asymmetric Ideal junction elements Relax the symmetry assumption and examine the resulting junction structure. For simplicity, consider two-port junction elements. As before, assume instantaneous power transmission between the ports without storage or dissipation of energy. Characterize the power flow in and out of a twoport junction structure using four real-valued wave-scattering variables. Using vector notation:
u1 u = u 2 v1 v = v 2
(A.1) (A.2)
The input and output power flows are the square of the length of these vectors, their inner products. Pin =
ui2 = utu
i=1
2
(A.3)
Pout =
vi2 = vtv
i=1
2
(A.4)
The constitutive equations of the junction structure may be written as follows. v = f(u) (A.5)
Geometrically, the requirement that power in equal power out means that the length of the vector v must equal the length of the vector u, i.e. their tips must lie on the perimeter of a circle (see figure A.1). For any two particular values of u and v, the algebraic relation f(.) is equivalent to a rotation operator. v = S(u) u where the square matrix S is known as a scattering matrix. (A.6)
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u2
v2
u u1 v1 v
S need not be a constant matrix, but may in general depend on the power flux through the junction, hence the notation S(u). However, S is subject to important restrictions. In particular, vtv = utStSu = utu (A.7)
S is an orthogonal matrix: the vectors formed by each of its rows (or columns) are (i) orthogonal and (ii) have unit magnitude; its transpose is its inverse. S tS = 1 This constrains the coefficients of the scattering matrix as follows.
a b S=c d
(A.8)
(A.9) (A.10) (A.11) (A.12) page 2 Neville Hogan
a2 + c2 = 1 ab + cd = 0 b2 + d2 = 1 Mod. Sim. Dyn. Sys.
As there are only three independent equations and four unknown quantities, we see that this junction is characterized by a single parameter. We may also write the orthogonality condition as SSt = 1 which yields the following equations. a2 + b2 = 1 ac + bd = 0 c2 + d2 = 1 (A.14) (A.15) (A.16) (A.13)
There are four possible solutions to these equations. Combining A.10 and A.16, a2 = 1 - c2 = d2. Thus a = d. If a = d then b =c = 1 - a2 . One solution Choosing the positive root yields one solution. Assuming the coefficient a to be the undetermined parameter,
a S= - 1 - a2
1 - a2 a
(A.17)
Rewrite in terms of effort and flow variables.
e1 e = e 2 f1 f = f 2
(A.18) (A.19)
The relation between efforts and wave-scattering variables is as follows. e = (u - v) c = c (1 - S) u (A.20)
where c is a scaling constant. The relation between flows and wave-scattering variables is as follows.
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f = (u + v)/c = 1/c (1 + S) u
(A.21)
If |a| 1 then 1 + S and 1 - S are nonsingular matrices and the input wave scattering variables u1 and u2 may be eliminated as follows. e = c2 (1 - S) (1 + S)-1 f
e1 2 =c e2
(A.22) - (1 - a)/(1 + a) f1 0
f2
0 (1 - a)/(1 + a)
(A.23)
Writing G = c2 (1 - a)/(1 + a) we obtain the equation for an ideal gyrator.
e1 0 -G f1 = G 0 f2 e2
(A.24)
Note that equations A.20 and A.21 imply a sign convention in effort-flow coordinates such that power is positive inwards on both ports. Pnet inwards = etf = utu - vtv (A.25)
To follow the more common sign convention we simply may change the sign of f2 in equation A.24. If a = 1, e is identically zero for all values of f. No energy is exchanged between the ports and the junction structure behaves like a dissipator with zero resistance. If a = -1, f is identically zero for all values of e. No energy is exchanged between the ports and the junction structure behaves like a dissipator with infinite resistance (zero conductance).
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A second solution Choosing a = d and using the negative root yields another solution. Again assuming the coefficient a to be the undetermined parameter,
S=
a 1 - a2
- 1 - a2 a
(A.26)
In this case the relation between efforts and flows is
e1 0 2 =c - (1 - a)/(1 + a) e2
(1 - a)/(1 + a) f1 0
f2
(A.27)
Again we obtain the equation for an ideal gyrator.
e1 0 G f1 = -G 0 f2 e2
(A.28)
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A third solution If a = - d, b = c = 1 - a2 . Using the positive root and assuming a to be the undetermined parameter
S=
a 1a2
1 - a2 -a
(A.29)
In this case the matrices 1 + S and 1 - S are singular for all values of the parameter a. However, equations A.20 and A.21 may be combined as follows:
1 1 - S e2/c = ----- u1 cf1 1 + Su2 cf2 1 1-a e2/c - 1 - a2 cf = 1 + a 1 cf2 1 - a2
e2/c - 1 - a2 = 1+a cf1 e1/c = cf2
e /c
(A.30)
e /c
- 1 - a2 1+a 1 - a2 1-a
u1 u2
(A.31)
If |a| 1, the 4 x 2 matrix relating efforts and flows to the input scattering variables contains two nonsingular 2 x 2 submatrices. 1 + a u1
1 - a2 u2 u2
(A.32)
1-a 1a2
- 1 - a2 u1 1-a
(A.33)
Solving the second of these for u and substituting into the first we obtain a relation between efforts and flows. Mod. Sim. Dyn. Sys. page 6 Neville Hogan
e2 - (1 + a)/(1 - a) = 0 f1
e1 (1 + a)/(1 - a) f2
0
(A.34)
Writing T = (1 + a)/(1 - a) we obtain the equation for an ideal transformer.
e2 -T 0 e1 = 0 T f2 f1
(A.35)
To follow the more common sign convention we may change the sign of e2. If the parameter a = 1, an argument similar to that used above shows that a degenerate case results in which no energy is exchanged between the ports.
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Final solution Choosing a = d and using the negative root we obtain the fourth solution.
a - 1 - a2 S= -a - 1 - a2
(A.36)
Once again, the matrices 1 + S and 1 - S are singular for all values of the parameter a, but by rearranging equations A.20 and A.21 as before the corresponding relation between efforts and flows is
e2 = f1
(1 + a)/(1 - a) 0
e1 - (1 + a)/(1 - a) f2
0
(A.37)
Again we obtain the equation for an ideal transformer
e2 T 0 e1 = 0 -T f2 f1
(A.38)
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Two-port junction elements There are only two possible power-continuous, asymmetric two-port junction elements, the gyrator and the transformer. Unlike the ideal symmetric junction elements (0 and 1) the ideal asymmetric junction elements may be nonlinear. The relation between efforts and flows must have a multiplicative form. The general asymmetric junction elements are a modulated gyrator (MGY) and a modulated transformer (MTF) respectively.
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