First Trans-Atlantic CableTrans-Atlantic Cable AnalysisLumped Circuit Properties of CableSectional Model (I)Sectional Model (II)Distributed ModelKCL and KVL for a small sectionDerivation of Wave EquationsThe Wave EquationWave Equation SolutionWave MotionWave SpeedCurrent / Voltage Relationship (I)Current / Voltage Relationship (II)Example: Step Into Infinite LineExample 1 (cont)Example 1 (cont)EECS 117Lecture 1: Transmission LinesProf. NiknejadUniversity of California, BerkeleyUniversity of California, Berkeley EECS 117 Lecture 1 – p. 1/18First Trans-Atlantic CableProblem: A long cable – the trans-atlantic telephonecable – is laid out connecting NY to London. We wouldlike analyze the electrical properties of this cable.For simplicity, assume the cable has a uniformcross-secitonal configuration (shown as two wires here)VNY(t)RNYRLondonUniversity of California, Berkeley EECS 117 Lecture 1 – p. 2/18Trans-Atlantic Cable AnalysisCan we do it with circuit theory?Fundamental problem with circuit theory is that itassumes that the speed of light is infinite. So all signalsare in phase: V (z) = V (z + ℓ)Consequently, all variations in space are ignored:∂∂z→ 0This allows the lumped circuit approximation.University of California, Berkeley EECS 117 Lecture 1 – p. 3/18Lumped Circuit Properties of CableShorted Line: The long loop has inductance since themagnetic flux ψ is not negligible (long cable) (ψ = LI)ψIIOpen Line: The cable also has substantial capacitance(Q = CV )++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++_________________________________________________________________________________++VQQ+_University of California, Berkeley EECS 117 Lecture 1 – p. 4/18Sectional Model (I)So do we model the cable as an inductor or as acapacitor? Or both? How?Try a distributed model: Inductance and capacitanceoccur together. They are intermingled.L L L LC C C C CCan add loss (series and shunt resistors) but let’s keepit simple for now.Add more sections and solution should convergeUniversity of California, Berkeley EECS 117 Lecture 1 – p. 5/18Sectional Model (II)More sections → The equiv LC circuit represents asmaller and smaller section and therefore lumpedcircuit approximation is more validThis is an easy problem to solve with SPICE.But the people 1866 didn’t have computers ... how didthey analyze a problem with hundreds of inductors andcapacitors?University of California, Berkeley EECS 117 Lecture 1 – p. 6/18Distributed ModelL′L′L′L′C′C′C′C′C′L = δzL′C = δzC′δzGo to a fully distributed model by letting the number ofsections go to infinityDefine inductance and capacitance per unit lengthL′= L/ℓ, C′= C/ℓFor an infinitesimal section of the line, circuit theoryapplies since signals travel instantly over aninfinitesimally small lengthUniversity of California, Berkeley EECS 117 Lecture 1 – p. 7/18KCL and KVL for a small sectionKCL: i(z) = δzC′∂v(z)∂t+ i(z + δz)KVL: v(z) = δzL′∂i(z+δz)∂t+ v(z + δz)Take limit as δz → 0We arrive at “Telegrapher’s Equatins”limδz→0i(z) − i(z + δz)δz= −∂i∂z= C′∂v∂tlimδz→0v(z) − v(z + δz)δz= −∂v∂z= L′∂i∂tUniversity of California, Berkeley EECS 117 Lecture 1 – p. 8/18Derivation of Wave EquationsWe have two coupled equations and two unkowns (iand v) ... can reduce it to two de-coupled equations:∂2i∂t∂z= − C′∂2v∂t2∂2v∂z2= − L′∂2i∂z∂tnote order of partials can be changed (at least in EE)∂2v∂z2= L′C′∂2v∂t2Same equation can be derived for current:∂2i∂z2= L′C′∂2i∂t2University of California, Berkeley EECS 117 Lecture 1 – p. 9/18The Wave EquationWe see that the currents and voltages on the transmissionline satisfy the one-dimensional wave equation. This is apartial differential equation. The solution depends onboundary conditions and the initial condition.∂2i∂z2= L′C′∂2i∂t2University of California, Berkeley EECS 117 Lecture 1 – p. 10/18Wave Equation SolutionConsider the function f(z, t) = f(z ± vt) = f(u):∂f∂z=∂f∂u∂u∂z=∂f∂u∂2f∂2z=∂2f∂u2∂f∂t=∂f∂u∂u∂t= ±v∂f∂u∂2f∂t2= ±v∂∂u∂f∂t= v2∂2f∂u2∂2f∂z2=1v2∂2f∂t2It satisfies the wave equation!University of California, Berkeley EECS 117 Lecture 1 – p. 11/18Wave Motionf(z − vt)zzf(z + vt)General voltage solution: v(z, t) = f+(z − vt) + f−(z + vt)Where v =q1LCUniversity of California, Berkeley EECS 117 Lecture 1 – p. 12/18Wave SpeedSpeed of motion can be deduced if we observe thespeed of a point on the aveformz ± vt = constantTo follow this point as time elapses, we must move the zcoordinate in step. This point moves with velocitydzdt± v = 0This is the speed at which we move with speeddzdt= ± vv is the velocity of wave propagationUniversity of California, Berkeley EECS 117 Lecture 1 – p. 13/18Current / Voltage Relationship (I)Since the current also satisfies the wave equationi(z, t) = g+(z − vt) + g−(z + vt)Recall that on a transmission line, current and voltageare related by∂i∂z= −C′∂v∂tFor the general function this gives∂g+∂u+∂g−∂u= −C′−v∂f+∂u+ v∂f−∂uUniversity of California, Berkeley EECS 117 Lecture 1 – p. 14/18Current / Voltage Relationship (II)Since the forward waves are independent of the reversewaves∂g+∂u= C′v∂f+∂u∂g−∂u= − C′v∂f−∂uWithin a constant we haveg+=f+Z0g−= −f−Z0Where Z0=qL′C′is the “Characteristic Impedance” ofthe lineUniversity of California, Berkeley EECS 117 Lecture 1 – p. 15/18Example: Step Into Infinite LineExcite a step function onto a transmission lineThe line is assumped uncharged: Q(z, 0) = 0,ψ(z, 0) = 0 or equivalently v(z, 0) = 0 and i(z, 0) = 0By physical intuiition, we would only expect a forwardtraveling wave since the line is infinite in extentThe general form of current and voltage on the line isgiven byv(z, t) = v+(z − vt)i(z, t) = i+(z − vt) =v+(z − vt)Z0The T-line looks like a resistor of Z0ohms!University of California, Berkeley EECS 117 Lecture 1 – p. 16/18Example 1 (cont)We may therefore model the line with the followingsimple equivalent circuitRsZ0isi+=v+Z0VsSince is= i+, the excited voltage wave has anamplitude ofv+=Z0Z0+ RsVsIt’s surprising that the voltage on the line is not equal tothe source voltageUniversity of California, Berkeley EECS 117 Lecture 1 –