Transmission Lines (Time Domain) Ch2

Circuit Model i ( z, t ) v ( z, t ) + - z ∆z Circuit Model: R ∆z Note: The order of the elements is not important, as long as we keep the R and L as series elements and the G and C as parallel elements. L ∆z G ∆z C ∆z z ∆z 13 Coaxial Cable σ ma Example (coaxial cable) Usually µ m = µ0 σ mb (cooper or aluminum) Usually εr µ d = µ0 (e.g., Teflon) z 2 ωµmσ m (skin depth of metal) σm = conductivity of metal [S/m] µm = permeability of metal [H/m] b 2πε 0ε r F/m ] C= [ b ln   a µ0  b  L= ln   [ H/m ] 2π  a  δ= σd = conductivity of dielectric [S/m] a δ a ,b = 2πσ d S/m ] [ b ln   a   1 1 + R=   π a δ σ π b δ σ 2 2 a ma b mb   G= 2 ωµma ,mbσ ma ,mb [Ω /m] Note: The derivation of R involves knowledge of the skin depth concept, and this will be given later. 14 Coaxial Cable (cont.) Overview of derivation: capacitance per unit length E - ρ sa - Assume static fields + + + + - + + - ρl - Q 1 ρ C = = V ∆z V − ρl Gauss’s law ρ = ρ sa ( 2π a ) b b  ρ  ρ b = = ln V ∫= Eρ d ρ ∫  d ρ    πε ε ρ πε ε 2 2 a 0 r 0 r  a a More details can be found in ECE 3318, Notes 25. 2πε 0ε r C= b ln   a [ F/m] 15 Coaxial Cable (cont.) Overview of derivation: inductance per unit length y × Assume static fields J sa × × a × × ψ Bφ × × b × I = J sza ( 2π a ) (current flowing in z direction on inner conductor) More details can be found in ECE 3318, Notes 31. L= x ψ 1 I ∆z Ampere’s law b b a a  I  dρ  2πρ  ψ= ∆z ∫ Bφ d ρ = ∆z ∫ µ0   I = ∆z µ0   2π ( µ = µ0 )  b  ln    a µ0  b  L= ln   2π  a  [ H/m] 16 Coaxial Cable (cont.) Overview of derivation: conductance per unit length RC Analogy (ECE 3318): a b σd More details can be found in ECE 3318, Notes 27. ε →σd C →G 2πε C= b ln   a 2πσ d G= b ln   a [S/m] 17 Coaxial Cable (cont.) Relation Between L and C: µ0 µ r  b  2πε 0ε r = C = L F/m ] ln   [ a 2π b  ln   a [ H/m] = LC µ= µε 0 µ r ε 0ε r cd Speed of light in dielectric medium: = 1 = µε 1 µ 0ε 0 1 = µr ε r c µr ε r c ≡ 2.99792458 ×108 [ m/s ] Hence: 1 LC = 2 cd This is true for ALL transmission lines. ( A proof will be seen later.) 18 Telegrapher’s Equations  These are a fundamental set of differential equations that describe how voltage and current propagate (travel) on a transmission line.  The derivation holds for any type of transmission line.  The equations are exact, showing how any type of signal propagates on a transmission line.  We only have an exact solution for the telegrapher’s equations in the time domain for a lossless transmission line. 19 Telegrapher’s Equations (cont.) Apply KVL and KCL laws to a small ∆z slice of line: R ∆z i ( z, t ) L ∆z i ( z + ∆z , t ) + + v ( z, t ) G ∆z C ∆z v ( z + ∆z , t ) - - z z z + ∆z ∂i ( z, t ) KVL : v ( z,= t ) v ( z + ∆z, t ) + i ( z, t ) R∆z + L∆z ∂t ∂v ( z + ∆z, t ) KCL : i ( z, t= ) i ( z + ∆z , t ) + v ( z + ∆z , t ) G ∆z + C ∆z ∂t 20 Telegrapher’s Equations (cont.) Hence, we have: v ( z + ∆z, t ) − v ( z, t ) ∂i ( z, t ) = − Ri ( z, t ) − L ∆z ∂t i ( z + ∆z , t ) − i ( z , t ) ∂v ( z + ∆z, t ) = −Gv ( z + ∆z, t ) − C ∆z ∂t “Telegrapher’s Equations (TEs)” Now let ∆z → 0: ∂v ∂i = − Ri − L ∂z ∂t ∂i ∂v = − Gv − C ∂z ∂t 21 Telegrapher’s Equations (cont.) To combine take therespect derivative  Take the derivative of thethese, first TE with to z.of the first one with respect to z:  Substitute in from the second TE. ∂i ∂v = − Gv − C ∂z ∂t ∂ 2v ∂i ∂  ∂i  = −R − L   2 ∂z ∂z ∂z  ∂t  ∂i ∂  ∂i  = −R − L   ∂z ∂t  ∂z   ∂v  ∂v ∂ 2v   = − R  −Gv − C  − L  −G − C 2  ∂t  ∂t ∂t    22 Telegrapher’s Equations (cont.)  ∂ 2v ∂v  ∂v ∂ 2v   = − R  −Gv − C  − L  −G − C 2  2 ∂z ∂t  ∂t ∂t    Hence, we have:  ∂ 2v  ∂ 2v ∂v − ( RG ) v − ( RC + LG ) − LC  2  = 0 2 ∂z ∂t  ∂t  There is no general exact solution to this differential equation, except for the lossless case. Hence, we will assume lossless transmission lines when we are in the time domain. Note: The current satisfies the same differential equation. 23 Telegrapher’s Equations (cont.) Lossless case: R= G= 0 ∂v ∂v ∂ v − ( RG ) v − ( RC + LG ) − LC  0 = ∂z ∂t  ∂t  2 2 2 2 ∂v ∂ v − LC  0 = ∂z  ∂t  2 2 2 Recall : LC = µε = 2 1 cd2 24