Application of the Two Port Networks to the Ladder Network Synthesis
Introduction
In electrical network analysis and design, two-port networks provide a powerful framework for representing complex circuits using standardized parameters. One of the most important applications of two-port network theory is in the synthesis of ladder networks, which are widely used in filter design, impedance matching, and communication systems. Ladder network synthesis involves constructing a network of inductors and capacitors arranged in a ladder-like structure to realize a desired impedance or transfer function. Two-port network parameters make this process systematic, modular, and mathematically tractable.Two-Port Network Concept
A two-port network is an electrical network with two pairs of terminals: an input port and an output port. The behavior of the network is described by relating input voltages and currents to output voltages and currents using parameter sets. Common two-port parameters include:- Z-parameters (impedance parameters)
- Y-parameters (admittance parameters)
- ABCD (transmission) parameters
- h-parameters (hybrid parameters)
Ladder Network Overview
A ladder network is formed by alternating series and shunt elements, typically inductors and capacitors. The structure resembles a ladder, where:- Series elements form the “rails”
- Shunt elements form the “rungs”
- Use passive components only
- Are easy to manufacture
- Provide stable and predictable frequency responses
Role of Two-Port Networks in Ladder Synthesis
Ladder network synthesis relies on breaking a complex network into a cascade of simpler two-port sections. Each section consists of either a series impedance or a shunt admittance, which can be modeled as a two-port network. By representing each ladder element as a two-port network and combining them using transmission parameters, the overall network behavior can be derived efficiently.ABCD Parameters and Cascading
The ABCD parameters (Hybrid Paramenters) relate input and output variables as: $$ \begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_2 \\ I_2 \end{bmatrix} $$ For ladder networks, this form is ideal because:- Cascaded two-port networks multiply their ABCD matrices
- Overall system behavior is obtained by simple matrix multiplication
- Each ladder section contributes directly to the final transfer function
Two-Port Representation of Ladder Elements
A ladder network consists of two basic building blocks:Series Impedance Element
For a series element ( Z ): $$ \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} 1 & Z \\ 0 & 1 \end{bmatrix} $$Shunt Admittance Element
For a shunt element ( Y ): $$ \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ Y & 1 \end{bmatrix} $$ These simple forms make ladder synthesis straightforward.Ladder Network Synthesis Using Two-Port Theory
The synthesis process generally follows these steps:- 1. Start with a desired driving-point impedance or transfer function.
- 2. Express the function in a continued-fraction or polynomial form.
- 3. Extract series and shunt elements sequentially.
- 4. Represent each extracted element as a two-port section.
- 5. Cascade all sections using ABCD matrices.
- 6. Verify that the final network meets the required specifications.
Application in Filter Design
Two-port ladder synthesis is extensively used in the design of:- Low-pass filters
- High-pass filters
- Band-pass filters
- Band-stop filters
- Low sensitivity to component variations
- Good impedance matching
- High power handling capability
Application of two-port parameters in Low-pass filters
Another application of two-port parameters is the synthesis (or building) of ladder networks which are found frequently in practice and have particular use in designing passive lowpass filters. Figure 1(a) shows an $ L C $ ladder network with an odd number of elements (to realize an odd-order filter), while Fig. 1(b) shows one with an even number of elements (for realizing an even-order filter).

Fig. 1: LC ladder networks for lowpass filters of: (a) odd order, (b) even order.

Fig. 2: LC ladder network with terminating impedances.
$$\mathbf{I}_{1}=\mathbf{y}_{11} \mathbf{V}_{1}+\mathbf{y}_{12} \mathbf{V}_{2} \tag{1.a}$$
$$\mathbf{I}_{2}=\mathbf{y}_{21} \mathbf{V}_{1}+\mathbf{y}_{22} \mathbf{V}_{2} \tag{1.b}$$
$$-\mathbf{V}_{o} \mathbf{Y}_{L}=\mathbf{y}_{21} \mathbf{V}_{s}+\mathbf{y}_{22} \mathbf{V}_{o}$$
$$\mathbf{H}(s)=\frac{\mathbf{V}_{o}}{\mathbf{V}_{s}}=\frac{-\mathbf{y}_{21}}{\mathbf{Y}_{L}+\mathbf{y}_{22}}$$
$$\mathbf{H}(s)=-\frac{\mathbf{y}_{21} / \mathbf{Y}_{L}}{1+\mathbf{y}_{22} / \mathbf{Y}_{L}} \tag{2}$$
$$\mathbf{H}(s)=\frac{\mathbf{N}(s)}{\mathbf{D}(s)}=\frac{\mathbf{N}_{o}+\mathbf{N}_{e}}{\mathbf{D}_{o}+\mathbf{D}_{e}} \tag{3}$$
$$\mathbf{H}(s)=\left\{\begin{array}{ll}\frac{\mathbf{N}_{o}}{\mathbf{D}_{o}+\mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{N}_{e}}{\mathbf{D}_{o}+\mathbf{D}_{e}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right. \tag{4}$$
$$\mathbf{H}(s)=\left\{\begin{array}{ll}\frac{\mathbf{N}_{o} / \mathbf{D}_{e}}{1+\mathbf{D}_{o} / \mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{N}_{e} / \mathbf{D}_{o}}{1+\mathbf{D}_{e} / \mathbf{D}_{o}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right. \tag{5}$$
$$\frac{\mathbf{y}_{21}}{\mathbf{Y}_{L}}=\left\{\begin{array}{ll}\frac{\mathbf{N}_{o}}{\mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{N}_{e}}{\mathbf{D}_{o}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right.$$
$$\frac{\mathbf{y}_{22}}{\mathbf{Y}_{L}}=\left\{\begin{array}{ll}\frac{\mathbf{D}_{o}}{\mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{D}_{e}}{\mathbf{D}_{o}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right.$$
Example 1: Design the $ L C $ ladder network terminated with a $ 1\Omega $ resistor that has the normalized transfer function
(This transfer function is for a Butterworth lowpass filter.)
Solution:
The denominator shows that this is a third-order network, so that the $ L C $ ladder network is shown in Fig. 3(a), with two inductors and one capacitor.


Our goal is to determine the values of the inductors and capacitor. To achieve this, we group the terms in the denominator into odd or even parts:
so that
Divide the numerator and denominator by the odd part of the denominator to get
From Eq. (2), when $\mathbf{Y}_L=1$,
Any realization of $y_{22}$ will automatically realize $y_{21}$, since $y_{22}$ is the output driving-point admittance, that is, the output admittance of the network with the input port short-circuited. We determine the values of $L$ and $C$ in Fig. 3(a) that will give us $\mathbf{y}_{22}$. Recall that $\mathbf{y}_{22}$ is the short-circuit output admittance. So we short-circuit the input port as shown in Fig. 3(b). First we get $L_3$ by letting
By long division,
Comparing Eqs. (1.3) and (1.4) shows that
Next, we seek to get $C_2$ as in Fig. 3(c) and let
from which $C_2=1.33 \mathrm{~F}$ and
Thus, the $L C$ ladder network in Fig. 3(a) with $L_1=1.5 \mathrm{H}, C_2=$ $1.333 \mathrm{~F}$, and $L_3=0.5 \mathrm{H}$ has been synthesized to provide the given transfer function $\mathbf{H}(s)$. This result can be confirmed by finding $\mathbf{H}(s)=\mathbf{V}_2 / \mathbf{V}_1$ in Fig. 3(a) or by confirming the required $y_{21}$.
$$\mathbf{H}(s)=\frac{1}{s^{3}+2 s^{2}+2 s+1}$$



Fig. 3: For Example 1.
$$\mathbf{D}(s)=\left(s^{3}+2 s\right)+\left(2 s^{2}+1\right)$$
$$\mathbf{H}(s)=\frac{1}{\left(s^{3}+2 s\right)+\left(2 s^{2}+1\right)}$$
$$\mathbf{H}(s)=\frac{\frac{1}{s^{3}+2 s}}{1+\frac{2 s^{2}+1}{s^{3}+2 s}} \tag{1.1}$$
$$\mathbf{H}(s)=\frac{-y_{21}}{1+y_{22}} \tag{1.2}$$
Comparing Eqs. (1.1) and (1.2), we obtain
$$
\mathbf{y}_{21}=-\frac{1}{s^3+2 s}, \quad \mathbf{y}_{22}=\frac{2 s^2+1}{s^3+2 s}
$$
$$
\mathbf{Z}_A=\frac{1}{\mathbf{y}_{22}}=\frac{s^3+2 s}{2 s^2+1}=s L_3+\mathbf{Z}_B \tag{1.3}
$$
$$
\mathbf{Z}_A=0.5 s+\frac{1.5 s}{2 s^2+1} \tag{1.4}
$$
$$
L_3=0.5 \mathrm{H}, \quad \mathbf{Z}_B=\frac{1.5 s}{2 s^2+1}
$$
$$
\mathbf{Y}_B=\frac{1}{\mathbf{Z}_B}=\frac{2 s^2+1}{1.5 s}=1.333 s+\frac{1}{1.5 s}=s C_2+Y_C
$$
$$
\mathbf{Y}_C=\frac{1}{1.5 s}=\frac{1}{s L_1} \quad \Longrightarrow \quad L_1=1.5 \mathrm{H}
$$
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