---
title: "Transmission coefficient; realize 1/(s²+√2s+1)"
subject: "Filter Design"
chapter: 5
key: "semester_5/filter_design:transmission-coefficient-ladder-realization"
source: https://www.bctengineer.com/q/semester_5/filter_design/transmission-coefficient-ladder-realization
site: BCT Engineering
---

# Transmission coefficient; realize 1/(s²+√2s+1)

Filter Design · Chapter 5: Design of Resistively-Terminated Lossless Filters

- Asked in: 2080 Baisakh, 2+5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/transmission-coefficient-ladder-realization
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-5

## Question

What information does the transmission coefficient indicate? Realize the following transfer function using LC Ladder with equal termination of R1 = R2 = 1Ω. T(s) = 1/(s² + √2 s + 1)

## Short answer

The transmission coefficient t(s) indicates the fraction of the source’s maximum available power that reaches the load, thus defining the passband/stopband magnitude response and transmission zeros. It relates to the reflection coefficient by |t|²+|ρ|²=1. Realization of T(s)=1/(s²+√2 s+1) as LC ladder (1 Ω terminations): series L₁=1.414 H, shunt C₂=1.414 F, R₁=R₂=1 Ω.

## Answer

#### Information indicated by the transmission coefficient

$t(s) = 2\sqrt{R_1/R_2}\,V_2/V_1$, and $|t(j\omega)|^2 = P_2/P_{\max}$ is the fraction of the source's maximum available power ($|V_1|^2/4R_1$) that reaches the load. Hence:

- it is the **magnitude (power) response** of the filter: passband where $|t| \approx 1$, stopband where $|t| \approx 0$;
- $|t| = 1$ means perfect matching and maximum power transfer (no reflection, $\rho = 0$);
- $t = 0$ at a frequency means a **transmission zero**;
- $|t| \le 1$ always, which limits the gain of a passive ladder (e.g. $V_2/V_1 \le 0.5$ for equal terminations);
- with $|\rho|^2 = 1 - |t|^2$ it leads to the reflection coefficient used for synthesis.

#### Realization of $T(s) = \dfrac{1}{s^2+\sqrt2 s+1}$ with $R_1 = R_2 = 1\ \Omega$

**Step 1: Transmission coefficient.** For $n = 2$ (Table 1), $B(s) = s^2 + \sqrt2 s + 1$. With equal terminations $|t(0)|$ can be 1:

$$
t(s) = \frac{1}{s^2+\sqrt2 s+1}, \qquad |t(j\omega)|^2 = \frac{1}{1+\omega^4}
$$

(The actual voltage gain is $V_2/V_1 = \tfrac12 t(s)$, since $t = 2\sqrt{R_1/R_2}\,V_2/V_1$.)

**Step 2: Reflection coefficient.**

$$
|\rho(j\omega)|^2 = 1 - \frac{1}{1+\omega^4} = \frac{\omega^4}{1+\omega^4}
\;\Rightarrow\;
\rho(s)\rho(-s) = \frac{s^4}{B(s)B(-s)}, \quad \rho(s) = \frac{s^2}{s^2+\sqrt2 s+1}
$$

**Step 3: Input impedance** ($R_1 = 1$):

$$
Z_1(s) = \frac{1+\rho}{1-\rho} = \frac{2s^2+\sqrt2 s+1}{\sqrt2 s+1}
$$

**Step 4: Continued fraction.**

$$
\begin{aligned}
Z_1 &= \sqrt2 s + \frac{1}{\sqrt2 s + 1} \\
&= 1.414\,s + \cfrac{1}{1.414\,s + 1}
\end{aligned}
$$

So series $L_1 = 1.414$ H, then a shunt $C_2 = 1.414$ F, and the last "1" is $R_2 = 1\ \Omega$.

**Step 5: Circuit.**

```text
  R1=1  L1=1.414 H
 o-/\/\---UUU----+------+
 |               |      |
(~) V1     C2=1.414 F  R2=1
 |               |      |
 o---------------+------+
```

**Answer:** $R_1 = 1\ \Omega$, $L_1 = 1.414$ H, $C_2 = 1.414$ F, $R_2 = 1\ \Omega$ (normalized, $\omega_c = 1$ rad/s). The dual form $Z_1 = (1-\rho)/(1+\rho)$ gives shunt $C_1 = 1.414$ F and series $L_2 = 1.414$ H. These agree with the standard table values (1.414, 1.414).

## Check yourself

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

1. With equal terminations, the passive ladder's V2/V1 cannot exceed:
   A. 1
   B. 0.5
   C. 2
   D. 0.25
   Answer: B. |t| <= 1 and t = 2 V2/V1 for R1 = R2, so V2/V1 <= 0.5. (Key point: Gain limit)

2. With rho = s^2/(s^2+sqrt2 s+1), Z1 = (1+rho)/(1-rho) equals:
   A. (2s^2+1.414s+1)/(1.414s+1)
   B. (s^2+1)/(1.414s)
   C. (1.414s+1)/(2s^2+1.414s+1)
   D. (s^2+1.414s+1)/s^2
   Answer: A. 1+rho = (2s^2+sqrt2 s+1)/B and 1-rho = (sqrt2 s+1)/B. (Key point: Z1 from rho)

3. Using Z1 = (1-rho)/(1+rho) instead gives:
   A. Series L1 = 1.414 H then shunt C2 = 1.414 F
   B. Shunt C1 = 1.414 F then series L2 = 1.414 H
   C. A high-pass ladder
   D. A ladder with R2 = 2 ohm
   Answer: B. The dual circuit has shunt C1 and series L2, with the same values. (Key point: Dual realization)

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