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MIT 2 141 - Ideal asymmetric junction elements

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Ideal asymmetric 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 A 1 v1 v v A 2 2 2 The input and output power flows are the square of the length of these vectors their inner products Pin 2 ui2 utu A 3 i 1 2 vi2 vtv Pout A 4 i 1 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 A 6 where the square matrix S is known as a scattering matrix December 6 1994 page 1 Neville Hogan 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 ortho normal 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 A 8 This constrains the coefficients of the scattering matrix as follows a b S c d A 9 a2 c2 1 A 10 ab cd 0 A 11 b2 d2 1 A 12 December 6 1994 page 2 Neville Hogan 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 A 13 which yields the following equations a2 b2 1 A 14 ac bd 0 A 15 c2 d2 1 A 16 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 A 18 f1 f f A 19 2 2 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 December 6 1994 page 3 Neville Hogan 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 0 A 22 1 a 1 a f1 1 a 1 a 0 f2 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 may simply 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 December 6 1994 page 4 Neville Hogan 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 December 6 1994 A 28 page 5 Neville Hogan A third solution If a d b c 1 a2 Using the positive root and assuming a to be the undetermined parameter S 1 a2 a 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 e c 1 1 S e2 c u1 cf1 1 S u2 cf2 e c 1 1 a e2 c 1 a2 cf 1 a 1 cf2 1 a2 A 30 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 e2 c 1 a2 1 a cf1 e1 c cf2 1 a 1 a2 1 a u1 1 a2 u2 1 a2 u1 1 a u2 A 32 A 33 Solving the second of these for u and substituting into the first we obtain a relation between efforts and flows December 6 1994 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 December 6 1994 page 7 Neville Hogan 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 December 6 1994 A 38 page 8 Neville Hogan 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 …


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MIT 2 141 - Ideal asymmetric junction elements

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