# Supplementary Material to: Passive Controller Realization of a Biquadratic
Impedance with Double Poles and Zeros as a Seven-Element Series-Parallel
Network for Effective Mechanical Control

Kai Wang1, Michael Z. Q. Chen2,∗, Chanying Li3,4, and Guanrong Chen5 1Key
Laboratory of Advanced Process Control for Light Industry (Ministry of
Education), School of Internet of Things Engineering, Jiangnan University,
Wuxi, Jiangsu 214122, P. R. China.2School of Automation, Nanjing University of
Science and Technology, Nanjing, Jiangsu 210094, P. R. China.3Key Laboratory
of Systems and Control, Academy of Mathematics and Systems Science, Chinese
Academy of Sciences, Beijing 100190, China.4School of Mathematical Sciences,
University of Chinese Academy of Sciences, Beijing 100049, China.5Department
of Electronic Engineering, City University of Hong Kong, Tat Chee Avenue,
Kowloon, Hong Kong.This work is supported by the National Natural Science
Foundation of China under grants 61703184 and 61374053.Correspondence: MZQ
Chen<EMAIL_ADDRESS>
###### Abstract

This report presents some supplementary material to the paper entitled
“Passive controller realization of a biquadratic impedance with double poles
and zeros as a seven-element Series-parallel network for effective mechanical
control” [1].

Keywords: Passive mechanical control, passive network synthesis, inerters,
biquadratic impedances, series-parallel networks.

##  I Introduction

This report presents some supplementary material to the paper entitled
“Passive controller realization of a biquadratic impedance with double poles
and zeros as a seven-element Series-parallel network for effective mechanical
control” [1], which are omitted from the paper for brevity.

##  II Preliminaries of Passive Network Synthesis

A two-terminal electrical network is defined to be passive if
$\int_{-\infty}^{T}i(t)v(t)dt\geq 0$ for all $T$ and for all admissible
current $i(t)$ and voltage $v(t)$ [2]. A real-rational function $F(s)$ is
positive real if $F(s)$ is analytic and $\mathfrak{R}(F(s))\geq 0$ for any
$\mathfrak{R}(s)>0$ [9]. An impedance (resp. admittance) is defined to be
$Z(s)=V(s)/I(s)$ (resp. $Y(s)=I(s)/V(s)$), where $V(s)$ and $I(s)$ are
voltages and currents of the port of a two-terminal network, where the network
is said to realize (or is a realization of) its impedance (resp. admittance).
A two-terminal electrical network is passive if and only if its impedance
(resp. admittance) is positive real, and any positive-real impedance (resp.
admittance) is realizable as a two-terminal passive $RLC$ network [9]. Any
positive-real impedance (resp. admittance) is realizable as a two-terminal
passive $RLC$ network. A reactive element is an inductor or capacitor, and a
resistor is also called as a resistive element.

##  III Definitions of the network duality and the frequency inverse

Any two-terminal passive $RLC$ network $N$ can be regarded as a one-terminal-
pair labeled graph $\mathcal{N}$ with two distinguished terminal vertices (see
[13, pg. 14]), in which the labels designate passive circuit elements
regardless of element values, namely resistors, capacitors, and inductors,
which are labeled as $R_{i}$, $C_{i}$, and $L_{i}$, respectively.

Two natural maps acting on the labeled graph are defined as follows:

  1. 1.

$\text{GDu}:=$ Graph duality, which takes the one-terminal-pair graph into its
dual (see [13, Definition 3-12]) while preserving the labeling.

  2. 2.

$\text{Inv}:=$ Inversion, which preserves the graph but interchanges the
reactive elements, that is, capacitors to inductors and inductors to
capacitors with their labels $C_{i}$ to $L_{i}$ and $L_{i}$ to $C_{i}$.

Consequently, one defines

$\text{Dual}:=\text{network duality of one-terminal-pair labeled
graph}:=\text{GDu}\circ\text{Inv}=\text{Inv}\circ\text{GDu}.$

Consider a network $N$ whose one-terminal-pair labeled graph is $\mathcal{N}$.
Denote $\text{Inv}(N)$ as the network whose one-terminal-pair labeled graph is
$\text{Inv}(\mathcal{N})$, resistors are of the same values as those of $N$,
and inductors (resp. capacitors) are replaced by capacitors (resp. inductors)
with reciprocal values, which is called the frequency inverse network of $N$.
Denote $\text{GDu}(N)$ as the network whose one-terminal-pair labeled graph is
$\text{GDu}(\mathcal{N})$ and elements are of the reciprocal values to those
of $N$, which is called the frequency inverse dual network of $N$. Denote
$\text{Dual}(N)$ as the network whose one-terminal-pair labeled graph is
$\text{Dual}(\mathcal{N})$, resistors are of reciprocal values to those of
$N$, and inductors (resp. capacitors) are replaced by capacitors (resp.
inductors) with same values, which is called the dual network of $N$.

It can be proved that $Z(s)$ (resp. $Y(s)$) is realizable as the impedance
(resp. admittance) of a network $N$ whose one-terminal-pair labeled graph is
$\mathcal{N}$, if and only if $Z(s^{-1})$ (resp. $Y(s^{-1})$) is realizable as
the impedance (resp. admittance) of $\text{Inv}(N)$ whose one-terminal-pair
labeled graph is $\text{Inv}(\mathcal{N})$, if and only if $Z(s^{-1})$ (resp.
$Y(s^{-1})$) is realizable as the admittance (resp. impedance) of
$\text{GDu}(N)$ whose one-terminal-pair labeled graph is
$\text{GDu}(\mathcal{N})$, and if and only if it is realizable as the
admittance (resp. impedance) of $\text{Dual}(N)$ whose one-terminal-pair
labeled graph is $\text{Dual}(\mathcal{N})$. Therefore, if a necessary and
sufficient condition is derived for
$H(s)=\sum_{k=0}^{m}a_{k}s^{k}/\sum_{k=0}^{m}b_{k}s^{k}$ to be realizable as
the impedance (resp. admittance) of a two-terminal network whose one-terminal-
pair labeled graph is $\mathcal{N}$, then the corresponding condition for
$\text{Inv}(\mathcal{N})$ can be obtained from that for $\mathcal{N}$ through
conversion $a_{k}\leftrightarrow a_{m-k}$ and $b_{k}\leftrightarrow b_{m-k}$
for $k=0,1,...,\lfloor m/2\rfloor$ (the principle of frequency inversion). The
corresponding condition for $\text{GDu}(\mathcal{N})$ can be obtained from
that for $\mathcal{N}$ through conversion $a_{k}\leftrightarrow b_{m-k}$ for
$k=0,1,...,m$ (the principle of frequency-inverse duality). Furthermore, the
corresponding condition for $\text{Dual}(\mathcal{N})$ can be obtained from
that for $\mathcal{N}$ through conversion $a_{k}\leftrightarrow b_{k}$ for
$k=0,1,...,m$ (the principle of duality).

Specifically, based on the principle of frequency inversion, a necessary and
sufficient condition for $Z(s)$ in the form of (1) with $k$, $z$, $p$ $>0$ to
be realizable as the impedance of a two-terminal network whose one-terminal-
pair labeled graph is $\text{Inv}(\mathcal{N})$ can be obtained from that for
$\mathcal{N}$ through $z\leftrightarrow z^{-1}$ and $p\leftrightarrow p^{-1}$.
Based on the principle of frequency-inversion duality, a necessary and
sufficient condition for $Z(s)$ in the form of (1) with $k$, $z$, $p$ $>0$ to
be realizable as the impedance of a two-terminal network whose one-terminal-
pair labeled graph is $\text{GDu}(\mathcal{N})$ can be obtained from that for
$\mathcal{N}$ through $p\leftrightarrow z^{-1}$ and $z\leftrightarrow p^{-1}$.
Based on the principle of duality, a necessary and sufficient condition for
$Z(s)$ in the form of (1) with $k$, $z$, $p$ $>0$ to be realizable as the
impedance of a two-terminal network whose one-terminal-pair labeled graph is
$\text{Dual}(\mathcal{N})$ can be obtained from that for $\mathcal{N}$ through
$p\leftrightarrow z$.

##  IV Realizability as Series-Parallel Networks with No More Than Five
Elements

###### Theorem IV.1

A biquadratic impedance $Z(s)$ in the form of (1), that is,

$Z(s)=k\frac{(s+z)^{2}}{(s+p)^{2}},$

where $k$, $z$, $p$ $>0$ and $p\neq z$, cannot be realized with fewer than
four elements.

###### Proof:

Let $A=kx$, $B=2kzx$, $C=kz^{2}x$, $D=x$, $E=2px$, and $F=p^{2}x$ for $x>0$.
This theorem can be proved from the realizability conditions of a general
biquadratic impedance in the form of (2), that is,

$Z(s)=\frac{As^{2}+Bs+C}{Ds^{2}+Es+F},$ (IV.1)

where $A$, $B$, $C$, $D$, $E$, $F$ $>0$, with at most three elements in [15].
∎

###### Theorem IV.2

A biquadratic impedance $Z(s)$ in the form of (1), where $k$, $z$, $p$ $>0$
and $p\neq z$, is realizable as a four-element series-parallel network if and
only if $p=z/3$ or $p=3z$.

###### Proof:

Let $A=kx$, $B=2kzx$, $C=kz^{2}x$, $D=x$, $E=2px$, and $F=p^{2}x$ for $x>0$.
This condition can be derived from the realizability conditions of a general
biquadratic impedance in the form of (2) where $A$, $B$, $C$, $D$, $E$, $F$
$>0$ as a four-element network in [15], where it is obvious that any four-
element network must be series-parallel. ∎

###### Theorem IV.3

A biquadratic impedance $Z(s)$ in the form of (1), where $k$, $z$, $p$ $>0$
and $p\neq z$, is realizable as a five-element series-parallel network if and
only if $p/z\in(1/3,3)$, $p=(2+\sqrt{2})z$, or $p=z/(2+\sqrt{2})$.

###### Proof:

Let $A=kx$, $B=2kzx$, $C=kz^{2}x$, $D=x$, $E=2px$, and $F=p^{2}x$ for $x>0$.
This condition can be derived from the realizability conditions of a general
biquadratic impedance in the form of (2) where $A$, $B$, $C$, $D$, $E$, $F$
$>0$ as a five-element series-parallel network in [10]. ∎

##  V Some Basic Lemmas

###### Lemma V.1

Consider a biquadratic impedance $F(s)$ in the form of

$F(s)=\frac{\alpha s^{2}+\beta s+\gamma}{(s+p)^{2}},$ (V.1)

where $\alpha$, $\beta$, $\gamma$ $\geq 0$, and $p>0$. Then, $F(s)$ is
realizable as a three-element series-parallel network if and only if at least
one of the following conditions holds: 1. $\alpha\gamma=0$; 2. $\beta=0$ and
$\alpha p^{2}-\gamma=0$; 3. $\gamma=0$ and $\alpha p-2\beta=0$; 4. $\alpha=0$
and $2\beta p-\gamma=0$; 5. $\alpha p^{2}-\beta p+\gamma=0$.

###### Proof:

Since it is obvious that $F(s)$ in the form of (V.1) is not a reactance
function [3, Definition 3.1], there is at least one resistor, which means that
there are at most two reactive elements.

When the number of reactive elements is at most one, the degree [2, Section
3.6] of $F(s)$ cannot exceed one by [2, Theorem 4.4.3]. Therefore, there must
exist at least one common factor between $(\alpha s^{2}+\beta s+\gamma)$ and
$(s+p)^{2}$, which holds if and only if Condition 5 is satisfied.

When the number of reactive elements is two, there is one resistor. Based on
the method of enumeration, one can obtain Conditions 1–4. ∎

###### Lemma V.2

Consider a biquadratic impedance $F(s)$ in the form of (V.1), where $\alpha$,
$\beta$, $\gamma$ $\geq 0$, $p>0$, assuming that the condition of Lemma V.1
does not hold. Then, $F(s)$ is realizable as a four-element series-parallel
network if and only if at least one of the six conditions holds: 1\.
$\alpha=0$ and $\gamma<2\beta p$; 2. $\gamma=0$ and $\alpha p<2\beta$; 3.
$\alpha$, $\beta$, $\gamma$ $>0$ and $\alpha p^{2}-\gamma=0$; 4. $\alpha$,
$\beta$, $\gamma$ $>0$, $\alpha p^{2}<\gamma$, and $3\alpha
p^{2}+\gamma-2\beta p=0$ or $\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma
p^{2}-2\beta\gamma p=0$ holds; 5. $\alpha$, $\beta$, $\gamma$ $>0$, $\alpha
p^{2}>\gamma$, and $\alpha p^{2}+3\gamma-2\beta p=0$ or
$\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma=0$ holds; 6. $\alpha$,
$\beta$, $\gamma$ $>0$ and $\alpha^{2}p^{4}-2\alpha\beta p^{3}+6\alpha\gamma
p^{2}-2\beta\gamma p+\gamma^{2}=0$. Moreover, if one of the above six
conditions holds, then $F(s)$ is realizable as a two-reactive four-element
series-parallel network.

###### Proof:

Let $A=\alpha x$, $B=\beta x$, $C=\gamma x$, $D=x$, $E=2px$, and $F=p^{2}x$
for any $x>0$. A necessary and sufficient condition for $Z(s)$ in the form of
(2) with $A$, $B$, $C$, $D$, $E$, $F$ $\geq 0$ to be positive real is
$(\sqrt{AF}-\sqrt{CD})^{2}\leq BE$ [4, 7]. Since any positive-real biquadratic
impedance with zero coeffients is realizable as a two-reactive four-element
series-parallel network [10, Lemma 8], one obtains Conditions 1 and 2 together
with Lemma V.1. Now, it remains to considering the case of $\alpha$, $\beta$,
$\gamma$ $>0$. By [15, Theorem 5], one obtains Conditions 3–6.

By the covering configurations in [15, Figs. 4–6], the number of reactive
elements for realizations is two. ∎

###### Lemma V.3

Consider a biquadratic impedance $F(s)$ in the form of (V.1), where $\alpha$,
$\beta$, $\gamma$, $p>0$, and neither the condition of Lemmas V.1 nor the
condition of Lemma V.2 holds. Then, $F(s)$ is realizable as a two-reactive
five-element series-parallel network if and only if at least one of the
following conditions holds: 1. $\alpha p^{2}>\gamma$ and $\alpha
p^{2}+3\gamma-2\beta p<0$; 2. $\alpha p^{2}>\gamma$ and
$\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma<0$; 3. $\alpha
p^{2}<\gamma$ and $3\alpha p^{2}+\gamma-2\beta p<0$; 4. $\alpha p^{2}<\gamma$
and $\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p<0$.

###### Proof:

A necessary and sufficient condition for $Z(s)$ in the form of (2) where $A$,
$B$, $C$, $D$, $E$, $F$ $>0$ to be realizable as a two-reactive five-element
series-parallel network is presented in [10, Theorem 1]. Letting $A=\alpha x$,
$B=\beta x$, $C=\gamma x$, $D=x$, $E=2px$, and $F=p^{2}x$ for any $x>0$, this
lemma can be proved together with the assumption that neither the condition of
Lemma V.1 nor the condition of Lemma does not hold. ∎

##  VI Supplementary Lemmas of Three-Reactive Seven-Element Series-Parallel
Realizations for the Proof of Lemma 2

###### Lemma VI.1

Consider a biquadratic impedance $Z(s)$ in the form of (IV.1) with $A$, $B$,
$C$, $D$, $E$, $F$ $>0$ that cannot be realized as a series-parallel network
containing fewer than seven elements. If $Z(s)$ is realizable as a three-
reactive seven-element series-parallel network as shown in Fig. 2(a), where
$N_{1}$ is a three-element series-parallel network and $N_{2}$ is a four-
element series-parallel network, then $Z(s)$ is also realizable as shown in
Fig. VI.1, where $N_{b}$ is a two-reactive five-element series-parallel
network.

Figure VI.1: A class of three-reactive seven-element series-parallel networks,
where $N_{b}$ is a two-reactive five-element series-parallel network (Fig. 7).

###### Proof:

By [15, Lemma 2], $Z(s)$ cannot be realized as the series connection of two
networks, one of which only contains reactive elements. Therefore, $N_{1}$
must contain one or two reactive elements. If $N_{1}$ contains two reactive
elements, then $N_{2}$ contains one reactive element. Therefore, the degree
[2, Section 3.6] of $Z_{2}(s)$ cannot exceed one by [2, Theorem 4.4.3], where
$Z_{2}(s)$ denotes the impedance of $N_{2}$. By the discussion in [15],
$Z_{2}(s)$ is realizable as a one-reactive three-element series-parallel
network, which implies that $Z(s)$ is realizable with a series-parallel
network containing fewer than seven elements. This contradicts the assumption.
If $N_{1}$ contains one reactive element, then the degree [2, Section 3.6] of
$Z_{1}(s)$ cannot exceed one by [2, Theorem 4.4.3], where $Z_{1}(s)$ denotes
the impedance of $N_{1}$. By the discussion in [15], $Z_{1}(s)$ is realizable
as a one-reactive three-element series-parallel network, which is equivalent
to one of configurations in Fig. VI.2. Regarding the series connection of
$R_{2}$ and $N_{2}$ as $N_{b}$, this lemma is proved. ∎

Figure VI.2: Configurations for $N_{1}$ mentioned in the proof of Lemma VI.1.

##  VII Complete Proof of Lemma 2

Assume that $Z(s)$ is realizable as in Fig. VI.1(a), where $N_{b}$ is a two-
reactive five-element series-parallel network. Let $Z(s)=Z_{a}(s)+Z_{b}(s)$,
where $Z_{a}(s)$ is the impedance of the parallel connection of $R_{1}$ and
$C_{1}$ and $Z_{b}(s)$ is the impedance of $N_{b}$. It is clear that
$Z_{a}(s)$ can be written in the form of

$Z_{a}(s)=\frac{m}{s+p},$

where $m>0$. Since the condition of Theorem 1 does not hold, $N_{b}$ cannot be
equivalent to a network containing fewer than five elements. Therefore,
$Z_{b}(s)$ can be expressed in the form of

$Z_{b}(s)=\frac{\alpha s^{2}+\beta s+\gamma}{(s+p)^{2}},$

where $\alpha$, $\beta$, $\gamma$ $>0$, and the condition of Lemma V.3 holds.
Since

$Z_{a}(s)+Z_{b}(s)=\frac{\alpha s^{2}+(\beta+m)s+(\gamma+mp)}{(s+p)^{2}},$

it follows that $\alpha s^{2}+(\beta+m)s+(\gamma+mp)=k(s+z)^{2}$. Therefore,

$\beta=2\alpha z-m,~{}~{}~{}~{}~{}\gamma=\alpha z^{2}-mp.$ (VII.1)

Since $\beta$, $\gamma$ $>0$, it follows from (VII.1) that

$\alpha>\max\left\\{\frac{m}{2z},\frac{mp}{z^{2}}\right\\}.$ (VII.2)

If $Z_{b}(s)$ satisfies Condition 1 of Lemma V.3, then $\alpha p^{2}>\gamma$
and $\alpha p^{2}+3\gamma-2\beta p<0$. Together with (VII.1), one obtains

$\displaystyle(p-z)(p+z)\alpha+mp$ $\displaystyle>0,$ (VII.3)
$\displaystyle(p-z)(p-3z)\alpha-mp$ $\displaystyle<0.$ (VII.4)

If $z<p\leq 3z$, then it is obvious that the condition of Theorem 1 holds. If
$p>3z$ or $p<z$, then it follows from (VII.4) that $\alpha<mp/((p-z)(p-3z))$.
Together with (VII.2), one obtains $mp/z^{2}<mp/((p-z)(p-3z))$, which implies
$(2-\sqrt{2})z<p<(2+\sqrt{2})z$. Therefore, the condition of Theorem 1 holds.

If $Z_{b}(s)$ satisfies Condition 2 of Lemma V.3, then $\alpha p^{2}>\gamma$
and $\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma<0$. Together with
(VII.1), one obtains (VII.3) and

$(p-z)(p-3z)\alpha^{2}+m(3p-4z)\alpha+m^{2}<0.$ (VII.5)

It follows from (VII.5) that $p<3z$. If $z<p<3z$, then it is obvious that the
condition of Theorem 1 holds. If $p<z$, then (VII.3) implies that

$\alpha<-\frac{mp}{(p-z)(p+z)}.$ (VII.6)

From (VII.2) and (VII.6), it follows that $m/(2z)<-mp/((p-z)(p+z))$, which
implies $p>z/(1+\sqrt{2})$. Therefore, the condition of Theorem 1 holds.

If $Z_{b}(s)$ satisfies Condition 3 of Lemma V.3, then $\alpha p^{2}<\gamma$
and $3\alpha p^{2}+\gamma-2\beta p<0$. Together with (VII.1), one obtains

$\displaystyle(p-z)(p+z)\alpha+mp$ $\displaystyle<0,$ (VII.7)
$\displaystyle(3p-z)(p-z)\alpha+mp$ $\displaystyle<0.$ (VII.8)

It follows from (VII.8) that $z/3<p<z$. Therefore, the condition of Theorem 1
holds.

If $Z_{b}(s)$ satisfies Condition 4 of Lemma V.3, then $\alpha p^{2}<\gamma$
and $\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p<0$. Together
with (VII.1), one obtains (VII.7) and

$z^{2}(3p-z)(p-z)\alpha^{2}+p^{3}m\alpha<0.$ (VII.9)

It follows from (VII.9) that $z/3<p<z$. Therefore, the condition of Theorem 1
holds.

This means that $Z(s)$ cannot be realized as in Fig. VI.1(a), where $N_{b}$ is
a two-reactive five-element series-parallel network. It is clear that any
network in Fig. VI.1(b) can be a frequency inverse network of another one in
Fig. VI.1(a). Therefore, by the principle of frequency inverse (Section III),
$Z(s)$ cannot be realized as in Fig. VI.1(b), where $N_{b}$ is a two-reactive
five-element series-parallel network.

By Lemma V.3, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as a three-
reactive seven-element series-parallel network as shown in Fig. 2(a), where
$N_{1}$ is a three-element series-parallel network and $N_{2}$ is a four-
element series-parallel network. Since any network in Fig. 2(b), where $N_{1}$
is a three-element series-parallel network and $N_{2}$ is a four-element
series-parallel network, can be a dual network of the case of Fig. 2(a), by
the principle of duality (Section III), this lemma can be proved.

##  VIII Supplementary Lemmas of Four-Reactive Seven-Element Series-Parallel
Realizations for the Proof of Lemma 3

###### Lemma VIII.1

Consider the four-reactive seven-element series-parallel network in Fig. 2,
realizing a biquadratic impedance $Z(s)$ in the form of (IV.1) with $A$, $B$,
$C$, $D$, $E$, $F$ $>0$, where $N_{1}$ is a two-reactive three-element series-
parallel network and $N_{2}$ is a two-reactive four-element series-parallel
network. If $Z(s)$ cannot be realized as a series-parallel network containing
fewer than seven elements, then $N_{1}$ must be one of the configurations in
Fig. VIII.1.

Figure VIII.1: Two-reactive three-element configurations for the $N_{1}$
mentioned in Lemma VIII.1 (Fig. 8).

###### Proof:

Let $\mathcal{C}(a,a^{\prime})$ denote the cut-set [13, pg. 28] separating a
one-terminal-pair labeled graph of a network into two connected subgraphs
containing two terminal vertices $a$ and $a^{\prime}$, respectively. By [15,
Lemma 1], for any realization of $Z(s)$ there is no cut-set
$\mathcal{C}(a,a^{\prime})$ corresponding to only one kind of reactive
elements, where $a$ and $a^{\prime}$ denote two terminals. Since $N_{1}$
contains three elements, all the possible network graphs [5] of $N_{1}$ are
listed as in Fig. VIII.2. By the method of enumeration, this lemma can be
proved. ∎

Figure VIII.2: Possible network graphs for three-element networks.

###### Lemma VIII.2

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in
Fig. 2(a), where $N_{1}$ is the configuration in Fig. VIII.1(a) and $N_{2}$ is
a two-reactive four-element series-parallel network.

###### Proof:

By calculation, the impedance of the configuration in Fig. VIII.1(a) is
obtained as

$Z_{1}(s)=\frac{R_{1}L_{1}s}{R_{1}L_{1}C_{1}s^{2}+L_{1}s+R_{1}}.$

Since by assumption the condition of Theorem 1 does not hold, $N_{2}$ cannot
be equivalent to a series-parallel network containing fewer than four
elements. If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(a) and $N_{2}$ is a two-reactive four-element
series-parallel network, then the impedance of $N_{1}$ is in the form of

$Z_{1}(s)=\frac{ms}{(s+p)^{2}},$

where $m>0$, and the impedance of $N_{2}$ is in the form of

$Z_{2}(s)=\frac{\alpha s^{2}+\beta s+\gamma}{(s+p)^{2}},$ (VIII.1)

where $\alpha$, $\beta$, $\gamma$ $>0$, and moreover the condition of Lemma
V.2 holds. Since $\alpha s^{2}+(\beta+m)s+\gamma=k(s+z)^{2}$, it follows that

$\displaystyle\beta+m$ $\displaystyle=2z\alpha,$ (VIII.2)
$\displaystyle\gamma$ $\displaystyle=z^{2}\alpha.$ (VIII.3)

Since $\alpha$, $\gamma$ $>0$, $Z_{2}(s)$ satisfies neither Condition 1 nor
Condition 2 of Lemma V.2.

If $Z_{2}(s)$ satisfies Condition 3 of Lemma V.2, then $\alpha
p^{2}-\gamma=0$. Together with (VIII.3), it follows that $p=z$, which
contradicts the assumption.

If $Z_{2}(s)$ satisfies Condition 4 of Lemma V.2, then $\alpha p^{2}<\gamma$
and either $3\alpha p^{2}+\gamma-2\beta p=0$ or
$\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p=0$ holds. For the
case of $3\alpha p^{2}+\gamma-2\beta p=0$, together with (VIII.2) and
(VIII.3), one obtains $\alpha=-2mp/((3p-z)(p-z))$,
$\beta=-m(3p^{2}+z^{2})/((3p-z)(p-z))$, and $\gamma=-2mz^{2}p/((3p-z)(p-z))$,
which implies $z/3<p<z$ by $\alpha$, $\beta$, $\gamma$ $>0$. Thus, the
condition of Theorem 1 holds. For the case of
$\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p=0$, together with
(VIII.2) and (VIII.3), one obtains

$\alpha=\frac{mp}{z(3p-z)},~{}~{}~{}\beta=-\frac{m(p-z)}{3p-z},~{}~{}~{}\gamma=\frac{mzp}{3p-z},$
(VIII.4)

or

$\alpha=\frac{mp}{z(p-z)},~{}~{}~{}\beta=\frac{m(p+z)}{p-z},~{}~{}~{}\gamma=\frac{mzp}{p-z}.$
(VIII.5)

Because $\alpha$, $\beta$, $\gamma$ $>0$, it follows from (VIII.4) that
$z/3<p<z$, which satisfies the condition of Theorem 1. Substituting (VIII.5)
into $\alpha p^{2}<\gamma$ yields $(p+z)pm/z<0$, which is impossible.

If $Z_{2}(s)$ satisfies Condition 5 of Lemma V.2, then $\alpha p^{2}>\gamma$
and either $\alpha p^{2}+3\gamma-2\beta p=0$ or
$\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma=0$ holds. For the case
of $\alpha p^{2}+3\gamma-2\beta p=0$, together with (VIII.2) and (VIII.3), one
obtains $\alpha=-2mp/((p-z)(p-3z))$, $\beta=-m(p^{2}+3z^{2})/((p-z)(p-3z))$,
and $\gamma=-2mz^{2}p/((p-z)(p-3z))$, which implies $z<p<3z$ because $\alpha$,
$\beta$, $\gamma$ $>0$. Thus, the condition of Theorem 1 holds. For the case
of $\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma=0$, together with
(VIII.2) and (VIII.3), one obtains

$\alpha=-\frac{m}{p-3z},~{}~{}~{}\beta=-\frac{m(p-z)}{p-3z},~{}~{}~{}\gamma=-\frac{mz^{2}}{p-3z},$
(VIII.6)

or

$\alpha=-\frac{m}{p-z},~{}~{}~{}\beta=-\frac{m(p+z)}{p-z},~{}~{}~{}\gamma=-\frac{mz^{2}}{p-z}.$
(VIII.7)

It follows from (VIII.6) that $z<p<3z$ by $\alpha$, $\beta$, $\gamma$ $>0$.
Thus, the condition of Theorem 1 holds. Substituting (VIII.7) into $\alpha
p^{2}>\gamma$ yields $-m(p+z)>0$, which is impossible.

If $Z_{2}(s)$ satisfies Condition 6 of Lemma V.2, then
$\alpha^{2}p^{4}+6\alpha\gamma p^{2}+\gamma^{2}-2\alpha\beta
p^{3}-2\beta\gamma p=0$. Together with (VIII.2) and (VIII.3), one obtains
$\alpha=0$, $\beta=-m$, and $\gamma=0$ or
$\alpha=-2mp(z^{2}+p^{2})/(p-z)^{4}$,
$\beta=-m(p^{4}+6z^{2}p^{2}+z^{4})/(p-z)^{4}$, and
$\gamma=-2mz^{2}p(z^{2}+p^{2})/(p-z)^{4}$, which contradicts the assumption
that $\alpha$, $\beta$, $\gamma$ $>0$.

As a conclusion, this lemma is proved. ∎

###### Lemma VIII.3

If a biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ is realizable as in
Fig. 2(a), where $N_{1}$ is one of the configurations in Figs. VIII.1(b),
VIII.1(c), and VIII.1(d), and $N_{2}$ is a two-reactive four-element series-
parallel network, then the condition of Lemma 1 holds.

###### Proof:

First, the case where $N_{1}$ is a configuration in Fig. VIII.1(b) and $N_{2}$
is a two-reactive four-element series-parallel network will be discussed.

It is calculated that the impedance of the configuration in Fig. VIII.1(b) is
in the form of

$Z_{1}(s)=\frac{R_{1}L_{1}C_{1}s^{2}+R_{1}}{L_{1}C_{1}s^{2}+R_{1}C_{1}s+1}.$
(VIII.8)

Since it is assumed that the condition of Theorem 1 does not hold, $N_{2}$
cannot be equivalent to a series-parallel network containing fewer than four
elements. If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(b) and $N_{2}$ is a two-reactive four-element
series-parallel network, then the impedance of $N_{1}$ is in the form of

$Z_{1}(s)=\frac{m(s^{2}+p^{2})}{(s+p)^{2}},$

where $m>0$, and the impedance of $N_{2}$ is in the form of (VIII.1), where
$\beta>0$, $\alpha$, $\gamma$ $\geq 0$, and the condition of Lemma V.2 holds.
Since $(m+\alpha)s^{2}+\beta s+(mp^{2}+\gamma)=k(s+z)^{2}$, it follows that

$\displaystyle\beta$ $\displaystyle=2z(m+\alpha),$ (VIII.9) $\displaystyle
mp^{2}+\gamma$ $\displaystyle=z^{2}(m+\alpha).$ (VIII.10)

If $Z_{2}(s)$ satisfies Condition 1 of Lemma V.2, then $\beta$, $\gamma$ $>0$,
$\alpha=0$, and $\gamma<2\beta p$. Together with (VIII.9) and (VIII.10), one
obtains

$\beta=2mz,~{}~{}~{}\gamma=-m(p-z)(p+z),$ (VIII.11)

which implies $p<z$ by $\beta$, $\gamma$ $>0$. Substituting (VIII.11) into
$\gamma<2\beta p$ yields $m(p^{2}+4zp-z^{2})>0$. This implies
$z/(2+\sqrt{5})<p<z$, which satisfies the condition of Lemma 1.

If $Z_{2}(s)$ satisfies Condition 2 of Lemma V.2, then $\alpha$, $\beta$ $>0$,
$\gamma=0$, and $\alpha p<2\beta$. Together with (VIII.9) and (VIII.10), one
obtains

$\alpha=\frac{m(p-z)(p+z)}{z^{2}},~{}~{}~{}\beta=\frac{2mp^{2}}{z},$ (VIII.12)

which implies $p>z$ by $\alpha$, $\beta$ $>0$. Substituting (VIII.12) into
$\alpha p<2\beta$ yields $mp(p^{2}-4zp-z^{2})/z^{2}<0$. This further implies
$z<p<(2+\sqrt{5})z$. Therefore, the condition of Lemma 1 holds.

If $Z_{2}(s)$ satisfies Condition 3 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$ and $\alpha p^{2}-\gamma=0$. Together with (VIII.9) and
(VIII.10), one obtains $\alpha=-m$, $\beta=0$, and $\gamma=-mp^{2}$, which
contradicts the assumption.

If $Z_{2}(s)$ satisfies Condition 4 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$, $\alpha p^{2}<\gamma$, and either $3\alpha p^{2}+\gamma-2\beta
p=0$ or $\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p=0$. For
the case of $3\alpha p^{2}+\gamma-2\beta p=0$, together with (VIII.9) and
(VIII.10), one obtains

$\alpha=\frac{m(p^{2}+4zp-z^{2})}{(3p-z)(p-z)},~{}~{}~{}\beta=\frac{8mzp^{2}}{(3p-z)(p-z)},~{}~{}~{}\gamma=-\frac{mp^{2}(3p^{2}-4zp-3z^{2})}{(3p-z)(p-z)}.$
(VIII.13)

Substituting (VIII.13) into $\alpha p^{2}<\gamma$ yields
$4mp^{2}(p+z)/(3p-z)<0$, which implies $z/(2+\sqrt{5})<p<z/3$ together with
(VIII.13) by $\alpha$, $\beta$, $\gamma$ $>0$. Thus, the condition of Lemma 1
holds. For the case of $\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma
p^{2}-2\beta\gamma p=0$, together with (VIII.9) and (VIII.10), one obtains

$\alpha=-m,~{}~{}~{}\beta=0,~{}~{}~{}\gamma=-mp^{2},$ (VIII.14)

or

$\alpha=-\frac{m(p^{2}+2zp-z^{2})^{2}}{z^{2}(3p-z)(p-z)},~{}~{}~{}\beta=-\frac{2mp^{2}(p^{2}+4zp-z^{2})}{z(3p-z)(p-z)},~{}~{}~{}\gamma=-\frac{4mp^{4}}{(3p-z)(p-z)}.$
(VIII.15)

By $\alpha$, $\beta$, $\gamma$ $>0$, (VIII.14) is impossible. It is implied
from (VIII.15) that $z/3<p<z$, which satisfies the condition of Lemma 1.

If $Z_{2}(s)$ satisfies Condition 5 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$, $\alpha p^{2}>\gamma$, and either $\alpha p^{2}+3\gamma-2\beta
p=0$ or $\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma=0$. For the
case of $\alpha p^{2}+3\gamma-2\beta p=0$, together with (VIII.9) and
(VIII.10), one obtains

$\alpha=\frac{m(3p^{2}+4zp-3z^{2})}{(p-z)(p-3z)},~{}~{}~{}\beta=\frac{8mzp^{2}}{(p-z)(p-3z)},~{}~{}~{}\gamma=-\frac{mp^{2}(p^{2}-4zp-z^{2})}{(p-z)(p-3z)}.$
(VIII.16)

Substituting (VIII.16) into $\alpha p^{2}>\gamma$ yields
$4mp^{2}(p+z)/(p-3z)>0$, which implies $3z<p<(2+\sqrt{5})z$ together with
(VIII.16) by $\alpha$, $\beta$, $\gamma$ $>0$. For the case of
$\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma=0$, together with
(VIII.9) and (VIII.10), one obtains

$\alpha=-m,~{}~{}~{}\beta=0,~{}~{}~{}\gamma=-mp^{2},$ (VIII.17)

or

$\alpha=-\frac{4mz^{2}}{(p-z)(p-3z)},~{}~{}~{}\beta=\frac{2mz(p^{2}-4zp-z^{2})}{(p-z)(p-3z)},~{}~{}~{}\gamma=-\frac{m(p^{2}-2zp-z^{2})^{2}}{(p-z)(p-3z)}.$
(VIII.18)

It is obvious that (VIII.17) contradicts the assumption that $\alpha$,
$\beta$, $\gamma$ $>0$. Moreover, it is implied from (VIII.18) that $z<p<3z$,
which satisfies the condition of Theorem 1.

If $Z_{2}(s)$ satisfies Condition 6 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$ and $\alpha^{2}p^{4}+6\alpha\gamma p^{2}+\gamma^{2}-2\alpha\beta
p^{3}-2\beta\gamma p=0$. Together with (VIII.9) and (VIII.10), one obtains

$\begin{split}\alpha&=\frac{m(3p^{4}-2z^{2}p^{2}+4z^{3}p-z^{4}+2p^{2}\sqrt{2p^{4}+2z^{4}})}{(p-z)^{4}},\\\
\beta&=\frac{4mzp^{2}(2p^{2}-2zp+2z^{2}+\sqrt{2p^{4}+2z^{4}})}{(p-z)^{4}},\\\
\gamma&=\frac{mp^{2}(-p^{4}+4zp^{3}-2z^{2}p^{2}+3z^{4}+2z^{2}\sqrt{2p^{4}+2z^{4}})}{(p-z)^{4}},\end{split}$
(VIII.19)

or

$\begin{split}\alpha&=\frac{m(3p^{4}-2z^{2}p^{2}+4z^{3}p-z^{4}-2p^{2}\sqrt{2p^{4}+2z^{4}})}{(p-z)^{4}},\\\
\beta&=\frac{4mzp^{2}(2p^{2}-2zp+2z^{2}-\sqrt{2p^{4}+2z^{4}})}{(p-z)^{4}},\\\
\gamma&=\frac{mp^{2}(-p^{4}+4zp^{3}-2z^{2}p^{2}+3z^{4}-2z^{2}\sqrt{2p^{4}+2z^{4}})}{(p-z)^{4}}.\end{split}$
(VIII.20)

Consider the solutions in (VIII.19). Assume that $p\geq(2+\sqrt{5})z$. Then,
$\gamma<0$ since $-p^{4}+4zp^{3}-2z^{2}p^{2}+3z^{4}<0$ and
$(2z^{2}\sqrt{2p^{4}+2z^{4}})^{2}-(-p^{4}+4zp^{3}-2z^{2}p^{2}+3z^{4})^{2}=-(p+z)(p^{2}-4zp-z^{2})(p-z)^{5}\leq
0$. This contradicts the assumption. Assume that $p\leq z/(2+\sqrt{5})$. Then,
$\alpha<0$ since $3p^{4}-2z^{2}p^{2}+4z^{3}p-z^{4}<0$ and
$(2p^{2}\sqrt{2p^{4}+2z^{4}})^{2}-(3p^{4}-2z^{2}p^{2}+4z^{3}p-z^{4})^{2}=-(p+z)(p^{2}+4zp-z^{2})(p-z)^{5}\leq
0$. This also contradicts the assumption. Consider the solutions in (VIII.20).
Assume that $p\geq(2+\sqrt{5})z$. Then, $\gamma<0$ because of
$-p^{4}+4zp^{3}-2z^{2}p^{2}+3z^{4}<0$. This contradicts the assumption. Assume
that $p\leq z/(2+\sqrt{5})$. Then, $\alpha<0$ because of
$3p^{4}-2z^{2}p^{2}+4z^{3}p-z^{4}<0$. This also contradicts the assumption.

Therefore, it can be proved that if a biquadratic impedance
$Z(s)\in\mathcal{Z}_{p2,z2}$ is realizable as in Fig. 2(a), where $N_{1}$ is
one of the configurations in Fig. VIII.1(b) and $N_{2}$ is a two-reactive
four-element series-parallel network, then the condition of Lemma 1 holds.

Then, it turns to the case where $N_{1}$ is a configurations in Figs.
VIII.1(c) and $N_{2}$ is a two-reactive four-element series-parallel network.

It is calculated that the impedance of the configuration in Fig. VIII.1(c) is
in the form of

$Z_{1}(s)=\frac{s(R_{1}L_{1}C_{1}s+L_{1})}{L_{1}C_{1}s^{2}+R_{1}C_{1}s+1}.$
(VIII.21)

Since it is assumed that the condition of Theorem 1 does not hold, $N_{2}$
cannot be equivalent to a series-parallel network containing fewer than four
elements. If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(c) and $N_{2}$ is a two-reactive four-element
series-parallel network, then the impedance of $N_{1}$ is in the form of

$Z_{1}(s)=\frac{ms(s+p/2)}{(s+p)^{2}},$

where $m>0$, and the impedance of $N_{2}$ is in the form of (VIII.1), where
$\beta$, $\gamma$ $>0$, $\alpha$ $\geq 0$, and the condition of Lemma V.2
holds. Since $(m+\alpha)s^{2}+(mp/2+\beta)s+\gamma=k(s+z)^{2}$, it follows
that

$\displaystyle\frac{mp}{2}+\beta$ $\displaystyle=2(m+\alpha)z,$ (VIII.22)
$\displaystyle\gamma$ $\displaystyle=(m+\alpha)z^{2}.$ (VIII.23)

If $Z_{2}(s)$ satisfies Condition 1 of Lemma V.2, then $\beta$, $\gamma$ $>0$,
$\alpha=0$, and $\gamma<2\beta p$. Together with (VIII.22) and (VIII.23), one
obtains

$\beta=-\frac{1}{2}m(p-4z),~{}~{}~{}\gamma=mz^{2},$ (VIII.24)

which implies $p<4z$ by $\beta>0$. Substituting (VIII.24) into $\gamma<2\beta
p$ yields $m(p^{2}-4zp+z^{2})<0$, which further implies
$z/(2+\sqrt{3})<p<(2+\sqrt{3})z$ together with (VIII.24). Thus, the condition
of Lemma 1 holds.

Since $\gamma>0$, $Z_{2}(s)$ cannot satisfy Condition 2 of Lemma V.2.

If $Z_{2}(s)$ satisfies Condition 3 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$ and $\alpha p^{2}-\gamma=0$. Together with (VIII.22) and
(VIII.23), one obtains

$\alpha=\frac{mz^{2}}{(p-z)(p+z)},~{}~{}~{}\beta=-\frac{mp(p^{2}-4zp-z^{2})}{2(p-z)(p+z)},~{}~{}~{}\gamma=\frac{mz^{2}p^{2}}{(p-z)(p+z)},$

which implies $z<p<(2+\sqrt{5})z$ by $\alpha$, $\beta$, $\gamma$ $>0$. Thus,
the condition of Lemma 1 holds.

If $Z_{2}(s)$ satisfies Condition 4 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$, $\alpha p^{2}<\gamma$, and either $3\alpha p^{2}+\gamma-2\beta
p=0$ or $\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p=0$. For
the case of $3\alpha p^{2}+\gamma-2\beta p=0$, together with (VIII.22) and
(VIII.23), one obtains

$\alpha=-\frac{m(p^{2}-4zp+z^{2})}{(3p-z)(p-z)},~{}~{}~{}\beta=-\frac{mp(3p^{2}-12zp+z^{2})}{2(3p-z)(p-z)},~{}~{}~{}\gamma=\frac{2mz^{2}p^{2}}{(3p-z)(p-z)},$

which implies $z/(2+\sqrt{3})<p<z/3$ or $z<p<(2+\sqrt{3})z$ by $\alpha$,
$\beta$, $\gamma$ $>0$. Thus, the condition of Lemma 1 holds. For the case of
$\beta^{2}p^{2}+\gamma^{2}-\alpha\gamma p^{2}-2\beta\gamma p=0$, together with
(VIII.22) and (VIII.23), one obtains

$\begin{split}\alpha&=\frac{m(2p^{3}-8zp^{2}+8z^{2}p-2z^{3}+p^{2}\sqrt{(p-z)(p-3z)})}{2z(3p-z)(p-z)},\\\
\beta&=\frac{mp(p^{2}-z^{2}+2p\sqrt{(p-z)(p-3z)})}{2(3p-z)(p-z)},\\\
\gamma&=\frac{mzp^{2}(2p-2z+\sqrt{(p-z)(p-3z)})}{2(3p-z)(p-z)},\end{split}$
(VIII.25)

or

$\begin{split}\alpha&=\frac{m(2p^{3}-8zp^{2}+8z^{2}p-2z^{3}-p^{2}\sqrt{(p-z)(p-3z)})}{2z(3p-z)(p-z)},\\\
\beta&=\frac{mp(p^{2}-z^{2}-2p\sqrt{(p-z)(p-3z)})}{2(3p-z)(p-z)},\\\
\gamma&=\frac{2p-2z-\sqrt{(p-z)(p-3z)}}{2(3p-z)(p-z)},\end{split}$ (VIII.26)

where $p<z$ or $p>3z$ must hold to guarantee the existence of the solutions.
Consider the solutions in (VIII.25). Substituting (VIII.25) into $\alpha
p^{2}<\gamma$ yields

$\frac{mp^{2}(2p(p-3z)+(p+z)\sqrt{(p-z)(p-3z)})}{2(3p-z)z}<0.$ (VIII.27)

It is further implied from (VIII.27) that $p<z$. Moreover, since
$((p+z)\sqrt{(p-z)(p-3z)})^{2}-(2p^{2}-6zp)^{2}=-(3p-z)(p-3z)(p^{2}-4zp-z^{2})$,
it is implied that $z/3<p<z$, which satisfies the condition of Theorem 1.
Consider the solutions in (VIII.26). The assumption that $\beta$, $\gamma$
$>0$ implies $z/3<p<z$ or $p>3z$. Assume that $p\geq(2+\sqrt{5})z$. Then,
$\beta\leq 0$ since $p^{2}-z^{2}>0$ and
$(p^{2}-z^{2})^{2}-(2p\sqrt{(p-z)(p-3z)})^{2}=-(p-z)(3p-z)(p^{2}-4zp-z^{2})\leq
0$. This contradicts the assumption.

If $Z_{2}(s)$ satisfies Condition 5 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$, $\alpha p^{2}>\gamma$, and either $\alpha p^{2}+3\gamma-2\beta
p=0$ or $\alpha^{2}p^{2}+\beta^{2}-2\alpha\beta p-\alpha\gamma=0$. For the
case of $\alpha p^{2}+3\gamma-2\beta p=0$, together with (VIII.22) and
(VIII.23), one obtains

$\begin{split}\alpha&=\frac{m(-p^{2}+6zp-7z^{2}+\sqrt{-z^{2}(p^{2}-4zp-z^{2})})}{2(p-z)(p-3z)},\\\
\beta&=\frac{m(-p^{3}+6zp^{2}-7z^{2}p-2z^{3}+2z\sqrt{-z^{2}(p^{2}-4zp-z^{2})})}{2(p-z)(p-3z)},\\\
\gamma&=\frac{mz^{2}(p^{2}-2zp-z^{2}+\sqrt{-z^{2}(p^{2}-4zp-z^{2})})}{2(p-z)(p-3z)},\end{split}$
(VIII.28)

or

$\begin{split}\alpha&=\frac{m(-p^{2}+6zp-7z^{2}-\sqrt{-z^{2}(p^{2}-4zp-z^{2})})}{2(p-z)(p-3z)},\\\
\beta&=\frac{m(-p^{3}+6zp^{2}-7z^{2}p-2z^{3}-2z\sqrt{-z^{2}(p^{2}-4zp-z^{2})})}{2(p-z)(p-3z)},\\\
\gamma&=\frac{mz^{2}(p^{2}-2zp-z^{2}-\sqrt{-z^{2}(p^{2}-4zp-z^{2})})}{2(p-z)(p-3z)},\end{split}$
(VIII.29)

where $p\leq(2+\sqrt{5})z$ must hold to guarantee the existence of the
solutions. Assume that $p=(2+\sqrt{5})z$. It is implied from (VIII.28) and
(VIII.29) that $\beta=0$, which contradicts the assumption. Assume that $p\leq
z/(2+\sqrt{5})$. For (VIII.28), it is derived that $\alpha<0$ since
$(-z^{2}(p^{2}-4zp-z^{2}))^{2}-(-p^{2}+6zp-7z^{2})^{2}=-(p-z)(p-3z)(p-4z)^{2}<0$.
This contradicts the assumption. For (VIII.29), it is derived that $\alpha$,
$\beta$, $\gamma$ $<0$ since $-p^{2}+6zp-7z^{2}<0$,
$-p^{3}+6zp^{2}-7z^{2}p-2z^{3}<0$, and $p^{2}-2zp-z^{2}<0$. This also
contradicts the assumption.

If $Z_{2}(s)$ satisfies Condition 6 of Lemma V.2, then $\alpha$, $\beta$,
$\gamma$ $>0$ and $\alpha^{2}p^{4}+6\alpha\gamma p^{2}+\gamma^{2}-2\alpha\beta
p^{3}-2\beta\gamma p=0$. Together with (VIII.22) and (VIII.23), one obtains

$\alpha=-m,~{}~{}~{}\beta=-\frac{mp}{2},~{}~{}~{}\gamma=0,$ (VIII.30)

or

$\begin{split}\alpha&=-\frac{mz^{2}(p^{2}-4zp+z^{2})}{(p-z)^{4}},\\\
\beta&=-\frac{mp(p^{4}-8zp^{3}+22z^{2}p^{2}-24z^{3}p+z^{4})}{2(p-z)^{4}},\\\
\gamma&=\frac{mz^{2}p^{2}(p^{2}-4zp+5z^{2})}{(p-z)^{4}}.\end{split}$ (VIII.31)

It is obvious that (VIII.30) is impossible. For (VIII.31), one implies
$z/(2+\sqrt{3})<p<(2+\sqrt{3})z$ by $\alpha$, $\beta$, $\gamma$ $>0$. Thus,
the condition of Lemma 1 holds.

Therefore, it can be proved that if a biquadratic impedance
$Z(s)\in\mathcal{Z}_{p2,z2}$ is realizable as in Fig. 2(a), where $N_{1}$ is
one of the configurations in Fig. VIII.1(c) and $N_{2}$ is a two-reactive
four-element series-parallel network, then the condition of Lemma 1 holds.

It is clear that any network in Fig. 2(a), where $N_{1}$ is a configuration in
Fig. VIII.1(d) and $N_{2}$ is a two-reactive four-element series-parallel
network can be a frequency inverse network of the case where $N_{1}$ is a
configuration in Fig. VIII.1(c). By the principle of frequency inverse, if a
biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ is realizable as in Fig.
2(a), where $N_{1}$ is one of the configurations in Fig. VIII.1(d) and $N_{2}$
is a two-reactive four-element series-parallel network, then the condition of
Lemma 1 holds. ∎

###### Lemma VIII.4

Consider the four-reactive seven-element series-parallel network in Fig. 2(a),
realizing a biquadratic impedance $Z(s)$ in the form of (IV.1) with $A$, $B$,
$C$, $D$, $E$, $F$ $>0$, where $N_{1}$ is a one-reactive three-element series-
parallel network and $N_{2}$ is a three-reactive four-element series-parallel
network. If $Z(s)$ cannot be realized as a series-parallel network containing
fewer than seven elements, then $N_{1}$ will be equivalent to one of the
configurations in Fig. VIII.3, and $N_{2}$ will be equivalent to one of the
configurations in Fig. VIII.4.

###### Proof:

For any realization of $Z(s)$, there is no cut-set $\mathcal{C}(a,a^{\prime})$
corresponding to one kind of reactive elements, where $a$ and $a^{\prime}$
denote two terminal vertices, by [15, Lemma 1]. The possible network graphs
for subnetworks $N_{1}$ and $N_{2}$ are listed as in Figs. VIII.2 and VIII.5,
respectively. Based on the method of enumeration and the equivalence in [10,
Lemma 11], $N_{1}$ can be equivalent to one of configurations in Fig. VIII.3,
and $N_{2}$ can be equivalent to one of configurations in Fig. VIII.4. ∎

Figure VIII.3: One-reactive three-element series-parallel configurations for
the $N_{1}$ mentioned in Lemma VIII.4.

Figure VIII.4: Three-reactive four-element series-parallel configurations for
the $N_{2}$ mentioned in Lemma VIII.4.

Figure VIII.5: Possible network graphs for four-element networks.

###### Lemma VIII.5

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in
Fig. 2(a), where $N_{1}$ is the configuration in Fig. VIII.3(a), and $N_{2}$
is one of the configurations in Figs. VIII.4(a) and VIII.4(c)–VIII.4(e).

###### Proof:

By calculation, the impedance of the configuration in Fig. VIII.4(a) is
obtained as

$Z_{2}(s)=\frac{R_{21}L_{21}C_{22}s^{2}+R_{21}}{R_{21}L_{21}C_{21}C_{22}s^{3}+L_{21}C_{22}s^{2}+R_{21}(C_{21}+C_{22})s+1}.$
(VIII.32)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(a), then the
impedance of $N_{1}$ is in the form of

$Z_{1}(s)=\frac{ms+q}{s+p_{1}},$ (VIII.33)

where $m$, $q$, $p_{1}$ $>0$ and

$q-mp_{1}>0$ (VIII.34)

holds, and moreover the impedance of $N_{2}$ is in the form of

$Z_{2}(s)=\frac{\alpha s^{2}+\gamma}{(s+p_{1})(s+p)^{2}},$ (VIII.35)

where $\alpha$, $\gamma$ $>0$. Then, it follows from (VIII.32) and (VIII.35)
that

$\displaystyle\frac{1}{C_{21}}$ $\displaystyle=\alpha,$ (VIII.36)
$\displaystyle\frac{1}{L_{21}C_{21}C_{22}}$ $\displaystyle=\gamma,$ (VIII.37)
$\displaystyle\frac{1}{R_{21}C_{21}}$ $\displaystyle=2p+p_{1},$ (VIII.38)
$\displaystyle\frac{C_{21}+C_{22}}{L_{21}C_{21}C_{22}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.39)
$\displaystyle\frac{1}{R_{21}L_{21}C_{21}C_{22}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.40)

From (VIII.36), one obtains

$C_{21}=\frac{1}{\alpha}.$ (VIII.41)

It follows from (VIII.38) and (VIII.41) that

$R_{21}=\frac{\alpha}{2p+p_{1}}.$ (VIII.42)

By (VIII.37), (VIII.39), and (VIII.41), one obtains

$C_{22}=\frac{\alpha p^{2}+2\alpha p_{1}p-\gamma}{\alpha\gamma}.$ (VIII.43)

By (VIII.37), (VIII.41), and (VIII.43), one obtains

$L_{21}=\frac{\alpha^{2}}{\alpha p^{2}+2\alpha p_{1}p-\gamma}.$ (VIII.44)

Substituting (VIII.41)–(VIII.44) into (VIII.40) gives

$\alpha p_{1}p^{2}-2\gamma p-\gamma p_{1}=0.$ (VIII.45)

The assumption that $C_{22}>0$ and $L_{21}>0$ gives

$\alpha p^{2}+2\alpha p_{1}p-\gamma>0.$ (VIII.46)

Based on (VIII.33) and (VIII.35), calculation yields

$\begin{split}&Z(s)=Z_{1}(s)+Z_{2}(s)=\\\
&\frac{ms^{3}+(2mp+\alpha+q)s^{2}+p(mp+2q)s+(qp^{2}+\gamma)}{(s+p_{1})(s+p)^{2}}.\end{split}$
(VIII.47)

Comparing (1) with (VIII.47), one obtains

$\displaystyle m$ $\displaystyle=k,$ (VIII.48) $\displaystyle 2mp+\alpha+q$
$\displaystyle=k(p_{1}+2z),$ (VIII.49) $\displaystyle p(mp+2q)$
$\displaystyle=kz(z+2p_{1}),$ (VIII.50) $\displaystyle qp^{2}+\gamma$
$\displaystyle=kp_{1}z^{2}.$ (VIII.51)

Then, (VIII.48) and (VIII.50) together yield

$q=\frac{-k(p^{2}-2p_{1}z-z^{2})}{2p}.$ (VIII.52)

By (VIII.48), (VIII.49), and (VIII.52), one obtains

$\alpha=\frac{-k(p-z)(3p-z-2p_{1})}{2p}.$ (VIII.53)

By (VIII.48) and (VIII.51), one obtains

$\gamma=\frac{1}{2}k(p-z)(p^{2}+zp-2zp_{1}).$ (VIII.54)

Together with (VIII.48) and (VIII.52)–(VIII.54), condition (VIII.34) is
equivalent to $p<z$, and (VIII.45) and (VIII.46) are equivalent to

$\displaystyle(p+z)p_{1}^{2}-2p(p-z)p_{1}-p^{2}(p+z)$ $\displaystyle=0,$
(VIII.55) $\displaystyle p^{2}+p_{1}p-(p_{1}^{2}+zp_{1})$ $\displaystyle>0,$
(VIII.56)

respectively. Then, it follows from (VIII.55) and (VIII.56) that
$-p_{1}(p-z)^{2}/(p+z)>0$, which is impossible.

Therefore, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in Fig. 2(a),
where $N_{1}$ is the configuration in Fig. VIII.3(a), and $N_{2}$ is the
configurations in Fig. VIII.4(a).

It is calculated that the impedance of the configuration in Fig. VIII.4(c) is

$Z_{2}(s)=\frac{R_{21}L_{21}C_{22}s^{2}+L_{21}s}{R_{21}L_{21}C_{21}C_{22}s^{3}+L_{21}(C_{21}+C_{22})s^{2}+R_{21}C_{22}s+1}.$
(VIII.57)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(c), then the
impedance of $N_{1}$ is in the form of (VIII.33), where $m$, $q$, $p_{1}$ $>0$
and (VIII.34) holds, and the impedance of $N_{2}$ is in the form of

$Z_{2}(s)=\frac{\alpha s^{2}+\beta s}{(s+p_{1})(s+p)^{2}},$ (VIII.58)

where $\alpha$, $\beta$ $>0$. Then, it follows from (VIII.57) and (VIII.58)
that

$\displaystyle\frac{1}{C_{21}}$ $\displaystyle=\alpha,$ (VIII.59)
$\displaystyle\frac{1}{R_{21}C_{21}C_{22}}$ $\displaystyle=\beta,$ (VIII.60)
$\displaystyle\frac{C_{21}+C_{22}}{R_{21}C_{21}C_{22}}$
$\displaystyle=2p+p_{1},$ (VIII.61) $\displaystyle\frac{1}{L_{21}C_{21}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.62)
$\displaystyle\frac{1}{R_{21}L_{21}C_{21}C_{22}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.63)

It follows from (VIII.59) that

$C_{21}=\frac{1}{\alpha}.$ (VIII.64)

By (VIII.62) and (VIII.64), one obtains

$L_{21}=\frac{\alpha}{p(p+2p_{1})}.$ (VIII.65)

By (VIII.60), (VIII.61), and (VIII.64), one obtains

$C_{22}=\frac{2\alpha p+\alpha p_{1}-\beta}{\alpha\beta}.$ (VIII.66)

Then, (VIII.60), (VIII.64), and (VIII.66) yield

$R_{21}=\frac{\alpha^{2}}{2\alpha p+\alpha p_{1}-\beta}.$ (VIII.67)

Then, it follows from (VIII.63)–(VIII.67) that

$(\beta-\alpha p_{1})p+2\beta p_{1}=0.$ (VIII.68)

The assumption that $R_{21}>0$ and $C_{22}>0$ gives

$2\alpha p+\alpha p_{1}-\beta>0.$ (VIII.69)

By (VIII.33) and (VIII.58), $Z(s)$ is calculated as

$Z(s)=Z_{1}(s)+Z_{2}(s)=\frac{ms^{3}+(2mp+\alpha+q)s^{2}+(mp^{2}+2qp+\beta)s+qp^{2}}{(s+p_{1})(s+p)^{2}}.$
(VIII.70)

Comparing (1) with (VIII.70), one obtains

$\displaystyle m$ $\displaystyle=k,$ (VIII.71) $\displaystyle 2mp+\alpha+q$
$\displaystyle=k(p_{1}+2z),$ (VIII.72) $\displaystyle mp^{2}+2qp+\beta$
$\displaystyle=kz(z+2p_{1}),$ (VIII.73) $\displaystyle qp^{2}$
$\displaystyle=kp_{1}z^{2}.$ (VIII.74)

Then, (VIII.74) yields

$q=\frac{kp_{1}z^{2}}{p^{2}}.$ (VIII.75)

By (VIII.71), (VIII.72), and (VIII.75), one obtains

$\alpha=-\frac{k(p-z)(2p^{2}-p_{1}(p+z))}{p^{2}}.$ (VIII.76)

It follows from (VIII.71), (VIII.73), and (VIII.75) that

$\beta=-\frac{k(p-z)(p^{2}+z(p-2p_{1}))}{p}.$ (VIII.77)

Together with (VIII.71) and (VIII.75)–(VIII.77), condition (VIII.34) is
equivalent to $p<z$, and (VIII.68) and $\alpha>0$ are equivalent to

$(p-3z)p_{1}^{2}+p^{2}(p+z)=0$ (VIII.78)

and

$p_{1}<\frac{2p^{2}}{p+z},$ (VIII.79)

respectively. Combining (VIII.78) and (VIII.79), one obtains
$p^{2}(5p+z)(p-z)^{2}/(p+z)^{2}<0$, which is impossible.

Therefore, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in Fig. 2(a),
where $N_{1}$ is the configuration in Fig. VIII.3(a), and $N_{2}$ is the
configurations in Fig. VIII.4(c).

It is calculated that the impedance of the configuration in Fig. VIII.4(d) is

$Z_{2}(s)=\frac{L_{21}L_{22}s^{2}+R_{21}L_{21}s}{L_{21}L_{22}C_{21}s^{3}+R_{21}L_{21}C_{21}s^{2}+(L_{21}+L_{22})s+R_{21}}.$
(VIII.80)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(d), then the
impedance of $N_{1}$ is in the form of (VIII.33), where $m$, $q$, $p_{1}$ $>0$
and (VIII.34) holds, and the impedance of $N_{2}$ is in the form of (VIII.58)
where $\alpha$, $\beta$ $>0$. Then, it follows from (VIII.58) and (VIII.80)
that

$\displaystyle\frac{1}{C_{21}}$ $\displaystyle=\alpha,$ (VIII.81)
$\displaystyle\frac{R_{21}}{L_{22}C_{21}}$ $\displaystyle=\beta,$ (VIII.82)
$\displaystyle\frac{R_{21}}{L_{22}}$ $\displaystyle=2p+p_{1},$ (VIII.83)
$\displaystyle\frac{L_{21}+L_{22}}{L_{21}L_{22}C_{21}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.84)
$\displaystyle\frac{R_{21}}{L_{21}L_{22}C_{21}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.85)

It follows from (VIII.81) that

$C_{21}=\frac{1}{\alpha}.$ (VIII.86)

By (VIII.82), (VIII.83), and (VIII.86), one obtains

$2\alpha p+\alpha p_{1}-\beta=0.$ (VIII.87)

Then, (VIII.83), (VIII.85), and (VIII.86) yield

$L_{21}=\frac{\alpha(2p+p_{1})}{p_{1}p^{2}}.$ (VIII.88)

It follows from (VIII.84), (VIII.86), and (VIII.88) that

$L_{22}=\frac{\alpha(2p+p_{1})}{2p(p+p_{1})^{2}}.$ (VIII.89)

By (VIII.83) and (VIII.89), one obtains

$R_{21}=\frac{\alpha(2p+p_{1})^{2}}{2p(p+p_{1})^{2}}.$ (VIII.90)

By (VIII.33) and (VIII.58), $Z(s)$ is calculated as (VIII.70). Then, one
obtains (VIII.71)–(VIII.77). Furthermore, it follows from conditions (VIII.34)
and (VIII.87) that $z/3<p<z$, which satisfies the condition of Theorem 1.

Therefore, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in Fig. 2(a),
where $N_{1}$ is the configuration in Fig. VIII.3(a), and $N_{2}$ is the
configurations in Fig. VIII.4(d).

It is calculated that the impedance of the configuration in Fig. VIII.4(e) is

$Z_{2}(s)=\frac{R_{21}L_{21}C_{22}s^{2}+L_{21}s+R_{21}}{R_{21}L_{21}C_{21}C_{22}s^{3}+L_{21}(C_{21}+C_{22})s^{2}+R_{21}C_{21}s+1}.$
(VIII.91)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(e), then the
impedance of $N_{1}$ is in the form of (VIII.33), where $m$, $q$, $p_{1}$ $>0$
and (VIII.34) holds, and the impedance of $N_{2}$ is in the form of

$Z_{2}(s)=\frac{\alpha s^{2}+\beta s+\gamma}{(s+p_{1})(s+p)^{2}},$ (VIII.92)

where $\alpha$, $\beta$, $\gamma$ $>0$. Then, it follows from (VIII.91) and
(VIII.92) that

$\displaystyle\frac{1}{C_{21}}$ $\displaystyle=\alpha,$ (VIII.93)
$\displaystyle\frac{1}{R_{21}C_{21}C_{22}}$ $\displaystyle=\beta,$ (VIII.94)
$\displaystyle\frac{1}{L_{21}C_{21}C_{22}}$ $\displaystyle=\gamma,$ (VIII.95)
$\displaystyle\frac{C_{21}+C_{22}}{R_{21}C_{21}C_{22}}$
$\displaystyle=2p+p_{1},$ (VIII.96) $\displaystyle\frac{1}{L_{21}C_{22}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.97)
$\displaystyle\frac{1}{R_{21}L_{21}C_{21}C_{22}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.98)

It follows from (VIII.93) that

$C_{21}=\frac{1}{\alpha}.$ (VIII.99)

By (VIII.94), (VIII.96), and (VIII.99), one obtains

$C_{22}=\frac{2\alpha p+\alpha p_{1}-\beta}{\alpha\beta}.$ (VIII.100)

By (VIII.94), (VIII.99), and (VIII.100), one obtains

$R_{21}=\frac{\alpha^{2}}{2\alpha p+\alpha p_{1}-\beta}.$ (VIII.101)

Then, (VIII.95), (VIII.99), and (VIII.100) yield

$L_{21}=\frac{\alpha^{2}\beta}{\gamma(2\alpha p+\alpha p_{1}-\beta)}.$
(VIII.102)

It follows from (VIII.97), (VIII.100), and (VIII.102) that

$\alpha p^{2}+2\alpha p_{1}p-\gamma=0.$ (VIII.103)

Moreover, substituting (VIII.99)–(VIII.102) into (VIII.98) yields

$\alpha^{2}p_{1}p^{2}-2\alpha\gamma p-\gamma(\alpha p_{1}-\beta)=0.$
(VIII.104)

The assumption that $R_{21}>0$, $L_{21}>0$, and $C_{22}>0$ gives

$2\alpha p+\alpha p_{1}-\beta>0.$ (VIII.105)

By (28) and (VIII.92), $Z(s)$ is calculated as

$Z(s)=Z_{1}(s)+Z_{2}(s)=\frac{ms^{3}+(2mp+\alpha+q)s^{2}+(mp^{2}+2qp+\beta)s+(qp^{2}+\gamma)}{(s+p_{1})(s+p)^{2}}.$
(VIII.106)

Comparing (1) with (VIII.106), one obtains

$\displaystyle m$ $\displaystyle=k,$ (VIII.107) $\displaystyle 2mp+\alpha+q$
$\displaystyle=k(p_{1}+2z),$ (VIII.108) $\displaystyle mp^{2}+2qp+\beta$
$\displaystyle=kz(z+2p_{1}),$ (VIII.109) $\displaystyle qp^{2}+\gamma$
$\displaystyle=kp_{1}z^{2}.$ (VIII.110)

Then, (VIII.107) and (VIII.108) yield

$\alpha+q=k(p_{1}+2z-2p).$ (VIII.111)

By (VIII.103) and (VIII.110), one obtains

$(p^{2}+2p_{1}p)\alpha+p^{2}q=kp_{1}z^{2}.$ (VIII.112)

By (VIII.111) and (VIII.112), one obtains

$\alpha=\frac{k(p-z)(2p^{2}-p_{1}p-p_{1}z)}{2p_{1}p}$ (VIII.113)

and

$q=-\frac{k(2p^{3}+(3p_{1}-2z)p^{2}-(2p_{1}^{2}+4p_{1}z)p+p_{1}z^{2})}{2p_{1}p}.$
(VIII.114)

Then, (VIII.107), (VIII.109), and (VIII.114) yield

$\beta=\frac{2k(p-z)(p^{2}+p_{1}p-p_{1}z-p_{1}^{2})}{p_{1}}.$ (VIII.115)

It follows from (VIII.110) and (VIII.114) that

$\gamma=\frac{k(p+2p_{1})(p-z)(2p^{2}-p_{1}p-p_{1}z)}{2p_{1}}.$ (VIII.116)

Based on (VIII.107) and (VIII.114), one implies that (VIII.1) is equivalent to

$(p-z)(2p^{2}+3p_{1}p-p_{1}z)<0,$ (VIII.117)

which implies $p<z$. Since the condition of Theorem 1 does not hold, one only
needs to consider to the case of $p\leq z/(2+\sqrt{5})$. Furthermore,
(VIII.105) is equivalent to

$(p-z)(4p^{2}-(3p_{1}+2z)p+p_{1}z)<0,$ (VIII.118)

by (VIII.113) and (VIII.115). Combining (VIII.117) and (VIII.118), one obtains

$\frac{2zp-4p^{2}}{z-3p}<p_{1}<\frac{2p^{2}}{z-3p},$

which implies $p>z/3$. Therefore, the condition of Theorem 1 holds.

Therefore, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in Fig. 2(a),
where $N_{1}$ is the configuration in Fig. VIII.3(a), and $N_{2}$ is the
configurations in Fig. VIII.4(e). ∎

###### Lemma VIII.6

If a biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ is realizable as in
Fig. 2(a), where $N_{1}$ is the configuration in Fig. VIII.3(a) and $N_{2}$ is
one of the configurations in Figs. VIII.4(b) and VIII.4(f), then the condition
of Lemma 1 holds.

###### Proof:

It is calculated that the impedance of the configuration in Fig. VIII.4(b) is

$Z_{2}(s)=\frac{R_{21}L_{21}L_{22}C_{21}s^{3}+R_{21}L_{21}s}{L_{21}L_{22}C_{21}s^{3}+R_{21}C_{21}(L_{21}+L_{22})s^{2}+L_{21}s+R_{21}}.$
(VIII.119)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(b), then the
impedance of $N_{1}$ is in the form of (VIII.33), where $m$, $q$, $p_{1}$ $>0$
and (VIII.34) holds, and the impedance of $N_{2}$ is in the form of

$Z_{2}(s)=\frac{\alpha s^{3}+\gamma s}{(s+p_{1})(s+p)^{2}},$ (VIII.120)

where $\alpha$, $\gamma$ $>0$. Then, it follows from (VIII.119) and (VIII.120)
that

$\displaystyle R_{21}$ $\displaystyle=\alpha,$ (VIII.121)
$\displaystyle\frac{R_{21}}{L_{22}C_{21}}$ $\displaystyle=\gamma,$ (VIII.122)
$\displaystyle\frac{R_{21}(L_{21}+L_{22})}{L_{21}L_{22}}$
$\displaystyle=2p+p_{1},$ (VIII.123) $\displaystyle\frac{1}{L_{22}C_{21}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.124)
$\displaystyle\frac{R_{21}}{L_{21}L_{22}C_{21}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.125)

By (VIII.121), (VIII.124), and (VIII.125), one obtains

$L_{21}=\frac{\alpha(p+2p_{1})}{p_{1}p}.$ (VIII.126)

Then, (VIII.121), (VIII.123), and (VIII.126) yield

$L_{22}=\frac{\alpha(p+2p_{1})}{2(p+p_{1})^{2}}.$ (VIII.127)

By (VIII.124) and (VIII.127), one obtains

$C_{21}=\frac{2(p+p_{1})^{2}}{\alpha p(p+2p_{1})^{2}}.$ (VIII.128)

Then, (VIII.121), (VIII.122), (VIII.127), and (VIII.128) yield

$\alpha p^{2}+2\alpha p_{1}p-\gamma=0.$ (VIII.129)

By (VIII.33) and (VIII.120), $Z(s)$ is calculated as

$Z(s)=Z_{1}(s)+Z_{2}(s)=\frac{(m+\alpha)s^{3}+(2mp+q)s^{2}+(mp^{2}+2qp+\gamma)s+qp^{2}}{(s+p_{1})(s+p)^{2}}.$
(VIII.130)

Comparing (1) with (VIII.130), one obtains

$\displaystyle m+\alpha$ $\displaystyle=k,$ (VIII.131) $\displaystyle 2mp+q$
$\displaystyle=k(p_{1}+2z),$ (VIII.132) $\displaystyle mp^{2}+2qp+\gamma$
$\displaystyle=kz(z+2p_{1}),$ (VIII.133) $\displaystyle qp^{2}$
$\displaystyle=kp_{1}z^{2}.$ (VIII.134)

By (VIII.134), one obtains

$q=\frac{kp_{1}z^{2}}{p^{2}}.$ (VIII.135)

Then, (VIII.132) and (VIII.135) yield

$m=\frac{k(p_{1}p^{2}+2zp^{2}-p_{1}z^{2})}{2p^{3}}.$ (VIII.136)

By (VIII.131) and (VIII.136), one obtains

$\alpha=\frac{k(p-z)(2p^{2}-p_{1}p-p_{1}z)}{2p^{3}}.$ (VIII.137)

It follows from (VIII.133), (VIII.135), and (VIII.136) that

$\gamma=\frac{-k(p-z)(p_{1}p+2zp-3zp_{1})}{2p}.$ (VIII.138)

Together with (VIII.135)–(VIII.138), condition (VIII.34) is equivalent to
$p<z$, and (VIII.129) and $m>0$ are equivalent to

$(p+z)p_{1}^{2}-2p(p-z)p_{1}-p^{2}(p+z)=0$ (VIII.139)

and

$p_{1}<\frac{2zp^{2}}{z^{2}-p^{2}},$ (VIII.140)

respectively. Combining (VIII.139) and (VIII.140), one obtains
$z/(2+\sqrt{3})<p<z$, which satisfies the condition of Lemma 1.

Therefore, if a biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ is
realizable as in Fig. 2(a), where $N_{1}$ is the configuration in Fig.
VIII.3(a) and $N_{2}$ is one of the configurations in Figs. VIII.4(b), then
the condition of Lemma 1 holds.

It is calculated that the impedance of the configuration in Fig. VIII.4(f) is

$Z_{2}(s)=\frac{R_{21}L_{21}C_{22}s^{2}+L_{21}s+R_{21}}{R_{21}L_{21}C_{21}C_{22}s^{3}+L_{21}C_{21}s^{2}+R_{21}(C_{21}+C_{22})s+1}.$
(VIII.141)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(f), then the
impedance of $N_{1}$ is in the form of (VIII.33), where $m$, $q$, $p_{1}$ $>0$
and (VIII.34) holds, and the impedance of $N_{2}$ is in the form of (VIII.92),
where $\alpha$, $\beta$, $\gamma$ $>0$. Then, it follows from (VIII.92) and
(VIII.141) that

$\displaystyle\frac{1}{C_{21}}$ $\displaystyle=\alpha,$ (VIII.142)
$\displaystyle\frac{1}{R_{21}C_{21}C_{22}}$ $\displaystyle=\beta,$ (VIII.143)
$\displaystyle\frac{1}{L_{21}C_{21}C_{22}}$ $\displaystyle=\gamma,$ (VIII.144)
$\displaystyle\frac{1}{R_{21}C_{22}}$ $\displaystyle=2p+p_{1},$ (VIII.145)
$\displaystyle\frac{C_{21}+C_{22}}{L_{21}C_{21}C_{22}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.146)
$\displaystyle\frac{1}{R_{21}L_{21}C_{21}C_{22}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.147)

It follows from (VIII.142) that

$C_{21}=\frac{1}{\alpha}.$ (VIII.148)

Then, (VIII.144), (VIII.146), and (VIII.148) yield

$C_{22}=\frac{\alpha p^{2}+2\alpha p_{1}p-\gamma}{\alpha\gamma}.$ (VIII.149)

By (VIII.145) and (VIII.149), one obtains

$R_{21}=\frac{\alpha\gamma}{(\alpha p^{2}+2\alpha p_{1}p-\gamma)(2p+p_{1})}.$
(VIII.150)

Then, it follows from (VIII.144), (VIII.148), and (VIII.149) that

$L_{21}=\frac{\alpha^{2}}{\alpha p^{2}+2\alpha p_{1}p-\gamma}.$ (VIII.151)

By (VIII.143) and (VIII.148)–(VIII.150), one obtains

$2\alpha p-\beta+\alpha p_{1}=0.$ (VIII.152)

The assumption that $R_{21}>0$, $L_{21}>0$, and $C_{22}>0$ gives

$\alpha p^{2}+2\alpha p_{1}p-\gamma>0.$ (VIII.153)

Then, (VIII.147)–(VIII.151) yield

$2\alpha pp_{1}^{2}+(4\alpha p^{2}-\gamma)p_{1}+2p(\alpha p^{2}-\gamma)=0.$
(VIII.154)

By (28) and (VIII.92), $Z(s)$ is calculated as

$Z(s)=Z_{1}(s)+Z_{2}(s)=\frac{ms^{3}+(2mp+\alpha+q)s^{2}+(mp^{2}+2qp+\beta)s+(qp^{2}+\gamma)}{(s+p_{1})(s+p)^{2}}.$
(VIII.155)

Comparing (1) with (VIII.155), one obtains

$\displaystyle m$ $\displaystyle=k,$ (VIII.156) $\displaystyle 2mp+\alpha+q$
$\displaystyle=k(p_{1}+2z),$ (VIII.157) $\displaystyle mp^{2}+2qp+\beta$
$\displaystyle=kz(z+2p_{1}),$ (VIII.158) $\displaystyle qp^{2}+\gamma$
$\displaystyle=kp_{1}z^{2}.$ (VIII.159)

Then, it follows from (VIII.152) and (VIII.156)–(VIII.158) that

$\displaystyle\alpha$ $\displaystyle=\frac{k(p-z)(3p-z-2p_{1})}{p_{1}},$
(VIII.160) $\displaystyle q$ $\displaystyle=-\frac{k(3p^{2}-4zp-
p_{1}^{2}+z^{2})}{p_{1}}.$ (VIII.161)

By (VIII.159) and (VIII.161), one obtains

$\gamma=\frac{k(p-z)((3p-z)p^{2}-(p+z)p_{1}^{2})}{p_{1}}.$ (VIII.162)

Then, (VIII.152) and (VIII.160) yield

$\beta=\frac{k(p-z)(3p-z-2p_{1})(2p+p_{1})}{p_{1}}.$ (VIII.163)

Together with (VIII.156) and (VIII.161), condition (VIII.34) is equivalent to
$z/3<p<z$, which satisfies the condition of Lemma 1.

Therefore, if a biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ is
realizable as in Fig. 2(a), where $N_{1}$ is the configuration in Fig.
VIII.3(a) and $N_{2}$ is one of the configurations in Figs. VIII.4(f), then
the condition of Lemma 1 holds. ∎

###### Lemma VIII.7

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ can be realized as the
configuration in Fig. 2(a), where $N_{1}$ is the configuration in Fig.
VIII.3(a) and $N_{2}$ is the configurations in Figs. VIII.4(g) (that is, the
configuration in Fig. 3(a)), if and only if

$\displaystyle(p-z)(p-3z)$ $\displaystyle>0,$ (VIII.164) $\displaystyle
p^{4}-6zp^{3}+6z^{2}p^{2}-14z^{3}p+5z^{4}$ $\displaystyle<0.$ (VIII.165)

###### Proof:

The realizability condition of a general biquadratic impedance $Z(s)$ in the
form of (IV.1) as a configuration that is equivalent to Fig. 3(a) is available
in [11, Table I]. Letting $A=kx$, $B=2kzx$, $C=kz^{2}x$, $D=x$, $E=2px$, and
$F=p^{2}x$ for $x>0$, the realizability condition for such a specific
biquadratic impedance can be derived.

The element values can be derived as $R_{1}=q/p_{1}$, $R_{2}=mq/(q-mp_{1})$,
$C_{1}=(q-mp_{1})/q^{2}$, $R_{21}=\alpha$, $L_{21}=\alpha/(2p+p_{1})$,
$L_{22}=\alpha\beta/\gamma$, and $C_{21}=1/\beta$, where
$\alpha=k(p-z)(2p+p_{1})(p^{2}+zp-2zp_{1})/(2p^{4})$,
$\beta=(2k(p-z)(-zp_{1}^{2}+p(p-z)p_{1}+zp^{2}))/p^{3}$,
$\gamma=(kp_{1}(p-z)(p^{2}+zp-2zp_{1}))/(2p^{2})$, and $p_{1}$ is a positive
root of $(3z-p)p_{1}^{2}-2p(p-z)p_{1}+p^{2}(p-3z)=0$. ∎

###### Lemma VIII.8

If a biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ can be realized as in
Fig. 2(a), where $N_{1}$ is the configuration in Fig. VIII.3(a) and $N_{2}$ is
the configuration in Fig. VIII.4(h), then the condition of Lemma VIII.7 holds.

###### Proof:

By calculation, the impedance of the configuration in Fig. VIII.4(h) is
obtained as

$Z_{2}(s)=\frac{s(R_{21}L_{21}L_{22}C_{21}s^{2}+L_{21}L_{22}s+R_{21}L_{21})}{L_{21}L_{22}C_{21}s^{3}+R_{21}(L_{21}+L_{22})C_{21}s^{2}+L_{22}s+R_{21}}.$
(VIII.166)

If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.3(a) and $N_{2}$ is the configuration in Fig. VIII.4(h), then the
impedance of $N_{1}$ is in the form of (VIII.33), where $m$, $q$, $p_{1}$ $>0$
and (VIII.34) holds, and moreover the impedance of $N_{2}$ is in the form of

$Z_{2}(s)=\frac{s(\alpha s^{2}+\beta s+\gamma)}{(s+p_{1})(s+p)^{2}},$
(VIII.167)

where $\alpha$, $\beta$, $\gamma$ $>0$. Consequently, it follows from
(VIII.166) and (VIII.167) that

$\displaystyle R_{21}$ $\displaystyle=\alpha,$ (VIII.168)
$\displaystyle\frac{1}{C_{21}}$ $\displaystyle=\beta,$ (VIII.169)
$\displaystyle\frac{R_{21}}{L_{22}C_{21}}$ $\displaystyle=\gamma,$ (VIII.170)
$\displaystyle\frac{R_{21}(L_{21}+L_{22})}{L_{21}L_{22}}$
$\displaystyle=2p+p_{1},$ (VIII.171) $\displaystyle\frac{1}{L_{21}C_{21}}$
$\displaystyle=p(p+2p_{1}),$ (VIII.172)
$\displaystyle\frac{R_{21}}{L_{21}L_{22}C_{21}}$ $\displaystyle=p_{1}p^{2}.$
(VIII.173)

Thus, (VIII.169) yields

$C_{21}=\frac{1}{\beta}.$ (VIII.174)

By (VIII.170) and (VIII.173), one obtains

$L_{21}=\frac{\gamma}{p_{1}p^{2}}.$ (VIII.175)

It follows from (VIII.168), (VIII.170), and (VIII.174) that

$L_{22}=\frac{\alpha\beta}{\gamma}.$ (VIII.176)

By (VIII.168), (VIII.171), (VIII.175), and (VIII.176), one obtains

$(\alpha p^{2}-\gamma)\beta p_{1}+\gamma(\gamma-2\beta p)=0.$ (VIII.177)

It follows from (VIII.172), (VIII.174), and (VIII.175) that

$(\beta p-2\gamma)p_{1}-\gamma p=0.$ (VIII.178)

Based on (VIII.33) and (VIII.167), calculation yields

$\begin{split}&Z(s)=Z_{1}(s)+Z_{2}(s)=\\\
&\frac{(m+\alpha)s^{3}+(2mp+\beta+q)s^{2}+(mp^{2}+2qp+\gamma)s+qp^{2}}{s^{3}+(2p+p_{1})s^{2}+p(p+2p_{1})s+p_{1}p^{2}}.\end{split}$
(VIII.179)

Comparing (1) with (VIII.179), one obtains

$\displaystyle m+\alpha$ $\displaystyle=k,$ (VIII.180) $\displaystyle
2mp+\beta+q$ $\displaystyle=k(p_{1}+2z),$ (VIII.181) $\displaystyle
mp^{2}+2qp+\gamma$ $\displaystyle=kz(z+2p_{1}),$ (VIII.182) $\displaystyle
qp^{2}$ $\displaystyle=kp_{1}z^{2}.$ (VIII.183)

It follows from (VIII.183) that

$q=\frac{kz^{2}p_{1}}{p^{2}}.$ (VIII.184)

Thus, (VIII.178), (VIII.181), (VIII.182), and (VIII.184) together yield

$\beta=\frac{k(p-z)(p+2p_{1})((p-3z)p_{1}+2zp)}{p^{3}},$ (VIII.185)

and

$m=\frac{-k((p-z)(p-3z)p_{1}^{2}-z^{2}p^{2})}{p^{4}}.$ (VIII.186)

By (VIII.180) and (VIII.186), one obtains

$\alpha=\frac{k(p-z)((p-3z)p_{1}^{2}+p^{2}(p+z))}{p^{4}}.$ (VIII.187)

It follows from (VIII.178) and (VIII.185) that

$\gamma=\frac{kp_{1}(p-z)((p-3z)p_{1}+2zp)}{p^{2}}.$ (VIII.188)

Together with (VIII.184) and (VIII.186), condition (VIII.34) is equivalent to
(VIII.164). Together with (VIII.185), (VIII.187), and (VIII.188), condition
(VIII.177) is equivalent to

$(5z-3p)p_{1}^{2}+(p-3z)p^{2}=0.$ (VIII.189)

From (VIII.186), it follows that $m>0$ is equivalent to
$p_{1}^{2}<z^{2}p^{2}/((p-z)(p-3z))$, which, in turn, is equivalent to
$p^{2}-5zp+2z^{2}<0$, together with (VIII.164) and (VIII.189). Therefore, the
condition of Lemma VIII.7 must hold. ∎

##  IX Supplementary Lemmas of Five-Reactive Seven-Element Series-Parallel
Realizations for the Proof of Lemma 4

###### Lemma IX.1

Consider the five-reactive seven-element series-parallel network in Fig. 2(a),
realizing a biquadratic impedance $Z(s)$ in the form of (IV.1) with $A$, $B$,
$C$, $D$, $E$, $F$ $>0$, where $N_{1}$ is a three-element series-parallel
network and $N_{2}$ is a four-element series-parallel network. If $Z(s)$
cannot be realized as a series-parallel network containing fewer than seven
elements, then $N_{1}$ is one of the configurations in Figs.
VIII.1(b)–VIII.1(d) and $N_{2}$ will be equivalent to one of the
configurations in Fig. VIII.4.

###### Proof:

By [15, Lemma 2], $Z(s)$ cannot be realized as the series connection of two
networks, one of which contains only reactive elements. Therefore, $N_{1}$ can
only contain two reactive elements. For any realization of $Z(s)$, there is no
cut-set $\mathcal{C}(a,a^{\prime})$ corresponding to one kind of reactive
elements, where $a$ and $a^{\prime}$ denote two terminals, by [15, Lemma 1].
The possible network graphs for subnetworks $N_{1}$ and $N_{2}$ are listed in
Figs. VIII.2 and VIII.5, respectively. Based on the method of enumeration and
the equivalence in [10, Lemma 11], $N_{1}$ is one of the configurations in
Figs. VIII.1(b)–VIII.1(d) and $N_{2}$ can be equivalent to one of the
configurations in Fig. VIII.4. ∎

###### Lemma IX.2

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ not satisfying the
condition of Lemma 3 cannot be realized as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(b) and $N_{2}$ is one of the configurations in
Figs. VIII.4(a)–VIII.4(d), VIII.4(f), and VIII.4(h).

###### Proof:

It has been shown that the impedance of the configuration in Fig. VIII.1(b) is
in the form of

$Z_{1}(s)=\frac{R_{1}L_{1}C_{1}s^{2}+R_{1}}{L_{1}C_{1}s^{2}+R_{1}C_{1}s+1},$
(IX.1)

and the impedance of the configuration in Fig. VIII.4(a) is in the form of
(VIII.32). Since it is assumed that the condition of Lemma 3 does not hold,
$Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized with fewer than five reactive
elements. If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig.
VIII.4(a), then the impedance of $N_{1}$ is of degree two and is in the form
of

$Z_{1}(s)=\frac{m(s^{2}+pp_{1})}{(s+p_{1})(s+p)},$ (IX.2)

where $m$, $p_{1}$, $p$ $>0$, and the impedance of $N_{2}$ is of degree three
and is in the form of (VIII.35), where $\alpha$, $\gamma$ $>0$, and (VIII.45)
and (VIII.46) hold. Furthermore,

$\begin{split}&Z(s)=Z_{1}(s)+Z_{2}(s)=\\\
&\frac{ms^{3}+(mp+\alpha)s^{2}+mp_{1}ps+(mp_{1}p^{2}+\gamma)}{(s+p_{1})(s+p)^{2}}.\end{split}$
(IX.3)

Comparing (1) with (IX.3), one obtains

$\displaystyle m$ $\displaystyle=k,$ (IX.4) $\displaystyle mp+\alpha$
$\displaystyle=k(p_{1}+2z),$ (IX.5) $\displaystyle mp_{1}p$
$\displaystyle=kz(2p_{1}+z),$ (IX.6) $\displaystyle mp_{1}p^{2}+\gamma$
$\displaystyle=kz^{2}p_{1}.$ (IX.7)

By (IX.4)–(IX.6), one obtains

$p_{1}=\frac{z^{2}}{p-2z}.$ (IX.8)

By (IX.4), (IX.7), and (IX.8), one obtains

$\gamma=\frac{-kz^{2}(p-z)(p+z)}{p-2z}.$ (IX.9)

It follows from (IX.8) and $p_{1}>0$ that $p>2z$, which further implies that
$\gamma<0$ by (IX.9). This is impossible.

Therefore, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in Fig. 2(a),
where $N_{1}$ is the configuration in Fig. VIII.1(b) and $N_{2}$ is the
configuration in Fig. VIII.4(a).

It is clear that any network in Fig. 2(a), where $N_{1}$ is the configuration
in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig. VIII.4(b), can be a
frequency inverse dual network of another one in Fig. 2(a), where $N_{1}$ is
the configuration in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig.
VIII.4(a). Based on the principle of frequency inverse,
$Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as such a network.

It has been shown that the impedance of the configuration in Fig. VIII.1(b) is
in the form of (IX.1), and the impedance of the configuration in Fig.
VIII.4(c) is in the form of
$Z(s)=(R_{21}L_{21}C_{22}s^{2}+L_{21}s)/(R_{21}L_{21}C_{21}C_{22}s^{3}+L_{21}(C_{21}+C_{22})s^{2}+R_{21}C_{22}s+1)$.
If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the configuration in
Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig. VIII.4(c), then the
impedance of $N_{1}$ is of degree two and is in the form of (IX.2), where $m$,
$p_{1}$, $p$ $>0$, and the impedance of $N_{2}$ is of degree three and is in
the form of $Z(s)=(\alpha s^{2}+\beta s)/((s+p_{1})(s+p)^{2})$, where
$\alpha$, $\beta$ $>0$. Furthermore,

$\begin{split}&Z(s)=Z_{1}(s)+Z_{2}(s)=\\\
&\frac{ms^{3}+(mp+\alpha)s^{2}+(mp_{1}p+\beta)s+mp_{1}p^{2}}{(s+p_{1})(s+p)^{2}}.\end{split}$
(IX.10)

Comparing (1) with (IX.10), one obtains $m=k$ and $mp_{1}p^{2}=kz^{2}p_{1}$,
which further implies $p=z$. This contradicts the assumption.

Therefore, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as in Fig. 2(a),
where $N_{1}$ is the configuration in Fig. VIII.1(b) and $N_{2}$ is the
configuration in Fig. VIII.4(c).

It is clear that any network in Fig. 2(a), where $N_{1}$ is the configuration
in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig. VIII.4(d), can be a
frequency inverse dual network of another one in Fig. 2(a), where $N_{1}$ is
the configuration in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig.
VIII.4(c). Based on the principle of frequency inverse,
$Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as such a network.

It has been shown that the impedance of the configuration in Fig. VIII.1(b) is
in the form of (IX.1), and the impedance of the configuration in Fig. 8(f) is
in the form of (VIII.141). If $Z(s)$ is realizable as in Fig. 2(a), where
$N_{1}$ is the configuration in Fig. VIII.1(b) and $N_{2}$ is the
configuration in Fig. 8(f), then the impedance of $N_{1}$ is of degree two and
is in the form of (IX.2), where $m$, $p_{1}$, $p$ $>0$ and the impedance of
$N_{2}$ is degree three and is in the form of (VIII.92), where $\alpha$,
$\beta$, $\gamma$ $>0$ and (VIII.152)–(VIII.154) hold. Furthermore, one
obtains (IX.18)–(IX.25). Substituting (IX.19) and (IX.23)–(IX.25) into
(VIII.152)–(VIII.154) yields

$\displaystyle p_{1}^{2}+2pp_{1}-(2p^{2}-4zp+z^{2})=0,$ (IX.11) $\displaystyle
2pp_{1}^{2}+z(4p-z)p_{1}-p^{2}(p-2z)>0,$ (IX.12) $\displaystyle
2pp_{1}^{3}+(3p^{2}+4zp-z^{2})p_{1}^{2}+2zp(4p-z)p_{1}-2p^{3}(p-2z)=0,$
(IX.13)

respectively. By (IX.11), (IX.13) can be further equivalent to

$(p^{2}-4zp+z^{2})p_{1}^{2}-4p^{3}p_{1}+2p^{3}(p-2z)=0.$ (IX.14)

It is calculated that the resultant of (IX.11) and (IX.14) in $p_{1}$ is
$8p^{8}+48zp^{7}-312z^{2}p^{6}+624z^{3}p^{5}-617z^{4}p^{4}+336z^{5}p^{3}-102z^{6}p^{2}+16z^{7}p-z^{8}$.
Since the condition of Lemma 3 does not hold, there exists at least one common
root between (IX.11) and (IX.14) in $p_{1}$ if and only if

$\begin{split}8p^{8}+48zp^{7}&-312z^{2}p^{6}+624z^{3}p^{5}-617z^{4}p^{4}\\\
&+336z^{5}p^{3}-102z^{6}p^{2}+16z^{7}p-z^{8}=0\end{split}$ (IX.15)

holds with $p<z/(2+\sqrt{5})$, which implies that the condition of Lemma 3
holds. One further implies $p_{1}>0$. By (IX.11), it is implied that (IX.12)
is equivalent to

$p_{1}<\frac{p(3p^{2}-6zp+2z^{2})}{(2p-z)^{2}}.$ (IX.16)

By (IX.14), (IX.16) is further equivalent to
$(p-z)(7p^{5}+63zp^{4}-174z^{2}p^{3}+134z^{3}p^{2}-40z^{4}p+4z^{5})<0$, which
is verified to be satisfied.

It is clear that any network in Fig. 2(a), where $N_{1}$ is the configuration
in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig. VIII.4(h), can be a
frequency inverse dual network of another one in Fig. 2(a), where $N_{1}$ is
the configuration in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig.
VIII.4(f). Based on the principle of frequency inverse,
$Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized as such a network either. ∎

###### Lemma IX.3

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ not satisfying the
condition of Lemma 3 is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(b) and $N_{2}$ is the configuration in Fig.
VIII.4(e) (that is, the configuration in Fig. 4(a) whose one-terminal-pair
labeled graph is $N_{4a}$), if and only if

$16p^{4}-40zp^{3}+31z^{2}p^{2}-10z^{3}p+z^{4}=0$ (IX.17)

and $p<z/(2+\sqrt{5})$ (it can be verified that there is only one distinct
root of the equation $16\eta^{4}-40\eta^{3}+31\eta^{2}-10\eta+1=0$ for
$\eta\in(0,1/(2+\sqrt{5}))$).

###### Proof:

Necessity. It has been shown that the impedance of the configuration in Fig.
VIII.1(b) is in the form of (IX.1), and the impedance of the configuration in
Fig. VIII.4(e) is in the form of (VIII.91). Since it is assumed that the
condition of Lemma 3 does not hold, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be
realized with fewer than five reactive elements. If $Z(s)$ is realizable as in
Fig. 2(a), where $N_{1}$ is the configuration in Fig. VIII.1(b) and $N_{2}$ is
the configuration in Fig. VIII.4(e), then the impedance of $N_{1}$ is of
degree two and is in the form of (IX.2), where $m$, $p_{1}$, $p$ $>0$, and the
impedance of $N_{2}$ is of degree three and is in the form of (VIII.92), where
$\alpha$, $\beta$ $>0$ and (VIII.103)–(VIII.105) hold. Furthermore, one
obtains

$Z(s)=Z_{1}(s)+Z_{2}(s)=\frac{ms^{3}+(mp+\alpha)s^{2}+(mp_{1}p+\beta)s+(mp_{1}p^{2}+\gamma)}{(s+p_{1})(s+p)^{2}}.$
(IX.18)

Comparing (1) with (IX.18), one obtains

$\displaystyle m$ $\displaystyle=k,$ (IX.19) $\displaystyle mp+\alpha$
$\displaystyle=k(p_{1}+2z),$ (IX.20) $\displaystyle mp_{1}p+\beta$
$\displaystyle=kz(2p_{1}+z),$ (IX.21) $\displaystyle mp_{1}p^{2}+\gamma$
$\displaystyle=kz^{2}p_{1}.$ (IX.22)

Then, (IX.19) and (IX.20) yield

$\alpha=-k(p-p_{1}-2z).$ (IX.23)

By (IX.19) and (IX.21), one obtains

$\beta=-k(p_{1}p-2zp_{1}-z^{2}).$ (IX.24)

It follows from (IX.19) and (IX.22) that

$\gamma=-kp_{1}(p-z)(p+z).$ (IX.25)

Substituting (IX.19) and (IX.23)–(IX.25) into (VIII.103)–(VIII.105) yields

$2pp_{1}^{2}+z(4p-z)p_{1}-p^{2}(p-2z)=0,$ (IX.26)
$\begin{split}(2p^{2}-z^{2})p_{1}^{2}&+2pz(2p-z)p_{1}\\\
&-(p^{4}-5z^{2}p^{2}+4z^{3}p-z^{4})=0,\end{split}$ (IX.27)
$p_{1}^{2}+2pp_{1}-(2p^{2}-4zp+z^{2})>0,$ (IX.28)

respectively. It is calculated that the resultant of (IX.26) and (IX.27) in
$p_{1}$ is
$z^{2}(p+z)(p-z)^{3}(16p^{4}-40zp^{3}+31z^{2}p^{2}-10z^{3}p+z^{4})$. Since the
condition of Lemma 3 does not hold, there exists at least one common root
between (IX.26) and (IX.27) in $p_{1}$ if and only if (6) holds with
$p<z/(2+\sqrt{5})$. One implies $p_{1}>0$. By (IX.26), it is implied that
(IX.28) is equivalent to

$p_{1}>\frac{p(3p^{2}-6zp+2z^{2})}{(2p-z)^{2}}.$ (IX.29)

Based on (IX.27), one implies that (IX.29) is equivalent to
$(p^{2}+2zp-z^{2})(p-z)^{3}(2p^{3}+10zp^{2}-7z^{2}p+z^{3})>0$, which can be
verified to be satisfied.

Sufficiency. Based on the discussion in the necessity part, there exists
$p_{1}>0$ such that (IX.26)–(IX.28) hold. Let $m$, $\alpha$, $\beta$, and
$\gamma$ satisfy (IX.19) and (IX.23)–(IX.25), which obviously implies that
$\alpha$, $\beta$, $\gamma$, $m$ $>0$. Therefore, (VIII.103)–(VIII.105) and
(IX.20)–(IX.22) hold. Therefore, $Z(s)$ can be written in the form of (IX.18).
Decompose $Z(s)$ as $Z(s)=Z_{1}(s)+Z_{2}(s)$, where $Z_{1}(s)$ is in the form
of (IX.2) where $m$, $p_{1}$, $p$ $>0$ and $Z_{2}(s)$ is in the form of
(VIII.92) where $\alpha$, $\beta$ $>0$. By letting $R_{1}=m$,
$L_{1}=m/(p+p_{1})$, and $C_{1}=(p+p_{1})/(mpp_{1})$, $Z_{1}(s)$ is realizable
as in Fig. VIII.1(b). Let $C_{21}$, $C_{22}$, $R_{21}$, and $L_{21}$ satisfy
(VIII.99)–(VIII.102). Since (VIII.105) holds, $C_{21}$, $C_{22}$, $R_{21}$,
$L_{21}$ $>0$. Based on the discussion in the proof of Lemma 11, it follows
that (VIII.93)–(VIII.98) hold because (VIII.103) and (VIII.104) are satisfied.
Therefore, $Z_{2}(s)$ can be realized as in Fig. VIII.4(e). The sufficiency
part is proved. ∎

###### Lemma IX.4

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ not satisfying the
condition of Lemma 3 cannot be realized as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(c) and $N_{2}$ is one of the configurations in
Fig. VIII.4(a) and VIII.4(f).

###### Proof:

It has been shown that the impedance of the configuration in Fig. VIII.1(c) is
in the form of
$Z_{1}(s)=s(R_{1}L_{1}C_{1}s+L_{1})/(L_{1}C_{1}s^{2}+R_{1}C_{1}s+1)$, and the
impedance of the configuration in Fig. VIII.4(a) is in the form of (VIII.32).
Since it is assumed that the condition of Lemma 3 does not hold,
$Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be realized with fewer than five reactive
elements. If $Z(s)$ is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(c) and $N_{2}$ is the configuration in Fig.
VIII.4(a), then the impedance of $N_{1}$ is in the form of

$Z_{1}(s)=\frac{ms(s+q)}{(s+p_{1})(s+p)},$ (IX.30)

where $m$, $p_{1}$, $p$ $>0$ and $q=p_{1}p/(p_{1}+p)$, and the impedance of
$N_{2}$ is in the form of (VIII.35), where $\alpha$, $\gamma$ $>0$, and
moreover (VIII.45) and (VIII.46) hold. Furthermore,

$\begin{split}&Z(s)=Z_{1}(s)+Z_{2}(s)=\\\
&\frac{ms^{3}+\frac{mp^{2}+(2mp_{1}+\alpha)p+\alpha
p_{1}}{p_{1}+p}s^{2}+\frac{mp_{1}p^{2}}{p_{1}+p}s+\gamma}{(s+p_{1})(s+p)^{2}}.\end{split}$
(IX.31)

Comparing (1) with (IX.31), one obtains

$\displaystyle m$ $\displaystyle=k,$ (IX.32)
$\displaystyle\frac{mp^{2}+(2mp_{1}+\alpha)p+\alpha p_{1}}{p+p_{1}}$
$\displaystyle=k(p_{1}+2z),$ (IX.33)
$\displaystyle\frac{mp_{1}p^{2}}{p+p_{1}}$ $\displaystyle=kz(2p_{1}+z),$
(IX.34) $\displaystyle\gamma$ $\displaystyle=kp_{1}z^{2}.$ (IX.35)

Then, it follows from (IX.32) and (IX.33) that

$\alpha=\frac{k(p_{1}^{2}-(p-2z)p_{1}-p(p-2z))}{p+p_{1}}.$ (IX.36)

Substituting (IX.32), (IX.35), and (IX.36) into (VIII.45), (VIII.46), and
(IX.34) yields

$\begin{split}2pp_{1}^{3}&-(p^{2}-4zp+z^{2})p_{1}^{2}\\\
&-p(3p^{2}-6zp+z^{2})p_{1}-p^{3}(p-2z)>0,\end{split}$ (IX.37)
$\begin{split}(p-z)(p+z)p_{1}^{2}-p(p^{2}&-2zp+3z^{2})p_{1}\\\
&-p^{2}(p^{2}-2zp+2z^{2})=0,\end{split}$ (IX.38)
$\begin{split}2zp_{1}^{2}-(p^{2}-2zp-z^{2})p_{1}+z^{2}p=0,\end{split}$ (IX.39)

respectively. By (IX.36), the assumption of $\alpha>0$ implies

$p_{1}^{2}-(p-2z)p_{1}-p(p-2z)>0.$ (IX.40)

By calculation, the resultant of (IX.38) and (IX.39) in $p_{1}$ is
$-p^{4}(p^{6}-8zp^{5}+20z^{2}p^{4}-28z^{3}p^{3}+21z^{4}p^{2}-12z^{5}p+2z^{6})$.
Since the condition of Lemma 3 does not hold, there exists at least one common
root in $p_{1}$ between (IX.38) and (IX.39) if and only if

$p^{6}-8zp^{5}+20z^{2}p^{4}-28z^{3}p^{3}+21z^{4}p^{2}-12z^{5}p+2z^{6}=0$
(IX.41)

holds with $p>(2+\sqrt{5})z$, which implies that the condition of Lemma 3
holds. This further implies that $p_{1}>0$. From (IX.39), it follows that
(IX.37) is equivalent to

$p_{1}>\frac{zp(2p-3z)}{(p-z)(p-3z)}.$ (IX.42)

By (IX.38), one has that (IX.42) is further equivalent to
$p^{10}-12zp^{9}+54z^{2}p^{8}-114z^{3}p^{7}+100z^{4}p^{6}+40z^{5}p^{5}-164z^{6}p^{4}+142z^{7}p^{3}-65z^{8}p^{2}+16z^{9}p-2z^{10}>0$,
which can indeed be verified to be true.

It has been shown that the impedance of the configuration in Fig. VIII.1(c) is
in the form of (VIII.21), and the impedance of the configuration in Fig. 8(f)
is in the form of (VIII.141). If $Z(s)$ is realizable as in Fig. 2(a), where
$N_{1}$ is the configuration in Fig. VIII.1(c) and $N_{2}$ is the
configuration in Fig. VIII.4(f), then the impedance of $N_{1}$ is in the form
of (IX.30), where $m$, $p_{1}$, $p$ $>0$ and $q=p_{1}p/(p_{1}+p)$, and the
impedance of $N_{2}$ is in the form of (VIII.92), where $\alpha$, $\beta$,
$\gamma$ $>0$ and (VIII.152)–(VIII.154) hold. Furthermore, one obtains
(IX.50)–(IX.54). It follows from (VIII.154) and (IX.54) that

$\alpha=\frac{kp_{1}z^{2}(p_{1}+2p)}{2p(p+p_{1})^{2}}.$ (IX.43)

Then, (VIII.152) and (IX.43) yield

$\beta=\frac{kp_{1}z^{2}(p_{1}+2p)^{2}}{2p(p+p_{1})^{2}}.$ (IX.44)

Substituting (IX.51), (IX.54), (IX.55), and (IX.44) into (VIII.153), (IX.52),
and (IX.53) gives

$\displaystyle\frac{kp_{1}^{2}z^{2}p}{2(p+p_{1})^{2}}$ $\displaystyle>0,$
(IX.45) $\displaystyle
2pp_{1}^{3}+z(4p-z)p_{1}^{2}-2p(2p^{2}-4zp+z^{2})p_{1}-2p^{3}(p-2z)$
$\displaystyle=0,$ (IX.46) $\displaystyle
z(4p-z)p_{1}^{3}-2p(p^{2}-4zp+z^{2})p_{1}^{2}-2p^{3}(p-2z)p_{1}+2z^{2}p^{3}$
$\displaystyle=0,$ (IX.47)

respectively. It is obvious that (IX.45) holds. It is calculated that the
resultant of (IX.46) and (IX.47) in $p_{1}$ is
$-4p^{6}z^{4}(2p^{4}-12zp^{3}+18z^{2}p^{2}-8z^{3}p+z^{4})^{2}$. Since the
condition of Lemma 3 does not hold, it is implied that there exists at least
one common root between (IX.46) and (IX.47) if and only if

$2p^{4}-12zp^{3}+18z^{2}p^{2}-8z^{3}p+z^{4}=0$ (IX.48)

holds with $p<z/(2+\sqrt{5})$, which implies that the condition of Lemma 3
holds. This further implies that $p_{1}>0$. ∎

###### Lemma IX.5

A biquadratic impedance $Z(s)\in\mathcal{Z}_{p2,z2}$ not satisfying the
condition of Lemma 3 is realizable as in Fig. 2(a), where $N_{1}$ is the
configuration in Fig. VIII.1(c) and $N_{2}$ is the configuration in Fig.
VIII.4(e) (that is, the configuration in Fig. 5(a) whose one-terminal-pair
labeled graph is $N_{5a}$), if and only if

$\begin{split}p^{10}&-16zp^{9}+118z^{2}p^{8}-476z^{3}p^{7}\\\
&+1066z^{4}p^{6}-1372z^{5}p^{5}+1064z^{6}p^{4}\\\
&-524z^{7}p^{3}+161z^{8}p^{2}-28z^{9}p+2z^{10}=0\end{split}$ (IX.49)

and $p<z/(2+\sqrt{5})$ (it can be verified that the equation
$\eta^{10}-16\eta^{9}+118\eta^{8}-476\eta^{7}+1066\eta^{6}-1372\eta^{5}+1064\eta^{4}-524\eta^{3}+161\eta^{2}-28\eta+2=0$
has only one distinct root for $\eta\in(0,1/(2+\sqrt{5}))$).

###### Proof:

Necessity. It has been shown that the impedance of the configuration in Fig.
VIII.1(c) is in the form of (VIII.21), and the impedance of the configuration
in Fig. VIII.4(e) is in the form of (VIII.91). Since it is assumed that the
condition of Lemma 3 does not hold, $Z(s)\in\mathcal{Z}_{p2,z2}$ cannot be
realized with fewer than five reactive elements. If $Z(s)$ is realizable as in
Fig. 2(a), where $N_{1}$ is the configuration in Fig. VIII.1(c) and $N_{2}$ is
the configuration in Fig. VIII.4(e), then the impedance of $N_{1}$ is in the
form of (IX.30), where $m$, $p_{1}$, $p$ $>0$ and $q=p_{1}p/(p_{1}+p)$, and
the impedance of $N_{2}$ is in the form of (VIII.92), where $\alpha$, $\beta$,
$\gamma$ $>0$ and (VIII.103)–(VIII.105) hold. Furthermore, one obtains

$Z(s)=Z_{1}(s)+Z_{2}(s)=\frac{ms^{3}+\frac{mp^{2}+(2mp_{1}+\alpha)p+\alpha
p_{1}}{p_{1}+p}s^{2}+\frac{mp_{1}p^{2}+\beta p+\beta
p_{1}}{p_{1}+p}s+\gamma}{(s+p_{1})(s+p)^{2}}.$ (IX.50)

Combining (1) with (IX.50), one obtains

$\displaystyle m$ $\displaystyle=k,$ (IX.51)
$\displaystyle\frac{mp^{2}+(2mp_{1}+\alpha)p+\alpha p_{1}}{p+p_{1}}$
$\displaystyle=k(p_{1}+2z),$ (IX.52) $\displaystyle\frac{mp_{1}p^{2}+\beta
p+\beta p_{1}}{p+p_{1}}$ $\displaystyle=kz(2p_{1}+z),$ (IX.53)
$\displaystyle\gamma$ $\displaystyle=kp_{1}z^{2}.$ (IX.54)

It follows from (VIII.103) and (IX.54) that

$\alpha=\frac{kp_{1}z^{2}}{p(p+2p_{1})}.$ (IX.55)

By (VIII.104), (IX.54), and (IX.55), one obtains

$\beta=\frac{2kp_{1}z^{2}(p+p_{1})^{2}}{(p+2p_{1})^{2}p}.$ (IX.56)

Substituting (IX.51) and (IX.54)–(IX.56) into (VIII.105), (IX.52), and (IX.53)
gives

$\displaystyle\frac{kp_{1}^{2}z^{2}}{(p+2p_{1})^{2}}$ $\displaystyle>0,$
(IX.57) $\displaystyle
2pp_{1}^{3}-(p^{2}-4zp+z^{2})p_{1}^{2}-p(3p^{2}-6zp+z^{2})p_{1}-p^{3}(p-2z)$
$\displaystyle=0,$ (IX.58)

and

$\begin{split}2z(4p-z)p_{1}^{4}-2p(2p^{2}-8zp+z^{2})p_{1}^{3}&-2p^{2}(2p^{2}-5zp-z^{2})p_{1}^{2}\\\
&-p^{3}(p+z)(p-3z)p_{1}+z^{2}p^{4}=0,\end{split}$ (IX.59)

respectively. It is obvious that (IX.57) holds. It is calculated that the
resultant of (IX.58) and (IX.59) in $p_{1}$ is
$-4z^{3}p^{10}(4p-z)(p^{10}-16zp^{9}+118z^{2}p^{8}-476z^{3}p^{7}+1066z^{4}p^{6}-1372z^{5}p^{5}+1064z^{6}p^{4}-524z^{7}p^{3}+161z^{8}p^{2}-28z^{9}p+2z^{10})$.
Since the condition of Lemma 3 does not hold, it is implied that there exists
at least one common root between (IX.58) and (IX.59) if and only if (IX.49)
holds with $p<z/(2+\sqrt{5})$. This further implies that $p_{1}>0$.

Sufficiency. Based on the discussion in the necessity part, there exists
$p_{1}>0$ such that (IX.57)–(IX.59) hold. Let $m$, $\gamma$, $\alpha$, and
$\beta$ satisfy (IX.51) and (IX.54)–(IX.56), which implies $\alpha$, $\beta$,
$\gamma$, $m$ $>0$. Therefore, (VIII.103)–(VIII.105), (IX.52), and (IX.53)
hold. Therefore, $Z(s)$ can be written in the form of (IX.50). Decompose
$Z(s)$ as $Z(s)=Z_{1}(s)+Z_{2}(s)$, where $Z_{1}(s)$ is in the form of (IX.30)
with $m$, $p_{1}$, $p$ $>0$ and $q=p_{1}p/(p_{1}+p)$, and $Z_{2}(s)$ is in the
form of (VIII.92) with $\alpha$, $\beta$, $\gamma$ $>0$. By letting $R_{1}=m$,
$C_{1}=1/(mq)$, and $L_{1}=m/(p_{1}+p)$, $Z_{1}(s)$ is realizable as in Fig.
VIII.1(c). Let $C_{21}$, $C_{22}$, $R_{21}$, and $L_{21}$ satisfy
(VIII.99)–(VIII.102). Since (VIII.105) hold, $C_{21}$, $C_{22}$, $R_{21}$,
$L_{21}$ $>0$. Based on the discussion in the proof of Lemma 11,
(VIII.93)–(VIII.98) hold. Therefore, $Z_{2}(s)$ can be realized as the
configuration in Fig. VIII.4(e). The sufficiency part is proved. ∎

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