ECE 6130: LECTURE 2 FIELD ANALYSIS OF TRANSMISSION LINES Text Section 2.2 Portfolio: 1) How do you calculate the RLGC parameters for a transmission line configuration that isn't in the standard table? See Chapter 2, problem 3. Field Analysis of Transmission Lines Role of E and H in R,L,G,C: Transmission Lines in ECE 3170 used V and I as the waveform variables, R,L,G,C to represent the properties of the line. R,L,G,C were given in tables as a function of the size and shape of the transmission line. These tables are derived from the E and H inside the transmission line. “Unit Length” Section of TL – E and H Field Patterns Given a section of 2-conductor transmission line that is “one unit length long” (like 1 meter): The two conductors have opposite charges, which makes E lines between them, and H lines around them. SEE FIELD PATTERNS FOR TL LINES How do you find these fields? Analytical (Maxwell's equations): In ECE3170, we learned how to find them for coax, parallel plate, single wire, etc. (symmetrical systems) using Maxwell's equations. Analytical (Boundary Value): You can also find the field distributions by applying Laplace's equation ∇2 V =0 and electric field boundary conditions (tangential E is constant across a boundary, normal D is constant across the boundary), and solve this analytically. This is called a "boundary value problem". You can find out more about solving these in… Computational Electromagnetics: We will solve for the fields on a microstrip and/or stripline using the Finite Difference method, which is one way of solving boundary value problems.For all of these methods, we assume: The voltage between the conductors is Vo e±jβz and the current is Io e±jβz . Z is direction of propagation, so this represents + and – traveling waves. Inductance and Capacitance (L and C) from E and H The time-averaged stored magnetic energy is given by (See Section 1.6 and circuit theory): So Inductance per unit length : The time-averaged stored electric energy is given by (See Section 1.6 and circuit theory): So Capacitance per unit length : Power Loss due to losses in the metal conductors (See equation 1.131, surface impedance section): Power Loss PER UNIT LENGTH WILWdsHHWSomm∫==•= 4/||42*μWVCWdsEEWSoee∫==•= 4/||42*εmFdsEEVCSo/*2∫•=εmHdsHHILSo/*2∫•=μ[]WdzdlHHRdsHHRPCsSsc∫∫∫•=•=**22mWdlHHRPCsc/2*∫•=Notes: (1) 2 comes from RMS. H is given as PEAK value, so RMS value = Peak / sqrt(2), and then you are squaring H. (2) * represents CONJUGATE (change sign on imaginary part) (3) C is the total contour of both conductors , C = C1 + C2 (4) Rs is the surface resistance of the conductors = 1/ (σδs) where δs (meters) is the “skin depth” (how far field penetrates before being reduced to 1/e, σ is the conductivity (S/m) From circuit theory: Pc = R |Io|2 / 2 So, series resistance PER UNIT LENGTH : When there is resistance between the conductors (in the substrate or insulating material) PER UNIT LENGTH: unit length is Where ε’’ = imaginary part of permittivity = σ / ω From circuit theory Pd = G|Vo|2 /2, so the shunt conductance per unit length is: Telegrapher’s Equations Lumped element model represents voltage and current on transmission line. Text derives these for field, I will derive them for voltage and current. ∫+Ω•=21/||*2CCsmdlHHIoRRdsEEPSd•=∫2''ωε∫•=SmSdsEEVoG /||''2ωεKirchoff Loop Equation: V(z,t) – R I(z,t) – L di(z,t)/dt – v(z+Δz,t) = 0 R = R’Δz, L = L’ Δz Divide by Δz, combine V terms, and rearrange: V(z,t) / Δz – R’ I(z,t) – L’ dI(z,t) /dt – V(z+Δz,t) / Δz = 0 -[ V(z+Δz,t) - V(z,t) ] / Δz = R’ I(z,t) – L’ dI(z,t)/dt Limit as Δz Æ 0 (difference goes to differential): -dV(z,t)/dz = R’ I(z,t) – L’ dI(z,t)/dt Kirchoff Node Equation: I(z,t) – G v(z+Δz,t) – C dv(z+Δz,t)/dt – I(z+Δz,t) = 0 I(z,t) – G’Δz v(z+Δz,t) – C’Δz dv(z+Δz,t)/dt – I(z+Δz,t) = 0 Divide by Δz and rearrange : -[ I(z+Δz,t) - I(z,t)] / Δz = G’ v(z+Δz,t) + C’ dv(z+Δz,t)/dt Limit as Δz Æ 0 : -dI(z,t) / dt = G’ v(z+Δz,t) + C’ dv(z+Δz,t)/dt Telegraphers equations: -dV(z,t)/dz = R’ I(z,t) + L’ dI(z,t)/dt -dI(z,t) / dt = G’ v(z+Δz,t) + C’ dv(z+Δz,t)/dt Sinusoidal Steady-State Form: Convert to Phasor form (d/dt Æ jω). -dV(z)/dz = (R’ + jωL’) Ι(z) -dI(z)/dz = (G’ + jωC’) V(z) What is difference between this and “regular” circuit equations?? They are a function of z…. distance down the transmission line. Wave Equation form: Take d/dz of one equation, and substitute it into the other equation. (Doesn’t matter which one). See text and copy of Chapter 2 in the IEEE room.-d2 I(z)/dz2 –γ2Ι(z) = 0 Alternate form : -d2 V(z)/dz2 –γ2V(z) =