1ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityLecture 24Time Domain Analysis of Transmission LinesIn this lecture you will learn:• Time domain analysis of transmission lines• Transients in transmission linesECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityoZ0=z()tVsl−=zTime Domain Analysis - Basics Question: How does one handle transmission lines for signals that are NOT time harmonic and when one is NOT dealing with the sinusoidal steady state?a) First thing to realize is that the notion of complex impedance has meaning only for the sinusoidal steady stateb) For an arbitrary source voltage Vs(t ), one needs to work in the time domain and start from the basic time-domain equations:sRLR() ()ttzILztzV∂∂−=∂∂ ,,() ()ttzVCztzI∂∂−=∂∂ ,,()()22222,1,ttzVvztzV∂∂=∂∂()()22222,1,ttzIvztzI∂∂=∂∂LCv1=2ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityTime Domain Analysis - BasicsoZThe equation:()()22222,1,ttzVvztzV∂∂=∂∂Has forward moving solutions of the form:()()vtzVtzV−=+,And backward moving solutions of the form:()()vtzVtzV+=−,Examples:()vtzV −+zv()vtzV+−zvLCv1=ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityVoltages and CurrentsoZ()vtzV −+zvVoltage:()vtzI −+zvCorresponding Current:()()ttzILztzV∂∂−=∂∂,,The current is related to the voltage and satisfies:And this:()()ttzVCztzI∂∂−=∂∂ ,,()()()()ooZvtzVvtzIZvtzVvtzI+−=+−=−−−++andThe solution is:++++++------Current is proportional to voltage since higher voltage means more surface charges and more surface charges mean more current flowv3ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityVisualizing PropagationoZ0=zl−=zt1Forward moving solutions are of the form:()vtzV−+And backward moving solutions are of the form:()vtzV+−So suppose somebody tells you that at z = -ℓ : ()=−+tV ,lThen what is the forward moving voltage on the line at t = ℓ / 2v ?T0=zl−=z()vtzV 2, l=+vt2l=vTECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityLoad End Boundary ConditionoZ0=z()tVsl−=zsRLRLoad end boundary condition:()()LRtzItzV ,0,0 ===()()( )tzVtzVtzV ,0,0,0 =+===−+()()()()()ooZtzVZtzVtzItzItzI,0,0 ,0,0,0=−===+===−+−+()()LtzVtzVΓ===⇒+−,0,011+−=ΓoLoLLZRZR+-()tzV ,0=()tzI ,0=For all time t we must have:Substitute these in this to get:4ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversitySource End Boundary ConditionoZ0=z()tVsl−=zsRLR+-Source end boundary condition:() ( )()tzVRtzItVss,, ll−=+−==()()( )tzVtzVtzV ,,, lll −=+−==−=−+()()( )()()ooZtzVZtzVtzItzItzI,, ,,,lllll−=−−==−=+−==−=−+−+()()()osossZRZtVtzVtzV++Γ−==−=⇒−+,, ll11+−=ΓosossZRZR()tzV ,l−=()tzI ,l−=For all time t we must have:Substitute these in this to get:ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On Transient - IΩ=50oZ0=z()tVsl−=zΩ= 50sRΩ= 150LRSuppose the source voltage is a step function:()()tutVs4=t421=ΓL0=Γs()()()()20, ,,=>−=⇒++Γ−==−=+−+tzVZRZtVtzVtzVososslll0=zl−=z()tzV ,+()tzV ,−()tzV ,0=t5ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On Transient - IIΩ=50oZ0=z()tVsl−=zΩ= 50sRΩ= 150LRSuppose the source voltage is a step function:()()tutVs4=t40=zl−=z()tzV ,+()tzV ,−()tzV ,220=zl−=z()tzV ,+()tzV ,−()tzV ,22vt2l=vt23l=21=ΓL130=ΓsECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityΩ=50oZ0=z()tVsl−=zΩ= 50sRΩ= 150LR0=zl−=z()tzV ,+()tzV ,−()tzV ,23vtl2≥21=ΓL1• The wave reflected back from the load end does not suffer a reflection at the source end since the source impedance is matched to the line impedance• After a time greater than 2ℓ/v the line voltage is at constant 3 Volts Step Voltage Source: Turn-On Transient - III0=Γs()()()osossZRZtVtzVtzV++Γ−==−=−+ ,, ll6ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-Off Transient - IΩ=50oZ0=z()tVsl−=zΩ= 50sRΩ= 150LRSuppose the source voltage has been at 4 Volts for a long long time – but is shut off at time t = 0() ()[]tutVs−= 14t40=zl−=z()tzV ,+()tzV ,−()tzV ,2vt2l=21=ΓL130=zl−=z()tzV ,+()tzV ,−()tzV ,230=t10=Γs1ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-Off Transient - IIΩ=50oZ0=z()tVsl−=zΩ= 50sRΩ= 150LRt0=zl−=z()tzV ,+()tzV ,−()tzV ,vtl2≥21=ΓL0=zl−=z()tzV ,+()tzV ,−()tzV ,vt23l=10=Γs1• After a time greater than 2ℓ/v the line voltage is at constant 0 Volts7ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On Transient – General CaseoZ0=z()tVsl−=zsRLRLΓsΓSuppose the source voltage is a step function:()()tuVtVos=toV0=zl−=z()tzV ,+()tzV ,−()tzV ,0=zl−=z()tzV ,+()tzV ,−()tzV ,vt2l=vt23l=osooZRZV+osooZRZV+LosooZRZV Γ+osooZRZV+()LosooZRZV Γ++1osooZRZV+ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On Transient – General CaseoZ()tVssRLRLΓsΓ0=zl−=z()tzV ,+()tzV ,−()tzV ,vt25l=osooZRZV+LosooZRZV Γ+()LsosooZRZV ΓΓ++1()⎥⎦⎤⎢⎣⎡Γ+ΓΓ++LLsosooZRZV1()LosooZRZV Γ++1• The waves will keep bouncing forever• But the net voltage on the line will slowly converge to the value one would expect in DC operation()()()[]()⎥⎦⎤⎢⎣⎡ΓΓ−+=∞+ΓΓ+ΓΓ+ΓΓ++=∞++LsosooLsLsLsosooZRZVzVZRZVzV11,........1,32()()()[]()⎥⎦⎤⎢⎣⎡ΓΓ−Γ+=∞+ΓΓ+ΓΓ+ΓΓ+Γ+=∞−−LsLosooLsLsLsLosooZRZVzVZRZVzV1,........1,32()()()LsLoLsLosooRRRVZRZVzVzVzV+=ΓΓ−Γ++=∞+∞=∞−+11,,,8ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On TransientΩ=50oZ0=z()tVsl−=zΩ= 150sRΩ= 150LR21=ΓL21=ΓsSource voltage is a step function:()()tutVs1=t1()tzV ,0=zl−=zECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On Transient – Capacitive Load - IΩ=50oZ0=z()tVsl−=zΩ= 50sR0=ΓsLCSource voltage is a step function:()()tutVs1=t1()tzV ,0=zl−=z9ECE 303 – Fall 2007 – Farhan Rana – Cornell UniversityStep Voltage Source: Turn-On Transient in a RC Circuit()tVsRC+-()tVCSource voltage is a step function:() ()tutVs1=t1()tVCSolution for is:t1()tVCtime constantCR==τ()tVs() ()tuetVtC⎟⎟⎠⎞⎜⎜⎝⎛−=−τ1At short times the capacitor acts like a shortAt long times the capacitor acts like an openECE 303 – Fall 2007 – Farhan Rana – Cornell