Electromagnetic Waves: Transmission Line Bounce Diagram Explained

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Learn how bounce diagrams are used to find step responses on terminated lossless lines, understand the concept through detailed diagrams, the steady-state solution, and practical examples in applied electromagnetic waves with Professor David R. Jackson. Explore the practical application of bounce diagrams for analyzing voltage waves and traces on transmission lines.

  • Electromagnetic Waves
  • Transmission Lines
  • Bounce Diagrams
  • Step Response
  • David R. Jackson
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  1. ECE 3317 Applied Electromagnetic Waves Prof. David R. Jackson Fall 2024 Notes 8 Transmission Lines (Bounce Diagram) 1

  2. Step Response The concept of the bounce diagram is useful to find a step response on a terminated lossless line: t = 0 Lossless line R R L + - g ( ) t Z gv 0 0V V = z L z = 0 u(t) = unit step function V 0 Note: ( ) t gv ( ) ( ) The bounce diagram is useful if the source is a step function or a rectangular pulse (discussed later in these notes). If the source is something else, it is better to use the general theory presented in Notes 7. = gv t V u t 0 t Generator voltage 2

  3. Step Response (cont.) The voltage wave is shown approaching the load: = t = = t t 0 t t dc 2 1 V+ t = 0 R R L + - g ( ) t Z gv 0 1 V V = 0 = z L z = 0 Z + + = 0 V V (from voltage divider) 0 R Z 0 g 3

  4. Bounce Diagram t = 0 R R L + - g ( ) t Z g v 0 1 V V = 0 = z L z = 0 Basic constants g L L c z t = = 0 T 0 V+ d V+ T Z + LV+ + = 0 V V t LV+ The purple values give the total voltage in each triangular region. 0 + (1 ) R Z 2T 0 g LV+ g LV+ + + (1 ) 3T L g + R R Z Z 2 LV+ + + + (1 g = 0 L 2 LV+ ) L 4T L g L g 0 L 2 LV+ 2 g 2 LV+ + + 2 g (1 ) 5T + R R Z Z 3 LV+ 2 g 0 g = g 3 LV+ + + 2 g (1 ) 6T 0 g 4

  5. Steady-State Solution Adding all infinite number of bounces (t = ), we have: + + = + + + + + + + + + 2 g 2 L 3 g 3 L 2 g 2 L 3 g 3 L ( , ) V z (1 ) (1 ) V V g L L g L Sum of all right-traveling waves Sum of all left-traveling waves After some math, we have (please see the Appendix): R + Note: = ( , ) V z V L The steady-state solution does not depend on the transmission line length or the characteristic impedance! 0 R R L g This is the DC circuit-theory voltage divider equation! 5

  6. Example R = t = 225 0 g + - ( ) t Z = R = 75 gv 25 T = V = 1 ns 4 V 0 L 0 = z L z = 0 1 2 1 2 = = g L 0 0 [m] z 1[V] 1[V] Z + 1 1[V] 2 + = = 1 [V] V V 0 0 R Z 0 g 0.5[V] 2 1[V] 4 1[V] 8 1[V] 16 1[V] 32 [ns] t + 1 2 R R Z Z 0.25[V] 3 = = 0 L L + 0 L 0.375[V] 4 + + R R Z Z 1 2 0 g = = 0.4375[V] 5 g 0 g 0.40625[V] 6 1[V] 64 0.390625[V] 6

  7. Example (cont.) The bounce diagram can be used to get an oscilloscope trace of the voltage at any point on the line. 3 4 3 4 = z L = z L 1 2 1 2 = = g L 0 0 [m] z 1[V] 0.75 [ns] 1[V] 1 1[V] 2 1.25 [ns] oscilloscope trace 4( , ) ( ) v L t 3 0.5[V] 2 [ns] t 1[V] 4 1[V] 8 1[V] 16 1[V] 32 2.75 [ns] [V] 1[V] 0.25[V] 3 + 3.25 [ns] 0.375[V] 0.5[V] 4 0.4375[V] 0.375[V] + 4.75 [ns] 0.4375[V] 5 0.25[V] 2 3 0.40625[V] 1 4 5 6 1[V] 64 [ns] t 0.390625[V] R + Steady state voltage: = = ( , ) v z 0.400 [V] V L 0 R R L g 7

  8. Example (cont.) 3 4 = Final oscilloscope trace of the voltage at z L t = 0 + ( ) t Z gv R V = 4 V 0 L 0 - = z L = z = 0.75 z L 0 1[V] 0.5[V] 0.375[V] 0.4375[V] 0.25[V] 2 3 1 4 5 [ns] t 8

  9. Example (cont.) The bounce diagram can also be used to get a snapshot of the line voltage at any point in time. t= 3.75 ns ( , 3.75[ns]) v z ( ) snapshot [V] Wavefront is moving to the left L/4 0.375[V] t= 3.75[ns] 0.25[V] L L 3 L L 4 2 4 [m] z 9

  10. Example (cont.) This set of snapshots shows the waves bouncing back and forth. Initial wave First reflected wave 1 2 = L 1.0 V 1.0 V 0.5 V 1.0 V 1.0 V 0.5 V 1.0 V 1.0 V 0.5 V 1.0 V 0.5 V 10

  11. Bounce Diagram for Current = I We just change the signs of the reflection coefficients, as shown. 1 2 1 2 = = I g I L Normalized Current 0 ( ( ) ) , , i i L t L t 0 1 = [m] z I L + 1 1 1 2 Z + + = = 1 [V] 0 V V ( )( )( , , L t ) , 1/ 1/ v L t L t Z 0 1.5 R Z 2 = 0 0 g [ns] t 1 4 ( ) + , v Z 0 1.25 + 3 V A 1 8 += ( ( ) ) I v v L t Z = 4 1.125 0 + 1 16 1.1875 5 + + + = = = = N 1 [V] I Z I V 1 32 L 0 0 1.21875 6 1 64 1.203125 Note: The normalized current iN(z,t) is defined as Z0i(z,t). 11

  12. Example for Current Oscilloscope Trace for Current 3 4 = z L Normalized Current 3 4 1 2 1 2 = = = I g z L I L 4( , ) Z i L t 3 0 ( 0 oscilloscope trace of current) 0 1 0.75 [ns] [m] z 1 1 1 2 1.25 [ns] 1.5 1.5 2 [ns] t 1 4 1.25 1 1.125 1.1875 1.25 3 2.75 [ns] (units are volts) 1 8 3.25 [ns] 4 1.125 1 16 1.1875 4.75 [ns] 5 2 3 1 4 5 1 32 [ ] t ns 1.21875 6 1 64 1.203125 V + ( )( ) = = Steady state current: 0( , ) Z i z = = ( , ) i z 0.016 [A] 0 0.016 75 1.20 V R R L g 12

  13. Example for Current (cont.) t= 3.75 ns Snapshot for Current 0( , 3.75[ns]) Z i z (snapshot of current) 1 2 1 2 = = I g I L Normalized Current 0 Wavefront is moving to the left 0 1 [m] z 1 1 1.25 1 2 1.125 1.5 2 [ns] t 1 4 (units are volts) 1.25 3 1 8 L/4 4 1.125 L L 3 L 1 t= 3.75[ns] L 16 4 2 4 1.1875 5 1 32 [m] z 1.21875 6 1 64 1.203125 13

  14. Two Transmission Lines Here we have a reflection and transmission coefficient at the junction between two lines. Junction R = 225 t = 0 T = T = g 1 ns 1 ns + - ( ) t Z = R = V = Z = 150 50 gv 4 V 75 0 L 0 0 z = = 0 z L 75 150 225 1 3 150 75 225 1 3 J T+ + = = = = Note: J J + When the wave first hits the junction, it sees the characteristic impedance of the line on the other side as a load. J 4 3 2 3 + + J T = + = = + = 1 T 1 T J J J J J KVL: TJ = 1 + J (This follows from the fact that voltage must be continuous across the junction.) 14

  15. Two Transmission Lines (cont.) Bounce Diagram for Two Lines Junction R = 225 t = 0 T = T = g 1 ns 1 ns + - ( ) t Z = R = V = Z = 150 gv 50 4 V 75 0 0 L 0 = z L z = 0 1 3 2 3 4 3 + T+ = = J J 1 3 = = JT J 1 2 1 2 = = g L 0 1[V] 0[V] 0[V] [m] z 1[V] 1 0.3333[V] 1.3333[V] Note: 1.3333[V] 1.3333[V] 2 The magnitude of a transmission coefficient can be larger than 1! [ns] t 0.6667 [V] 0.1667[V] 1.5000[V] 0.6667 [V] 3 0.2222 [V] 0.2222 [V] 0.4444 [V] 0.0555[V] -0.4444 [V] -0.3888 [V] 4 1.1111[V] 1.1111[V] 15

  16. Pulse Response Superposition can be used to get the response due to a rectangular pulse. R g + - ( ) t ( ) t Z R +- gv gv 0 L z = = 0 z L ( ) ( ) t ( ) ( ) = V gv V u t u t W 0 0 ( ) t u(t) = unit step function gv 1.0 ( ) u t t t W We thus subtract two bounce diagrams, with the second one being a shifted version of the first one. 16

  17. Example: Pulse Find an oscilloscope trace of the voltage at z = 0.75 L R = Oscilloscope trace 225 g ( ) t R = 25 Z = +- 75 T = 1 ns gv L 0 = z L z = 0 = 0.75 z L Given pulse: V+ = = 4 V 1[V] V V = 4 V 0 0 ( ) gv t Z + W = 0.25 ns + = = Recall : 1 [V] V V 0 0 R Z 0 g t W 17

  18. Example (cont.) Oscilloscope trace for original step function ( ) 3 4 = z L [V] 1[V] 1 2 1 2 = = g L 0.5[V] 0 0.4375[V] 0.375[V] 0 [m] z 1[V] 0.75 [ns] 1[V] 1 1[V] 2 0.25[V] 1.25 [ns] [ns] t 2 3 1 4 5 0.5[V] 2 [ns] t 1[V] 4 1[V] 8 1[V] 16 1[V] 32 2.75 [ns] [V] 0.25[V] 3 + 3.25 [ns] Subtract 0.375[V] 4 Shifted trace + 4.75 [ns] [V] 1[V] 0.4375[V] 5 (shifted by W =0.25 [ns]) 0.5[V] 0.4375[V] 0.40625[V] 6 0.375[V] 1[V] 64 0.390625[V] 0.25[V] [ns] t 2 3 1 4 5 18

  19. Example: Pulse (cont.) V V = 4 V 0 0 ( ) gv t W = 0.25 ns t W R = Oscilloscope trace 225 g ( ) t R = 25 Z = +- 75 T = 1 ns gv L 0 = z L z = 0 = 0.75 z L 4( , ) v L t 3 1[V] Oscilloscope trace of voltage Incident pulse sweeping past the oscilloscope 0.125[V] 0.0625[V] Reflected pulse sweeping past the oscilloscope [ns] t 2 3 1 4 5 0.03125[V] Re-reflected pulse (reflected from the generator) sweeping past the oscilloscope 0.25[V] 0.5[V] 19

  20. Example (cont.) Final oscilloscope trace of the voltage at 3 4 = z L 5 + ( ) R Z gv t - L 0 z = 0 = z L = 0.75 z L 1[V] 4( , ) v L t 3 0.125[V] 0.0625[V] [ns] t 2 3 1 4 5 0.03125[V] 0.25[V] 0.5[V] 20

  21. Example: Pulse (cont.) Find a snapshot of the voltage at t = 1.5 [ns]. R = 225 g ( ) t R = 25 Z = +- 75 T = 1 ns gv L 0 = z L z = 0 t = 1.5 ns V = 4 V V+= 1[V] 0 ( ) gv t W = 0.25 ns t W 21

  22. Example: Pulse (cont.) Shifted numbers Original step Delayed step 1 2 1 2 = = 1 2 1 2 L = = g L g 0 0.25 nS 0.25 0 1[V] 0 [m] z 1[V] 1[V] 1 1[V] 2 L / 2 1[V] 1.25 1[V] 2 3L / 4 0.5[V] 2 0.5[V] 2.25 1[V] 4 1[V] 8 1[V] 16 1[V] 32 [ns] t 1[V] 4 1[V] 8 1[V] 16 1[V] 32 0.25[V] 3 0.25[V] 3.25 + + 0.375[V] 4 0.375[V] 4.25 + + 0.4375[V] 5 0.4375[V] 5.25 0.40625[V] 6 1[V] 64 0.40625[V] 6.25 1[V] 64 0.390625[V] 0.390625[V] 22

  23. Example: Pulse (cont.) Snapshot from original step function V 1.0 0.5 [m] z 0.25L 0.5L L 0.75L Subtract V Snapshot from shifted step function 1.0 0.5 [m] z 0.25L 0.5L L 0.75L 23

  24. Example: Pulse (cont.) V V = 4 V 0 Snapshot 0 ( ) gv t W = 0.25 ns t W R = 225 g ( ) t R = 25 Z = +- 75 T = 1 ns gv L 0 = z L z = 0 Snapshot of voltage ( ,1.5[ns]) v z 0.25L 0.5L 0.75L L [m] z Pulse is moving to the left (after being reflected from the generator). 0.5[V] 24

  25. Capacitive Load = R Z t = 0 0 g C L + - ( ) t Z gv 0V V 0 = z = z L 0 Note: The generator is assumed to be matched to the transmission line for convenience (we wish to focus on the effects of the capacitive load). = 0 Hence g The load reflection coefficient L is now a function of time. 25

  26. Capacitive Load (cont.) ( )/ = = ( L z t c d d dt ) dt delay = R Z t = 0 0 g C L + - ( ) Z gv t 0V V 0 = z = z L 0 z (time = t) ( ) = 0 Z + Lt + = g V V 0 z 0 R Z t = 0 0 g 0 V+ = 0/2 V V+ T t ( ) t t V+ L d 2T += 0/ 2 V V ( ) ( ) V+ + t t 1 L d 3T 26

  27. Capacitive Load (cont.) = R Z t = 0 0 g C L + - ( ) Z gv t 0V V 0 = z = z L 0 ( ) = At t = T: Thecapacitor acts as a short circuit: 1 LT ( ) 1 = At t = : Thecapacitor acts as an open circuit: L Between t= Tandt = , there is an exponential time-constant behavior. General time-constant formula: Hence, we have: ( ) ( ) + ( ) ( ) ( )/ t T = ( ) ( )/ F t F F T F e t T 1 2 = , Lt e t T t T = Z C 0 L 27

  28. Capacitive Load (cont.) += 0/ 2 V V Assume z = 0 = R Z t = 0 0 g C + L + - ( ) ( ) 0, v t Z gv t 0V V 0 - = z = z L 0 ( ) = 0 V 0 Lt Steady-state (v(t) = V0) ( V e g t = 0 ) 0 V+ z ( ) 2 / t T 01 V+ ( ) T 0, v t t ( ) Lt T V+ 0/2 V 2T ( ) ( ) + + t T 1 V L 3T z = 0 t T 2T ( ) ( ) ( ) ( ) t ( ) ( ) ( ) ( ) ( ) + / 2 / 2 / t T t T t T 1 2 = + t T = + t T = , 1 2 1 , 1 1 e e V V e 0 L L L 28

  29. Inductive Load = R Z t = 0 0 g + - ( ) t L L Z gv 0V V 0 = z = z L 0 ( ) 1 = At t = T: inductor acts as an open circuit: LT ( ) L = At t = : inductor acts as a short circuit: 1 Between t= Tandt = , there is an exponential time-constant behavior. ( ) ( )/ t T = + 1 2 , Lt e t T Recall: ( ) ( ) + ( ) ( ) ( )/ t T = F t F F T F e = / L L Z 0 t T 29

  30. Inductive Load (cont.) += 0/2 V V Assume z = 0 = R Z t = 0 0 g + + - ( ) t ( ) L L Z gv 0, v t 0V V 0 - = z = z L 0 0 V ( ) = 0 Lt g t = 0 ( ) 2 / t T V e 0 V+ ( ) z 0 0, v t V+ 0/2 V T t ( ) Lt T V+ 2T ( ) ( ) V+ + 1 Lt T 3T Steady-state (v(t) = 0) z = 0 t T 2T ( ) ( ) t ( ) ( ) ( ) ( ) ( ) + / 2 / 2 / t T t T t T = + + t T = + t T = 1 2 , 1 2 , 1 e e V V e 0 L L L 30

  31. Time-Domain Reflectometer (TDR) This is a device that is used to look at reflections on a line, to look for potential problems such as breaks of other types of faults on the line. The 20/20 Step Time Domain Reflectometer (TDR) was designed to provide the clearest picture of coaxial or twisted pair cable lengths and to pin-point cable faults. AEA Technology, Inc. 31

  32. Time-Domain Reflectometer (cont.) = dt 2 / t z c round - trip time down to fault) ( d F d = R Z t = 0 = 0 g z z F + - ( ) t = R Z gv Z 0V V Z 0 L 0 0 = z L z = 0 Fault = The fault is modeled as a load resistor at : R z z R F F F The fault might be acting as a parallel resistor or a series resistor. R Z R Z or = 0 0 F F 2 / t z c d F d = R Z t = 0 = 0 g z z F + - ( ) t gv Z 0V V R 0 F z = 0 32

  33. Time-Domain Reflectometer (cont.) dt = 2 / t z c round - trip time down to fault) z ( = d F d R Z t = 0 = 0 g z F + - ( ) t = R Z gv Z 0V V Z 0 L 0 0 = z L z = 0 Fault = The fault is modeled as a load resistor at . R z z = = Short on line: Break on line: 0, R R F F F F ( ) ( ) 0, v t 0, v t = t t d = t t The time tdindicates where the fault is. d t t Resistive load, RF > Z0 Resistive load, RF < Z0 33

  34. Time-Domain Reflectometer (cont.) The TDR can also tell us what kind of a load we have at the end of a line. = R Z t = 0 0 g + - ( ) t Load gv Z 0V V 0 = z L z = 0 ( ) 0, v t ( ) 0, v t t t Inductive load Capacitive load 34

  35. Appendix: Steady-State Solution Adding all infinite number of bounces (t = ) for the step response, we have: + + = + + + + + + + + + 2 g 2 L 3 g 3 L 2 g 2 L 3 g 3 L ( , ) V z (1 ) (1 ) V V g L L g L Sum of all right-traveling waves V + Sum of all left-traveling waves + + + (1 1 ) V + = = V L L Note: We have used 1 1 g L g L g L 1 + + R R Z Z Z Z = n + z 1 0 L 1 1 z Z + = 0 n = 0 L V 0 ( ) 0 + R Z R R + z R R Z Z 0 g 0 g 1 0 L 0 0 g L + R R Z Z + ( )( ) + + + 1 R Z R Z 0 L 0 0 g L Z + = 0 L V 0 ( )( ) ( )( ) 0 R Z R Z R Z R Z R Z 0 g 0 0 0 0 g L g L 35

  36. Appendix: Steady-State Solution (cont.) Simplifying, we have: + + R R Z Z + ( )( ) + + + 1 R Z R Z 0 L 0 0 g L Z + = 0 L ( , ) V z 0 V ( )( ) ( )( ) 0 R Z R Z R Z R Z R Z 0 g 0 0 0 0 g L g L 2 R + ( )( ) + + R Z R Z L Z R 0 0 g L R Z + = 0 L 0 V ( )( ) ( ( ) ( 0 2 L R Z V Z ) ( 0 )( ) 0 + + R Z R Z Z R Z R Z 0 g 0 0 0 0 g L g L ) + 2 + R R Z Z + 0 L g = 0 V ( )( )( ) 0 + R Z R Z R Z R Z R Z 0 g 0 0 0 g L g L = 0 0 ( ) )( ( ) + + R Z R R Z R Z 0 0 0 g L g L 2 R Z V = 0 R R 0 L + + + + + 2 0 2 0 R R Z R Z R Z Z R Z R Z 0 0 0 0 g L L g g L L g 2 R Z V R Z =+ 0 0 L + + + R Z R Z R Z 0 0 0 0 L g L g 36

  37. Appendix: Steady-State Solution (cont.) Continuing with the simplification: 2 R Z V + = ( , ) V z 0 0 R Z L + + R Z R Z R Z 0 2 0 0 0 L g L g R Z V R Z = 0 0 R Z + L ( ) 2 0 0 L R Z V R Z g = 0 0 R Z + L ( ) 0 0 L R V R g = 0 R L + ( ) L g Hence we finally have: R + = ( , ) V z V L 0 R R L g 37

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