Supplementary Material to: Realizations of a Special Class of Admittances with Strictly Lower Complexity than Canonical Forms

0.14 MB·English

by  Michael Z. Q. Chen

1
Supplementary Material to: Realizations of a
Special Class of Admittances with Strictly Lower
Complexity than Canonical Forms
Michael Z. Q. Chen, Kai Wang, Zhan Shu, and Chanying Li
5
1
0 Abstract—This is supplementary material to “Realizations of d = 0 and d = 0, implying that a −d ≥ 0 must hold. If
0 1 1 1
2 aspecialclassofadmittanceswithstrictlylowercomplexitythan thenumberofzerocoefficientsisexactlythree,thentoensure
canonical forms” [1], whichpresentsthedetailed proofsof some
n the positive-realness only the cases when (1) a =0, d =0,
results. For more background information, refer to [2]–[22] and 0 0
a and d =0;(2) a =0, d =0, and d =0 are possible. It is
J references therein. 1 1 0 1
obviousthatthetwocasesmustalwaysbepositive-real,which

9 Index Terms—Network synthesis, transformerless synthesis,
implies a −d ≥0. If the four coefficients are all zero, then
1 canonical realization. 1 1
Y(s) is always positive-real, and a −d ≥ 0 always holds.
1 1
] Summarizing all the above discussion, the two cases in the
C I. INTRODUCTION theorem are obtained.
O This reportpresents the proofsof some results in the paper
. “Realizations of a special class of admittances with strictly III. PROOF OFTHEOREM1
h
lowercomplexitythancanonicalforms”[1].Itisassumedthat
t Proof: Based on Lemma 2, it is obvious that there are
a the numbering of lemmas, theorems, corollaries, and figures
m twopossiblecases.ForCase1,sinced0 =0anda1−d1 ≥0,
in this document agrees with that in the original paper.
Y(s) can be written as
[

1 II. PROOF OF LEMMA2 Y(s)=ka0s2+a1s+1,
v Proof: Sufficiency. It suffices to prove that the three d1s2+s
8
8 inequalities of Lemma 1 hold. For Case 1, d0 = 0 makes andRk =a0(a0+d21−a1d1).IfRk >0,thenY(s)iswritten
4 the three inequalities be equivalent to a0d1 ≥ 0, a0 ≥ 0, as
4 and a1 −d1 ≥ 0, which are obviously satisfied because of Y(s)=k/s+(1/(ka /d )
0 1
0 a −d ≥ 0 and the assumption that all the four coefficients
1. be1 non1negative. For Case 2, a1 = 0, d1 = 0 make the three +1/ ka30s/Rk+ka20(a1−d1)/Rk −1,
0 inequalities be equivalent to a0 −d0 ≥ 0, which obviously which is realizable(cid:0)with one inductor, one capac(cid:1)i(cid:1)tor, and at
5 holds. most two resistors. If Rk <0, then Y(s) can be written as
1 Necessity. Suppose that Y(s) is positive-real with at least
v: one of the four coefficients being zero. If the number of zero Y(s)=k/s+ka0/d1+(−a0d1/(kRk)−a0d21s/(kRk))−1,
i coefficientsisexactlyone,thentoensurethethreeinequalities
X which is realizable with two inductors and two resistors. For
ofLemma1tohold,itisnotedthatonlyd =0ispossible.In
0 Case 2, since a =0, d =0, and a −d ≥0, Y(s) can be
r 1 1 0 0
a this case, a necessary and sufficient condition for Y(s) to be written as
positive-realisa −d ≥0. Ifthe numberof zerocoefficients
1 1 a s2+1 k k(a −d )s
is exactlytwo,then to guaranteethepositive-realnessofY(s) Y(s)=k 0 = + 0 0
s(d s2+1) s d s2+1
only the cases when (1) a = 0 and d = 0; (2) a = 0 0 0
0 0 1
and d1 = 0; (3) d0 = 0 and d1 = 0 are possible. Y(s) is and Rk = (a0 −d0)2. Since it is assumed that Rk 6= 0, we
positive-real if and only if a1 −d1 ≥ 0 when a0 = 0 and concludethatRk >0becauseofthepositive-realnessofY(s).
d0 =0. Y(s) is positive-realif and only if a0−d0 ≥0 when Y(s)isrealizablewithtwoinductorsandonecapacitor,which
a1 = 0 and d1 = 0. Y(s) must always be positive-real when is of strictly lower complexity than the canonicalnetwork.
ThisworkissupportedbyHKUCRCG201111159110,NNSFC61004093,
“973 Program” 2012CB720200, and the “Innovative research projects for IV. PROOF OFLEMMA 3
graduate students inuniversities ofJiangsuprovince” CXLX12 0200. Proof: Suppose that Y(s) can be realized by the lossless
M.Z.Q. Chen is with the Department of Mechanical Engineering, The
University ofHongKong,PokfulamRoad,HongKong;[email protected]. network,thenby[23]theevenpartofY−1(s)isequaltozero,
K.WangiswiththeSchoolofAutomation,NanjingUniversityofScience that is, Ev Y−1(s)=0. It therefore follows that
andTechnology, Nanjing, P.R.China; [email protected].
Z. Shu is with the Electro-Mechanical Engineering Group, Faculty Ev Y−1(s)= 1 Y−1(s)+Y−1(−s)
of Engineering and the Environment, University of Southampton, UK; 2
[email protected]. (cid:0) (cid:1)
2s2((a d −a d )s2+(d −a ))
C.LiiswiththeAcademyofMathematics andSystemsScience, Chinese = 0 1 1 0 1 1 =0.
AcademyofSciences, Beijing, P.R.China; [email protected]. k(a s2+a s+1)(a s2−a s+1)
0 1 0 1
2
Thus, 2s2((a d −a d )s2 +(d −a )) = 0 holds for all its frequency-inversedual as shown in Fig. 5(c). In summary,
0 1 1 0 1 1
s. Then we have a −d = 0 and a d −a d = 0, which Y(s) can be realized as the admittance of at least one of the
1 1 0 1 1 0
indicates that Y(s)=k/s. networks shown in Fig. 5. By calculating their admittances,
we see that Rk =0.
V. PROOF OF LEMMA4
Proof: It is obvious that Z(s) = Y−1(s) has a pole at
s=∞ and a zero at s=0. By [24, Theorem 2], this lemma 1 1 2
a a' a a'
can be easily proven.
(a) (b)
VI. PROOF OF LEMMA 5
3
Proof: For the network in Fig. 3(b), we see that there
must exist poles on the imaginary axis s=jω0 with ω0 6=0, a 1 2 3 a' a 1 a'
which contradicts the fact that all the coefficients be positive. 2
Since any network in the form of Fig. 3(a) is the frequency-
(c) (d)
inverse dual of a network in Fig. 3(b), then it cannot realize
this class of admittances, either. Fig.S1. Onehalfofthenetworkgraphsofthetwo-terminal networks with
atmostthreeelements.
VII. PROOF OF LEMMA 7
Proof: It has been discussed that Y1(s) can IX. PROOF OF THEOREM2
be written as Y (s) = Y−1(s−1) = k′(a′s2 +
1 0 Proof: Sufficiency. For the network shown in Fig. 7(a),
a′s + 1)/ s(d′s2+d′s+1) with a′ = 1/d ,
1 0 1 0 0 its admittance is calculated as
a′1 = d1(cid:0)/d0, d′0 = 1/(cid:1)a0, and d′1 = a1/a0. L C s2+R C s+1
Calculating the corresponding Rk1, we obtain Y(s)= 2 1 1 1 ,
Rk1 = (a0−d0)2−(a1−d1)(a0d1−a1d0) /(a20d20) = s(L1L2C1s2+R1C1(L1+L2)s+L1+L2)
Rk/(a20d20(cid:0)). Thus, this completes the proof. (cid:1) which can be expressed in the form of (VIII.1), where a0 =
L C > 0, a = R C > 0, d = L L C /(L +L ) >
2 1 1 1 1 0 1 2 1 1 2
VIII. PROOF OF LEMMA8 0, d1 = R1C1 > 0, and k = 1/(L1 + L2) > 0. Then it
Sufficiency. Since Rk = 0, there must be at least one is calculated that Rk = C12L42/(L1 + L2)2 6= 0. So is the
commonfactorbetween(a s2+a s+1)and(d s2+d s+1). admittance of its frequency-inverse dual shown in Fig. 7(b)
0 1 0 1
Therefore, Y(s) must be of the form by Lemma 7, and hence the equivalent networks in Fig. 7(c)
and Fig. 7(d). Thus, the sufficiency part of this theorem is
(As+1)(Bs+1) As+1
Y(s)=k =k , proven.
s(Cs+1)(Bs+1) s(Cs+1) Necessity. Suppose that Y(s) can be realized with at most
where A, B, C >0. Comparing the above equation with four elements whose values are positive and finite. Based on
a s2+a s+1 Lemma8,Rk 6=0guaranteesthatweonlyneedtoconsiderthe
Y(s)=k 0 1 , (VIII.1) irreduciblefour-elementnetwork.Themethodofenumeration
s(d s2+d s+1)
0 1 is used here for the proof. One half of the network graphs of
where a , a , d , d ≥ 0 and k > 0, we have the following the two-terminal four-element networks are listed in Fig. S2,
0 1 0 1
relations:a =AB, a =A+B,d =BC,andd =B+C. and other possible graphs are dual with them.
0 1 0 1
Then we obtain a /d =A/C. Since Y(s) is positive-real, it If a four-element network can always be equivalent to one
0 0
isthereforeimpliedthata0−d0 ≥0,whichindicatesthatA≥ withlesselements,thenbyLemma8weknowRk =0,contra-
C. If A=C, then Y(s) reduces to Y(s)=k/s, realizable as dictingwiththeassumption.IthasbeenstatedinLemma4that
just an inductor. Otherwise, A>C leads to theremustbe a pathP(a,a′) anda cut-setC(a,a′)consisting
of onlyinductorsfor the possible realizations.If the networks
As+1 k 1
Y(s)=ks(Cs+1) = s + Cs + 1 , whose graphs in Fig. S2(a), Fig. S2(b), and Fig. S2(e) satisfy
k(A−C) k(A−C) this property, then they must be equivalent to the ones with
which is realizable with three elements. at most three elements. Therefore, all these network graphs
Necessity. By the method of enumeration, one half of shouldbeeliminated.ForFig.S2(c),Edge1andEdge4must
network graphs of two-terminal networks with at most three be inductorsby Lemma 4 and Lemma8. Since Rk 6=0, Y(s)
elements are shown in Fig. S1. Lemma 4 implies that the cannot be written as Y(s) = k/s, which means that lossless
networkwith the networkgraphshownin Fig. S1(a)can only networks do not need to be considered by Lemma 3. Then
realize the network in Fig. 5(a). If we disregard the networks there must be exactly one resistor either on Edge 2 or on
which can always be reduced to those with less elements, Edge 3. If the other element is the inductor, then we obtain
then graphs in Fig. S1(b) and Fig. S1(c) are immediately the network shown in Fig. 6, which by Lemma 9 means that
eliminated,becausealltheelementscouldonlybeinductorsby Rk = 0. Then, it is eliminated. Therefore, only the network
Lemma4.ThenbasedonLemma5,weconcludethatonlythe in Fig. 7(a) is possible for the graph in Fig. S2(c). Using a
networkshown in Fig. 5(b)is possible, which is equivalentto similar argument, we have that only the network in Fig. 7(d)
3
is possible for graph in Fig. S2(d). Using the frequency- From (X.8) and (X.10), it is obvious that (X.2) holds. From
inversedualoperation,the networksin Fig. 7(b)andFig. 7(c) (X.7) and (X.9), we obtain
are obtained. It is noted that by Lemma 6 the networks in
a L +L
0 1 2
Fig. 7(a)andFig. 7(b)are equivalentto thosein Fig. 7(c)and = ,
d L
0 1
Fig. 7(d), respectively. So far, all the possible networks have
from which and together with (X.11), we obtain that L can
been discovered. 1
be expressed as (X.4). Substituting L into (X.11), L is
1 1
4 yielded in the form of (X.5), from which we indicate (X.1).
SubstitutingL into(X.7),wehave(X.6).Finally,from(X.8),
a 1 2 3 4 a' a 1 2 a' 2
3 we obtain (X.3).
(a) (b) Sufficiency. Consider an admittance Y(s) in the form of
4 2 (VIII.1), where a0, a1, d0, d1, k > 0, Rk 6= 0, and (X.1)
and(X.2)hold.Calculate R ,L ,L ,andC by(X.3)–(X.6).
3 1 1 2 1
1 1 Since (X.1) holds, it can be verified that all the values of ele-
a a' a a'
2 3 4 ments are positive and finite. Since (X.2) holds, (X.7)–(X.11)
(c) (d) must be satisfied. Therefore,the admittance of the networkin
1 3 Fig. 7(a) equals (VIII.1), which implies the sufficiency.
a 2 4 a' XI. PROOF OF THEOREM4
(e) Proof:Sufficiency.ByLemma8,therealizationofCase1
can obviously be satisfied when Rk = 0. It is not difficult to
Fig.S2. Onehalfofthenetworkgraphsofthetwo-terminal networkswith
see that the condition of Case 2 is equivalent to a −d >0,
fourelements. 0 0
andeithera −d =0ora d −a d =0.Whena −d >0
1 1 0 1 1 0 0 0
and a − d = 0, Y(s) can be realized as in Fig. 7(a) by
1 1
X. PROOF OF THEOREM3 Theorem 3, which is a four-element network. Through the
Theorem 3: Consider an admittance Y(s) in the form of use of frequency-inverse dual, the case can be proven when
(VIII.1) with a0, a1, d0, d1, k > 0 and Rk 6= 0. Then it can a0−d0 >0 and a0d1−a1d0 =0.
be realized by the network shown in Fig. 7(a) with all the Necessity. Suppose that admittance Y(s) in the form of
elements positive and finite if and only if (VIII.1) with a , a , d , d , k > 0 can be realized with at
0 1 0 1
most four elements, whose values are positive and finite. We
a −d >0, (X.1)
0 0
thendivideitintotwocases:Rk =0andRk 6=0.Itisknown
a −d =0. (X.2)
1 1 fromLemma8thatthecasewhenRk =0alwaysholds.When
Moreover, if the condition is satisfied, then the values of Rk 6=0,itisseenfromTheorem2thatY(s)istheadmittance
elements are expressed as ofthenetworkshowninFig.7.Togetherwiththerealizability
a (a −d ) conditions derived above, we obtain a0−d0 > 0, and either
R1 = 1 k0a2 0 , (X.3) a1−d1 =0 or a0d1−a1d0 =0, implying Case 2.
0
d
L1 = 0 , (X.4) XII. PROOF OF THEOREM5
ka
0
a −d Proof:Necessity.SinceRk 6=0,thenY(s)mustbeanRL
L2 = 0 0, (X.5) (SD) admittance of McMillan degree three. By [22, Theorem
ka
0
ka2 4], the coefficients must satisfy Rk <0.
C1 = a −0d . (X.6) Sufficiency. Since Rk :=(a0−d0)2−(a0d1−a1d0)(a1−
0 0 d )<0,thenwehavea d −a d >0.Inaddition,weknow
1 0 1 1 0
Proof: Necessity. The admittance of the network shown
d ,d >0, then it is known from [22, Theorem 4] that Y(s)
0 1
in Fig. 7(a) can be calculated as
is an RL (SD) admittance of McMillan degree three. From
L2C1s2+R1C1s+1 [22], [25], Y(s) can be of the form
Y(s)= .
s(L L C s2+R C (L +L )s+L +L )
1 2 1 1 1 1 2 1 2 (As+1)(Cs+1)
Y(s)=k ,
Suppose that it can realize admittance Y(s) in the form of s(Bs+1)(Ds+1)
(VIII.1) with a , a , d , d , k >0, then we obtain
0 1 0 1 where A > B > C > D > 0 and k > 0. Therefore, it is the
a =L C , (X.7) admittance of a network with two resistors and an arbitrary
0 2 1
a =R C , (X.8) number of inductors. Furthermore, Y(s) can be written as
1 1 1
d0 = L1L2C1, (X.9) Y(s)= k + k(A−B)(B−C) + k(A−D)(C−D),
L1+L2 s (B−D)(Bs+1) (B−D)(Ds+1)
d =R C , (X.10)
1 1 1 whichisobviouslytheadmittanceofthenetworkinFig.8with
1 the values satisfying L = 1/k, L = B(B −D)/(k(A−
k = . (X.11) 1 2
L1+L2 B)(B − C)), L3 = D(B − D)/(k(A − D)(C − D)),
4
R1 =(B−D)/(k(A−B)(B−C)), R2 =(B−D)/(k(A− XIV. PROOF OF LEMMA12
D)(C − D)), where A,C = a1± a21−4a0 /2 and Lemma 12: Consider a positive-real function Y(s). Then it
(cid:16) p (cid:17) can be realized as the admittance of the network shown in
B,D = d ± d2−4d /2.Sincea =AC, a =A+C,
1 1 0 0 1 Fig. 12 with the values of the elements being positive and
d =BD(cid:16), andpd =B+D(cid:17), then by solving them we obtain
0 1 finite, if and only if Y(s) can be expressed as
the expressions of A, B, C, and D as stated in this theorem.
α s3+α s2+α s+1
Y(s)= 3 2 1 , (XIV.1)
β s4+β s3+β s2+β s
4 3 2 1
where α , α , α , β , β , β , β > 0, W , W , W , W −
XIII. PROOF OF LEMMA 10 1 2 2 1 2 3 4 1 2 3
2α W >0, and the following two equations hold
2 3
Proof: For Fig. 9(a), the admittance can be calculated W2−4W W W =0, (XIV.2)
1 2 3
as Y(s) = (C L L s3 + R C L s2 + (L + L )s +
1 2 3 1 1 3 2 3 β +α β +α β −α β =0. (XIV.3)
R )/(s(C L L L s3 + R C L L s2 + (L L + L L + 4 1 3 3 1 2 2
1 1 1 2 3 1 1 1 3 1 2 2 3
L1L3)s+R1(L1 +L3))). Since Rk 6= 0, then Y(s) can be Moreover, if the conditions hold, then the values of the
realized as in Fig. 9(a), if and only if there exists T >0 such network can be expressed as follows
that the following equations hold
(α α −α )β2
R = 1 2 3 1, (XIV.4)
a T = C1L2L3, a +a T =C L , a +T = L2+L3, 1 α21(α2β1−β3)
0 R1 0 1 1 3 1 R1 L = (α1α2β1−α3β1−α1β3)β1, (XIV.5)
(XIII.1) 1 α (α β −β )
1 2 1 3
d0T = RC11(LL11L+2LL33), d0+d1T = LC11L+1LL33, L2 = α1(αα2β3β112−β3), (XIV.6)
(XIII.2)
β β
L = 1 3 , (XIV.7)
L1L2+L2L3+L1L3 1 3 α β −β
d +T = , k = . 2 1 3
1
R1(L1+L3) L1+L(X3III.3) C1 = α2ββ12−β3. (XIV.8)
1
Therefore,it sufficesto show thatT >0 doesnotexist. After Proof:Necessity.Calculatetheadmittanceofthenetwork
a series of calculations, it is verified that (XIII.1)–(XIII.3) are in Fig. 12, then we obtain Y(s) = (C1L2(L1 + L3)s3 +
equivalent to R1C1(L1+L2+L3)s2+(L1+L3)s+R1)/(s(C1L1L2L3s3+
R C L (L +L )s2+(L L +L L +L L )s+R (L +
1 1 3 1 2 1 2 2 3 1 3 1 1
L ))). Therefore, Y(s) can be expressed as (XIV.1) with the
2
(a0+a1T)(a0−d0) coefficients satisfying
R =
1 ka a (T2+a T +a )
0 1 1 0 C L (L +L )
= (a0−d0)(a1d0T2+(a20+a21d0)T +a0a1d0), α3 = 1 2 R11 3 , α2 =C1(L1+L2+L3), (XIV.9)
ka2a (d +T)(T2+a T +a )
0 1 1 1 0 L1+L3 C1L1L2L3
(XIII.4) α1 = , β4 = , β3 =C1L3(L1+L2),
R R
1 1
(XIV.10)
L L +L L +L L
L1 =kda0 , L2 = ka (T(a20+−ad0T)T+a ), L3 = a0k−ad0, β2 = 1 2 R213 1 3, β1 =L1+L2. (XIV.11)
0 1 1 0 0
It is obvious that α , α , α , β , β , β , β > 0. After a
1 2 2 1 2 3 4
(XIII.5) series of calculations, the above equations are equivalent to
(XIV.4)–(XIV.8) and the following two equations
α α2β β +α β β +α2β β +α α β2
ka (a +a T) ka2(d +d T) 1 2 1 4 3 3 4 3 1 3 1 3 3
C1 = 0a 0−d 1 = d0(a0 −d1) . (XIII.6) =α2α3β1β4+α1α2β3β4+α1α2α3β1β3,
0 0 0 0 0 (XIV.12)
From(XIII.6),weobtaina0−d0 >0anda1d0−a0d1 =0.The α α2β β +α β β +α2β2+α α β β +α2β2
secondequalityof(XIII.4)isequivalenttoa (a −d )T2 =0, 1 2 1 2 3 2 3 3 1 1 3 1 3 1 3
1 0 0 =α α β β +α α β β +α α α β2+α2α β β .
which obviously cannot hold for any T >0. Therefore, there 2 3 1 2 1 2 2 3 1 2 3 1 1 2 1 3
(XIV.13)
does not exist any T > 0 such that (XIII.1)–(XIII.3) hold
simultaneously. Furthermore, (XIV.13) is equivalent to α α α β β −
1 2 3 1 3
Using a similar argument,the conclusion of the network in α2β β − α α β2 = (α α2β + α β − α α β −
3 1 3 1 3 3 1 2 1 3 3 2 3 1
Fig. 9(c) can also be proven.Furthermore,since the networks α α β )(α β −α β −α β ). Since all the values of the
1 2 3 2 2 1 3 3 1
in Fig. 9(b) and Fig. 9(d) are the frequency-inverse duals of elements are positive and finite, then we can calculate that
those in Fig. 9(a) and Fig. 9(c), respectively, they cannot be W = α α − α = C (L + L )2/R1 > 0, W =
1 1 2 3 1 1 3 2
realized, either. α β −β = C (L +L )2 > 0, W = L2/R > 0, and
2 1 3 1 1 2 3 1 1
5
W −2α W = 2C L L L /R > 0. Since W , W > 0, where
2 3 1 1 2 3 1 1 2
then T = a0d1−a1d0. (XV.7)
α α2β +α β −α α β −α α β r a1−d1
1 2 1 3 3 2 3 1 1 2 3
Proof: Necessity. Suppose that Y(s) in the form of
=(α α −α )(α β −β )=W W >0.
1 2 3 2 1 3 1 2
(VIII.1)withallthecoefficientspositivecanberealizedasthe
(XIV.14)
admittance of the network shown in Fig. 12. It then follows
Therefore, it follows from (XIV.12) that from Lemma 12 that Y(s) can be expressed as (XIV.1) with
α α α β β −α2β β −α α β2 all the coefficients positive and satisfying the condition of
β4 =α α12β2 +3 α1 3β −α3 1α 3β −1α 3α 3β Lemma12.SinceRk 6=0,theonlywaytoexpress(VIII.1)as
1 2 1 3 3 2 3 1 1 2 3
(XIV.1) where α , α , α , β , β , β , β > 0 is to multiply
(α α2β +α β −α α β −α α β ) 1 2 2 1 2 3 4
= 1 2 1 3 3 2 3 1 1 2 3 (XIV.15) the numerator and denominator with (Ts+1) where T > 0.
α α2β +α β −α α β −α α β
1 2 1 3 3 2 3 1 1 2 3 Consequently, it follows that
×(α β −α β −α β )
2 2 1 3 3 1
α =a T, α =a +a T, α =a +T,
=α β −α β −α β , 3 0 2 0 1 1 1
2 2 1 3 3 1
d T d +d T d +T 1
0 0 1 1
β = , β = , β = , β = .
which implies (XIV.3). Substituting the above equation into 4 3 2 1
k k k k
W, we obtain W =2(α α β −α β −α β ), implying (XV.8)
1 2 1 1 3 3 1
4(α1α3β1β3−α21α2β1β3−α1α2α3β12+α21β32+α23β12 It is obvious that the condition that α1, α2, α2, β1, β2, β3,
+α α2β β −α α β β −α α β β +α β β ) β4 > 0 always holds. Furthermore, other conditions can be
1 2 1 2 2 3 1 2 1 2 2 3 3 2 3 presented as follows
=4(α α β −α β −α β )2
1 2 1 3 1 1 3
W =α α −α =a T2+a2T +a a >0, (XV.9)
−4(α1α2−α3)(α2β1−β3)(α1β1−β2) 1 1 2 3 1 1 0 1
(a −d )T +(a −d )
=W2−4W1W2W3. W2 =α2β1−β3 = 1 1 0 0 >0, (XV.10)
k
(XIV.16)
a −d
W =α β −β = 1 1 >0, (XV.11)
Now, (XIV.13) and (XIV.16) yield (XIV.2). 3 1 1 2 k
Sufficiency.LetthevaluesofR1,L1,L2,L3,andC1 satisfy W −2α W = (a1−d1)T2+(a0d1−a1d0) >0,
(XIV.4)–(XIV.8). W , W >0 indicates that R , L , L , C 2 3
1 2 1 2 3 1 k
> 0. Then substituting β obtained from (XIV.3) into W, (XV.12)
4
we obtain W − 2α2W3 = 2(α2β2 − α1β3 − α3β1). Since (a −d )T2−(a d −a d ) 2
W −2α W >0, it follows that L >0. Substituting β ob- W2−4W W W = 1 1 0 1 1 0 =0,
2 3 1 4 1 2 3 (cid:0) k2 (cid:1)
tainedfrom(XIV.3)into(XIV.2),weobtainEquation(XIV.13)
(XV.13)
immediately.Togetherwith (XIV.14) and (XIV.15), we obtain
Equation (XIV.12). It is known in the necessity part that β4+α1β3+α3β1−α2β2
(XIV.9)–(XIV.11) must hold. Now, the sufficiency is proven. −(a −d )T2+2d T −(a d −a d )
1 1 0 0 1 1 0
= =0.
k
(XV.14)
XV. PROOF OF LEMMA 13 Then (XV.11) leads to a −d > 0. Then T can be solved
1 1
Lemma 13: Consider any positive-realfunctionY(s) in the from (XV.13) as (XV.7). The constraint that T > 0 yields
formof(VIII.1)witha0,a1,d0,d1,k >0,andRk 6=0.Then a0d1−a1d0 >0. Substituting the solved T into (XV.14), we
it can be realized as the admittance of the network as shown obtain (XV.1).
in Fig. 12 with the values of the elements being positive and Sufficiency. Suppose that (a d −a d )(a −d )−d2 =0
0 1 1 0 1 1 0
finite if and only if holds, then d >0 and the positive-realness of Y(s) leads to
0
a d −a d >0 and a −d >0. Then we can let
(a d −a d )(a −d )−d2 =0. (XV.1) 0 1 1 0 1 1
0 1 1 0 1 1 0
a d −a d
0 1 1 0
Moreover, the values of the elements can be expressed as T = >0.
r a −d
1 1
R = a1(T2+a1T +a0) , (XV.2) Furthermore, multiplying the numerator and denominator of
1 k(a1+T)2((a1−d1)T +(a0−d0)) Y(s) with the factor (Ts+1), then we can express Y(s) in
(a −d )T2+(a2−a d −d )T +a (a −d ) the form of (XIV.1) with the coefficients satisfying (XV.8). It
L = 1 1 1 1 1 0 1 0 0 ,
1 k(a1+T)((a1−d1)T +(a0−d0)) is obvious that α1, α2, α2, β1, β2, β3, β4 > 0 and (XV.9)–
(XV.3) (XV.13) hold. Since (a0d1 −a1d0)(a1−d1)−d20 = 0, then
(XV.14) is also satisfied. It is concluded that the condition
a T
L2 = k(a +T)((a −0d )T +(a −d )), (XV.4) of Lemma 12 holds, therefore Y(s) can be realized as the
1 1 1 0 0
admittance of the network as shown in Fig. 12 with the
d T +d
L = 1 0 , (XV.5) elements positive and finite. The expressions for the values
3 k((a −d )T +(a −d ))
1 1 0 0 of the elements are presented in (XV.2)–(XV.6) derived from
C =k((a −d )T +(a −d )), (XV.6)
1 1 1 0 0 (XIV.4)–(XIV.8) with the relation (XV.8).
6
XVI. PROOF OF LEMMA 14 [10] M.Z.Q.Chen,K.Wang,M.Yin,C.Li,Z.Zuo,andG.Chen,“Realiz-
ability ofn-portresistive networks with 2nterminals,” inProceedings
Proof: Calculating the admittance of the network shown
ofthe9thAsianControlConference, 2013,pp.1–6.
in Fig. 13(a) yields Yd(s) = (C1L2(L1 + L3)s3 + [11] M. Z. Q. Chen, K. Wang, M. Yin, C. Li, Z. Zuo, and G. Chen,
R C (L +L )s2+(L +L +L )s+R )/(s(C L L L s3+ “Synthesis of n-port resistive networks containing 2n terminals,” In-
1 1 1 3 1 2 3 1 1 1 2 3
ternational Journal ofCircuitTheoryandApplications, inpress(DOI:
R C (L L +L L +L L )s2+L (L +L )s+R (L +
1 1 1 2 2 3 1 3 3 1 2 1 1 10.1002/cta.1951).
L2))), which is obviously in the form of (XIV.1) with α3 = [12] M. Z. Q. Chen, “The classical n-port resistive synthesis problem,” in
C L (L +L3)/R , α = C (L +L ), α = (L +L + Workshop on “Dynamics and Control in Networks”, Lund University,
1 2 1 1 2 1 1 3 1 1 2
2014(http://www.lccc.lth.se/media/2014/malcolm3.pdf,lastaccessedon
L )/R , β = C L L L /R , β = C (L L + L L +
3 1 4 1 1 2 3 1 3 1 1 2 2 3 19/01/2015).
L1L3), β2 = L3(L1 +L2)/R1, and β1 = L1 +L2. It can [13] M.Z.Q.Chen, K.Wang,Y.Zou,andG.Chen, “Realization ofthree-
be calculatedthatthe coefficientsmustsatisfy W−2α W = portspringnetworkswithinerterforeffectivemechanicalcontrol,”IEEE
2 3
Trans.AutomaticControl, inpress.
−2C L (L +L )(L +L )/R <0. Similarly, for network
1 2 1 2 1 3 1 [14] M.Z.Q.Chen, Y.Hu, andB. Du,“Suspension performance with one
in Fig. 13(b), its admittance can also be in the form of damper and one inerter,” in Proceedings of the 24th Chinese Control
(XIV.1) where α , α , α , β , β , β , β > 0 and satisfy andDecisionConference, Taiyuan,China,2012,pp.3534–3539.
1 2 2 1 2 3 4
[15] M. Z. Q. Chen, Y. Hu, L. Huang, and G. Chen, “Influence of inerter
W −2α W < 0. Assume that Y(s) in the form of (VIII.1)
2 3 on natural frequencies of vibration systems,” Journal of Sound and
with a0, a1, d0, d1, k > 0 and Rk 6= 0 can be realized by Vibration, vol.333,no.7,pp.1874–1887, 2014.
a network as in Fig. 13(a) or Fig. 13(b). Consequently, Y(s) [16] M. Z. Q. Chen, Y. Hu, C. Li, and G. Chen, “Performance benefits of
using inerter in semiactive suspensions,” IEEETrans. Control Systems
must be able to be expressed in the form of (XIV.1) with α ,
1 Technology, inpress(DOI:10.1109/TCST.2014.2364954).
α2, α2, β1, β2, β3, β4 > 0, implying that the coefficients [17] Y. Hu, M. Z. Q. Chen, and Z. Shu, “Passive vehicle suspensions
satisfy (XV.8). In the proof of Lemma 13, it is seen that employing inerters with multiple performance requirements,” Journal
ofSoundandVibration, vol.333,no.8,pp.2212–2225, 2014.
(a −d )T2+(a d −a d ) [18] K. Wang and M. Z. Q. Chen, “Realization of biquadratic impedances
W −2α2W3 = 1 1 0 1 1 0 ≥0, withatmostfourelements,”inProceedingsofthe24thChineseControl
k
andDecisionConference, 2012,pp.2900–2905.
which contradicts with the hypothesis. Thus this lemma is [19] K.WangandM.Z.Q.Chen,“Generalizedseries-parallelRLCsynthesis
withoutminimizationforbiquadraticimpedances,”IEEETrans.Circuits
proven.
andSystemsII:ExpressBriefs,vol.59,no.11,pp.766–770,2012.
[20] K. Wang, M. Z. Q. Chen, and Y. Hu, “Synthesis of biquadratic
impedanceswithatmostfourpassiveelements,”JournaloftheFranklin
XVII. CONCLUSION
Institute, vol.351,no.3,pp.1251–1267, 2014.
In this report, the proofs of some results in the original [21] K.WangandM.Z.Q.Chen,“Minimalrealizationsofthree-portresistive
networks,”IEEETrans.CircuitsandSystemsI:RegularPapers,inpress
paper [1] have been presented.
(10.1109/TCSI.2015.2390560).
[22] M. C. Smith, “Synthesis of mechanical networks: the inerter,” IEEE
ACKNOWLEDGMENT Trans.AutomaticControl, vol.47,no.10,pp.1648–1662, 2002.
[23] H.Baher, Synthesis ofElectrical Networks. NewYork:Wiley,1984.
The authors are grateful to the Associate Editor and the [24] S. Seshu, “Minimal realization of the biquadratic minimum function,”
reviewers for their insightful suggestions. IRETrans.CircuitTheory,vol.6,no.4,pp.345–350,1959.
[25] E. A. Guillemin, Synthesis of Passive Networks, John Wiley & Sons,
1957.
REFERENCES
[1] M.Z.Q.Chen,K.Wang,Z.Shu,andC.Li,“Realizations ofaspecial
class of admittances with strictly lower complexity than canonical
forms,” IEEE Trans. Circuits and Systems I: Regular Papers, vol. 60,
no.9,pp.2465–2473,2013.
[2] M. Z. Q. Chen, Passive Network Synthesis of Restricted Complexity,
Ph.D.Thesis,CambridgeUniv.Eng.Dept.,U.K.,2007.
[3] M. Z. Q. Chen and M. C. Smith, “Electrical and mechanical passive
networksynthesis,”inRecentAdvancesinLearningandControl,V.D.
Blondel,S.P.Boyd,andH.Kimura(Eds.),NewYork:Springer-Verlag,
2008,LNCIS,vol.371,pp.35–50.
[4] M. Z. Q. Chen, “A note on PIN polynomials and PRIN rational
functions,”IEEETrans.CircuitsandSystemsII:ExpressBriefs,vol.55,
no.5,pp.462–463,2008.
[5] M. Z. Q. Chen and M. C. Smith, “Restricted complexity network
realizations for passive mechanical control,” IEEE Trans. Automatic
Control, vol.54,no.10,pp.2290–2301, 2009.
[6] M. Z. Q. Chen and M. C. Smith, “A note on tests for positive-real
functions,”IEEETrans.AutomaticControl,vol.54,no.2,pp.390–393,
2009.
[7] M. Z. Q. Chen, C. Papageorgiou, F. Scheibe, F.-C. Wang, and M. C.
Smith, “The missing mechanical circuit element,” IEEE Circuits Syst.
Mag.,vol.9,no.1,pp.10–26,2009.
[8] M.Z.Q.Chen,K.Wang,Y.Zou,andJ.Lam,“Realization ofaspecial
class ofadmittances withonedamperandoneinerter,” InProceedings
ofthe51stIEEEConferenceonDecisionandControl,2012,pp.3845–
3850.
[9] M.Z.Q.Chen,K.Wang,Y.Zou,andJ.Lam,“Realization ofaspecial
class of admittances with one damper and one inerter for mechanical
control,”IEEETrans.AutomaticControl,vol.58,no.7,pp.1841–1846,
2013.