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EECS 117 Lecture 3: Transmission Line Junctions / Time Harmonic Excitation Prof. Niknejad University of California, Berkeley University of California, Berkeley EECS 117 Lecture 3 – p. 1/23
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  • EECS 117Lecture 3: Transmission Line Junctions / Time

    Harmonic Excitation

    Prof. Niknejad

    University of California, Berkeley

    University of California, Berkeley EECS 117 Lecture 3 p. 1/23

  • Transmission Line Menagerie

    striplinemicrostripline coplanar

    rectangular

    waveguide

    coaxialtwo wires

    T-Lines come in many shapes and sizes

    Coaxial usually 75 or 50 (cable TV, Internet)

    Microstrip lines are common on printed circuit boards(PCB) and integrated circuit (ICs)

    Coplanar also common on PCB and ICs

    Twisted pairs is almost a T-line, ubiquitous forphones/Ethernet

    University of California, Berkeley EECS 117 Lecture 3 p. 2/23

  • Waveguides and Transmission Lines

    The transmission lines weve been considering havebeen propagating the TEM mode or TransverseElectro-Magnetic. Later well see that they can alsopropagation other modes

    Waveguides cannot propagate TEM but propagationTM (Transverse Magnetic) and TE (Transverse Electric)

    In general, any set of more than one losslessconductors with uniform cross-section can transmitTEM waves. Low loss conductors are commonlyapproximated as lossless.

    University of California, Berkeley EECS 117 Lecture 3 p. 3/23

  • Cascade of T-Lines (I)

    Z01 Z02

    z = 0

    i1 i2

    v1 v2

    Consider the junction between two transmission linesZ01 and Z02At the interface z = 0, the boundary conditions are thatthe voltage/current has to be continuous

    v+1

    + v1

    = v+2

    (v+1 v

    1)/Z01 = v

    +

    2/Z02

    University of California, Berkeley EECS 117 Lecture 3 p. 4/23

  • Cascade of T-Lines (II)

    Solve these equations in terms of v+1

    The reflection coefficient has the same form (easy toremember)

    =v1

    v+1

    =Z02 Z01Z01 + Z02

    The second line looks like a load impedance of valueZ02

    Z01 Z02

    z = 0

    i1

    +v1

    University of California, Berkeley EECS 117 Lecture 3 p. 5/23

  • Transmission Coefficient

    The wave launched on the new transmission line at theinterface is given by

    v+2

    = v+1

    + v1

    = v+1

    (1 + ) = v+1

    This transmitted wave has a coefficient

    = 1 + =2Z02

    Z01 + Z02

    Note the incoming wave carries a power

    Pin =|v+

    1|2

    2Z01

    University of California, Berkeley EECS 117 Lecture 3 p. 6/23

  • Conservation of Energy

    The reflected and transmitted waves likewise carry apower of

    Pref =|v

    1|2

    2Z01= ||2 |v

    +

    1|2

    2Z01Ptran =

    |v+2|2

    2Z02= | |2 |v

    +

    1|2

    2Z02

    By conservation of energy, it follows that

    Pin = Pref + Ptran

    1

    Z022 +

    1

    Z012 =

    1

    Z01

    You can verify that this relation holds!

    University of California, Berkeley EECS 117 Lecture 3 p. 7/23

  • Bounce Diagram

    Consider the bouncediagram for the follow-ing arrangement

    Z01 Z02

    Rs

    RL

    `1 `2

    Ti

    me

    Space

    td

    2td

    3td

    4td

    5td

    6td

    v+1

    1v+1

    L1v+

    1

    L12v+

    1

    sL12v+1

    jv+

    1

    jsv+

    1

    s2jv

    +

    1

    `1 `1 + `2

    td1

    University of California, Berkeley EECS 117 Lecture 3 p. 8/23

  • Junction of Parallel T-Lines

    Z01Z

    02

    z = 0

    Z03

    Again invoke voltage/current continuity at the interface

    v+1

    + v1

    = v+2

    = v+3

    v+1 v

    1

    Z01=

    v+2

    Z02+

    v+3

    Z02

    But v+2

    = v+3

    , so the interface just looks like the case oftwo transmission lines Z01 and a new line with char.impedance Z01||Z02.

    University of California, Berkeley EECS 117 Lecture 3 p. 9/23

  • Reactive Terminations (I)

    Rs

    Vs

    `

    Z0, td L

    Lets analyze the problem intuitively first

    When a pulse first sees the inductance at the load, itlooks like an open so 0 = +1

    As time progresses, the inductor looks more and morelike a short! So = 1

    University of California, Berkeley EECS 117 Lecture 3 p. 10/23

  • Reactive Terminations (II)

    So intuitively we might expect the reflection coefficientto look like this:

    1 2 3 4 5

    -1

    -0.5

    0.5

    1

    t/

    The graph starts at +1 and ends at 1. In between wellsee that it goes through exponential decay (1st orderODE)

    University of California, Berkeley EECS 117 Lecture 3 p. 11/23

  • Reactive Terminations (III)

    Do equations confirm our intuition?

    vL = Ldi

    dt= L

    d

    dt

    (

    v+

    Z0 v

    Z0

    )

    And the voltage at the load is given by v+ + v

    v +L

    Z0

    dv

    dt=

    L

    Z0

    dv+

    dt v+

    The right hand side is known, its the incomingwaveform

    University of California, Berkeley EECS 117 Lecture 3 p. 12/23

  • Solution for Reactive Term

    For the step response, the derivative term on the RHSis zero at the load

    v+ =Z0

    Z0 + RsVs

    So we have a simpler case dv+

    dt = 0

    We must solve the following equation

    v +L

    Z0

    dv

    dt= v+

    For simplicity, assume at t = 0 the wave v+ arrives atload

    University of California, Berkeley EECS 117 Lecture 3 p. 13/23

  • Laplace Domain Solution I

    In the Laplace domain

    V (s) +sL

    Z0V (s) L

    Z0v(0) = v+/s

    Solve for reflection V (s)

    V (s) =v(0)L/Z01 + sL/Z0

    v+

    s(1 + sL/Z0)

    Break this into basic terms using partial fractionexpansion

    1s(1 + sL/Z0)

    =1

    1 + sL/Z0+

    L/Z01 + sL/Z0

    University of California, Berkeley EECS 117 Lecture 3 p. 14/23

  • Laplace Domain Solution (II)

    Invert the equations to get back to time domain t > 0

    v(t) = (v(0) + v+)et/ v+

    Note that v(0) = v+ since initially the inductor is anopen

    So the reflection coefficient is

    (t) = 2et/ 1

    The reflection coefficient decays with time constantL/Z0

    University of California, Berkeley EECS 117 Lecture 3 p. 15/23

  • Time Harmonic Steady-State

    Compared with general transient case, sinusoidal caseis very easy t jSinusoidal steady state has many importantapplications for RF/microwave circuits

    At high frequency, T-lines are like interconnect fordistances on the order of

    Shorted or open T-lines are good resonators

    T-lines are useful for impedance matching

    University of California, Berkeley EECS 117 Lecture 3 p. 16/23

  • Why Sinusoidal Steady-State?

    Typical RF system modulates a sinusoidal carrier(either frequency or phase)

    If the modulation bandwidth is much smaller than thecarrier, the system looks like its excited by a puresinusoid

    Cell phones are a good example. The carrier frequencyis about 1 GHz and the voice digital modulation is about200 kHz(GSM) or 1.25 MHz(CDMA), less than a 0.1% ofthe bandwidth/carrier

    University of California, Berkeley EECS 117 Lecture 3 p. 17/23

  • Generalized Distributed Circuit Model

    Z Z Z Z Z

    Y Y Y Y Y

    Z : impedance per unit length (e.g. Z = jL + R)

    Y : admittance per unit length (e.g. Y = jC + G)

    A lossy T-line might have the following form (but wellanalyze the general case)

    L

    C

    R

    G

    L

    C

    R

    G

    L

    C

    R

    G

    L

    C

    R

    G

    University of California, Berkeley EECS 117 Lecture 3 p. 18/23

  • Time Harmonic Telegraphers Equations

    Applying KCL and KVL to a infinitesimal section

    v(z + z) v(z) = Z zi(z)

    i(z + z) i(z) = Y zv(z)Taking the limit as before (z 0)

    dv

    dz= Zi(z)

    di

    dz= Y v(z)

    University of California, Berkeley EECS 117 Lecture 3 p. 19/23

  • Sin. Steady-State (SSS) Voltage/Current

    Taking derivatives (notice z is the only variable) wearrive at

    d2v

    dz2= Z di

    dz= Y Zv(z) = 2v(z)

    d2i

    dz2= Y dv

    dz= Y Zi(z) = 2i(z)

    Where the propagation constant is a complex function

    = + j =

    (R + jL)(G + jC )

    The general solution to D2G 2G = 0 is ez

    University of California, Berkeley EECS 117 Lecture 3 p. 20/23

  • Lossless Line for SSS

    The voltage and current are related (just as before, butnow easier to derive)

    v(z) = V +ez + V ez

    i(z) =V +

    Z0ez V

    Z0ez

    Where Z0 =

    Z

    Y is the characteristic impedance of the

    line (function of frequency with loss)

    For a lossless line we discussed before, Z = jL andY = jC

    Propagation constant is imaginary

    =

    jLjC = j

    LC University of California, Berkeley EECS 117 Lecture 3 p. 21/23

  • Back to Time-Domain

    Recall that the real voltages and currents are the < and= parts of

    v(z, t) = ezejt = ejtz

    Thus the voltage/current waveforms are sinusoidal inspace and time

    Sinusoidal source voltage is transmitted unaltered ontoT-line (with delay)

    If there is loss, then has a real part , and the wavedecays or grows on the T-line

    ez = ezejz

    The first term represents amplitude response of theT-line

    University of California, Berkeley EECS 117 Lecture 3 p. 22/23

  • Passive T-Line/Wave Speed

    For a passive line, we expect the amplitude to decaydue to loss on the line

    The speed of the wave is derived as before. In order tofollow a constant point on the wavefront, you have tomove with velocity

    d

    dt(t z = constant)

    Or, v = dzdt = =

    1

    LC

    University of California, Berkeley EECS 117 Lecture 3 p. 23/23

    Transmission Line MenagerieWaveguides and Transmission LinesCascade of T-Lines (I)Cascade of T-Lines (II)Transmission CoefficientConservation of EnergyBounce DiagramJunction of Parallel T-LinesReactive Terminations (I)Reactive Terminations (II)Reactive Terminations (III)Solution for Reactive TermLaplace Domain Solution ILaplace Domain Solution (II)Time Harmonic Steady-State Why Sinusoidal Steady-State?Generalized Distributed Circuit ModelTime Harmonic Telegrapher's EquationsSin. Steady-State (SSS)Voltage/Current Lossless Line for SSSBack to Time-Domain Passive T-Line/Wave Speed