notes 8 transmission lines (bounce diagram)courses.egr.uh.edu/ece/ece3317/sectionjackson/class...
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Notes 8 Transmission Lines (Bounce Diagram)
1
ECE 3317 Applied Electromagnetic Waves
Prof. David R. Jackson Fall 2018
Step Response
The concept of the bounce diagram is illustrated for a step response on a terminated line:
2
( )gv t( ) ( )0gv t V u t=
0V
t
Generator voltage
0Z( )gv t+ -
0z =[ ]0 VV
0t =
LR
z L=
gR
Step Response (cont.)
The wave is shown approaching the load.
00
0g
ZV VR Z
+
= + (from voltage divider)
3
dc0t = 1t t= 2t t=V +
0Z( )gv t+ -
0z =[ ]0 1 VV =
0t =
LR
z L=
gR
Bounce Diagram
d
LTc
=
00
0g
ZV VR Z
+
= +
0
0
gg
g
R ZR Z
−Γ = +
0
0
LL
L
R ZR Z
−Γ = +
t↓
4
0t =
T
2T
3T
4T
5T
6T
LΓgΓ
V +
L V +Γ
g L V +Γ Γ
2g L V +Γ Γ
2 2g L V +Γ Γ
2 3g L V +Γ Γ
0
V +
(1 )L V ++ Γ
(1 )L g L V ++ Γ +Γ Γ
2(1 )L g L g L V ++ Γ +Γ Γ +Γ Γ
2 2(1 )g L V ++ + Γ Γ
2 3(1 )g L V ++ + Γ Γ
z
0Z( )gv t+-
0z =[ ]0 VV
0t =
LR
z L=
gR
The purple values give the total voltage in each region.
Steady-State Solution
2 2 3 3 2 2 3 3
Sum of all right-traveling waves Sum of all left-traveling waves
0
( , ) (1 ) (1 )
(1 )1 1 1
1
g L g L g L L g L g L g L
L L
g L g L g L
L
L
V z V V
VV V
R ZR
+ +
+++
∞ = +Γ Γ +Γ Γ +Γ Γ + + Γ +Γ Γ +Γ Γ +Γ Γ +
Γ +Γ= + =
−Γ Γ −Γ Γ −Γ Γ
−+
=
( )( )
( )( ) ( )( )
0 00
00 0
0 0
00 0
0 00
00 0 0 0
1
1
gg L
g L
Lg L
L
gg L g L
Z Z VR ZR Z R Z
R Z R Z
R Z R Z R ZR Z Z V
R ZR Z R Z R Z R Z
+ + − − − + +
−+ + + + = ++ + − − −
Adding all infinite number of bounces (t = ∞), we have:
( )0
111
n
nz
zz
∞
=
=−
<
∑
Note: We have used
5
Steady-State Solution (cont.)
( )( )
( )( ) ( )( )
( )( )
( )( ) ( )( )( )
( )( ) ( )( )
( )
00 0
0 00
00 0 0 0
0 00 0
000 0 0 0
0 00
00 0 0 0
0 0
0
1( , )
2
2
2
Lg L
L
gg L g L
Lg L
L
gg L g L
L g
gg L g L
L
g L
R Z R Z R ZR Z ZV z V
R ZR Z R Z R Z R Z
R R Z R ZR Z Z V
R ZR Z R Z R Z R Z
R R Z Z VR ZR Z R Z R Z R Z
R Z VR Z R Z
−+ + + + ∞ = ++ + − − −
+ + + = ++ + − − −
+ = ++ + − − −
=+ +( ) ( )( )0 0 0
0 02 20 0 0 0 0 0
0 0
0 0 0 0
2
2
g L
L
g L L g g L L g
L
L g L g
R Z R Z
R Z VR R Z R Z R Z R R Z R Z R Z
R Z VR Z R Z R Z R Z
− − −
=+ + + − − + +
=+ + + +
Simplifying, we have:
6
( )
( )
( )
0 0
0 0 0 0
0 0
0 0
0 0
0 0
0
2( , )
22
L
L g L g
L
L g
L
L g
L
L g
R Z VV zR Z R Z R Z R Z
R Z VR Z R Z
R Z VR Z R Z
R VR R
∞ =+ + +
=+
=+
=+
0( , ) L
L g
RV z VR R
∞ = +
Hence we finally have:
This is the DC circuit-theory voltage divider equation!
Continuing with the simplification:
Note: The steady-state solution does not depend on the transmission
line length or characteristic impedance!
7
Steady-State Solution (cont.)
Example
0
1
2
3
4
5
6
12LΓ = −
12gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t↓
[m]z0
00
1 [V]g
ZV VR Z
+ = = +
0
0
12
gg
g
R ZR Z
−Γ = = +
0
0
12
LL
L
R ZR Z
−Γ = = − +
8
+
- ( )gv t
0z = z L=
0t =
[ ]0 4 VV =
[ ]225gR = Ω
[ ]25LR = Ω[ ]0 75Z = Ω [ ]1 nsT =
Example (cont.) The bounce diagram can be used to get an “oscilloscope trace” of the voltage at any point on the line.
Steady state voltage: 0( , ) 0.400 [V]L
L g
Rv z VR R
∞ = = +
9
[ns]t
1 2 3 4 5
1 [V]
0.5 [V] 0.375 [V] 0.4375 [V]
0.25 [V]
34( , ) ( )v L t oscilloscope trace
[V]
34
z L=34
z L=
0.75 [ns]
1.25 [ns]
2.75 [ns]
3.25 [ns]
0
1
2
3
4
5
6
12LΓ = −
12gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t↓
[m]z
The bounce diagram can also be used to get a “snapshot” of the line voltage at any point in time.
Example (cont.)
10
[m]z4L
0.375 [V]0.25 [V]
( , 3.75 [ns]) ( )v z snapshot
2L 3
4L L
Wavefront is moving to the left
[V]
[ ]3.75 nst =
3.75 [ns]t =
L/4
0
1
2
3
4
5
6
12LΓ = −
12gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t↓
[m]z
To obtain a current bounce diagram from the voltage diagram, multiply forward-traveling voltages by 1/Z0, backward-traveling voltages by -1/Z0.
0
1
2
3
4
5
6
1
12
14
−
116
132
0
1
1.5
1.25
1.125
1.1875
164
−1.203125
[ns]t↓ 1
8−
1.21875
0
1
2
3
4
5
6
12LΓ = −1
2gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t↓
[m]z0
1
2
3
4
5
6
12LΓ = −1
2gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t↓
[m]z
Note: This diagram is for the normalized current,
defined as Z0 i (z,t).
[m]z
Voltage Normalized Current
Example (cont.)
11
Note: We can also just change the signs of the reflection coefficients, as shown.
Note: This diagram is for the normalized current, defined as Z0 i (z,t).
IΓ = −Γ
0
1
2
3
4
5
6
1
12
14
−
116
132
0
1
1.5
1.25
1.125
1.1875
164
−1.203125
[ns]t↓ 1
8−
1.21875
12
IgΓ = − 1
2ILΓ =
[m]z
Normalized Current
Example (cont.)
12
( )( )( ) ( )( ) ( )( )( )
0
0
,,
, 1 /, 1 /
,,
IL
L
i L ti L t
v L t Zv L t Z
v L tv L t
−
+
−
+
−
+
Γ =
−=
= −
= −Γ
Example (cont.)
Steady state current: 0( , ) 0.016 [A]L g
Vi zR R
∞ = = +
( ) ( )0 ( , ) 0.016 75 1.20Z i z ∞ = =
1 2 3 4 5
1
1.5
1.125 1.18751.25
30 4( , )
(Z i L t
oscilloscope trace of current)
[ ]t ns
13
0
1
2
3
4
5
6
1
12
14
−
116
132
0
1
1.5
1.25
1.125
1.1875
164
−1.203125
[ns]t↓ 1
8−
1.21875
12
IgΓ = − 1
2ILΓ =
[m]z
Normalized Current
2.75 [ns]
3.25 [ns]
0.75 [ns]
1.25 [ns]
34
z L=
(units are volts)
34
z L=
Example (cont.)
0
1
2
3
4
5
6
1
12
14
−
116
132
0
1
1.5
1.25
1.125
1.1875
164
−1.203125
[ns]t↓ 1
8−
1.21875
12
IgΓ = − 1
2ILΓ =
[m]z
Normalized Current
3.75 [ns]t =
L/4
[m]z4L
1.1251.25
0 ( , 3.75 [ns])Z i z(snapshot of current)
2L 3
4L L
14
Wavefront is moving to the left
(units are volts)
[ ]3.75 nst =
Example Reflection and Transmission Coefficient at
Junction Between Two Lines
KVL: TJ = 1 + ΓJ (This follows from the fact that voltage must be continuous across the junction.)
15
[ ]0 4 VV =+
- ( )gv t
150 75 1225 3
413
J
J JT
+
+ +
−Γ = =
= +Γ =
75 150 1225 3
213
J
J JT
−
− −
−Γ = = −
= +Γ =
Junction
J+Γ
JT +
JT −
J−Γ
0t = [ ]225gR = Ω [ ]1 nsT = [ ]1 nsT =
[ ]50LR = Ω[ ]0 75Z = Ω [ ]0 150Z = Ω
0z = z L=
Example (cont.)
1323
J
JT
+
−
Γ =
=
43
13
J
J
T +
−
=
Γ = −
Bounce Diagram for Cascaded Lines
0
1
2
3
4
12gΓ =
1 [V]
0.3333 [V]
0.1667[V]
0[V]1 [V]
1.3333 [V]
1.5000 [V][ns]t↓
12LΓ = −
1.3333 [V]
0.6667 [V]−
0 [V]
1.3333 [V]
0.6667 [V]-0.4444 [V] 0.0555 [V] -0.3888 [V]
1.1111 [V] 1.1111 [V]
0.2222 [V] 0.2222 [V] 0.4444 [V]
[m]z
16
+
- ( )gv t
Junction 0t =
[ ]0 4 VV =
[ ]225gR = Ω [ ]1 nsT = [ ]1 nsT =
[ ]50LR = Ω[ ]0 75Z = Ω [ ]0 150Z = Ω
0z = z L=
Pulse Response
Superposition can be used to get the response due to a pulse.
( ) ( ) ( )( )0gv t V u t u t W= − −
We thus subtract two bounce diagrams, with the second one being a shifted version of the first one.
17
+
- ( )gv t+ - ( )gv t
gR
0z = z L=
0Z LR
( )gv t
0V
Wt
Example: Pulse
1 [V]V + =
18
+ -
Oscilloscope trace
( )gv t
[ ]225gR = Ω
[ ]0 75Z = Ω [ ]1 nsT = [ ]25LR = Ω
0z = z L=
0.75z L=
( )gv t
0V
Wt
[ ]0 4 VV =
[ ]0.25 nsW =
Example: Pulse (cont.)
Subtract
19
( ) ( ) ( )( )0gv t V u t u t W= − −
0
1
2
3
4
5
6
12LΓ = −1
2gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0 [V]
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t
0.75 [ns]
1.25 [ns]
2.75 [ns]
3.25 [ns]
4.75 [ns]
5.25 [ns]
1.25
2.25
3.25
4.25
5.25
6.25
12LΓ = −1
2gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0 [V]
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[m]zW
0.25
1.00 [ns]
1.50 [ns]
3.00[ns]
3.50[ns]
5.00 [ns]
5.50 [ns]
34
z L= 34
z L=[ ]0.25 nsW =
Oscilloscope trace 0.75z L=
Example: Pulse (cont.)
Oscilloscope trace of voltage
[ns]t1 2 3 4 5
1 [V]
0.5 [V]−
0.125 [V]
0.25 [V]−
34( , )v L t
0.0625 [V]
0.03125 [V]−
20
( )gv t
0V
Wt
[ ]0 4 VV =
[ ]0.25 nsW =
+ -
Oscilloscope trace
( )gv t
[ ]225gR = Ω
[ ]0 75Z = Ω [ ]1 nsT = [ ]25LR = Ω
0z = z L=
0.75z L=
1 [V]V + =
21
+ -
Snapshot
( )gv t
[ ]225gR = Ω
[ ]0 75Z = Ω [ ]1 nsT = [ ]25LR = Ω
0z = z L=
[ ]1.5 nst =
( )gv t
0V
Wt
[ ]0 4 VV =
[ ]0.25 nsW =
Example: Pulse (cont.)
Example: Pulse (cont.)
12LΓ = −1
2gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[m]zW
3L / 4
0
1
2
3
4
5
6
12LΓ = −1
2gΓ =
1 [V]
1 [V]2
−
1 [V]4
−
1 [V]8
+
1 [V]16
+
1 [V]32
−
0
1 [V]
0.5 [V]
0.25 [V]
0.375 [V]
0.4375 [V]
0.40625 [V]1 [V]64
−0.390625 [V]
[ns]t
1.25
2.25
3.25
4.25
5.25
6.25
0.25
L / 2
22
subtract
W = 0.25 [ns]
Snapshot ( ) ( ) ( )( )0gV t V u t u t W= − −[ ]1.5 nst =
Example: Pulse (cont.)
Snapshot of voltage
[m]z
0.5 [V]−
( , 1.5 [ns])v z
L0.5L 0.75L0.25L
23
Pulse is moving to the left
( )gv t
0V
Wt
[ ]0 4 VV =
[ ]0.25 nsW =
+ -
Snapshot
( )gv t
[ ]225gR = Ω
[ ]0 75Z = Ω [ ]1 nsT = [ ]25LR = Ω
0z = z L=
[ ]1.5 nst =
Capacitive Load
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).
0gΓ =Hence
The reflection coefficient is now a function of time.
24
+ -
( )gv t
0z = z L=
0ZLC
0gR Z=
[ ]0 VV
0t =
Capacitive Load (cont.)
0t =
T
2T
3T
( )L tΓ0gΓ =
V +
( )L dt t V +Γ −
0V +
( )( )1 L dt t V ++ Γ −
t
z 0
00 0
0 / 2
ZV VZ Z
V
+ = + =
25
( ) /d dt L z c= −dt
z
+ -
( )gv t
0z = z L=
0ZLC
0gR Z=
[ ]0 VV
0t =
Capacitive Load (cont.)
At t = T : The capacitor acts as a short circuit: ( ) 1L TΓ = −
At t = ∞: The capacitor acts as an open circuit: ( ) 1LΓ ∞ =
Between t = T and t = ∞, there is an exponential time-constant behavior.
( ) ( ) ( )/1 1 1 ,t TL t e t Tτ− − Γ = + − − ≥
0 LZ Cτ =
( ) ( ) ( ) ( ) ( )/t TF t F F T F e τ− − = ∞ + − ∞
General time-constant formula: Hence we have:
26
t T≥
( ) /d dt L z c= −dt
z
+ -
( )gv t
0z = z L=
0ZLC
0gR Z=
[ ]0 VV
0t =
Capacitive Load (cont.)
0t =
T
2T
3T
( )L tΓ0gΓ =
V +
( )L t T V +Γ −
0V +
( )( )1 L t T V ++ Γ −
t
z
27
( ) ( ) ( ) ( )( ) ( )( ) ( )( )/ / 2 /01 2 , 1 2 1 , 1 1t T t T t T
L L Lt e t e V t T V eτ τ τ− − − − − −+Γ = − +Γ = − +Γ − = −
Assume z = 0 0 / 2V V+ =
+ -
( )gv t
0z = z L=
0ZLC
0gR Z=
[ ]0 VV
0t =
steady-state
( )( )2 /0 1 t TV e τ− −−
( )0,v t
0V
0 / 2V
T 2T t
Inductive Load
At t = T: inductor as a open circuit: ( ) 1L TΓ =
At t = ∞: inductor acts as a short circuit: ( ) 1LΓ ∞ = −
Between t = T and t = ∞, there is an exponential time-constant behavior.
( ) ( ) ( )/1 1 1 ,t TL t e t Tτ− − Γ = − + − − ≥
0/LL Zτ =
28
+
- ( )gv t
0gR Z=
0z = z L=
0Z
0t =
[ ]0 VV LL
Inductive Load (cont.)
0t =
T
2T
3T
( )L tΓ0gΓ =
V +
( )L t T V +Γ −
0V +
( )( )1 L t T V ++ Γ −
t
z
29
( ) ( ) ( ) ( ) ( )( ) ( )/ / 2 /01 2 , 1 2 , 1t T t T t T
L L Lt e t e V t T V eτ τ τ− − − − − −+Γ = − + + Γ = + Γ − =
0 / 2V V+ =Assume z = 0
+
- ( )gv t
0gR Z=
0z = z L=
0Z
0t =
[ ]0 VV LL
steady-state
( )2 /0
t TV e τ− −
( )0,v t
0V
0 / 2V
T 2T t
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 on the line.
Resistive load, RF > Z0 Resistive load, RF < Z0
30
The time indicates where the break is.
dt t=
t
( )0,v t
+
- ( )gv t
(dt round - trip time down to fault)
z = zF
2 /d F dt Z c=0gR Z=0t =
[ ]0 VV 0Z
0z = z L=
LoadFault
F FR z z=The fault is modeled as a load resistor at
t
( )0,v tdt t=
Time-Domain Reflectometer (cont.)
Resistive load, RL > Z0 Resistive load, RL < Z0
31
The reflectometer can also tell us what kind of a load we have.
+
- ( )gv t
0gR Z=0t =
[ ]0 VV 0Z
0z = z L=
Load
t
( )0,v t
t
( )0,v t
Time-Domain Reflectometer (cont.)
Capacitive load Inductive load
32
The reflectometer can also tell us what kind of a load we have.
+
- ( )gv t
0gR Z=0t =
[ ]0 VV 0Z
0z = z L=
Load
( )0,v t
t t
( )0,v t
Example of a commercial product
“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.
Time-Domain Reflectometer (cont.)
33