ECE 6130: LECTURE 1 REVIEW TRANSMISSION LINES Text Section 2.1, Handout Portfolio: 1) Describe the lumped element transmission line model. 2) Describe how to calculate the lumped element values for known lines such as coaxial and microstripline (See Chapter 2 Problems 1,2) "Transmission Line" -- Any structure or media which guides EM waves from one location to another. Two-port circuit diagram: Effect of Transmission line (function of frequency and speed of wave): VAA' = Vg(t) = Vocos(ωt) volts VBB' = VAA'( t + tdelay) = Vocos(ω(t + length/speed)) volts Speed = vo = 2.996e8 m/s Example 1: f = 60 Hz, length = 1 meter, ω∗tdelay = .0000012, VBB' = VAA' Example 2: f = 1GHz, length = 1meter, ω∗tdelay = 20.94 radians ! 6.67 cycles, VBB' = -0.49*VAA' Example 3: f=10 GHz, length = 1cm, ω∗tdelay = 2.094 radians, VBB' =-.49*VAA' Reflections: With 6.67 cycles, there has been a lot of time for bounces (reflections) to get back to source. • Reflections add (you think your computer line = 0.0, but it actually = .5, which is enough to give a digital "1") • Reflections subtract (you think your power is 1 W, but it is actually .5 W) Power Loss• Reflections through lossy material • Multiple bounces through semi-lossy material Dispersive effects: • Many materials have different properties at different frequencies. All water-based materials, many semi-conductors, etc. Propagation Modes • TEM : Transverse Electromagnetic (** we are going to study these) Transverse = perpendicular E and H fields are both entirely perpendicular to the direction of propagation. Made up of two parallel conducting surfaces: Coaxial Line: See transparency Figure 2-5 • Higher - order transmission lines: E and H fields have at least one significant component in direction of propagation. Combination of TE and TM. Lumped –Element Model: TEM transmission lines (remember, no fields in the direction of propagation) can be represented by a lumped element model … • Parallel-wire equivalent • Represents motion of fields down the transmission line • Transverse field effects (all fields ARE transverse in TEM) are modeled by equivalent circuit elements … RLC • Accuracy of method depends on: • True TEM nature (non-transverse fields not properly modeled) • Correct calculation of circuit parameters (RLC) of line (done analytically, as we will do in a minute) “Lumped” Elements: R’ : Combined resistance of both conductors / unit length (ohm/meter) L’: Combined inductance of both conductors / unit length (H/m)G’ : Combined conductance of both conductors / unit length (S/m = 1/ (ohm-meter)) C’: Combined capacitance of both conductors / unit length (F/m) How do you get them? From derivations See in a minuteÆ Tabulated for standard transmission lines What effects do we see? R’ : Waves move down the transmission lines, but if the material has a resistance (anything except a perfect conductor), R’ .ne. 0. Resistance would be continuous, rather than discrete as shown … it is exact in the limit as Δz Æ 0 . Thus a “lumped” element, is generally an approximation. L’: Inductance represents magnetic flux generated by the current on the transmission line. Again, it would be continuous, but is represented by a small lumped element. G’: Conductance (1/R) . This represents coupling currents between the line. If the internal material has some conductivity, it can draw current from one line to the other. C’: Capacitance between the lines. Charges induced on the lines produce a voltage . Capacitance is charge/voltage. For all TEM lines (properties are of insulating material): L’C’ = μ ε G’/C’ = σ / ε Characteristic Impedance Zo = Vo+ / Io+ = Vo- / Io- = sqrt( (R+jωL)/(G+jωC)