Fundamentals of Electromagnetics: A Two-Week, 8-Day, Intensive Course for Training Faculty in Electrical-, Electronics-, Communication-, and Computer- Related Engineering Departments by Nannapaneni Narayana Rao Edward C. Jordan Professor Emeritus of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, India Amrita Viswa Vidya Peetham, Coimbatore August 11, 12, 13, 14, 18, 19, 20, and 21, 2008
7-1 Module 7 Transmission Line Analysis in Time Domain Line terminated by a resistive load Transmission-line discontinuity Lines with reactive terminations and discontinuities Lines with initial conditions
7-2 Instructional Objectives 30. Find the voltage and current variations at a location on a lossless transmission line as functions of time and at an instant of time as functions of distance, and the steady state values of the line voltage and current, for a line terminated by a resistive load and excited by turning on a constant voltage source, by using the bounce-diagram technique 31. Design a lossless transmission line system by determining its parameters from information specified concerning the voltage and/or current variations on the line 32. Design a system of three lines in cascade for achieving a specified unit impulse response
7-3 Instructional Objectives 33. Compute the reflected power for a wave incident on a junction of multiple lossless transmission lines from one of the lines and the values of power transmitted into each of the other lines, where the junction may consist of lines connected in series, parallel, or series-parallel, and include resistive elements 34. Find the solutions for voltage and current along a transmission-line system excited by a constant voltage source and having reactive elements as terminations/discontinuities 35. Find the voltage and current variations at a location on a lossless transmission-line system as functions of time and at an instant of time as functions of distance, for specified nonzero initial voltage and/or current distributions along the system
7-4 Line Terminated by Resistive Load (FEME, Sec. 7.4; EEE6E, Sec. 6.2)
7-5 Notation
7-6
7-7
7-8
7-9 Excitation by Constant Voltage Source Semi-infinite Line, No Source Resistance
7-10
7-11
7-12 Example: t, s
7-13 t = 1 s
7-14 Effect of Source Resistance B.C. (+) Wave
7-15
7-16 Line Terminated by Resistance
7-17 B.C:
7-18 Define Voltage Reflection Coefficient, Then, Current Reflection Coefficient
7-19
7-20
7-21
7-22
7-23
7-24 For constant voltage source, Actual Situation in the Steady State One (+) Wave and One (–) Wave
7-25 Four equations for the four unknowns
7-26 Example:
7-27 Solving, we obtain
7-28 Bounce Diagram Technique: Constant Voltage Source
7-29 Voltage
7-30 Current
7-31 VoltageCurrent
7-32 t, s
7-33 t, s
7-34 t, s
7-35
7-36 Rectangular Pulse Source Use superposition. Example t, s
7-37 t, s
7-38 Transmission-Line Discontinuity (EEE6E, Sec. 6.3)
7-39 Transmission-Line Discontinuity
7-40
7-41
7-42 Current Transmission Coefficient, Define Voltage Transmission Coefficient,
7-43 Note that
7-44 Three Lines in Cascade (t)(t)
7-45 t, s
7-46 (t)(t)
7-47
7-48
7-49
7-50
7-51 Junction of Three Lines
7-52
7-53
7-54
7-55 Lines with ReactiveTerminations and Discontinuities (EEE6E, Sec. 6.4)
7-56 Line Terminated by an Inductor t = T+
7-57 Using I.C.,
7-58
7-59 Voltage = 0
7-60 – = 0
7-61
7-62
7-63 Line Terminated by a Capacitor t = T+
7-64
7-65 Using I.C., t > T
7-66 Lines with Initial Conditions (FEME, Sec. 7.5; EEE6E, Secs. 6.5)
7-67 Line with Initial Conditions
7-68
7-69 Example:
7-70
7-71
7-72
7-73
7-74
7-75 Uniform Distribution
7-76
7-77 t = 0.5 mS t = 1.5 mS
7-78 t = 2.5 mS
7-79 Bounce Diagram Technique for Uniform Distribution
7-80
7-81
7-82 Energy Storage in Transmission Lines w e, Electric stored energy density = W e, Electric stored energy =
7-83 w m, Magnetic stored energy density = W m, Magnetic stored energy =
7-84 Check of Energy Balance Initial stored energy
7-85 Energy dissipated in R L
7-86 Another Example: System in steady state at t = 0–.
7-87 t = 0–: steady state
7-88 t = 0+:
7-89 Solving, we obtain
7-90 Voltage = 0 V = 1
7-91 Current = 0 = 0, C = 1 C eff = 0.5
7-92 t = 3 s + : New steady state
7-93 t = 3 s + :
The End