ UUU L | === C | o-------------o o-----+-----o ``` **Procedure (zero shifting by partial removal):** 1. From $t(s)$ find $\rho(s)$ and $Z_1(s) = R_1\dfrac{1+\rho}{1-\rho}$ as usual. The finite zeros of $t(s)$ are known, say $\pm j\omega_z$. 2. Expand $Z_1$ (or $Y_1$) step by step. Before each finite zero, **partially remove** a pole at infinity (a series $L$ or shunt $C$), choosing its value so that the remaining function has a zero exactly at $s = j\omega_z$: e.g. $C = Y(s)/s$ evaluated at $s^2 = -\omega_z^2$. 3. The reciprocal of the remainder now has a pole at $\pm j\omega_z$; remove it fully as a series tank (or shunt series-LC branch). This creates the transmission zero. 4. Continue with the next zero; the final remainder must equal the load $R_2$. **Example.** For $y = s(s^2+4)/(s^2+1)$ with a required zero at $\omega = 3$: remove shunt $C = y/s|_{s^2=-9} = 5/8$ F, leaving $y' = 3s(s^2+9)/[8(s^2+1)]$, which is zero at $j3$. Then $1/y'$ has a pole at $j3$, removed as a series tank $C = 27/64$ F, $L = 64/243$ H that blocks the signal at $\omega = 3$ rad/s. The same idea gives the elliptic (Cauer) ladder, where each series inductor has a capacitor in parallel to produce a finite stopband zero. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a lossless terminated ladder, how are t and rho related? A. |rho| + |t| = 1 B. |rho|^2 + |t|^2 = 1 C. |rho|^2 - |t|^2 = 1 D. |rho| x |t| = 1 Answer: B. Power not reflected is delivered, giving |rho|^2 + |t|^2 = 1 (Feldtkeller). (Key point: Feldtkeller relation) 2. Which arrangement produces a finite transmission zero at wz in a ladder? A. A series inductor alone B. A shunt capacitor alone C. A parallel LC tank in a series arm, or a series LC branch in a shunt arm D. A series resistor in the shunt arm Answer: C. A series-arm tank is open at resonance and a shunt series-LC branch is a short, so no signal reaches the load. (Key point: Ways to create a finite zero) 3. In zero shifting by partial removal, why is only part of a pole at infinity removed before a finite zero? A. To make the remainder have a zero at j*wz so its reciprocal has a pole there B. To reduce the number of elements C. To make the source resistance zero D. To keep the function a pure resistance Answer: A. Choosing the partial value so the remainder is zero at j*wz means its reciprocal has a pole at j*wz, which is then removed as a tank. (Key point: Zero shifting) 4. Where are the transmission zeros of Butterworth and Chebyshev ladders? A. At s = 0 B. At finite points on the jw axis C. At s = infinity D. In the right half plane Answer: C. All-pole filters have all zeros at infinity, made by series L and shunt C; inverse Chebyshev and elliptic need finite jw zeros. 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