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Berkeley ELENG 117 - Transmission Line Discontinuities

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Energy to ``Charge'' Transmission LineEnergy Stored in Inds and Caps (I)Energy Stored (II)Transmission Line TerminationReflectionsReflection CoefficientPropagation of Reflected Wave (I)Propagation of Reflected Wave (II)Propagation of Reflected Wave (III)Bounce DiagramFreeze timeFreeze SpaceSteady-State Voltage on Line (I)Steady-State Voltage on Line (II)What Happend to the T-Line?PCB Interconnect Example: Open Line (I)Example: Open Line (II)Example: Open Line RingingPhysical Intuition: Shorted Line (I)Physical Intuition: Shorted Line (II)EECS 117Lecture 2: Transmission Line DiscontinuitiesProf. NiknejadUniversity of California, BerkeleyUniversity of California, Berkeley EECS 117 Lecture 2 – p. 1/22Energy to “Charge” Transmission LineRsZ0i+=v+Z0+Vs−+v+−The power flow into the line is given byP+line= i+(0, t)v+(0, t) =v+(0, t)2Z0Or in terms of the source voltageP+line=Z0Z0+ Rs2V2sZ0=Z0(Z0+ Rs)2V2sUniversity of California, Berkeley EECS 117 Lecture 2 – p. 2/22Energy Stored in Inds and Caps (I)But where is the power going? The line is lossless!Energy stored by a cap/ind is12CV2/12LI2At time td, a length of ℓ = vtdhas been “charged”:12CV2=12ℓC′Z0Z0+ Rs2V2s12LI2=12ℓL′VsZ0+ Rs2The total energy is thus12LI2+12CV2=12ℓV2s(Z0+ Rs)2L′+ C′Z20University of California, Berkeley EECS 117 Lecture 2 – p. 3/22Energy Stored (II)Recall that Z0=pL′/C′. The total energy stored on theline at time td= ℓ/v:Eline(ℓ/v) = ℓL′V2s(Z0+ Rs)2And the power delivered onto the line in time td:Pline×ℓv=lvZ0V2s(Z0+ Rs)2= ℓrL′C′√L′C′V2s(Z0+ Rs)2As expected, the results match (conservation ofenergy).University of California, Berkeley EECS 117 Lecture 2 – p. 4/22Transmission Line TerminationRsi+=v+Z0+Vs−+v+−ℓZ0, tdi =vLRLConsider a finite transmission line with a terminationresistanceAt the load we know that Ohm’s law is valid: IL= VL/RLSo at time t = ℓ/v, our pulse reaches the load. Sincethe current on the T-line is i+= v+/Z0= Vs/(Z0+ Rs)and the current at the load is VL/RL, a discontinuity isproduced at the load.University of California, Berkeley EECS 117 Lecture 2 – p. 5/22ReflectionsThus a reflected wave is created at discontinuityVL(t) = v+(ℓ, t) + v−(ℓ, t)IL(t) =1Z0v+(ℓ, t) −1Z0v−(ℓ, t) = VL(t)/RLSolving for the forward and reflected waves2v+(ℓ, t) = VL(t)(1 + Z0/RL)2v−(ℓ, t) = VL(t)(1 − Z0/RL)University of California, Berkeley EECS 117 Lecture 2 – p. 6/22Reflection CoefficientAnd therefore the reflection from the load is given byΓL=V−(ℓ, t)V+(ℓ, t)=RL− Z0RL+ Z0Reflection coefficient is a very important concept fortransmisslin lines: −1 ≤ ΓL≤ 1ΓL= −1 for RL= 0 (short)ΓL= +1 for RL= ∞ (open)ΓL= 0 for RL= Z0(match)Impedance match is the proper termination if we don’twant any reflectionsUniversity of California, Berkeley EECS 117 Lecture 2 – p. 7/22Propagation of Reflected Wave (I)If ΓL6= 0, a new reflected wave travels toward thesource and unless Rs= Z0, another reflection alsooccurs at source!To see this consider the wave arriving at the source.Recall that since the wave PDE is linear, asuperposition of any number of solutins is also asolution.At the source end the boundary condition is as followsVs− IsRs= v+1+ v−1+ v+2The new term v+2is used to satisfy the boundaryconditionUniversity of California, Berkeley EECS 117 Lecture 2 – p. 8/22Propagation of Reflected Wave (II)The current continuity requires Is= i+1+ i−1+ i+2Vs= (v+1− v−1+ v+2)RsZ0+ v+1+ v−1+ v+2Solve for v+2in terms of known termsVs=1 +RsZ0(v+1+ v+2) +1 −RsZ0v−1+But v+1=Z0Rs+Z0VsVs=Rs+ Z0Z0Z0Rs+ Z0Vs+1 −RsZ0v−1+1 +RsZ0v+2University of California, Berkeley EECS 117 Lecture 2 – p. 9/22Propagation of Reflected Wave (III)So the source terms cancel out andv+2=Rs− Z0Z0+ Rsv−1= Γsv−1The reflected wave bounces off the source impedancewith a reflection coefficient given by the same equationas beforeΓ(R) =R − Z0R + Z0The source appears as a short for the incoming waveInvoke superposition! The term v+1took care of thesource boundary condition so our new v+2only neededto compensate for the v−1wave ... the reflected wave isonly a function of v−1University of California, Berkeley EECS 117 Lecture 2 – p. 10/22Bounce DiagramWe can track the multiple reflections with a “bouncediagram”TimeSpacetd2td3td4td5td6tdℓℓ/2ℓ/4 3ℓ/4v+1v−1= ΓLv+1v+2=Γsv−1=ΓsΓLv+1v−2= ΓLv+2= ΓsΓ2Lv+1v+3=Γsv−2=Γ2sΓ2Lv+1v−3= ΓLv+3=Γ2sΓ3Lv+1v+4= Γsv−3= Γ3sΓ3Lv+1University of California, Berkeley EECS 117 Lecture 2 – p. 11/22Freeze timeIf we freeze time and look at the line, using the bouncediagram we can figure out how many reflections haveoccurredFor instance, at time 2.5td= 2.5ℓ/v three waves havebeen excited (v+1,v−1, v+2), but v+2has only travelled adistance of ℓ/2To the left of ℓ/2, the voltage is a summation of threecomponents: v = v+1+ v−1+ v+2= v+1(1 + ΓL+ ΓLΓs).To the right of ℓ/2, the voltage has only twocomponents: v = v+1+ v−1= v+1(1 + ΓL).University of California, Berkeley EECS 117 Lecture 2 – p. 12/22Freeze SpaceWe can also pick at arbitrary point on the line and plotthe evolution of voltage as a function of timeFor instance, at the load, assuming RL> Z0andRS> Z0, so that Γs,L> 0, the voltage at the load will willincrease with each new arrival of a reflectionv+1= .4v−1= .2v+2= .04v−2= .02v+3= .004v−3= .002Rs= 75ΩRL= 150ΩΓs=0.2ΓL=0.5vss=2/3V.6.64.66.664.666td2td3td4td5td6tdvL(t)tUniversity of California, Berkeley EECS 117 Lecture 2 – p. 13/22Steady-State Voltage on Line (I)To find steady-state voltage on the line, we sum over allreflected waves:vss= v+1+ v−1+ v+2+ v−2+ v+3+ v−3+ v+4+ v−4+ ···Or in terms of the first wave on the linevss= v+1(1 + ΓL+ ΓLΓs+ Γ2LΓs+ Γ2LΓ2s+ Γ3LΓ2s+ Γ3LΓ3s+ ···Notice geometric sums of terms like ΓkLΓksand Γk+1LΓks.Let x = ΓLΓs:vss= v+1(1 + x + x2+ ··· + ΓL(1 + x + x2+ ···))University of California, Berkeley EECS 117 Lecture 2 – p. 14/22Steady-State Voltage on Line (II)The sums converge since x < 1vss= v+111 − ΓLΓs+ΓL1 − ΓLΓsOr more compactlyvss= v+11 + ΓL1 − ΓLΓsSubstituting for ΓLand Γsgivesvss= VsRLRL+ RsUniversity of California, Berkeley EECS 117 Lecture 2 – p. 15/22What Happend to the T-Line?For steady state, the equivalent circuit shows that thetransmission line has disappeared.This happens


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Berkeley ELENG 117 - Transmission Line Discontinuities

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