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Transmission Line Basics II - Class 6

Prerequisite Reading assignment: CH2




TARA SAIKUMAR
2

Real Computer Issues
                data
     Dev a                      Dev b

                           Signal
                           Measured
       Clk                 here              Switch
                                             Threshold


An engineer tells you the measured clock is non-monotonic
and because of this the flip flop internally may double clock
the data. The goal for this class is to by inspection
determine the cause and suggest whether this is a problem
or not.              Transmission Lines Class 6
3

Agenda
 The Transmission Line Concept
 Transmission line equivalent circuits
  and relevant equations
 Reflection diagram & equation
 Loading
 Termination methods and comparison
 Propagation delay
 Simple return path ( circuit theory,
  network theory come later)
                Transmission Lines Class 6
4

Two Transmission Line Viewpoints
 Steady state ( most historical view)
    Frequency domain
 Transient
    Time domain
    Not circuit element Why?
 We mix metaphors all the time
    Why convenience and history




                Transmission Lines Class 6
5


 Transmission Line Concept
Power Frequency (f) is @ 60 Hz                           Power
     Wavelength (λ ) is 5× 10 m  6
                                                         Plant
      ( Over 3,100 Miles)




         Consumer
           Home




                            Transmission Lines Class 6
PC Transmission Lines
                                                                                                    6




Signal Frequency (f) is
   approaching 10 GHz                                      Integrated Circuit
                                                                           Stripline
       Wavelength (λ ) is 1.5 cm
        ( 0.6 inches)        Microstrip                                   T

             PCB substrate
                                                                  Cross section view taken here
 Stripline
                                             W
                                                                    Via
                                                                                           Micro-
                                                                     FR4 Dielectric
   Copper Trace
                              Cross Section of Above PCB
                                                                                           Strip

                                                                              Signal (microstrip)

                                                                              Ground/Power
   T                 Copper Plane                                             Signal (stripline)
                                                                              Signal (stripline)
                                                                              Ground/Power
                                                                              Signal (microstrip)

                                     W
                                Transmission Lines Class 6
7


Key point about transmission line operation
  Voltage and current on a transmission line is
  a function of both time and position.
                                             I2

  V = f ( z, t )
                        I1




  I = f ( z, t )
                             V1              V2




                              dz
 The major deviation from circuit theory with
   transmission line, distributed networks is this
   positional dependence of voltage and current!
      Must think in terms of position and time to
      understand transmission line behavior
      This positional dependence is added when the
      assumption of the size of the circuit being
      small compared to the signaling wavelength
                    Transmission Lines Class 6
8


Examples of Transmission Line
Structures- I
 Cables and wires
   (a)   Coax cable
   (b)   Wire over ground
   (c)   Tri-lead wire
   (d)   Twisted pair (two-wire line)
 Long distance interconnects
                                                             +


             +                                          -
                                (a)                         (b)
             -




                  -   +     -                       +   -
                      (c)                         (d)
                                  Transmission Lines Class 6
9


Segment 2: Transmission line equivalent
circuits and relevant equations

 Physics of transmission line structures
 Basic transmission line equivalent circuit
 ?Equations for transmission line propagation




                   Transmission Lines Class 6
10


E & H Fields – Microstrip Case

       How does the signal move   Signal path

         from source to load?                   Y

                                                    Z (into the page)
                                                           X


                                                           Electric field

   Remember fields are setup given field
                                     Magnetic


     an applied forcing function.
              (Source)         Ground return path

  The signal is really the wave
       propagating between the
               conductors
               Transmission Lines Class 6
Transmission Line “Definition”
                                                               11




 General transmission line: a closed system in which
 power is transmitted from a source to a destination

 Our class: only TEM mode transmission lines
    A two conductor wire system with the wires in close
    proximity, providing relative impedance, velocity and
    closed current return path to the source.
    Characteristic impedance is the ratio of the voltage and
    current waves at any one position on the transmission
    line                        V
                           Z0 =
                                    I
    Propagation velocity is the speed with which signals are
    transmitted through the transmission line in its
    surrounding medium.         c
                            v=
                                    εr
                      Transmission Lines Class 6
12

Presence of Electric and Magnetic Fields
                                                                  H
       I                             I + ∆I             I               I + ∆I
               +   +       +   +

                       E
           V                       V + ∆V                   V         V + ∆V
                                                                  H
       I                             I + ∆I             I               I + ∆I
               -   -       -   -



  Both Electric and Magnetic fields are present in the
   transmission lines
       These fields are perpendicular to each other and to the direction of wave
       propagation for TEM mode waves, which is the simplest mode, and
       assumed for most simulators(except for microstrip lines which assume
       “quasi-TEM”, which is an approximated equivalent for transient response
       calculations).
  Electric field is established by a potential difference
   between two conductors.
       Implies equivalent circuit model must contain capacitor.
  Magnetic field induced by current flowing on the line
       Implies equivalent circuit model must contain inductor.
                                    Transmission Lines Class 6
13


T-Line Equivalent Circuit
  General Characteristics of Transmission
   Line
     Propagation delay per unit length (T0) { time/distance} [ps/in]
        Or Velocity (v0) {distance/ time} [in/ps]
     Characteristic Impedance (Z0)
     Per-unit-length Capacitance (C0) [pf/in]
     Per-unit-length Inductance (L0) [nf/in]
     Per-unit-length (Series) Resistance (R0) [Ω/in]
     Per-unit-length (Parallel) Conductance (G0) [S/in]

                 lR0           lL0
                                           lG0       lC0




                        Transmission Lines Class 6
14

Ideal T Line
 Ideal (lossless) Characteristics of
 Transmission Line                                 lL0
    Ideal TL assumes:
       Uniform line                                      lC0
       Perfect (lossless) conductor (R0→ 0)
       Perfect (lossless) dielectric (G0→ 0)
    We only consider T0, Z0 , C0, and L0.
 A transmission line can be represented by a
 cascaded network (subsections) of these
 equivalent models.
    The smaller the subsection the more accurate the model
    The delay for each subsection should be
    no larger than 1/10th the signal rise time.
                      Transmission Lines Class 6
Signal Frequency and Edge Rate
                                                              15




                vs.
       Lumped or Tline Models
In theory, all circuits that deliver transient power from
one point to another are transmission lines, but if the
signal frequency(s) is low compared to the size of the
circuit (small), a reasonable approximation can be
used to simplify the circuit for calculation of the circuit
transient (time vs. voltage or time vs. current)
response.




                     Transmission Lines Class 6
16

T Line Rules of Thumb
   So, what are the rules of thumb to use?


                        May treat as lumped Capacitance
                        Use this 10:1 ratio for accurate modeling
                        of transmission lines


    Td < .1 Tx

                          May treat as RC on-chip, and treat as LC
                          for PC board interconnect




    Td < .4 Tx
                 Transmission Lines Class 6
Other “Rules of Thumb”
                                                      17




 Frequency knee (Fknee) = 0.35/Tr (so if Tr is
    1nS, Fknee is 350MHz)
   This is the frequency at which most energy is
    below
   Tr is the 10-90% edge rate of the signal
   Assignment: At what frequency can your thumb be
    used to determine which elements are lumped?
       Assume 150 ps/in




                      Transmission Lines Class 6
When does a T-line become a T-Line?
                                                                                      18




                                                                Whether it is a
                                                                  bump or a
                                                                  mountain depends
                                                                  on the ratio of its
     When do we need to                                           size (tline) to the
     use transmission line                                        size of the vehicle
    analysis techniques vs.                                       (signal
        lumped circuit                                            wavelength)
           analysis?
                                                                Similarly, whether
                                                                 or not a line is to
                                                                 be considered as a
                                                                 transmission line
                                                                 depends on the
                                                                 ratio of length of
                                                                 the line (delay) to
                                                                 the wavelength of
         Wavelength/edge rate                          Tline     the applied
                                                                 frequency or the
                                                                 rise/fall edge of the
                              Transmission Lines Class 6         signal
Equations & Formulas

 How to model & explain
 transmission line behavior
20


Relevant Transmission Line Equations

 Propagation equation
   γ = ( R + jωL)(G + jωC ) = α + jβ
       α is the attenuation (loss) factor
       β is the phase (velocity) factor
 Characteristic Impedance equation
        ( R + j ωL )
   Z0 =
        (G + jωC )
 In class problem: Derive the high frequency, lossless
    approximation for Z0
                        Transmission Lines Class 6
21

Ideal Transmission Line Parameters
  Knowing any two out of Z0,
   Td, C0, and L0, the other two
                                                        L0
   can be calculated.                         Z0 =         ;     T d = L0 C0 ;
  C0 and L0 are reciprocal                             C0
   functions of the line cross-                    T0
   sectional dimensions and                   C0 =    ;         L0 = Z 0 T 0 ;
                                                   Z0
   are related by constant me.
                                                    1
  ε is electric permittivity                 v0 =      ;      C0 L0 = µε;
       ε 0= 8.85 X 10-12 F/m (free space)           µε
       ε ri s relative dielectric constant
                                               µ = µr µ0 ;     ε = εr ε0 .
  µ is magnetic permeability
       µ 0= 4p X 10-7 H/m (free space)
       µ r is relative permeability


  Don’t forget these relationships and what they mean!

                                Transmission Lines Class 6
Parallel Plate Approximation
                                                                                                   22




 Assumptions                                                                            TC

      TEM conditions                                     ε                             TD

      Uniform dielectric (ε )
      between conductors                                     WC
      TC<< TD; WC>> TD
                                                ε * PlateArea Base
 T-line characteristics are                 C=
  function of:                                        d       equation
      Material electric and                         WC  F                         WC  pF 
      magnetic properties                  C0    ε⋅   ⋅               8.85 ⋅ε r ⋅   ⋅ 
                                                    TD  m                         TD  m 
      Dielectric Thickness (TD)
                                                    TD  F                           T D  µH 
      Width of conductor (WC)
                                           L0    µ⋅   ⋅               0.4 ⋅π ⋅µ r ⋅     ⋅   
 Trade-off                                         WC  m                           WC  m 
      TD ; C0 , L0 , Z0                              TD       µr
                                           Z0    377 ⋅        ⋅        ⋅Ω
      WC ; C0 , L0 , Z0                              WC       εr

To a first order, t-line capacitance and inductance can
be approximated using the parallel plate approximation.
                               Transmission Lines Class 6
23


Improved Microstrip Formula
 Parallel Plate Assumptions +
                                                                     WC
         Large ground plane with
                                                                              TC
         zero thickness
                                                                 ε                         TD
    To accurately predict
     microstrip impedance, you
     must calculate the effective
     dielectric constant.                               From Hall, Hall & McCall:
                87        5.98TD 
     Z0 ≈              ln                             Valid when:
              εr + 1.41  0.8WC + TC                   0.1 < WC/TD < 2.0 and 1 <   r   < 15

         εr + 1       εr − 1                                TC
    εe =        +               + F − 0.217( εr − 1)
           2             12TD                              WCTD
                    2 1+
                          WC
                                2                                You can’t beat
                           WC 
            0.02(εr −1)1 −                                      a field solver
                                              WC
                                       for
                                              TD
                                                   <1
    F=                     TD 
            0                           for
                                              WC
                                                   >1
                                              TD

                                  Transmission Lines Class 6
24


Improved Stripline Formulas
 Same assumptions as                                        WC                TD1
    used for microstrip                               ε
                                                                      TC

    apply here                                                                 TD2

     From Hall, Hall & McCall:
Symmetric (balanced) Stripline Case TD1 = TD2

                      60    4(TD1 + TD1)     
          Z 0 sym ≈   ln                     
                         0.67π (0.8WC + TC ) 
                    εr                       
      Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25
                                                                     You can’t beat a
Offset (unbalanced) Stripline Case TD1 > TD2                         field solver
                   Z 0 sym(2 A, WC , TC , εr ) ⋅ Z 0 sym(2 B, WC , TC , εr )
     Z 0offset ≈ 2
                   Z 0 sym(2 A,WC , TC , εr ) + Z 0 sym(2 B,WC , TC , εr )

                                Transmission Lines Class 6
25

Refection coefficient
 Signal on a transmission line can be analyzed by
    keeping track of and adding reflections and
    transmissions from the “bumps” (discontinuities)
   Refection coefficient
       Amount of signal reflected from the “bump”
       Frequency domain ρ=sign(S11)*|S11|
       If at load or source the reflection may be called gamma (ΓL
       or Γs)
       Time domain ρ is only defined a location
         The “bump”
       Time domain analysis is causal.
       Frequency domain is for all time.
       We use similar terms – be careful
 Reflection diagrams – more later

                         Transmission Lines Class 6
Reflection and Transmission
                                                                                       26




          Incident                    1+ρ           Transmitted




          ρ    Reflected

Reflection Coeficient Transmission Coeffiecent
                                                                             Zt − Z0
      ρ
          Zt − Z0                    τ   (1 + ρ)        "" → ""     τ   1+
                                                                             Zt + Z0
          Zt + Z0
                                                           2⋅ Zt
                                                    τ
                                                          Zt + Z0



                       Transmission Lines Class 6
27

Special Cases to Remember
              A: Terminated in Zo
         Zs                                             −
                           Zo              Zo    ρ = Zo Zo = 0
    Vs                                               Zo + Zo



              B: Short Circuit
         Zs                                           −
                           Zo                   ρ = 0 Zo = −1
    Vs                                              0 + Zo



              C: Open Circuit
         Zs                                          ∞ − Zo
                           Zo                   ρ=          =1
    Vs                                               ∞ + Zo




              Transmission Lines Class 6
28


Assignment – Building the SI Tool Box
 Compare the parallel plate
 approximation to the improved
 microstrip and stripline formulas
 for the following cases:
 Microstrip:
 WC = 6 mils, TD = 4 mils, TC = 1 mil, εr = 4
 Symmetric Stripline:
 WC = 6 mils, TD1 = TD2 = 4 mils, TC = 1 mil, εr = 4
 Write Math Cad Program to calculate Z0, Td, L
 & C for each case.
 What factors cause the errors with the parallel
 plate approximation?
                        Transmission Lines Class 6
29


Transmission line equivalent circuits and
relevant equations

  Basic pulse launching onto transmission lines
  Calculation of near and far end waveforms for
   classic load conditions




                  Transmission Lines Class 6
Review: Voltage Divider Circuit
                                                                          30




 Consider the
  simple circuit that                           RS
  contains source
  voltage VS, source                                            RL
                            VS                                       VL
  resistance RS, and
  resistive load RL.
 The output
  voltage, VL is
                                                           RL
  easily calculated                            VL = VS
  from the source                                        RL + R S
  amplitude and the
  values of the two
  series resistors.
               Why do we care for?
                  Next page….
                  Transmission Lines Class 6
31


Solving Transmission Line Problems
The next slides will establish a procedure that
  will allow you to solve transmission line
  problems without the aid of a simulator. Here
  are the steps that will be presented:
 Determination of launch voltage &
  final “DC” or “t =0” voltage
 Calculation of load reflection coefficient and
  voltage delivered to the load
 Calculation of source reflection coefficient
  and resultant source voltage
       These are the steps for solving
            all t-line problems.
                   Transmission Lines Class 6
Determining Launch Voltage
                                                                    32




                                           TD
                       Rs A                                B
             Vs
                                          Zo
    0           Vs                                             Rt


        (initial voltage)
                 t=0, V=Vi

                             Z0                             Rt
         Vi = VS                                Vf = VS
                       Z 0 + RS                           Rt + RS

 Step 1 in calculating transmission line waveforms
   is to determine the launch voltage in the circuit.
  The behavior of transmission lines makes it
  easy to calculate the launch & final voltages –
  it is simply a voltage divider!
                         Transmission Lines Class 6
33



Voltage Delivered to the Load
                                                  TD
                    Vs        Rs A                                    B
                      Vs                         Zo                       Rt
       0

             (initial voltage)
                      t=0, V=Vi

                                                                      (signal is reflected)
                 t=2TD,

      ρΒ =          Rt − Zo
                 V=Vi + ρB(Vi) +ρA(ρB)(Vi )                   t=TD, V=Vi +ρB(Vi )
                                                        Vreflected = ρ Β (Vincident)
                     Rt + Zo                            VB = Vincident + Vreflected
  Step 2: Determine VB in the circuit at time t = TD
     The transient behavior of transmission line delays the
      arrival of launched voltage until time t = TD.
          VB at time 0 < t < TD is at quiescent voltage (0 in this case)
     Voltage wavefront will be reflected at the end of the t-line
          VB = Vincident + Vreflected at time t = TD
                               Transmission Lines Class 6
34


Voltage Reflected Back to the Source

                     Rs A                                         B
           Vs
                                         Zo
 0           Vs          ρA                                  ρB       Rt


                                        TD
     (initial voltage)
             t=0, V=Vi


                                                                      (signal is reflected)
         t=2TD,
         V=Vi + ρB (Vi) + ρA B )(Vi )
                            (ρ                             t=TD, V=Vi + ρB (Vi )




                              Transmission Lines Class 6
Voltage Reflected Back to the Source                                                35




                 − Zo                             Vreflected = ρ Α (Vincident)
        ρ Α = Rs
              Rs + Zo                        VA = Vlaunch + Vincident + Vreflected


Step 3: Determine VA in the circuit at time t = 2TD
   The transient behavior of transmission line delays the
    arrival of voltage reflected from the load until time t =
    2TD.
       VA at time 0 < t < 2TD is at launch voltage
   Voltage wavefront will be reflected at the source
       VA = Vlaunch + Vincident + Vreflected at time t = 2TD

In the steady state, the solution converges to
  VB = VS[Rt / (Rt + Rs)]

                                  Transmission Lines Class 6
36


Problems      Solved Homework

 Consider the circuit
  shown to the right
  with a resistive load,
  assume propagation
                                     RS      I1             I2
                                                  Z0 ,Τ 0

  delay = T, RS= Z0 .           VS           V1
                                                   l
                                                            V2   RL

  Calculate and show
  the wave forms of
  V1(t),I1(t),V2(t),
  and I2(t) for (a) RL=
  ∞ and (b) RL= 3Z0

                Transmission Lines Class 6
37


Step-Function into T-Line: Relationships
 Source matched case: RS= Z0
     V1(0) = 0.5VA, I1(0) = 0.5IA
    Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L)
 Uncharged line
    V2(0) = 0, I2(0) = 0
 Open circuit means RL= ∞
    Γ L = ∞ /∞ = 1
    V1(∞) = V2(∞) = 0.5VA(1+1) = VA
    I1(∞) = I2 (∞) = 0.5IA(1-1) = 0
                                                Solution
                   Transmission Lines Class 6
38


Step-Function into T-Line with Open Ckt
 At t = T, the voltage wave reaches load end
 and doubled wave travels back to source end
    V1(T) = 0.5VA, I1(T) = 0.5VA/Z0
    V2(T) = VA, I2 (T) = 0
 At t = 2T, the doubled wave reaches the
 source end and is not reflected
    V1(2T) = VA, I1(2T) = 0
    V2(2T) = VA, I2(2T) = 0


                                              Solution
                 Transmission Lines Class 6
39



Waveshape:
Step-Function into T-Line with Open Ckt
                                                 I1
                  IA
                                                 I2
                                                              RS   I1               I2
Current (A)




              0.75IA                                                     Z0 ,Τ 0

                                                                          l
               0.5I A                                    VS        V1              V2     Open

              0.25IA



                        0   Τ   2Τ      3Τ        4Τ Time (ns)

                  VA
                                                 V1                This is called
                                                 V2                “reflected wave
Voltage (V)




              0.75VA
                                                                   switching”
               0.5VA

              0.25VA

                                                                               Solution
                        0   Τ   2Τ      3Τ        4Τ Time (ns)
                                 Transmission Lines Class 6
40


Problem 1b: Relationships
 Source matched case: RS= Z0
     V1(0) = 0.5VA, I1(0) = 0.5IA
    Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L)
 Uncharged line
    V2(0) = 0, I2(0) = 0
 RL= 3Z0
    Γ L = (3Z0 -Z0) / (3Z0 +Z0) = 0.5
    V1(∞) = V2(∞) = 0.5VA(1+0.5) = 0.75VA
    I1(∞) = I2(∞) = 0.5IA(1-0.5) = 0.25IA
                                                Solution
                   Transmission Lines Class 6
41


Problem 1b: Solution
 At t = T, the voltage wave reaches load end
 and positive wave travels back to the source
    V1(T) = 0.5VA, I1(T) = 0.5IA
    V2(T) = 0.75VA , I2(T) = 0.25IA
 At t = 2T, the reflected wave reaches the
 source end and absorbed
    V1(2T) = 0.75VA , I1(2T) = 0.25IA
    V2(2T) = 0.75VA , I2(2T) = 0.25IA


                                              Solution
                 Transmission Lines Class 6
Waveshapes for Problem 1b
                                                                                                         42




                                                   I1
                     IA
                                                   I2                    I1                  I2
                                                                   RS
                 0.75IA                                                        Z0 ,Τ 0
 Current (A)




                                                                                l
                  0.5IA                                     VS           V1                 V2      RL

                 0.25IA



                          0   Τ   2Τ       3Τ       4Τ Time (ns)

                                                   I1
                     VA
                                                   I2
                                                                        Note that a
                 0.75VA                                                 properly terminated
   Voltage (V)




                  0.5VA
                                                                        wave settle out at
                                                                        0.5 V
                 0.25VA
                                                        Solution
                          0   Τ   2Τ       3Τ       4Τ Time (ns)
                                                                                         Solution
                                       Transmission Lines Class 6
Transmission line step response
                                                          43




    Introduction to lattice diagram analysis
    Calculation of near and far end waveforms for
       classic load impedances
      Solving multiple reflection problems


   Complex signal reflections at different types of
   transmission line “discontinuities” will be analyzed
   in this chapter. Lattice diagrams will be introduced
   as a solution tool.


                     Transmission Lines Class 6
44


Lattice Diagram Analysis – Key Concepts
                                                                      V(source)     Zo           V(load)
                                                             Vs      Rs
     The lattice diagram is a                        0                         TD = N ps
                                                                Vs                                  Rt
     tool/technique to simplify
     the accounting of                                                     ρsource               ρload
     reflections and waveforms                                                                      V(load)
                                                         Time V(source)
  Diagram shows the boundaries                          0                           a
     (x =0 and x=l) and the reflection                                                                   A’
     coefficients (GL and GL )
                                                         N ps        A
    Time (in T) axis shown                                                              b
     vertically
    Slope of the line should                            2N ps
                                                                                             c
                                                                                                         B’
     indicate flight time of signal
        Particularly important for multiple
        reflection problems using both                   3N ps       B
        microstrip and stripline mediums.                                                d
  Calculate voltage amplitude                                                                           C’
     for each successive reflected                       4N ps
                                                                                         e
     wave
    Total voltage at any point is the                   5N ps
     sum of all the waves that have
     reached that point
                                    Transmission Lines Class 6
Lattice Diagram Analysis – Detail                                                                                             45



                                           ρ                               ρ
                                               source                          load

                               V(source)                                          V(load)

      0                                                  Vlaunch
                                                                                      0

     Time                      Vlaunch                                                                                 N ps
                                                        Vlaunch ρload

                                                                                      Vlaunch(1+ρload)                Time
      2N ps
                                                   Vlaunch ρloadρsource

     Vlaunch(1+ρload +ρload ρsource)                                                                                   3N ps

                                                   Vlaunch ρ2loadρsource

                                                                                       Vlaunch(1+ρload+ρ2loadρsource+ ρ2loadρ2source)
      4N ps
                                                  Vlaunch ρ2loadρ2source
          V(source) Zo    V(load)
    Vs   Rs                                                                                                            5N ps
0     Vs        TD = N ps
                             Rt


                                           Transmission Lines Class 6
Transient Analysis – Over Damped                                                                                                  46



                 V(source)     Zo       V(load)        Assume Zs=75 ohms
        2v      Zs                                     Zo=50ohms
0                        TD = 250 ps                   Vs=0-2 volts
         Vs
                                                                                    Zo          50 
                                                           Vinitial = Vs                  = (2)          = 0.8
                     ρ source = 0.2      ρ load = 1                               Zs + Zo       75 + 50 
    Time V(source)                            V(load)                         Zs − Zo 75 − 50
    0
                                                           ρ source =                =        = 0.2
                              0.8v                                            Zs + Zo 75 + 50
                                                  0v
                                                                          Zl − Zo ∞ − 50
    500 ps    0.8v
                                                           ρ load =              =       =1
                                                                          Zl + Zo ∞ + 50
                                0.8v

    1000 ps                                       1.6v                                Response fr om lattice diagram
                                0.16v
                                                                        2.5
    1500 ps 1.76v                                                        2
                                0.16v
                                                                        1.5
                                                              V olt s
                                                  1.92v                  1                                       Sour ce
    2000 ps
                              0.032v                                    0.5
                                                                                                                 Load


                                                                         0
    2500 ps                                                                   0        2 50     500        750      1000   1250

                                                                                                  Tim e , ps


                                        Transmission Lines Class 6
Transient Analysis – Under Damped                                                                                                 47



                     V(source)      Zo      V(load)        Assume Zs=25 ohms
        2v      Zs                                         Zo =50ohms
0          Vs                 TD = 250 ps                  Vs=0-2 volts

                                                                                    Zo           50 
                                                               Vinitial =Vs               = (2)          =1.3333
                     ρsource = −0 . 3333     ρload = 1                            Zs + Zo        25 +50 
    Time      V(source)                            V(load)
                                                                                 Zs − Zo 25 −50
    0                                                          ρsource =                =        = −0.33333
                                 1.33v                                           Zs + Zo 25 + 50
                                                      0v
                                                                             Zl − Zo ∞ −50
    500 ps 1.33v                                               ρ =                  =      =1
                                  1.33v
                                                                load
                                                                             Zl + Zo ∞ +50

    1000 ps                                           2.66v                             Response from lattice diagram
                                  -0.443v

                                                                        3
    1500 ps 2.22v
                                  -0.443v                              2.5
                                                                        2
                                                               Volts
                                                      1.77v            1.5
    2000 ps
                                 0.148v                                 1                                          Source
                                                                       0.5                                        Load
    2500 ps     1.92                                                    0
                                  0.148v                                     0    250   500   750 1000 1250 1500 1750 2000 2250

                                                                                                   Time, ps
                                                      2.07

                                             Transmission Lines Class 6
Two Segment Transmission Line Structures
                                                                                                              48

                       X             X
            Rs
                      Zo1            Zo2           Rt
      Vs
                      TD             TD                 A= a                     A' = b + e
                            T3 T2                       B = a+c+d                B' = b + e + g + i
                 ρ1         ρ 2 ρ3            ρ4        C = A+ c+ d + f + h      C' = b + e + g + i + k + l
                       a                                                      a = vi
                                                                  Z o1
                                                        vi = Vs
      TD A             c             b                          Rs + Z o1     b = aT2

      2TD                                                      Rs − Z o1      c = aρ 2
                       d             e                  ρ1 =
                                                               Rs + Z o1
      3TD B                          g
                                                   A’                         d = cρ 1
                       f
                                                               Z o 2 − Z o1
      4TD
                                                        ρ2 =                  e = bρ 4
                       h             i                         Z o 2 + Z o1
                                                   B’                         f = dρ 2 + eT3
      5TD C            j             k                       Z − Zo2
                                                        ρ 3 = o1              g = eρ 3 + dT2
                                                             Z o1 + Z o 2
                                     l
                                                             Rt − Z o 2       h = fρ 1
                                                   C’   ρ4 =
                                                             Rt + Z o 2       i = gρ 4

                                                        T2 = 1 + ρ 2          j = hρ 2 + iT3

                                                        T3 = 1 + ρ 3          k = iρ 3 + hT2
                                         Transmission Lines Class 6
49



Assignment                       Previous examples are the
                                 preparation
 Consider the two segment
   transmission line shown to
   the right. Assume RS= 3Z01
   and Z02= 3Z01 . Use Lattice R             I1               I2                I3
                                         S

   diagram and calculate                          Z01 ,Τ 01        Z02 ,Τ 02

                                                    l1               l2
   reflection coefficients atV       S
                                             V1               V2               V3    Short


   the interfaces and show
   the wave forms of V1(t),
   V2(t), and V3(t).
 Check results with PSPICE


                  Transmission Lines Class 6

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Transmission Line Basics

  • 1. Transmission Line Basics II - Class 6 Prerequisite Reading assignment: CH2 TARA SAIKUMAR
  • 2. 2 Real Computer Issues data Dev a Dev b Signal Measured Clk here Switch Threshold An engineer tells you the measured clock is non-monotonic and because of this the flip flop internally may double clock the data. The goal for this class is to by inspection determine the cause and suggest whether this is a problem or not. Transmission Lines Class 6
  • 3. 3 Agenda  The Transmission Line Concept  Transmission line equivalent circuits and relevant equations  Reflection diagram & equation  Loading  Termination methods and comparison  Propagation delay  Simple return path ( circuit theory, network theory come later) Transmission Lines Class 6
  • 4. 4 Two Transmission Line Viewpoints  Steady state ( most historical view) Frequency domain  Transient Time domain Not circuit element Why?  We mix metaphors all the time Why convenience and history Transmission Lines Class 6
  • 5. 5 Transmission Line Concept Power Frequency (f) is @ 60 Hz Power Wavelength (λ ) is 5× 10 m 6 Plant ( Over 3,100 Miles) Consumer Home Transmission Lines Class 6
  • 6. PC Transmission Lines 6 Signal Frequency (f) is approaching 10 GHz Integrated Circuit Stripline Wavelength (λ ) is 1.5 cm ( 0.6 inches) Microstrip T PCB substrate Cross section view taken here Stripline W Via Micro- FR4 Dielectric Copper Trace Cross Section of Above PCB Strip Signal (microstrip) Ground/Power T Copper Plane Signal (stripline) Signal (stripline) Ground/Power Signal (microstrip) W Transmission Lines Class 6
  • 7. 7 Key point about transmission line operation Voltage and current on a transmission line is a function of both time and position. I2 V = f ( z, t ) I1 I = f ( z, t ) V1 V2 dz The major deviation from circuit theory with transmission line, distributed networks is this positional dependence of voltage and current! Must think in terms of position and time to understand transmission line behavior This positional dependence is added when the assumption of the size of the circuit being small compared to the signaling wavelength Transmission Lines Class 6
  • 8. 8 Examples of Transmission Line Structures- I  Cables and wires (a) Coax cable (b) Wire over ground (c) Tri-lead wire (d) Twisted pair (two-wire line)  Long distance interconnects + + - (a) (b) - - + - + - (c) (d) Transmission Lines Class 6
  • 9. 9 Segment 2: Transmission line equivalent circuits and relevant equations  Physics of transmission line structures  Basic transmission line equivalent circuit  ?Equations for transmission line propagation Transmission Lines Class 6
  • 10. 10 E & H Fields – Microstrip Case How does the signal move Signal path from source to load? Y Z (into the page) X Electric field Remember fields are setup given field Magnetic an applied forcing function. (Source) Ground return path The signal is really the wave propagating between the conductors Transmission Lines Class 6
  • 11. Transmission Line “Definition” 11  General transmission line: a closed system in which power is transmitted from a source to a destination  Our class: only TEM mode transmission lines A two conductor wire system with the wires in close proximity, providing relative impedance, velocity and closed current return path to the source. Characteristic impedance is the ratio of the voltage and current waves at any one position on the transmission line V Z0 = I Propagation velocity is the speed with which signals are transmitted through the transmission line in its surrounding medium. c v= εr Transmission Lines Class 6
  • 12. 12 Presence of Electric and Magnetic Fields H I I + ∆I I I + ∆I + + + + E V V + ∆V V V + ∆V H I I + ∆I I I + ∆I - - - -  Both Electric and Magnetic fields are present in the transmission lines These fields are perpendicular to each other and to the direction of wave propagation for TEM mode waves, which is the simplest mode, and assumed for most simulators(except for microstrip lines which assume “quasi-TEM”, which is an approximated equivalent for transient response calculations).  Electric field is established by a potential difference between two conductors. Implies equivalent circuit model must contain capacitor.  Magnetic field induced by current flowing on the line Implies equivalent circuit model must contain inductor. Transmission Lines Class 6
  • 13. 13 T-Line Equivalent Circuit  General Characteristics of Transmission Line Propagation delay per unit length (T0) { time/distance} [ps/in] Or Velocity (v0) {distance/ time} [in/ps] Characteristic Impedance (Z0) Per-unit-length Capacitance (C0) [pf/in] Per-unit-length Inductance (L0) [nf/in] Per-unit-length (Series) Resistance (R0) [Ω/in] Per-unit-length (Parallel) Conductance (G0) [S/in] lR0 lL0 lG0 lC0 Transmission Lines Class 6
  • 14. 14 Ideal T Line  Ideal (lossless) Characteristics of Transmission Line lL0 Ideal TL assumes: Uniform line lC0 Perfect (lossless) conductor (R0→ 0) Perfect (lossless) dielectric (G0→ 0) We only consider T0, Z0 , C0, and L0.  A transmission line can be represented by a cascaded network (subsections) of these equivalent models. The smaller the subsection the more accurate the model The delay for each subsection should be no larger than 1/10th the signal rise time. Transmission Lines Class 6
  • 15. Signal Frequency and Edge Rate 15 vs. Lumped or Tline Models In theory, all circuits that deliver transient power from one point to another are transmission lines, but if the signal frequency(s) is low compared to the size of the circuit (small), a reasonable approximation can be used to simplify the circuit for calculation of the circuit transient (time vs. voltage or time vs. current) response. Transmission Lines Class 6
  • 16. 16 T Line Rules of Thumb So, what are the rules of thumb to use? May treat as lumped Capacitance Use this 10:1 ratio for accurate modeling of transmission lines Td < .1 Tx May treat as RC on-chip, and treat as LC for PC board interconnect Td < .4 Tx Transmission Lines Class 6
  • 17. Other “Rules of Thumb” 17  Frequency knee (Fknee) = 0.35/Tr (so if Tr is 1nS, Fknee is 350MHz)  This is the frequency at which most energy is below  Tr is the 10-90% edge rate of the signal  Assignment: At what frequency can your thumb be used to determine which elements are lumped? Assume 150 ps/in Transmission Lines Class 6
  • 18. When does a T-line become a T-Line? 18  Whether it is a bump or a mountain depends on the ratio of its When do we need to size (tline) to the use transmission line size of the vehicle analysis techniques vs. (signal lumped circuit wavelength) analysis?  Similarly, whether or not a line is to be considered as a transmission line depends on the ratio of length of the line (delay) to the wavelength of Wavelength/edge rate Tline the applied frequency or the rise/fall edge of the Transmission Lines Class 6 signal
  • 19. Equations & Formulas How to model & explain transmission line behavior
  • 20. 20 Relevant Transmission Line Equations Propagation equation γ = ( R + jωL)(G + jωC ) = α + jβ α is the attenuation (loss) factor β is the phase (velocity) factor Characteristic Impedance equation ( R + j ωL ) Z0 = (G + jωC ) In class problem: Derive the high frequency, lossless approximation for Z0 Transmission Lines Class 6
  • 21. 21 Ideal Transmission Line Parameters  Knowing any two out of Z0, Td, C0, and L0, the other two L0 can be calculated. Z0 = ; T d = L0 C0 ;  C0 and L0 are reciprocal C0 functions of the line cross- T0 sectional dimensions and C0 = ; L0 = Z 0 T 0 ; Z0 are related by constant me. 1  ε is electric permittivity v0 = ; C0 L0 = µε; ε 0= 8.85 X 10-12 F/m (free space) µε ε ri s relative dielectric constant µ = µr µ0 ; ε = εr ε0 .  µ is magnetic permeability µ 0= 4p X 10-7 H/m (free space) µ r is relative permeability Don’t forget these relationships and what they mean! Transmission Lines Class 6
  • 22. Parallel Plate Approximation 22  Assumptions TC TEM conditions ε TD Uniform dielectric (ε ) between conductors WC TC<< TD; WC>> TD ε * PlateArea Base  T-line characteristics are C= function of: d equation Material electric and WC  F  WC  pF  magnetic properties C0 ε⋅ ⋅  8.85 ⋅ε r ⋅ ⋅  TD  m  TD  m  Dielectric Thickness (TD) TD  F  T D  µH  Width of conductor (WC) L0 µ⋅ ⋅  0.4 ⋅π ⋅µ r ⋅ ⋅   Trade-off WC  m  WC  m  TD ; C0 , L0 , Z0  TD µr Z0 377 ⋅ ⋅ ⋅Ω WC ; C0 , L0 , Z0  WC εr To a first order, t-line capacitance and inductance can be approximated using the parallel plate approximation. Transmission Lines Class 6
  • 23. 23 Improved Microstrip Formula  Parallel Plate Assumptions + WC Large ground plane with TC zero thickness ε TD  To accurately predict microstrip impedance, you must calculate the effective dielectric constant. From Hall, Hall & McCall: 87  5.98TD  Z0 ≈ ln  Valid when: εr + 1.41  0.8WC + TC  0.1 < WC/TD < 2.0 and 1 < r < 15 εr + 1 εr − 1 TC εe = + + F − 0.217( εr − 1) 2 12TD WCTD 2 1+ WC 2 You can’t beat  WC  0.02(εr −1)1 − a field solver WC  for TD <1 F=  TD  0 for WC >1 TD Transmission Lines Class 6
  • 24. 24 Improved Stripline Formulas  Same assumptions as WC TD1 used for microstrip ε TC apply here TD2 From Hall, Hall & McCall: Symmetric (balanced) Stripline Case TD1 = TD2 60 4(TD1 + TD1)  Z 0 sym ≈ ln   0.67π (0.8WC + TC )  εr   Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25 You can’t beat a Offset (unbalanced) Stripline Case TD1 > TD2 field solver Z 0 sym(2 A, WC , TC , εr ) ⋅ Z 0 sym(2 B, WC , TC , εr ) Z 0offset ≈ 2 Z 0 sym(2 A,WC , TC , εr ) + Z 0 sym(2 B,WC , TC , εr ) Transmission Lines Class 6
  • 25. 25 Refection coefficient  Signal on a transmission line can be analyzed by keeping track of and adding reflections and transmissions from the “bumps” (discontinuities)  Refection coefficient Amount of signal reflected from the “bump” Frequency domain ρ=sign(S11)*|S11| If at load or source the reflection may be called gamma (ΓL or Γs) Time domain ρ is only defined a location The “bump” Time domain analysis is causal. Frequency domain is for all time. We use similar terms – be careful  Reflection diagrams – more later Transmission Lines Class 6
  • 26. Reflection and Transmission 26 Incident 1+ρ Transmitted ρ Reflected Reflection Coeficient Transmission Coeffiecent Zt − Z0 ρ Zt − Z0 τ (1 + ρ) "" → "" τ 1+ Zt + Z0 Zt + Z0 2⋅ Zt τ Zt + Z0 Transmission Lines Class 6
  • 27. 27 Special Cases to Remember A: Terminated in Zo Zs − Zo Zo ρ = Zo Zo = 0 Vs Zo + Zo B: Short Circuit Zs − Zo ρ = 0 Zo = −1 Vs 0 + Zo C: Open Circuit Zs ∞ − Zo Zo ρ= =1 Vs ∞ + Zo Transmission Lines Class 6
  • 28. 28 Assignment – Building the SI Tool Box Compare the parallel plate approximation to the improved microstrip and stripline formulas for the following cases: Microstrip: WC = 6 mils, TD = 4 mils, TC = 1 mil, εr = 4 Symmetric Stripline: WC = 6 mils, TD1 = TD2 = 4 mils, TC = 1 mil, εr = 4 Write Math Cad Program to calculate Z0, Td, L & C for each case. What factors cause the errors with the parallel plate approximation? Transmission Lines Class 6
  • 29. 29 Transmission line equivalent circuits and relevant equations  Basic pulse launching onto transmission lines  Calculation of near and far end waveforms for classic load conditions Transmission Lines Class 6
  • 30. Review: Voltage Divider Circuit 30  Consider the simple circuit that RS contains source voltage VS, source RL VS VL resistance RS, and resistive load RL.  The output voltage, VL is RL easily calculated VL = VS from the source RL + R S amplitude and the values of the two series resistors. Why do we care for? Next page…. Transmission Lines Class 6
  • 31. 31 Solving Transmission Line Problems The next slides will establish a procedure that will allow you to solve transmission line problems without the aid of a simulator. Here are the steps that will be presented:  Determination of launch voltage & final “DC” or “t =0” voltage  Calculation of load reflection coefficient and voltage delivered to the load  Calculation of source reflection coefficient and resultant source voltage These are the steps for solving all t-line problems. Transmission Lines Class 6
  • 32. Determining Launch Voltage 32 TD Rs A B Vs Zo 0 Vs Rt (initial voltage) t=0, V=Vi Z0 Rt Vi = VS Vf = VS Z 0 + RS Rt + RS Step 1 in calculating transmission line waveforms is to determine the launch voltage in the circuit.  The behavior of transmission lines makes it easy to calculate the launch & final voltages – it is simply a voltage divider! Transmission Lines Class 6
  • 33. 33 Voltage Delivered to the Load TD Vs Rs A B Vs Zo Rt 0 (initial voltage) t=0, V=Vi (signal is reflected) t=2TD, ρΒ = Rt − Zo V=Vi + ρB(Vi) +ρA(ρB)(Vi ) t=TD, V=Vi +ρB(Vi ) Vreflected = ρ Β (Vincident) Rt + Zo VB = Vincident + Vreflected Step 2: Determine VB in the circuit at time t = TD  The transient behavior of transmission line delays the arrival of launched voltage until time t = TD.  VB at time 0 < t < TD is at quiescent voltage (0 in this case)  Voltage wavefront will be reflected at the end of the t-line  VB = Vincident + Vreflected at time t = TD Transmission Lines Class 6
  • 34. 34 Voltage Reflected Back to the Source Rs A B Vs Zo 0 Vs ρA ρB Rt TD (initial voltage) t=0, V=Vi (signal is reflected) t=2TD, V=Vi + ρB (Vi) + ρA B )(Vi ) (ρ t=TD, V=Vi + ρB (Vi ) Transmission Lines Class 6
  • 35. Voltage Reflected Back to the Source 35 − Zo Vreflected = ρ Α (Vincident) ρ Α = Rs Rs + Zo VA = Vlaunch + Vincident + Vreflected Step 3: Determine VA in the circuit at time t = 2TD  The transient behavior of transmission line delays the arrival of voltage reflected from the load until time t = 2TD.  VA at time 0 < t < 2TD is at launch voltage  Voltage wavefront will be reflected at the source  VA = Vlaunch + Vincident + Vreflected at time t = 2TD In the steady state, the solution converges to VB = VS[Rt / (Rt + Rs)] Transmission Lines Class 6
  • 36. 36 Problems Solved Homework  Consider the circuit shown to the right with a resistive load, assume propagation RS I1 I2 Z0 ,Τ 0 delay = T, RS= Z0 . VS V1 l V2 RL Calculate and show the wave forms of V1(t),I1(t),V2(t), and I2(t) for (a) RL= ∞ and (b) RL= 3Z0 Transmission Lines Class 6
  • 37. 37 Step-Function into T-Line: Relationships  Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L)  Uncharged line V2(0) = 0, I2(0) = 0  Open circuit means RL= ∞ Γ L = ∞ /∞ = 1 V1(∞) = V2(∞) = 0.5VA(1+1) = VA I1(∞) = I2 (∞) = 0.5IA(1-1) = 0 Solution Transmission Lines Class 6
  • 38. 38 Step-Function into T-Line with Open Ckt  At t = T, the voltage wave reaches load end and doubled wave travels back to source end V1(T) = 0.5VA, I1(T) = 0.5VA/Z0 V2(T) = VA, I2 (T) = 0  At t = 2T, the doubled wave reaches the source end and is not reflected V1(2T) = VA, I1(2T) = 0 V2(2T) = VA, I2(2T) = 0 Solution Transmission Lines Class 6
  • 39. 39 Waveshape: Step-Function into T-Line with Open Ckt I1 IA I2 RS I1 I2 Current (A) 0.75IA Z0 ,Τ 0 l 0.5I A VS V1 V2 Open 0.25IA 0 Τ 2Τ 3Τ 4Τ Time (ns) VA V1 This is called V2 “reflected wave Voltage (V) 0.75VA switching” 0.5VA 0.25VA Solution 0 Τ 2Τ 3Τ 4Τ Time (ns) Transmission Lines Class 6
  • 40. 40 Problem 1b: Relationships  Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L)  Uncharged line V2(0) = 0, I2(0) = 0  RL= 3Z0 Γ L = (3Z0 -Z0) / (3Z0 +Z0) = 0.5 V1(∞) = V2(∞) = 0.5VA(1+0.5) = 0.75VA I1(∞) = I2(∞) = 0.5IA(1-0.5) = 0.25IA Solution Transmission Lines Class 6
  • 41. 41 Problem 1b: Solution  At t = T, the voltage wave reaches load end and positive wave travels back to the source V1(T) = 0.5VA, I1(T) = 0.5IA V2(T) = 0.75VA , I2(T) = 0.25IA  At t = 2T, the reflected wave reaches the source end and absorbed V1(2T) = 0.75VA , I1(2T) = 0.25IA V2(2T) = 0.75VA , I2(2T) = 0.25IA Solution Transmission Lines Class 6
  • 42. Waveshapes for Problem 1b 42 I1 IA I2 I1 I2 RS 0.75IA Z0 ,Τ 0 Current (A) l 0.5IA VS V1 V2 RL 0.25IA 0 Τ 2Τ 3Τ 4Τ Time (ns) I1 VA I2 Note that a 0.75VA properly terminated Voltage (V) 0.5VA wave settle out at 0.5 V 0.25VA Solution 0 Τ 2Τ 3Τ 4Τ Time (ns) Solution Transmission Lines Class 6
  • 43. Transmission line step response 43  Introduction to lattice diagram analysis  Calculation of near and far end waveforms for classic load impedances  Solving multiple reflection problems Complex signal reflections at different types of transmission line “discontinuities” will be analyzed in this chapter. Lattice diagrams will be introduced as a solution tool. Transmission Lines Class 6
  • 44. 44 Lattice Diagram Analysis – Key Concepts V(source) Zo V(load) Vs Rs The lattice diagram is a 0 TD = N ps Vs Rt tool/technique to simplify the accounting of ρsource ρload reflections and waveforms V(load) Time V(source)  Diagram shows the boundaries 0 a (x =0 and x=l) and the reflection A’ coefficients (GL and GL ) N ps A  Time (in T) axis shown b vertically  Slope of the line should 2N ps c B’ indicate flight time of signal Particularly important for multiple reflection problems using both 3N ps B microstrip and stripline mediums. d  Calculate voltage amplitude C’ for each successive reflected 4N ps e wave  Total voltage at any point is the 5N ps sum of all the waves that have reached that point Transmission Lines Class 6
  • 45. Lattice Diagram Analysis – Detail 45 ρ ρ source load V(source) V(load) 0 Vlaunch 0 Time Vlaunch N ps Vlaunch ρload Vlaunch(1+ρload) Time 2N ps Vlaunch ρloadρsource Vlaunch(1+ρload +ρload ρsource) 3N ps Vlaunch ρ2loadρsource Vlaunch(1+ρload+ρ2loadρsource+ ρ2loadρ2source) 4N ps Vlaunch ρ2loadρ2source V(source) Zo V(load) Vs Rs 5N ps 0 Vs TD = N ps Rt Transmission Lines Class 6
  • 46. Transient Analysis – Over Damped 46 V(source) Zo V(load) Assume Zs=75 ohms 2v Zs Zo=50ohms 0 TD = 250 ps Vs=0-2 volts Vs Zo  50  Vinitial = Vs = (2)  = 0.8 ρ source = 0.2 ρ load = 1 Zs + Zo  75 + 50  Time V(source) V(load) Zs − Zo 75 − 50 0 ρ source = = = 0.2 0.8v Zs + Zo 75 + 50 0v Zl − Zo ∞ − 50 500 ps 0.8v ρ load = = =1 Zl + Zo ∞ + 50 0.8v 1000 ps 1.6v Response fr om lattice diagram 0.16v 2.5 1500 ps 1.76v 2 0.16v 1.5 V olt s 1.92v 1 Sour ce 2000 ps 0.032v 0.5 Load 0 2500 ps 0 2 50 500 750 1000 1250 Tim e , ps Transmission Lines Class 6
  • 47. Transient Analysis – Under Damped 47 V(source) Zo V(load) Assume Zs=25 ohms 2v Zs Zo =50ohms 0 Vs TD = 250 ps Vs=0-2 volts Zo  50  Vinitial =Vs = (2)   =1.3333 ρsource = −0 . 3333 ρload = 1 Zs + Zo  25 +50  Time V(source) V(load) Zs − Zo 25 −50 0 ρsource = = = −0.33333 1.33v Zs + Zo 25 + 50 0v Zl − Zo ∞ −50 500 ps 1.33v ρ = = =1 1.33v load Zl + Zo ∞ +50 1000 ps 2.66v Response from lattice diagram -0.443v 3 1500 ps 2.22v -0.443v 2.5 2 Volts 1.77v 1.5 2000 ps 0.148v 1 Source 0.5 Load 2500 ps 1.92 0 0.148v 0 250 500 750 1000 1250 1500 1750 2000 2250 Time, ps 2.07 Transmission Lines Class 6
  • 48. Two Segment Transmission Line Structures 48 X X Rs Zo1 Zo2 Rt Vs TD TD A= a A' = b + e T3 T2 B = a+c+d B' = b + e + g + i ρ1 ρ 2 ρ3 ρ4 C = A+ c+ d + f + h C' = b + e + g + i + k + l a a = vi Z o1 vi = Vs TD A c b Rs + Z o1 b = aT2 2TD Rs − Z o1 c = aρ 2 d e ρ1 = Rs + Z o1 3TD B g A’ d = cρ 1 f Z o 2 − Z o1 4TD ρ2 = e = bρ 4 h i Z o 2 + Z o1 B’ f = dρ 2 + eT3 5TD C j k Z − Zo2 ρ 3 = o1 g = eρ 3 + dT2 Z o1 + Z o 2 l Rt − Z o 2 h = fρ 1 C’ ρ4 = Rt + Z o 2 i = gρ 4 T2 = 1 + ρ 2 j = hρ 2 + iT3 T3 = 1 + ρ 3 k = iρ 3 + hT2 Transmission Lines Class 6
  • 49. 49 Assignment Previous examples are the preparation  Consider the two segment transmission line shown to the right. Assume RS= 3Z01 and Z02= 3Z01 . Use Lattice R I1 I2 I3 S diagram and calculate Z01 ,Τ 01 Z02 ,Τ 02 l1 l2 reflection coefficients atV S V1 V2 V3 Short the interfaces and show the wave forms of V1(t), V2(t), and V3(t).  Check results with PSPICE Transmission Lines Class 6