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On the theorem of Reichert
Systems & Control Letters 61 (2012) 1124–1131
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On the theorem of Reichert Jason Zheng Jiang, Malcolm C. Smith ∗ Department of Engineering, University of Cambridge, Cambridge, CB2 1PZ, UK
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Article history: Received 19 March 2012 Received in revised form 14 September 2012 Accepted 19 September 2012 Available online 9 November 2012
abstract This paper reworks and amplifies Reichert’s proof of his theorem (1969) which asserts that any impedance function of a one-port electrical network which can be realised with two reactive elements and an arbitrary number of resistors can be realised with two reactive elements and three resistors. © 2012 Elsevier B.V. All rights reserved.
Keywords: Electrical networks Circuit synthesis Positive-real functions
1. Introduction Reichert’s theorem [1] is one of the fundamental results at the boundary of circuit theory and systems theory. The theorem completely characterises all second-order transferfunctions which may be realised by one-port RLC-circuits with at most two energy-storage (reactive) elements. It is shown that no transfer-function requires more than three resistors for its realisation and that eight such circuits suffice to realise every transfer function in the class. A consequence of the result is that not all second-order positive-real transfer-functions (biquadratics) are realisable with just two energy-storage elements. Stated more broadly, non-minimality in the sense of modern systems theory is necessary for RLC-realisations of (some) positive-real functions. To date no proof of the theorem other than Reichert’s original German language publication [1] has appeared. Reichert’s proof makes use of a characterisation of Auth [2] of realisable biquadratic impedances, which itself relies on Cederbaum’s necessary condition for the realisation of a purely resistive nport [3]. The crucial part of Reichert’s proof is a topological argument that a certain system of polynomial equations has no solution in some region of its variable parameters. It is our view that some parts of this argument need to be expanded. It is the purpose of the present paper to provide a complete reworking of Reichert’s proof, including the parametrisation of Auth, and to provide new lemmas to clarify the main topological argument. The class of transfer-functions realisable by two reactive elements and three resistors was illuminated in [4] by the
∗
Corresponding author. Tel.: +44 1223 3 32745; fax: +44 1223 3 32662. E-mail addresses: [email protected], [email protected] (J.Z. Jiang), [email protected], [email protected] (M.C. Smith). 0167-6911/$ – see front matter © 2012 Elsevier B.V. All rights reserved. doi:10.1016/j.sysconle.2012.09.009
concept of a regular positive-real function. It was shown that series–parallel networks of this type can only realise regular biquadratics and all regular biquadratics can be realised in this way. The only networks capable of realising non-regular biquadratics are a pair of bridge networks studied by Foster and Ladenheim [5]. Reichert’s theorem shows that the class of realisable transfer-functions is expanded no further by the use of additional resistors. Aside from the fundamental nature of the questions involved, renewed interest in the subject of network synthesis has been provided recently by mechanical network problems making use of the inerter device [6]. Such problems highlight the importance of gaining a complete understanding of the efficient realisation of (low-order) impedance functions, e.g. biquadratics. For further general background on the subject of classical network synthesis, the reader is referred to [7–9]. In the next section we provide the background and notation needed for our version of Reichert’s proof. 2. Preliminaries We will restrict our attention to RLC one-port networks, i.e. networks with two external terminals and comprising resistors, inductors and capacitors but excluding transformers (see Fig. 1). The energy storage (reactive) elements are the inductor and capacitor. For the analogous two-terminal mechanical elements the reactive elements are the spring and inerter [6]. For a network as in Fig. 1 the driving-point impedance Z (s) is defined by Z (s) = vˆ (s) /ˆi (s) where ˆ denotes Laplace transformation and Y (s) = Z −1 (s) is the driving-point admittance. The immittance of a network is the impedance or admittance. We recall that a rational function Z (s) is defined to be positive-real if Z (s) is analytic and
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
′
Fig. 1. One-port electrical network with two external terminals 1 and 1 , terminal voltage v and terminal current i.
Re (Z (s)) ≥ 0 for Re (s) > 0 [9]. It is known that the driving-point immittance of a linear network is positive-real if and only if the network is passive [10]. Following [4] we define a positive-real function Z (s) to be regular if the smallest value of Re (Z (jω)) or Re Z −1 (jω) occurs at ω = 0 or ω = ∞. In this paper, we are concerned with the realisation of biquadratics Z (s) =
As2 + Bs + C Ds2 + Es + F
,
(1)
where A, B, C , D, E , F ≥ 0. It is well known [5,9,11] that Z (s) is positive real if and only if σ = BE −
√
√
AF −
CD
2
≥ 0. In [4]
it was shown that a positive-real biquadratic impedance (1) with any of the parameters A, B, C , D, E and F equal to zero is regular. Moreover, any such impedance can be realised by a series–parallel network with at most two reactive elements and two resistors. Accordingly we restrict attention to the case that A, B, C , D, E , F > 0. Using impedance and frequency scaling [4, Section III], Z (s) can be reduced to a canonical form Zc (s) =
s2 + 2U
√ Ws + W
√
s2 + 2V / W s + 1/W
,
(U , V , W > 0) .
(2)
See [4, Section V] and [1]. Any circuit that realises Zc (s) will also realise the corresponding Z (s) after appropriate scaling of the elements [4]. It can be checked that Zc (s) is positive real if and only if
σc = 4UV + 2 −
1 W
+W
≥ 0.
The resultant of the numerator and denominator of Zc (s) is 2
2
Kc = 4U + 4V − 4UV
1 W
+W
+
1 W
2 −W
.
(3)
For any rational function ρ (U , V , W ) we make use of the notations ρ ∗ (U , V , W ) = ρ U , V , W −1 and ρ Ď (U , V , W ) =
ρ (V , U , W ). We observe that σc∗ = σcĎ = σc and Kc∗ = KcĎ = Kc . Let
λc = 4UV − 4V 2 W −
1 W
−W .
Lemma 1 ([4]). The biquadratic Zc is regular if and only if at least one of the following five conditions is satisfied:
< 1 and λc ≥ 0, < 1 and λĎc ≥ 0, = 1, > 1 and λ∗c ≥ 0, ∗Ď Case 5. W > 1 and λc ≥ 0.
Case 1. Case 2. Case 3. Case 4.
W W W W
In [4] it was shown that any regular biquadratic could be realised by a series–parallel network with (at most) two reactive
1125
Fig. 2. The two-reactive five-element bridge networks that can realise non-regular impedances.
elements and three resistors. Six network structures suffice to realise the whole class. Among all the bridge networks with two reactive elements and three resistors only the pair shown in Fig. 2 can realise non-regular immittances. We now present the realisation condition for this pair of networks for the canonical form (2) as given in [4]. Let
1 γ1 = 4U 2 + 4V 2 + 4UV 3W − W 1 1 + −W − 9W ,
(4)
1 γ2 = −4U 2 − 4V 2 + 4UV W + W 1 1 − −W − 3W .
(5)
W
W
W
W
Lemma 2 ([5]). Zc (s) in the form of (2) can be realised by the network shown in Fig. 2(a) and (b) with positive element values if and only if W < 1, γ1 ≥ 0 (W > 1, γ1∗ ≥ 0) and the signs of
λc , λĎc , γ2 (λ∗c , λ∗c Ď , γ2∗ ) are not all the same. When one of the three
equals zero, the other two must have different algebraic signs or both equal zero. The regions in the (U , V )-plane for fixed W which are realisable according to Lemmas 1 and 2 are illustrated in Fig. 3 for two W Ď values. We call the region with λc ≥ 0 or λc ≥ 0 (resp. λ∗c ≥ 0 ∗Ď
or λc ≥ 0) for W < 1 (resp. W > 1) the regular region (see single-hatched region in Fig. 3(i) and (ii)). All other points satisfying σc ≥ 0 make up the non-regular region. It can be checked that the Ď region defined by γ1 ≥ 0, γ2 > 0 with λc < 0 and λc < 0 is non-empty if and only if W ∈ (1/3, 1), as pointed out in [1] (see cross-hatched region in Fig. 3(i)). This region defines the part of the non-regular region which can be realised by the networks of Fig. 2. For W > 1, we obtain plots which are visually identical with curves ∗Ď generated by λ∗c , λc , γ1∗ and γ2∗ . For any fixed W < 1 we define Γ to be the region of the (U , V )-plane for which σc ≥ 0, λc < 0, λĎc < 0 and either γ1 < 0 or γ2 < 0 (see light shaded region in Fig. 3(i) and (ii)). The region Γ represents the positive-real biquadratics which cannot be realised using two reactive elements and three resistors. The following is the main theorem which implies that Γ is the nonrealisable region even if additional resistors are allowed. Theorem (Reichert). Any biquadratic which can be realised using two reactive elements and an arbitrary number of resistors can be realised by two reactive elements and three resistors. 3. Necessary conditions for realisability Following [1] we consider a one-port network with a welldefined impedance consisting of one inductor, one capacitor and an arbitrary number of resistors. The network can be arranged in the form shown in Fig. 4 where N is a three-port network containing all the resistors and vˆ 2 = −sLˆi2 , ˆi3 = −sC vˆ 3 . Under mild conditions
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J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
Fig. 4. Three-port resistive network N terminated with one inductor and one capacitor.
s+
Z (s) = R s+
hs11
s+
L ho11
s+
L
hs22
C ho22
C
2 h s + ( 12LC ) 2 h o + ( 12LC )
,
(7)
where we can compute directly that detX
hs11 =
X1 X3 − X52 X1
hs22 =
X1 X3 − X52
(8)
,
(9)
X1 X6 − X4 X5
hs12 =
X1 X3 − X52 X2 X3 − X62
ho11 =
X3 1
ho22 =
X3 X6
ho12 = R=
,
X3
,
(10)
,
(11)
,
(12)
,
(13)
X1 X3 − X52
.
(14)
X3 It can be verified that
ho11 − hs11
hs22 − ho22 = ho12 − hs12
2
.
(15)
(We remark incidentally that the parameters hs11 , ho11 , . . . etc. are the hybrid parameters of ports 2 and 3 in Fig. 4, namely Fig. 3. The regions in the (U , V )-plane which are realisable according to Lemmas 1 and 2 for (i) W = 0.6, (ii) W = 0.25.
(see Section 5, note 1), N has a non-negative definite impedance matrix given by
v1 X1 v 2 = X4 v3 X5
X4 X2 X6
X5 X6 X3
i1 i2 i3
i1
=: X i2 .
(6)
i3
=
i3
hs11 L C
hs12
=
X3 s2 +
1 C
+
X1 C
+
detX L
X2 X3 −X62 L
s+
s+
ho11
X1 X2 −X42 LC
,
X2 LC
where detX = X1 X2 X3 − X1 X62 − X2 X52 − X3 X42 + 2X4 X5 X6 . Following Auth [2] we write Z (s) in the form
h12 h22
i2
v3
,
√ =
L ho22 C
ho12 LC
W (U − x) ,
(16)
√ =
W (U + x) ,
(17)
2 = W 1 − U 2 + x2 ,
LC
X1 X3 − X52 s2 +
h11 −h12
where the superscripts o and s refer to the cases when the first port is open-circuited or short-circuited, respectively.) We now consider the biquadratic canonical form (2). If it is possible for (7) to equal (2), we see that there must exist x and y such that
hs22
Eliminating i2 , v2 , i3 and v3 we find that
vˆ 1 (s) Z ( s) = ˆi1 (s)
v2
(18)
V +y = √ ,
(19)
V −y = √ ,
(20)
W W
2 =
1 − V 2 + y2 W
.
(21)
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
The nonnegativity of (16)–(21) implies that |x| ≤ U , |y| ≤ V , x2 ≥ U 2 − 1 and y2 ≥ V 2 − 1. (This is the parametrisation of Auth [2], albeit with x and y interchanged.) We now turn to the conditions which must be imposed on X in order that it is realisable as a purely resistive network. A real symmetric matrix is defined to be paramount if each principal minor of the matrix is not less than the absolute value of any minor built from the same rows. Lemma 3 (Cederbaum [3]). A necessary condition for a matrix to be an impedance or admittance of a purely resistive n-port is that it is paramount. Among the inequalities implied by Lemma 3 we will highlight only four: X62 ≤ X22 , X62 ≤ X32 ,
2 (X1 X6 − X4 X5 )2 ≤ X1 X2 − X42 , 2 (X1 X6 − X4 X5 )2 ≤ X1 X3 − X52 . These reduce to the following inequalities after making use of (8)– (13): ho12 ho12
ho11 ho22
|
|≤
|
| ≤ 1,
+
2 ho12
,
Using (16)–(21) these inequalities become: W 1−V +y 1 − U 2 + x2 W
2
≤ LC ≤
≤ LC ≤
W
,
(22)
.
(23)
1 − V 2 + y2 1
W 1 − U 2 + x2
(24a)
y2 ≤ V 2 ,
(24b)
− UV ≤ xy ≤ 2 1 − V 2 x2 + 1 − U 2 y2
3W + W −1 2
− UV ,
(24c)
Kc = 4 (U − V )2 + 4UV 2 − W − W −1 + W − W −1
W + W −1 2
− UV xy −
Kc 4
= 0,
(24d)
where Kc is defined in (3). Proof. The bounds (24a), (24b) have already been noted as following from the positivity of (16), (17), (19) and (20). Combining the upper inequality of (22) and lower inequality of (23), we obtain 1 − U 2 + x2
1 − V 2 + y2 ≤ W 2 .
(25)
Substituting (16)–(21) into (15) gives 1 − U 2 + x2
1 − V 2 + y2 =
W + W −1 2
2
where the last step makes use of the left hand inequality of (27). Using the expression (26), we define f1 (x, y) = 1 − U 2 + x2
−
1 − V 2 + y2
W + W −1 2
2 − UV − xy , (26)
after some manipulation, which simplifies to (24d). Combining (25) with (26) gives (24c).
2 − UV − xy . 2
Then f1 (U , V ) = f1 (−U , −V ) = 1 − 41 W + W −1 − 4UV , which is nonnegative because (27) holds, and f1 (U , −V ) =
2
f1 (−U , V ) = 1 − W + W −1 /2 , which is negative. Since f1 (x, y) is a continuous function, it must have an odd number of intersection points with each edge of the rectangle defined by (24a) and (24b). Since f1 (x, y) is a quadratic function in both x and y, it intersects each edge only once. Substituting x = α U and y = −α V into the left hand side of (24d), we obtain the expression
Proof. It can be checked that the discriminant of (24d) is equal to Kc , and hence (24d) defines a hyperbola if and only if Kc > 0. It can be calculated that
x2 ≤ U 2 , W −1 − W
(27)
2 ≥ −4UV (W − 1)2 W −1 + W 2 − 1 W −2 > 0
Lemma 4. A necessary condition that (2) is realisable in the form of Fig. 4 is that there exists real x and y satisfying
+2
Lemma 5. Suppose
For the following and all subsequent lemmas we will impose the condition 0 < W < 1.
This section reworks and amplifies the main part of the proof of Reichert’s theorem — the essential idea is to show that there are no solutions (x, y) satisfying (24a)–(24d) for (U , V ) ∈ Γ . Lemmas 5–7 establish some facts about (24d) in relation to the bounds (24a) and (24b). Lemmas 8 and 9 consider the points (x, y) which satisfy (24c) and (24d) with strict inequality — which are intervals of the curve (24d). Lemma 8 shows that such intervals deform continuously apart from an exceptional point which is dealt with in Lemma 9. Lemma 10 introduces the set Ω of points (U , V ) inside Γ where (24a)–(24d) can be satisfied for some (x, y) and establishes a preliminary fact about it. The proof of theorem will eventually show that Ω is empty. Next, the ‘‘extreme values’’ (termed Grenzwerte by Reichert) are introduced. Lemmas 11–14 all relate to properties of the extreme values and in several respects follow arguments provided in [1]. The ‘‘proof of theorem’’ is then presented which provides the main topological argument in a systematic manner drawing on all the lemmas in the section. The section concludes with a graphical illustration showing a typical case of existence/non-existence of solution to (24a)–(24d).
holds. Then (24d) defines a hyperbola. Moreover, one branch of the hyperbola always intersects the upper and right hand edges of the rectangle defined by (24a) and (24b). The other branch always intersects the lower and left hand edges (see Fig. 5).
|hs12 | ≤ 1.
2
4. Proof of Reichert’s theorem
− 2 < W + W −1 − 4UV ≤ 2
2 |hs12 | ≤ hs11 hs22 + hs12 ,
1127
f2 (α) = U 2 + V 2 − UV W + W −1
α2 −
Kc 4
.
Since f2 (1) = f2 (−1) = f1 (U , −V ) ≤ 0 and f2 (0) = −Kc /4 < 0, it can be seen that the two branches of the hyperbola defined by (24d) do not intersect the line (x, y) = α (U , −V ) for |α| ≤ 1. This proves the assertion on the intersection of the hyperbola with the edges of the rectangle. Lemma 6. The curve UV = δ is always in the regular region when
δ ≥ (1 + W ) / (4W ).
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J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
Proof. If σc = 0, 2UV + 1 = W + W −1 /2. Substituting into (26), which is equivalent to (24d), we obtain
1 − U 2 + x2
1 − V 2 + y2 = (UV − xy + 1)2
which can be rearranged as
(x + y)2 = (U + V )2 + (Vx − Uy)2 .
(28)
From (24a) and (24b), it can be seen that the LHS of (28) is no greater than the first term on the RHS. It follows that (28) can only hold when (x, y) = ± (U , V ). Substituting into (25), which is equivalent to (24c) when (24d) holds, we obtain W 2 ≥ 1, which is contradictory to our assumption. Hence, (24a)–(24d) cannot all be satisfied on σc = 0. We now define the extreme values of (24c) and (24d) to be the real pairs (x, y) which satisfy (24d) and one of the equalities in (24c). The following lemmas establish properties of the extreme values which are needed in the proof of theorem.
Fig. 5. The hyperbola defined by (24d) and its intersection with the rectangle defined by (24a) and (24b) for W = 1/3 and U = 0.7, V = 0.6.
Proof. Consider the line V = α U where α ≥ 1. Then UV = δ implies U 2 ≥ (1 + W ) / (4W α). Therefore, on this line,
λĎc = 4U 2 (α − W ) + W − W −1 ≥ 1 − α −1 (1 + W ) ≥ 0. A similar argument shows that λc ≥ 0 if 0 ≤ α ≤ 1.
2
xν,µ
√ −αν + (−1)µ Dν , = 2 V2 − 1
yν,µ =
Lemma 7. Condition (27) holds for any positive-real bi-quadratic in the canonical form (2) which is non-regular. I.e. all points in the nonregular region satisfy (27). Proof. The right hand side of condition (27) is equivalent to σc ≥ 0, which is the condition The left hand side is for positive-realness. equivalent to UV < W + W −1 + 2 /4. It can be calculated that W + W −1 + 2 /4 > (1 + W ) / (4W ). The result then follows from Lemma 6 since all (U , V ) satisfying UV ≥ W + W −1 + 2 /4 are in the regular region.
Lemma 11. The extreme values of (24c) and (24d) consist of at most four pairs (x, y) which satisfy
Lemma 8. Suppose (27) holds and (U , V ) ̸= (1, 1). Let (x, y) satisfy (24c) and (24d) with strict inequality. Suppose (U1 , V1 ) → (U , V ). Then there exists (x1 , y1 ) so that (i) (24d) holds with U , V , x, y replaced by U1 , V1 , x1 , y1 , (ii) (x1 , y1 ) → (x, y) as (U1 , V1 ) → (U , V ) and (iii) (24c) holds with U , V , x, y replaced by U1 , V1 , x1 , y1 for (U1 , V1 ) sufficiently close to (U , V ). Proof. Suppose U ̸= 1. Then, for the given x, we can find y as one of the roots of the quadratic (24d) in y, which are distinct by Lemma 5. We now define y1 to be the same expression with (U , V ) replaced by (U1 , V1 ). Then with (x1 , y1 ) chosen to be (x, y1 ), all the required properties are satisfied. The case where V ̸= 1 follows similarly by solving (24d) for x in terms of y. Lemma 9. The point (1, 1) ∈ Γ if and only if W < 1/3. With U = V = 1 and W < 1/3, there are no real solutions x and y to (24c) and (24d). Proof. With U = V = 1 it can be checked that γ1 is a perfect Ď square (and hence is always nonnegative) and λc , λc and γ2 are each negative if and only if W < 1/3. This proves the first statement. For U = V = 1, (24d) reduces to xy = (W − 1)2 /4W , which does not satisfy the left hand inequality in (24c). Lemma 10. Let Ω be the set of (U , V ) inside Γ where (24a)–(24d) can be satisfied for some (x, y). Then Ω does not intersect the positivereal boundary (σc = 0).
βν xν,µ
(29)
,
(30)
where ν = 1, 2, µ = 1, 2, xν,µ ≥ 0, W + W −1
+ (−1)ν W − UV , αν = W 2 − U 2 − 1 V 2 − 1 − βν2 , Dν = αν2 − 4 U 2 − 1 V 2 − 1 βν2 ,
βν =
2
together with their negatives (−x, −y). An explicit expression for y2ν,µ is: y2ν,µ =
√ −αν − (−1)µ Dν . 2 U2 − 1
Proof. Satisfaction of (24c) with equality is equivalent to xy = βν for ν = 1, 2. Eliminating y or x from (24d) with equality sign gives V 2 − 1 x4 + αν x2 + U 2 − 1 βν2 = 0,
(31)
U 2 − 1 y4 + αν y2 + V 2 − 1 βν2 = 0,
(32)
which proves the result.
Lemma 12. For (U , V ) ∈ Γ , it holds that βν ̸= 0 for ν = 1, 2. Proof. The condition βν = 0 is satisfied on the curves κν = 0 where
κ1 = UV − κ2 = UV −
W −1 − W 2
,
W −1 + 3W 2
(ν = 1) , ,
(33)
(ν = 2) .
(34)
It can be calculated that W −1 − W 2
−
1+W 4W
=
(1 + W ) (1 − 2W ) W
.
Based on Lemma 6, when W ∈ (0, 1/2], κ1 = 0 is in the regular Ď region. Now suppose W ∈ (1/2, 1). First note that λc = 0 and λc =
0 intersect at P1 = (U1 , V1 ), where U1 = V1 =
√
√
1 + W/ 2 W .
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
1129
y-value and hence does not generate an extreme value. It remains only to consider the case of Dν = 0. For ν = 1, Dν = 0 is equivalent to Kc = 0 or γ1 = 0. For ν = 2, Dν = 0 is equivalent to Kc = 0 or κ4 = 0, where
2 κ4 = 4U 2 + 4V 2 − 4UV 5W + W −1 + 5W + W −1 − 4. It can be checked that the curve Kc = 0 lies inside the regular region. We first verify that κ4 = 0 lies inside the regular region. It can be calculated that κ4 = 0 crosses λc = 0 only at one point, Ď P8 = (U6 , V6 ), and crosses λc = 0 only at one point, P9 = (V6 , U6 ), where
V6 =
√
2 + 14W 2 + 2 50W 4 + 12W 2 + 2 4W
.
Using the expression for V1 from the proof of Lemma 12, it can be checked that V62 − V12 =
50W 4 + 12W 2 + 2 − 2W + 5W 2 + 1 / 8W 2
> 0. Fig. 6. The curves κ1 = 0, κ3 = 0 and κ4 = 0 in the (U , V ) plane with W = 0.6. Ď
Also the curve γ1 = 0 crosses λc = 0 and λc √ = 0 at the points P2 = (U2 , V2 ) and P3 = (V2 , U2 ) where U2 = 2 − 2W 2 / (4W ). Ď Further, the curve κ1 = 0 crosses λc = 0 and λ√ c = 0 at the points P4 = (U3 , V3 ) and P5 = (V3 , U3 ), where U3 = 1 − W 2 / (2W ). It can be easily checked that U1 > U3 > U2 when W > 1/2. Hence P4 and P5 are always on the regular boundary curve between P1 , P2 and P1 , P3 respectively (see Fig. 6). Since the curve κ1 = 0 only has Ď one intersection with λc = 0 and λc = 0, and does not intersect with γ1 = 0 when W ∈ (1/3, 1], it is always outside of Γ for W > 1/2. The curve κ2 = 0 can be easily checked to be outside of Γ by Lemma 6. Lemma 13. For (U , V ) ∈ Γ , there are no extreme values which satisfy x2ν,µ = U 2 or y2ν,µ = V 2 . Ď
Proof. When ν = 1 the condition x2ν,µ = U 2 implies that λc = 0. When ν = 2 the condition implies that κ3 = 0, where
κ3 = V −
W 1 − W2 1 + 3W 2
U+
W −1 + 3W 1 4
U
.
It can be calculated that κ3 = 0 crosses λc = 0 only at one point, Ď P6 = (U4 , V4 ), and crosses λc = 0 only at one point, P7 = (U5 , V5 ), where
(1 + W ) W W 2 + 1 (W + 1) V4 = , 2W W 2 + 1 √ and U5 = 3W 2 + 1/ (2W ). Using the expression for V1 from the proof of Lemma 12, it can be checked that V42 − V12 = 2 (W + 1) / 2 W + 1 > 0. It is also clear that U5 > U1 . Hence κ3 = 0 can never cross the outside boundary of the whole regular region (see Fig. 6). The condition y2ν,µ = V 2 is analogous. Therefore,
the conditions on the extreme values cannot occur in the nonregular region, and hence not in Γ . Lemma 14. For (U , V ) ∈ Γ , there are no coincident extreme values which satisfy (24a) and (24b). Proof. The quadratic (31) in x2 has discriminant equal to Dν . The only possibility for repeated x-values occurs with Dν = 0 or x = 0 being a repeated root. From Lemma 12 the latter implies an infinite
Hence κ4 = 0 can never cross the outside boundary of the whole regular region (see Fig. 6). It remains only to consider the case of γ1 = 0. It can be checked that γ1 = 0 lies inside the regular region (see [4]) and hence outside of Γ when W ∈ (0, 1/3]. We consider the case where 1/3 < W < 1. For this range, the curve γ1 = 0 lies completely inside the region |U | ≤ 1, |V | ≤ 1 and touchesthe line U = 1 at
√
the point PB = 1, W −1 − 3W /2 . For W ∈ 1/ 3, 1 , PB lies
outside of Γ in the 4th quadrant. We now consider the segment of γ1 = 0 between the axis V = 0 and the point P3 (see Fig. 6). On the curve γ1 = 0, there is a pair of coincident extreme values given by
−α1
,
(35)
.
(36)
x21,µ = x21 =
2 V2 − 1
y21,µ = y21 =
2 U2 − 1
−α1
From Lemma 13 we note that y21 ̸= V 2 on γ1 = 0 in Γ . From Lemma 12 we note that y21 = 0 can only occur when x21 = ∞ which is impossible on this segment of γ1 = 0 because of (35). Now consider the point P0 , which is the intersection of γ1 = 0 with V = 0. At P0 we have U = y21 =
W −1 − W 3W − W −1
W −1 − W
9W − W −1 /2 and
.
√ 1/ 3, 1 , y21 < 0 for √ √ W ∈ 1/3, 1/ 3 and y21 = ∞ when W = 1/ 3. Since y1 varies continuously on γ1 = 0 (apart from the possible singularity at PB )
We see that y21 > V 2 = 0 for W ∈
it is therefore impossible for y1 to satisfy (24b) on this segment of γ1 = 0. The part of γ1 = 0 between the axis U = 0 and P2 follows by a dual argument. Proof of theorem. Let W < 1. Networks containing two reactive elements of the same kind can only realise (U , V ) with Kc < 0 [4]. Hence we only need to consider networks which take the form of Fig. 4. Suppose then that there is a region Ω inside Γ where (24a)–(24d) are satisfied. Consider a line ℓ within Γ which connects a point in the positive-real boundary σc = 0 with a point in Ω (see Fig. 7). By Lemma 9, (1, 1) ̸∈ Ω , so we can choose ℓ so that it does not pass through the point (1, 1). As (U , V ) moves
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J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
Fig. 7. Illustration of proof of theorem. The assumption of a region Ω inside Γ where (24a)–(24d) are satisfied is shown to lead to a contradiction.
along ℓ, we can find solutions (x, y) to (24d) at every point which also satisfy (24a) and (24b) by Lemmas 5 and 7. Inside Ω some of these solutions will also satisfy (24c) whereas none of them do outside of Ω . By Lemma 10 there are points on ℓ which are outside of Ω . Now for each fixed (U , V ) the set of (x, y) which satisfy (24c) and (24d) consists of intervals of the hyperbola (24d) which are bounded by extreme values. Lemma 8 shows that these intervals deform continuously with (U , V ). Lemma 14 shows that such intervals cannot vanish by virtue of extreme values becoming coincident. Hence, as (U , V ) moves along ℓ and traverses δ Ω , an extreme value must cross x2ν,µ = U 2 or y2ν,µ = V 2 . Lemma 13 shows that this cannot happen inside Γ . Hence there is no region Ω inside Γ where (24a)–(24d) are satisfied. The result now follows from Lemma 4. The case of W > 1 follows in general by duality, or directly by using Lemma 4 in dual form with W replaced by W −1 in (24c). Since Zc is regular when W = 1 there is nothing to prove in this case. To conclude the section we present a graphical illustration in Fig. 8 of one typical case where solutions to (24a)–(24d) are lost by virtue of extreme values becoming coincident, as (U , V ) crosses the boundary of Γ . For other boundary points of Γ , solutions may be lost due to extreme values crossing the boundary x2ν,µ = U 2 or y2ν,µ = V 2 . The theorem proves that such behaviour cannot occur inside Γ . 5. Notes We present some remarks to compare our paper with [1]. 1. Section 3 provides a simple derivation of the parametrisation of Auth [2], from which the system of equations (Lemma 4) is derived on which the proof of [1] is built. As in [1,2], the mild assumption of the existence of an impedance matrix (6) is made. In general, matrices A and B exist for which A(v1 , v2 , v3 )| = B(i1 , i2 , i3 )| with A + B non-singular, which implies the existence of the scattering matrix of N [12, Chapter 3] and also that any singular cases of the impedance matrix will be limiting cases of (6) in which some elements tend to ∞. These limiting cases include impedances (1) in which some coefficients may be zero, after appropriate scaling, which are
=
=
Fig. 8. The curves which satisfy (24a)–(24c) with equality and (24d) for W = 0.6 and the (U , V ) values marked in Fig. 7 ((i) U = V = 0.71, (ii) U = V = 0.65). In (i) the (x, y) values satisfying (24a)–(24d) lie on the solid curve between the extreme values A and B (A′ and B′ ), while no such (x, y) exist in (ii).
trivial cases that are ignored here without loss of generality (see Section 2). 2. The key part of Reichert’s original proof is the claim that, in the notation of the present paper, (24a)–(24d) have no solution in Γ . In [1, p.206, col.1, l.3–14] it is stated that this reduces to four cases (Fall 1–4) involving the ‘‘extreme values’’. The underlying topological argument is not provided. The ‘‘proof of theorem’’ above provides this argument which makes use of the lemmas developed. It is evident that some further cases need to be analysed which are not treated in [1]. 3. One possibility which lies outside Reichert’s ‘‘Fall 1–4’’ is the possibility that (24d) itself loses contact with the rectangle defined by (24a) and (24b). This is dealt with by Lemma 5. 4. There is a further logical possibility that falls outside of Reichert’s ‘‘Fall 1–4’’. This is the possibility that the intervals of (24d) between extreme values which satisfy (24c) could ‘‘flip’’ discontinuously even when the extreme values themselves are
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
continuous. We show that this cannot happen when (U , V ) ̸= (1, 1) in Lemma 8. On the other hand this behaviour does occur when U = V = 1. We deal with this case separately in Lemma 9. 5. Lemmas 12–14 are a reworking of several cases related to Reichert’s ‘‘Fall 1–4’’ in a form which can be explicitly used in the ‘‘proof of theorem’’. 6. Concluding remarks The statement of Reichert’s theorem is remarkable for its simplicity. This contrasts with Reichert’s proof which is far from simple. The purpose of this paper has been to provide a complete reworking of the proof which highlights all the main steps and to expand the reasoning for some parts of the topological argument. Acknowledgement This work was supported by the Engineering and Physical Sciences Research Council grant number EP/G066477/1.
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References [1] M. Reichert, Die kanonisch und übertragerfrei realisierbaren Zweipolfunktionen zweiten Grades (Transformerless and canonic realisation of biquadratic immittance functions), Arch. Elek. Übertragung 23 (1969) 201–208. [2] L.V. Auth, RLC biquadratic driving-point synthesis using the resistive threeport, IEEE Trans. on Circuit Theory (1964) 82–88. [3] I. Cederbaum, Conditions for the impedance and admittance matrices of nports without ideal transformers, Proc. IEE 105 (1958) 245–251. [4] J.Z. Jiang, M.C. Smith, Regular positive-real functions and five-element network synthesis for electrical and mechanical networks, IEEE Trans. Automat. Control 56 (6) (2011) 1275–1290. [5] R.M. Foster, E.L. Ladenheim, A class of biquadratic impedances, IEEE Trans. on Circuit Theory (1963) 262–265. [6] M.C. Smith, Synthesis of mechanical networks: the inerter, IEEE Trans. Automat. Control 47 (10) (2002) 1648–1662. [7] R.M. Foster, Academic and theoretical aspects of circuit theory, Proc. IRE 50 (1962) 866–871. [8] E.A. Guillemin, Synthesis of Passive Networks, John Wiley & Sons, 1957. [9] M.E.V. Valkenburg, Introduction to Modern Network Synthesis, John Wiley & Sons, 1960. [10] B.D.O. Anderson, S. Vongpanitlerd, Network Analysis and Synthesis, PrenticeHall, Upper Saddle River, NJ, 1973. [11] M.Z.Q. Chen, M.C. Smith, A note on tests for positive-real functions, IEEE Trans. Automat. Control 54 (2) (2009) 390–393. [12] R.W. Newcomb, Linear Multiport Synthesis, McGrawHill, New York, 1966.
Contents lists available at SciVerse ScienceDirect
Systems & Control Letters journal homepage: www.elsevier.com/locate/sysconle
On the theorem of Reichert Jason Zheng Jiang, Malcolm C. Smith ∗ Department of Engineering, University of Cambridge, Cambridge, CB2 1PZ, UK
article
info
Article history: Received 19 March 2012 Received in revised form 14 September 2012 Accepted 19 September 2012 Available online 9 November 2012
abstract This paper reworks and amplifies Reichert’s proof of his theorem (1969) which asserts that any impedance function of a one-port electrical network which can be realised with two reactive elements and an arbitrary number of resistors can be realised with two reactive elements and three resistors. © 2012 Elsevier B.V. All rights reserved.
Keywords: Electrical networks Circuit synthesis Positive-real functions
1. Introduction Reichert’s theorem [1] is one of the fundamental results at the boundary of circuit theory and systems theory. The theorem completely characterises all second-order transferfunctions which may be realised by one-port RLC-circuits with at most two energy-storage (reactive) elements. It is shown that no transfer-function requires more than three resistors for its realisation and that eight such circuits suffice to realise every transfer function in the class. A consequence of the result is that not all second-order positive-real transfer-functions (biquadratics) are realisable with just two energy-storage elements. Stated more broadly, non-minimality in the sense of modern systems theory is necessary for RLC-realisations of (some) positive-real functions. To date no proof of the theorem other than Reichert’s original German language publication [1] has appeared. Reichert’s proof makes use of a characterisation of Auth [2] of realisable biquadratic impedances, which itself relies on Cederbaum’s necessary condition for the realisation of a purely resistive nport [3]. The crucial part of Reichert’s proof is a topological argument that a certain system of polynomial equations has no solution in some region of its variable parameters. It is our view that some parts of this argument need to be expanded. It is the purpose of the present paper to provide a complete reworking of Reichert’s proof, including the parametrisation of Auth, and to provide new lemmas to clarify the main topological argument. The class of transfer-functions realisable by two reactive elements and three resistors was illuminated in [4] by the
∗
Corresponding author. Tel.: +44 1223 3 32745; fax: +44 1223 3 32662. E-mail addresses: [email protected], [email protected] (J.Z. Jiang), [email protected], [email protected] (M.C. Smith). 0167-6911/$ – see front matter © 2012 Elsevier B.V. All rights reserved. doi:10.1016/j.sysconle.2012.09.009
concept of a regular positive-real function. It was shown that series–parallel networks of this type can only realise regular biquadratics and all regular biquadratics can be realised in this way. The only networks capable of realising non-regular biquadratics are a pair of bridge networks studied by Foster and Ladenheim [5]. Reichert’s theorem shows that the class of realisable transfer-functions is expanded no further by the use of additional resistors. Aside from the fundamental nature of the questions involved, renewed interest in the subject of network synthesis has been provided recently by mechanical network problems making use of the inerter device [6]. Such problems highlight the importance of gaining a complete understanding of the efficient realisation of (low-order) impedance functions, e.g. biquadratics. For further general background on the subject of classical network synthesis, the reader is referred to [7–9]. In the next section we provide the background and notation needed for our version of Reichert’s proof. 2. Preliminaries We will restrict our attention to RLC one-port networks, i.e. networks with two external terminals and comprising resistors, inductors and capacitors but excluding transformers (see Fig. 1). The energy storage (reactive) elements are the inductor and capacitor. For the analogous two-terminal mechanical elements the reactive elements are the spring and inerter [6]. For a network as in Fig. 1 the driving-point impedance Z (s) is defined by Z (s) = vˆ (s) /ˆi (s) where ˆ denotes Laplace transformation and Y (s) = Z −1 (s) is the driving-point admittance. The immittance of a network is the impedance or admittance. We recall that a rational function Z (s) is defined to be positive-real if Z (s) is analytic and
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
′
Fig. 1. One-port electrical network with two external terminals 1 and 1 , terminal voltage v and terminal current i.
Re (Z (s)) ≥ 0 for Re (s) > 0 [9]. It is known that the driving-point immittance of a linear network is positive-real if and only if the network is passive [10]. Following [4] we define a positive-real function Z (s) to be regular if the smallest value of Re (Z (jω)) or Re Z −1 (jω) occurs at ω = 0 or ω = ∞. In this paper, we are concerned with the realisation of biquadratics Z (s) =
As2 + Bs + C Ds2 + Es + F
,
(1)
where A, B, C , D, E , F ≥ 0. It is well known [5,9,11] that Z (s) is positive real if and only if σ = BE −
√
√
AF −
CD
2
≥ 0. In [4]
it was shown that a positive-real biquadratic impedance (1) with any of the parameters A, B, C , D, E and F equal to zero is regular. Moreover, any such impedance can be realised by a series–parallel network with at most two reactive elements and two resistors. Accordingly we restrict attention to the case that A, B, C , D, E , F > 0. Using impedance and frequency scaling [4, Section III], Z (s) can be reduced to a canonical form Zc (s) =
s2 + 2U
√ Ws + W
√
s2 + 2V / W s + 1/W
,
(U , V , W > 0) .
(2)
See [4, Section V] and [1]. Any circuit that realises Zc (s) will also realise the corresponding Z (s) after appropriate scaling of the elements [4]. It can be checked that Zc (s) is positive real if and only if
σc = 4UV + 2 −
1 W
+W
≥ 0.
The resultant of the numerator and denominator of Zc (s) is 2
2
Kc = 4U + 4V − 4UV
1 W
+W
+
1 W
2 −W
.
(3)
For any rational function ρ (U , V , W ) we make use of the notations ρ ∗ (U , V , W ) = ρ U , V , W −1 and ρ Ď (U , V , W ) =
ρ (V , U , W ). We observe that σc∗ = σcĎ = σc and Kc∗ = KcĎ = Kc . Let
λc = 4UV − 4V 2 W −
1 W
−W .
Lemma 1 ([4]). The biquadratic Zc is regular if and only if at least one of the following five conditions is satisfied:
< 1 and λc ≥ 0, < 1 and λĎc ≥ 0, = 1, > 1 and λ∗c ≥ 0, ∗Ď Case 5. W > 1 and λc ≥ 0.
Case 1. Case 2. Case 3. Case 4.
W W W W
In [4] it was shown that any regular biquadratic could be realised by a series–parallel network with (at most) two reactive
1125
Fig. 2. The two-reactive five-element bridge networks that can realise non-regular impedances.
elements and three resistors. Six network structures suffice to realise the whole class. Among all the bridge networks with two reactive elements and three resistors only the pair shown in Fig. 2 can realise non-regular immittances. We now present the realisation condition for this pair of networks for the canonical form (2) as given in [4]. Let
1 γ1 = 4U 2 + 4V 2 + 4UV 3W − W 1 1 + −W − 9W ,
(4)
1 γ2 = −4U 2 − 4V 2 + 4UV W + W 1 1 − −W − 3W .
(5)
W
W
W
W
Lemma 2 ([5]). Zc (s) in the form of (2) can be realised by the network shown in Fig. 2(a) and (b) with positive element values if and only if W < 1, γ1 ≥ 0 (W > 1, γ1∗ ≥ 0) and the signs of
λc , λĎc , γ2 (λ∗c , λ∗c Ď , γ2∗ ) are not all the same. When one of the three
equals zero, the other two must have different algebraic signs or both equal zero. The regions in the (U , V )-plane for fixed W which are realisable according to Lemmas 1 and 2 are illustrated in Fig. 3 for two W Ď values. We call the region with λc ≥ 0 or λc ≥ 0 (resp. λ∗c ≥ 0 ∗Ď
or λc ≥ 0) for W < 1 (resp. W > 1) the regular region (see single-hatched region in Fig. 3(i) and (ii)). All other points satisfying σc ≥ 0 make up the non-regular region. It can be checked that the Ď region defined by γ1 ≥ 0, γ2 > 0 with λc < 0 and λc < 0 is non-empty if and only if W ∈ (1/3, 1), as pointed out in [1] (see cross-hatched region in Fig. 3(i)). This region defines the part of the non-regular region which can be realised by the networks of Fig. 2. For W > 1, we obtain plots which are visually identical with curves ∗Ď generated by λ∗c , λc , γ1∗ and γ2∗ . For any fixed W < 1 we define Γ to be the region of the (U , V )-plane for which σc ≥ 0, λc < 0, λĎc < 0 and either γ1 < 0 or γ2 < 0 (see light shaded region in Fig. 3(i) and (ii)). The region Γ represents the positive-real biquadratics which cannot be realised using two reactive elements and three resistors. The following is the main theorem which implies that Γ is the nonrealisable region even if additional resistors are allowed. Theorem (Reichert). Any biquadratic which can be realised using two reactive elements and an arbitrary number of resistors can be realised by two reactive elements and three resistors. 3. Necessary conditions for realisability Following [1] we consider a one-port network with a welldefined impedance consisting of one inductor, one capacitor and an arbitrary number of resistors. The network can be arranged in the form shown in Fig. 4 where N is a three-port network containing all the resistors and vˆ 2 = −sLˆi2 , ˆi3 = −sC vˆ 3 . Under mild conditions
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J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
Fig. 4. Three-port resistive network N terminated with one inductor and one capacitor.
s+
Z (s) = R s+
hs11
s+
L ho11
s+
L
hs22
C ho22
C
2 h s + ( 12LC ) 2 h o + ( 12LC )
,
(7)
where we can compute directly that detX
hs11 =
X1 X3 − X52 X1
hs22 =
X1 X3 − X52
(8)
,
(9)
X1 X6 − X4 X5
hs12 =
X1 X3 − X52 X2 X3 − X62
ho11 =
X3 1
ho22 =
X3 X6
ho12 = R=
,
X3
,
(10)
,
(11)
,
(12)
,
(13)
X1 X3 − X52
.
(14)
X3 It can be verified that
ho11 − hs11
hs22 − ho22 = ho12 − hs12
2
.
(15)
(We remark incidentally that the parameters hs11 , ho11 , . . . etc. are the hybrid parameters of ports 2 and 3 in Fig. 4, namely Fig. 3. The regions in the (U , V )-plane which are realisable according to Lemmas 1 and 2 for (i) W = 0.6, (ii) W = 0.25.
(see Section 5, note 1), N has a non-negative definite impedance matrix given by
v1 X1 v 2 = X4 v3 X5
X4 X2 X6
X5 X6 X3
i1 i2 i3
i1
=: X i2 .
(6)
i3
=
i3
hs11 L C
hs12
=
X3 s2 +
1 C
+
X1 C
+
detX L
X2 X3 −X62 L
s+
s+
ho11
X1 X2 −X42 LC
,
X2 LC
where detX = X1 X2 X3 − X1 X62 − X2 X52 − X3 X42 + 2X4 X5 X6 . Following Auth [2] we write Z (s) in the form
h12 h22
i2
v3
,
√ =
L ho22 C
ho12 LC
W (U − x) ,
(16)
√ =
W (U + x) ,
(17)
2 = W 1 − U 2 + x2 ,
LC
X1 X3 − X52 s2 +
h11 −h12
where the superscripts o and s refer to the cases when the first port is open-circuited or short-circuited, respectively.) We now consider the biquadratic canonical form (2). If it is possible for (7) to equal (2), we see that there must exist x and y such that
hs22
Eliminating i2 , v2 , i3 and v3 we find that
vˆ 1 (s) Z ( s) = ˆi1 (s)
v2
(18)
V +y = √ ,
(19)
V −y = √ ,
(20)
W W
2 =
1 − V 2 + y2 W
.
(21)
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
The nonnegativity of (16)–(21) implies that |x| ≤ U , |y| ≤ V , x2 ≥ U 2 − 1 and y2 ≥ V 2 − 1. (This is the parametrisation of Auth [2], albeit with x and y interchanged.) We now turn to the conditions which must be imposed on X in order that it is realisable as a purely resistive network. A real symmetric matrix is defined to be paramount if each principal minor of the matrix is not less than the absolute value of any minor built from the same rows. Lemma 3 (Cederbaum [3]). A necessary condition for a matrix to be an impedance or admittance of a purely resistive n-port is that it is paramount. Among the inequalities implied by Lemma 3 we will highlight only four: X62 ≤ X22 , X62 ≤ X32 ,
2 (X1 X6 − X4 X5 )2 ≤ X1 X2 − X42 , 2 (X1 X6 − X4 X5 )2 ≤ X1 X3 − X52 . These reduce to the following inequalities after making use of (8)– (13): ho12 ho12
ho11 ho22
|
|≤
|
| ≤ 1,
+
2 ho12
,
Using (16)–(21) these inequalities become: W 1−V +y 1 − U 2 + x2 W
2
≤ LC ≤
≤ LC ≤
W
,
(22)
.
(23)
1 − V 2 + y2 1
W 1 − U 2 + x2
(24a)
y2 ≤ V 2 ,
(24b)
− UV ≤ xy ≤ 2 1 − V 2 x2 + 1 − U 2 y2
3W + W −1 2
− UV ,
(24c)
Kc = 4 (U − V )2 + 4UV 2 − W − W −1 + W − W −1
W + W −1 2
− UV xy −
Kc 4
= 0,
(24d)
where Kc is defined in (3). Proof. The bounds (24a), (24b) have already been noted as following from the positivity of (16), (17), (19) and (20). Combining the upper inequality of (22) and lower inequality of (23), we obtain 1 − U 2 + x2
1 − V 2 + y2 ≤ W 2 .
(25)
Substituting (16)–(21) into (15) gives 1 − U 2 + x2
1 − V 2 + y2 =
W + W −1 2
2
where the last step makes use of the left hand inequality of (27). Using the expression (26), we define f1 (x, y) = 1 − U 2 + x2
−
1 − V 2 + y2
W + W −1 2
2 − UV − xy , (26)
after some manipulation, which simplifies to (24d). Combining (25) with (26) gives (24c).
2 − UV − xy . 2
Then f1 (U , V ) = f1 (−U , −V ) = 1 − 41 W + W −1 − 4UV , which is nonnegative because (27) holds, and f1 (U , −V ) =
2
f1 (−U , V ) = 1 − W + W −1 /2 , which is negative. Since f1 (x, y) is a continuous function, it must have an odd number of intersection points with each edge of the rectangle defined by (24a) and (24b). Since f1 (x, y) is a quadratic function in both x and y, it intersects each edge only once. Substituting x = α U and y = −α V into the left hand side of (24d), we obtain the expression
Proof. It can be checked that the discriminant of (24d) is equal to Kc , and hence (24d) defines a hyperbola if and only if Kc > 0. It can be calculated that
x2 ≤ U 2 , W −1 − W
(27)
2 ≥ −4UV (W − 1)2 W −1 + W 2 − 1 W −2 > 0
Lemma 4. A necessary condition that (2) is realisable in the form of Fig. 4 is that there exists real x and y satisfying
+2
Lemma 5. Suppose
For the following and all subsequent lemmas we will impose the condition 0 < W < 1.
This section reworks and amplifies the main part of the proof of Reichert’s theorem — the essential idea is to show that there are no solutions (x, y) satisfying (24a)–(24d) for (U , V ) ∈ Γ . Lemmas 5–7 establish some facts about (24d) in relation to the bounds (24a) and (24b). Lemmas 8 and 9 consider the points (x, y) which satisfy (24c) and (24d) with strict inequality — which are intervals of the curve (24d). Lemma 8 shows that such intervals deform continuously apart from an exceptional point which is dealt with in Lemma 9. Lemma 10 introduces the set Ω of points (U , V ) inside Γ where (24a)–(24d) can be satisfied for some (x, y) and establishes a preliminary fact about it. The proof of theorem will eventually show that Ω is empty. Next, the ‘‘extreme values’’ (termed Grenzwerte by Reichert) are introduced. Lemmas 11–14 all relate to properties of the extreme values and in several respects follow arguments provided in [1]. The ‘‘proof of theorem’’ is then presented which provides the main topological argument in a systematic manner drawing on all the lemmas in the section. The section concludes with a graphical illustration showing a typical case of existence/non-existence of solution to (24a)–(24d).
holds. Then (24d) defines a hyperbola. Moreover, one branch of the hyperbola always intersects the upper and right hand edges of the rectangle defined by (24a) and (24b). The other branch always intersects the lower and left hand edges (see Fig. 5).
|hs12 | ≤ 1.
2
4. Proof of Reichert’s theorem
− 2 < W + W −1 − 4UV ≤ 2
2 |hs12 | ≤ hs11 hs22 + hs12 ,
1127
f2 (α) = U 2 + V 2 − UV W + W −1
α2 −
Kc 4
.
Since f2 (1) = f2 (−1) = f1 (U , −V ) ≤ 0 and f2 (0) = −Kc /4 < 0, it can be seen that the two branches of the hyperbola defined by (24d) do not intersect the line (x, y) = α (U , −V ) for |α| ≤ 1. This proves the assertion on the intersection of the hyperbola with the edges of the rectangle. Lemma 6. The curve UV = δ is always in the regular region when
δ ≥ (1 + W ) / (4W ).
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J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
Proof. If σc = 0, 2UV + 1 = W + W −1 /2. Substituting into (26), which is equivalent to (24d), we obtain
1 − U 2 + x2
1 − V 2 + y2 = (UV − xy + 1)2
which can be rearranged as
(x + y)2 = (U + V )2 + (Vx − Uy)2 .
(28)
From (24a) and (24b), it can be seen that the LHS of (28) is no greater than the first term on the RHS. It follows that (28) can only hold when (x, y) = ± (U , V ). Substituting into (25), which is equivalent to (24c) when (24d) holds, we obtain W 2 ≥ 1, which is contradictory to our assumption. Hence, (24a)–(24d) cannot all be satisfied on σc = 0. We now define the extreme values of (24c) and (24d) to be the real pairs (x, y) which satisfy (24d) and one of the equalities in (24c). The following lemmas establish properties of the extreme values which are needed in the proof of theorem.
Fig. 5. The hyperbola defined by (24d) and its intersection with the rectangle defined by (24a) and (24b) for W = 1/3 and U = 0.7, V = 0.6.
Proof. Consider the line V = α U where α ≥ 1. Then UV = δ implies U 2 ≥ (1 + W ) / (4W α). Therefore, on this line,
λĎc = 4U 2 (α − W ) + W − W −1 ≥ 1 − α −1 (1 + W ) ≥ 0. A similar argument shows that λc ≥ 0 if 0 ≤ α ≤ 1.
2
xν,µ
√ −αν + (−1)µ Dν , = 2 V2 − 1
yν,µ =
Lemma 7. Condition (27) holds for any positive-real bi-quadratic in the canonical form (2) which is non-regular. I.e. all points in the nonregular region satisfy (27). Proof. The right hand side of condition (27) is equivalent to σc ≥ 0, which is the condition The left hand side is for positive-realness. equivalent to UV < W + W −1 + 2 /4. It can be calculated that W + W −1 + 2 /4 > (1 + W ) / (4W ). The result then follows from Lemma 6 since all (U , V ) satisfying UV ≥ W + W −1 + 2 /4 are in the regular region.
Lemma 11. The extreme values of (24c) and (24d) consist of at most four pairs (x, y) which satisfy
Lemma 8. Suppose (27) holds and (U , V ) ̸= (1, 1). Let (x, y) satisfy (24c) and (24d) with strict inequality. Suppose (U1 , V1 ) → (U , V ). Then there exists (x1 , y1 ) so that (i) (24d) holds with U , V , x, y replaced by U1 , V1 , x1 , y1 , (ii) (x1 , y1 ) → (x, y) as (U1 , V1 ) → (U , V ) and (iii) (24c) holds with U , V , x, y replaced by U1 , V1 , x1 , y1 for (U1 , V1 ) sufficiently close to (U , V ). Proof. Suppose U ̸= 1. Then, for the given x, we can find y as one of the roots of the quadratic (24d) in y, which are distinct by Lemma 5. We now define y1 to be the same expression with (U , V ) replaced by (U1 , V1 ). Then with (x1 , y1 ) chosen to be (x, y1 ), all the required properties are satisfied. The case where V ̸= 1 follows similarly by solving (24d) for x in terms of y. Lemma 9. The point (1, 1) ∈ Γ if and only if W < 1/3. With U = V = 1 and W < 1/3, there are no real solutions x and y to (24c) and (24d). Proof. With U = V = 1 it can be checked that γ1 is a perfect Ď square (and hence is always nonnegative) and λc , λc and γ2 are each negative if and only if W < 1/3. This proves the first statement. For U = V = 1, (24d) reduces to xy = (W − 1)2 /4W , which does not satisfy the left hand inequality in (24c). Lemma 10. Let Ω be the set of (U , V ) inside Γ where (24a)–(24d) can be satisfied for some (x, y). Then Ω does not intersect the positivereal boundary (σc = 0).
βν xν,µ
(29)
,
(30)
where ν = 1, 2, µ = 1, 2, xν,µ ≥ 0, W + W −1
+ (−1)ν W − UV , αν = W 2 − U 2 − 1 V 2 − 1 − βν2 , Dν = αν2 − 4 U 2 − 1 V 2 − 1 βν2 ,
βν =
2
together with their negatives (−x, −y). An explicit expression for y2ν,µ is: y2ν,µ =
√ −αν − (−1)µ Dν . 2 U2 − 1
Proof. Satisfaction of (24c) with equality is equivalent to xy = βν for ν = 1, 2. Eliminating y or x from (24d) with equality sign gives V 2 − 1 x4 + αν x2 + U 2 − 1 βν2 = 0,
(31)
U 2 − 1 y4 + αν y2 + V 2 − 1 βν2 = 0,
(32)
which proves the result.
Lemma 12. For (U , V ) ∈ Γ , it holds that βν ̸= 0 for ν = 1, 2. Proof. The condition βν = 0 is satisfied on the curves κν = 0 where
κ1 = UV − κ2 = UV −
W −1 − W 2
,
W −1 + 3W 2
(ν = 1) , ,
(33)
(ν = 2) .
(34)
It can be calculated that W −1 − W 2
−
1+W 4W
=
(1 + W ) (1 − 2W ) W
.
Based on Lemma 6, when W ∈ (0, 1/2], κ1 = 0 is in the regular Ď region. Now suppose W ∈ (1/2, 1). First note that λc = 0 and λc =
0 intersect at P1 = (U1 , V1 ), where U1 = V1 =
√
√
1 + W/ 2 W .
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
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y-value and hence does not generate an extreme value. It remains only to consider the case of Dν = 0. For ν = 1, Dν = 0 is equivalent to Kc = 0 or γ1 = 0. For ν = 2, Dν = 0 is equivalent to Kc = 0 or κ4 = 0, where
2 κ4 = 4U 2 + 4V 2 − 4UV 5W + W −1 + 5W + W −1 − 4. It can be checked that the curve Kc = 0 lies inside the regular region. We first verify that κ4 = 0 lies inside the regular region. It can be calculated that κ4 = 0 crosses λc = 0 only at one point, Ď P8 = (U6 , V6 ), and crosses λc = 0 only at one point, P9 = (V6 , U6 ), where
V6 =
√
2 + 14W 2 + 2 50W 4 + 12W 2 + 2 4W
.
Using the expression for V1 from the proof of Lemma 12, it can be checked that V62 − V12 =
50W 4 + 12W 2 + 2 − 2W + 5W 2 + 1 / 8W 2
> 0. Fig. 6. The curves κ1 = 0, κ3 = 0 and κ4 = 0 in the (U , V ) plane with W = 0.6. Ď
Also the curve γ1 = 0 crosses λc = 0 and λc √ = 0 at the points P2 = (U2 , V2 ) and P3 = (V2 , U2 ) where U2 = 2 − 2W 2 / (4W ). Ď Further, the curve κ1 = 0 crosses λc = 0 and λ√ c = 0 at the points P4 = (U3 , V3 ) and P5 = (V3 , U3 ), where U3 = 1 − W 2 / (2W ). It can be easily checked that U1 > U3 > U2 when W > 1/2. Hence P4 and P5 are always on the regular boundary curve between P1 , P2 and P1 , P3 respectively (see Fig. 6). Since the curve κ1 = 0 only has Ď one intersection with λc = 0 and λc = 0, and does not intersect with γ1 = 0 when W ∈ (1/3, 1], it is always outside of Γ for W > 1/2. The curve κ2 = 0 can be easily checked to be outside of Γ by Lemma 6. Lemma 13. For (U , V ) ∈ Γ , there are no extreme values which satisfy x2ν,µ = U 2 or y2ν,µ = V 2 . Ď
Proof. When ν = 1 the condition x2ν,µ = U 2 implies that λc = 0. When ν = 2 the condition implies that κ3 = 0, where
κ3 = V −
W 1 − W2 1 + 3W 2
U+
W −1 + 3W 1 4
U
.
It can be calculated that κ3 = 0 crosses λc = 0 only at one point, Ď P6 = (U4 , V4 ), and crosses λc = 0 only at one point, P7 = (U5 , V5 ), where
(1 + W ) W W 2 + 1 (W + 1) V4 = , 2W W 2 + 1 √ and U5 = 3W 2 + 1/ (2W ). Using the expression for V1 from the proof of Lemma 12, it can be checked that V42 − V12 = 2 (W + 1) / 2 W + 1 > 0. It is also clear that U5 > U1 . Hence κ3 = 0 can never cross the outside boundary of the whole regular region (see Fig. 6). The condition y2ν,µ = V 2 is analogous. Therefore,
the conditions on the extreme values cannot occur in the nonregular region, and hence not in Γ . Lemma 14. For (U , V ) ∈ Γ , there are no coincident extreme values which satisfy (24a) and (24b). Proof. The quadratic (31) in x2 has discriminant equal to Dν . The only possibility for repeated x-values occurs with Dν = 0 or x = 0 being a repeated root. From Lemma 12 the latter implies an infinite
Hence κ4 = 0 can never cross the outside boundary of the whole regular region (see Fig. 6). It remains only to consider the case of γ1 = 0. It can be checked that γ1 = 0 lies inside the regular region (see [4]) and hence outside of Γ when W ∈ (0, 1/3]. We consider the case where 1/3 < W < 1. For this range, the curve γ1 = 0 lies completely inside the region |U | ≤ 1, |V | ≤ 1 and touchesthe line U = 1 at
√
the point PB = 1, W −1 − 3W /2 . For W ∈ 1/ 3, 1 , PB lies
outside of Γ in the 4th quadrant. We now consider the segment of γ1 = 0 between the axis V = 0 and the point P3 (see Fig. 6). On the curve γ1 = 0, there is a pair of coincident extreme values given by
−α1
,
(35)
.
(36)
x21,µ = x21 =
2 V2 − 1
y21,µ = y21 =
2 U2 − 1
−α1
From Lemma 13 we note that y21 ̸= V 2 on γ1 = 0 in Γ . From Lemma 12 we note that y21 = 0 can only occur when x21 = ∞ which is impossible on this segment of γ1 = 0 because of (35). Now consider the point P0 , which is the intersection of γ1 = 0 with V = 0. At P0 we have U = y21 =
W −1 − W 3W − W −1
W −1 − W
9W − W −1 /2 and
.
√ 1/ 3, 1 , y21 < 0 for √ √ W ∈ 1/3, 1/ 3 and y21 = ∞ when W = 1/ 3. Since y1 varies continuously on γ1 = 0 (apart from the possible singularity at PB )
We see that y21 > V 2 = 0 for W ∈
it is therefore impossible for y1 to satisfy (24b) on this segment of γ1 = 0. The part of γ1 = 0 between the axis U = 0 and P2 follows by a dual argument. Proof of theorem. Let W < 1. Networks containing two reactive elements of the same kind can only realise (U , V ) with Kc < 0 [4]. Hence we only need to consider networks which take the form of Fig. 4. Suppose then that there is a region Ω inside Γ where (24a)–(24d) are satisfied. Consider a line ℓ within Γ which connects a point in the positive-real boundary σc = 0 with a point in Ω (see Fig. 7). By Lemma 9, (1, 1) ̸∈ Ω , so we can choose ℓ so that it does not pass through the point (1, 1). As (U , V ) moves
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Fig. 7. Illustration of proof of theorem. The assumption of a region Ω inside Γ where (24a)–(24d) are satisfied is shown to lead to a contradiction.
along ℓ, we can find solutions (x, y) to (24d) at every point which also satisfy (24a) and (24b) by Lemmas 5 and 7. Inside Ω some of these solutions will also satisfy (24c) whereas none of them do outside of Ω . By Lemma 10 there are points on ℓ which are outside of Ω . Now for each fixed (U , V ) the set of (x, y) which satisfy (24c) and (24d) consists of intervals of the hyperbola (24d) which are bounded by extreme values. Lemma 8 shows that these intervals deform continuously with (U , V ). Lemma 14 shows that such intervals cannot vanish by virtue of extreme values becoming coincident. Hence, as (U , V ) moves along ℓ and traverses δ Ω , an extreme value must cross x2ν,µ = U 2 or y2ν,µ = V 2 . Lemma 13 shows that this cannot happen inside Γ . Hence there is no region Ω inside Γ where (24a)–(24d) are satisfied. The result now follows from Lemma 4. The case of W > 1 follows in general by duality, or directly by using Lemma 4 in dual form with W replaced by W −1 in (24c). Since Zc is regular when W = 1 there is nothing to prove in this case. To conclude the section we present a graphical illustration in Fig. 8 of one typical case where solutions to (24a)–(24d) are lost by virtue of extreme values becoming coincident, as (U , V ) crosses the boundary of Γ . For other boundary points of Γ , solutions may be lost due to extreme values crossing the boundary x2ν,µ = U 2 or y2ν,µ = V 2 . The theorem proves that such behaviour cannot occur inside Γ . 5. Notes We present some remarks to compare our paper with [1]. 1. Section 3 provides a simple derivation of the parametrisation of Auth [2], from which the system of equations (Lemma 4) is derived on which the proof of [1] is built. As in [1,2], the mild assumption of the existence of an impedance matrix (6) is made. In general, matrices A and B exist for which A(v1 , v2 , v3 )| = B(i1 , i2 , i3 )| with A + B non-singular, which implies the existence of the scattering matrix of N [12, Chapter 3] and also that any singular cases of the impedance matrix will be limiting cases of (6) in which some elements tend to ∞. These limiting cases include impedances (1) in which some coefficients may be zero, after appropriate scaling, which are
=
=
Fig. 8. The curves which satisfy (24a)–(24c) with equality and (24d) for W = 0.6 and the (U , V ) values marked in Fig. 7 ((i) U = V = 0.71, (ii) U = V = 0.65). In (i) the (x, y) values satisfying (24a)–(24d) lie on the solid curve between the extreme values A and B (A′ and B′ ), while no such (x, y) exist in (ii).
trivial cases that are ignored here without loss of generality (see Section 2). 2. The key part of Reichert’s original proof is the claim that, in the notation of the present paper, (24a)–(24d) have no solution in Γ . In [1, p.206, col.1, l.3–14] it is stated that this reduces to four cases (Fall 1–4) involving the ‘‘extreme values’’. The underlying topological argument is not provided. The ‘‘proof of theorem’’ above provides this argument which makes use of the lemmas developed. It is evident that some further cases need to be analysed which are not treated in [1]. 3. One possibility which lies outside Reichert’s ‘‘Fall 1–4’’ is the possibility that (24d) itself loses contact with the rectangle defined by (24a) and (24b). This is dealt with by Lemma 5. 4. There is a further logical possibility that falls outside of Reichert’s ‘‘Fall 1–4’’. This is the possibility that the intervals of (24d) between extreme values which satisfy (24c) could ‘‘flip’’ discontinuously even when the extreme values themselves are
J.Z. Jiang, M.C. Smith / Systems & Control Letters 61 (2012) 1124–1131
continuous. We show that this cannot happen when (U , V ) ̸= (1, 1) in Lemma 8. On the other hand this behaviour does occur when U = V = 1. We deal with this case separately in Lemma 9. 5. Lemmas 12–14 are a reworking of several cases related to Reichert’s ‘‘Fall 1–4’’ in a form which can be explicitly used in the ‘‘proof of theorem’’. 6. Concluding remarks The statement of Reichert’s theorem is remarkable for its simplicity. This contrasts with Reichert’s proof which is far from simple. The purpose of this paper has been to provide a complete reworking of the proof which highlights all the main steps and to expand the reasoning for some parts of the topological argument. Acknowledgement This work was supported by the Engineering and Physical Sciences Research Council grant number EP/G066477/1.
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References [1] M. Reichert, Die kanonisch und übertragerfrei realisierbaren Zweipolfunktionen zweiten Grades (Transformerless and canonic realisation of biquadratic immittance functions), Arch. Elek. Übertragung 23 (1969) 201–208. [2] L.V. Auth, RLC biquadratic driving-point synthesis using the resistive threeport, IEEE Trans. on Circuit Theory (1964) 82–88. [3] I. Cederbaum, Conditions for the impedance and admittance matrices of nports without ideal transformers, Proc. IEE 105 (1958) 245–251. [4] J.Z. Jiang, M.C. Smith, Regular positive-real functions and five-element network synthesis for electrical and mechanical networks, IEEE Trans. Automat. Control 56 (6) (2011) 1275–1290. [5] R.M. Foster, E.L. Ladenheim, A class of biquadratic impedances, IEEE Trans. on Circuit Theory (1963) 262–265. [6] M.C. Smith, Synthesis of mechanical networks: the inerter, IEEE Trans. Automat. Control 47 (10) (2002) 1648–1662. [7] R.M. Foster, Academic and theoretical aspects of circuit theory, Proc. IRE 50 (1962) 866–871. [8] E.A. Guillemin, Synthesis of Passive Networks, John Wiley & Sons, 1957. [9] M.E.V. Valkenburg, Introduction to Modern Network Synthesis, John Wiley & Sons, 1960. [10] B.D.O. Anderson, S. Vongpanitlerd, Network Analysis and Synthesis, PrenticeHall, Upper Saddle River, NJ, 1973. [11] M.Z.Q. Chen, M.C. Smith, A note on tests for positive-real functions, IEEE Trans. Automat. Control 54 (2) (2009) 390–393. [12] R.W. Newcomb, Linear Multiport Synthesis, McGrawHill, New York, 1966.