Application of the Two Port Networks to the Ladder Network Synthesis

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.
When either network is terminated by the load impedance $ Z_{L} $ and the source impedance $ Z_{s} $, we obtain the structure in Fig. 2. To make the design less complicated, we will assume that $ Z_{s}=0 $.
Fig. 2: LC ladder network with terminating impedances.
Our goal is to synthesize the transfer function of the $ L C $ ladder network. We begin by characterizing the ladder network by its admittance parameters, namely,
$$\mathbf{I}_{1}=\mathbf{y}_{11} \mathbf{V}_{1}+\mathbf{y}_{12} \mathbf{V}_{2} \tag{1.a}$$
$$\mathbf{I}_{2}=\mathbf{y}_{21} \mathbf{V}_{1}+\mathbf{y}_{22} \mathbf{V}_{2} \tag{1.b}$$
(Of course, the impedance parameters could be used instead of the admittance parameters.) At the input port, $ \mathbf{V}_{1}=\mathbf{V}_{s} $ since $ \mathbf{Z}_{s}=0 $. At the output port, $ \mathbf{V}_{2}=\mathbf{V}_{o} $ and $ \mathbf{I}_{2}=-\mathbf{V}_{2} / \mathbf{Z}_{L}=-\mathbf{V}_{o} \mathbf{Y}_{L} $. Thus Eq. (1.b) becomes
$$-\mathbf{V}_{o} \mathbf{Y}_{L}=\mathbf{y}_{21} \mathbf{V}_{s}+\mathbf{y}_{22} \mathbf{V}_{o}$$
We can write this as
$$\mathbf{H}(s)=-\frac{\mathbf{y}_{21} / \mathbf{Y}_{L}}{1+\mathbf{y}_{22} / \mathbf{Y}_{L}} \tag{2}$$
We may ignore the negative sign in Eq. (2) because filter requirements are often stated in terms of the magnitude of the transfer function. The main objective in filter design is to select capacitors and inductors so that the parameters $ \mathbf{y}_{21} $ and $ \mathbf{y}_{22} $ are synthesized, thereby realizing the desired transfer function. To achieve this, we take advantage of an important property of the $ L C $ ladder network: all $ z $ and $ y $ parameters are ratios of polynomials that contain only even powers of $ s $ or odd powers of $ s $-that is, they are ratios of either $ \mathrm{Od}(s) / \operatorname{Ev}(s) $ or $ \operatorname{Ev}(s) / \operatorname{Od}(s) $, where $ \mathrm{Od} $ and $ \mathrm{Ev} $ are odd and even functions, respectively. Let
$$\mathbf{H}(s)=\frac{\mathbf{N}(s)}{\mathbf{D}(s)}=\frac{\mathbf{N}_{o}+\mathbf{N}_{e}}{\mathbf{D}_{o}+\mathbf{D}_{e}} \tag{3}$$
where $ \mathbf{N}(s) $ and $ \mathbf{D}(s) $ are the numerator and denominator of the transfer function $ \mathbf{H}(s) ; \mathbf{N}_{o} $ and $ \mathbf{N}_{e} $ are the odd and even parts of $ \mathbf{N} ; \mathbf{D}_{o} $ and $ \mathbf{D}_{e} $ are the odd and even parts of $ \mathbf{D} $. Since $ \mathbf{N}(s) $ must be either odd or even, we can write Eq. (3) as
$$\mathbf{H}(s)=\left\{\begin{array}{ll}\frac{\mathbf{N}_{o}}{\mathbf{D}_{o}+\mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{N}_{e}}{\mathbf{D}_{o}+\mathbf{D}_{e}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right. \tag{4}$$
and can rewrite this as
$$\mathbf{H}(s)=\left\{\begin{array}{ll}\frac{\mathbf{N}_{o} / \mathbf{D}_{e}}{1+\mathbf{D}_{o} / \mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{N}_{e} / \mathbf{D}_{o}}{1+\mathbf{D}_{e} / \mathbf{D}_{o}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right. \tag{5}$$
Comparing this with Eq. (2), we obtain the $ y $ parameters of the network as
$$\frac{\mathbf{y}_{21}}{\mathbf{Y}_{L}}=\left\{\begin{array}{ll}\frac{\mathbf{N}_{o}}{\mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{N}_{e}}{\mathbf{D}_{o}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right.$$
$$\frac{\mathbf{y}_{22}}{\mathbf{Y}_{L}}=\left\{\begin{array}{ll}\frac{\mathbf{D}_{o}}{\mathbf{D}_{e}}, & \left(\mathbf{N}_{e}=0\right) \\\frac{\mathbf{D}_{e}}{\mathbf{D}_{o}}, & \left(\mathbf{N}_{o}=0\right)\end{array}\right.$$
The following example illustrates the procedure.
Example 1: Design the $ L C $ ladder network terminated with a $ 1\Omega $ resistor that has the normalized transfer function
$$\mathbf{H}(s)=\frac{1}{s^{3}+2 s^{2}+2 s+1}$$
(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.
Fig. 3: For Example 1.
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:
$$\mathbf{D}(s)=\left(s^{3}+2 s\right)+\left(2 s^{2}+1\right)$$
so that
$$\mathbf{H}(s)=\frac{1}{\left(s^{3}+2 s\right)+\left(2 s^{2}+1\right)}$$
Divide the numerator and denominator by the odd part of the denominator to get
$$\mathbf{H}(s)=\frac{\frac{1}{s^{3}+2 s}}{1+\frac{2 s^{2}+1}{s^{3}+2 s}} \tag{1.1}$$
From Eq. (2), when $\mathbf{Y}_L=1$,
$$\mathbf{H}(s)=\frac{-y_{21}}{1+y_{22}} \tag{1.2}$$ Comparing Eqs. (1.1) and (1.2), we obtain
$$ \mathbf{y}_{21}=-\frac{1}{s^3+2 s}, \quad \mathbf{y}_{22}=\frac{2 s^2+1}{s^3+2 s} $$
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
$$ \mathbf{Z}_A=\frac{1}{\mathbf{y}_{22}}=\frac{s^3+2 s}{2 s^2+1}=s L_3+\mathbf{Z}_B \tag{1.3} $$
By long division,
$$ \mathbf{Z}_A=0.5 s+\frac{1.5 s}{2 s^2+1} \tag{1.4} $$
Comparing Eqs. (1.3) and (1.4) shows that
$$ L_3=0.5 \mathrm{H}, \quad \mathbf{Z}_B=\frac{1.5 s}{2 s^2+1} $$
Next, we seek to get $C_2$ as in Fig. 3(c) and let
$$ \mathbf{Y}_B=\frac{1}{\mathbf{Z}_B}=\frac{2 s^2+1}{1.5 s}=1.333 s+\frac{1}{1.5 s}=s C_2+Y_C $$
from which $C_2=1.33 \mathrm{~F}$ and
$$ \mathbf{Y}_C=\frac{1}{1.5 s}=\frac{1}{s L_1} \quad \Longrightarrow \quad L_1=1.5 \mathrm{H} $$
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 $y_{21}$.

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