On the notion of effective impedance via ordered fields Anna Muranova IRTG 2235, Bielefeld University, Germany February 25, 2019 Differential Operators on Graphs and Waveguides TU Graz, Austria Advisor: Alexander Grigoryan Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Overview and references Definition Let ( V , µ ) be a locally finite weighted graph without isolated points (weights are positive/non-negative). Then for any function f : V → R , the function, defined by 1 � ∆ µ f ( x ) = ( f ( y ) − f ( x )) µ xy , µ ( x ) y is called the (classical) Laplace operator on ( V , µ ). Alexander Grigoryan. Introduction to Analysis on Graphs , 2018. Fact The Laplace operator on graphs is related with electric networks with resistors (since resistors have positive resistance) and direct current. The concept of network and effective resistance of the network is introduced in this case. David A. Levin, Yuval Peres, Elizabeth L. Wilmer. Markov Chains and Mixing Times . 2009. Doyle P.G., Snell J.L. Random walks and electric networks . 1984. Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Overview and references Question What about electric networks with other elements (coils, capacitors, and resistors)? All together they are called impedances . Note, that current in this case should be alternating. R. P. Feynman. The Feynman lectures on physics, Volume 2: Mainly Electromagnetism and Matter . 1964 O. Brune. Synthesis of a finite two-terminal network whose driving-point impedance is a prescribed function of frequency . Thesis (Sc. D.). Massachusetts Institute of Technology, Dept. of Electrical Engineering, 1931. Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Electrical network Definition Network is a (finite) connected graph ( V , E ) where to each edge xy ∈ E the triple of non-negative numbers R xy (resistance), L xy (inductance), and C xy (capacitance) is associated; two fixed vertices a 0 , a 1 “are connected to a source of alternating current”. The impedance of the edge xy is 1 z xy = R xy + i ω L xy + i ω C xy , where ω is the frequency of the current. Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Example a 2 1 1 i ω C 1 i ω C 2 a 0 a 1 R i ω L 1 i ω L 2 a 3 Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Example a 2 1 1 i ω C 1 i ω C 2 a 0 a 1 R i ω L 1 i ω L 2 a 3 Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Dirichlet problem By Ohm’s complex law and Kirchhoff’s complex law, the voltage v ( x ) as a function on V satisfies the following Dirichlet problem : � ∆ ρ v ( x ) := � y : y ∼ x ( v ( x ) − v ( y )) ρ xy = 0 on V \ { a 0 , a 1 } , (1) v ( a 0 ) = 0 , v ( a 1 ) = 1 , 1 where ρ xy = z xy is the admittance of the edge xy . ρ xy is hence a complex-valued weight of xy . Operator ∆ ρ is called the Laplace operator . Note, that if | V | = n , then (1) is a n × n system of linear equations. Definition Effective impedance of the network is 1 1 Z eff = = P eff , � x : x ∼ a 0 v ( x ) ρ xa 0 where v ( x ) is a solution of the Dirichlet problem. P eff is called effective admittance and it is equal to the current through a 0 due to the unit voltage on the source. Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Dirichlet problem By Ohm’s complex law and Kirchhoff’s complex law, the voltage v ( x ) as a function on V satisfies the following Dirichlet problem : � ∆ ρ v ( x ) := � y : y ∼ x ( v ( x ) − v ( y )) ρ xy = 0 on V \ { a 0 , a 1 } , (1) v ( a 0 ) = 0 , v ( a 1 ) = 1 , 1 where ρ xy = z xy is the admittance of the edge xy . ρ xy is hence a complex-valued weight of xy . Operator ∆ ρ is called the Laplace operator . Note, that if | V | = n , then (1) is a n × n system of linear equations. Definition Effective impedance of the network is 1 1 Z eff = = P eff , � x : x ∼ a 0 v ( x ) ρ xa 0 where v ( x ) is a solution of the Dirichlet problem. P eff is called effective admittance and it is equal to the current through a 0 due to the unit voltage on the source. Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Dirichlet problem By Ohm’s complex law and Kirchhoff’s complex law, the voltage v ( x ) as a function on V satisfies the following Dirichlet problem : � ∆ ρ v ( x ) := � y : y ∼ x ( v ( x ) − v ( y )) ρ xy = 0 on V \ { a 0 , a 1 } , (1) v ( a 0 ) = 0 , v ( a 1 ) = 1 , 1 where ρ xy = z xy is the admittance of the edge xy . ρ xy is hence a complex-valued weight of xy . Operator ∆ ρ is called the Laplace operator . Note, that if | V | = n , then (1) is a n × n system of linear equations. Definition Effective impedance of the network is 1 1 Z eff = = P eff , � x : x ∼ a 0 v ( x ) ρ xa 0 where v ( x ) is a solution of the Dirichlet problem. P eff is called effective admittance and it is equal to the current through a 0 due to the unit voltage on the source. Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Main problem There are easy examples, when the Dirichlet problem has no solution or has multiple solutions 1 i ω L i ω C y z x 1 1 i ω C i ω C i ω L 1 i ω C a 1 a 0 i ω L + R � 2 In the case ω = LC the Dirichlet problem has infinitely many solutions ( v ( x ) , v ( y ) , v ( z )) = ( − 2 τ + 1 , 2 τ − 1 , τ ); � 1 In the case ω = 3 LC the Dirichlet problem has no solutions. Question How to define P eff in this cases? Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Main problem There are easy examples, when the Dirichlet problem has no solution or has multiple solutions 1 i ω L i ω C y z x 1 1 i ω C i ω C i ω L 1 i ω C a 1 a 0 i ω L + R � 2 In the case ω = LC the Dirichlet problem has infinitely many solutions ( v ( x ) , v ( y ) , v ( z )) = ( − 2 τ + 1 , 2 τ − 1 , τ ); � 1 In the case ω = 3 LC the Dirichlet problem has no solutions. Question How to define P eff in this cases? Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Main problem There are easy examples, when the Dirichlet problem has no solution or has multiple solutions 1 i ω L i ω C y z x 1 1 i ω C i ω C i ω L 1 i ω C a 1 a 0 i ω L + R � 2 In the case ω = LC the Dirichlet problem has infinitely many solutions ( v ( x ) , v ( y ) , v ( z )) = ( − 2 τ + 1 , 2 τ − 1 , τ ); � 1 In the case ω = 3 LC the Dirichlet problem has no solutions. Question How to define P eff in this cases? Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Main problem There are easy examples, when the Dirichlet problem has no solution or has multiple solutions 1 i ω L i ω C y z x 1 1 i ω C i ω C i ω L 1 i ω C a 1 a 0 i ω L + R � 2 In the case ω = LC the Dirichlet problem has infinitely many solutions ( v ( x ) , v ( y ) , v ( z )) = ( − 2 τ + 1 , 2 τ − 1 , τ ); � 1 In the case ω = 3 LC the Dirichlet problem has no solutions. Question How to define P eff in this cases? Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
P eff is well defined Theorem 1 In the case of existence of multiple solutions of the Dirichlet problem (1) the value of P eff does not depend on the choice of solution. Proof. If one writes the Dirichlet problem (1) in the form � A ˆ v = b , v ( a 0 ) = 0 , v ( a 1 ) = 1 , where A is a symmetric ( n − 2) × ( n − 2) matrix, then P eff = ( Ae − b ) T ˆ v + ρ a 1 a 0 . And for any two solutions ˆ v 1 , ˆ v 2 , we have: ( Ae − b ) T ˆ v 1 = e T A T ˆ v 1 − b T ˆ v 2 ) T ˆ 2 A T ˆ v 1 = e T A ˆ v 1 = e T b − ˆ v T v 1 − ( A ˆ v 1 v 2 = e T A T ˆ 2 A T ˆ = e T A ˆ v T v 1 = e T A ˆ v T v T v 2 − ˆ 2 A ˆ v 2 − ˆ 2 A ˆ v 2 − ˆ v 2 = e T A T ˆ v 2 ) T ˆ v 2 = e T A T ˆ v 2 − b T ˆ v 2 = ( Ae − b ) T ˆ v 2 − ( A ˆ v 2 . In the case of lack of solution we set by definition P eff = + ∞ . Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
Conservation of the complex power Note that for 1 z xy = R xy + i ω L xy + i ω C we have Re z xy ≥ 0. Also, Re ρ xy ≥ 0. By physical meaning, Re Z eff and Re P eff should be non-negative. Theorem 2 P eff = 1 � | v ( x ) − v ( y ) | 2 ρ xy . 2 x ∼ y Key point of the proof (Green’s formula) For any two functions f , g on V ∆ z f ( x ) g ( x ) = 1 � � ( ∇ xy f )( ∇ xy g ) ρ xy , (2) 2 x ∈ V x , y ∈ V where g ( x ) is a complex conjugation of g ( x ). Corollary Re P eff ≥ 0 . Anna MuranovaIRTG 2235, Bielefeld University, Germany On the notion of effective impedance via ordered fields
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