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Prof. David R. Jackson
Notes 8 Transmission Lines(Bounce Diagram)
ECE 3317
Spring 2013
2
Step Response
The concept of the bounce diagram is illustrated for a unit step response on a terminated line.
RL
z = 0 z = L
V0 [V]
t = 0
+
-
Rg
gV t Z0
t
gV t
0gV t V u t
0V
3
Step Response (cont.)
The wave is shown approaching the load.
RL
z = 0 z = L
V0 [V]
t = 0
+
-
Rg
gV t Z0
dct = 0 t = t1 t = t2 V +
00
0g
ZV V
R Z
(from voltage divider)
4
Bounce Diagram
d
LT
c
00
0g
ZV V
R Z
0
0
gg
g
R Z
R Z
0
0
LL
L
R Z
R Z
0t
T
2T
3T
4T
5T
6T
Lg
V
L V
g L V
2g L V
2 2g L V
2 3g L V
0
V
(1 )L V
(1 )L g L V
2(1 )L g L g L V
2 2(1 )g L V
2 3(1 )g L V
z
t
z = 0
RL
z = L
V0 [V]
t = 0
+
-
Rg
gV t Z0
5
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
VVV
R Z
R
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 ZV
R ZR Z R Z
R Z R Z
R ZR Z R Z
R Z ZV
R ZR Z R Z R Z R Z
Adding all infinite number of bounces, we have:
0
1
1
1
n
n
zz
z
Note: We have used
6
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 ZR Z R Z
R Z ZV z V
R ZR Z R Z R Z R Z
RR Z R Z
R Z ZV
R ZR Z R Z R Z R Z
R R Z ZV
R ZR Z R Z R Z R Z
R Z V
R 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 V
R R Z R Z R Z R R Z R Z R Z
R Z V
R Z R Z R Z R Z
Simplifying, we have:
7
0 0
0 0 0 0
0 0
0 0
0 0
0 0
0
2( , )
2
2
L
L g L g
L
L g
L
L g
L
L g
R Z VV z
R Z R Z R Z R Z
R Z V
R Z R Z
R Z V
R Z R Z
R V
R R
0( , ) L
L g
RV z V
R 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!
Steady-State Solution (cont.)
8
Example
z = 0
RL = 25 []
z = L
V0 = 4 [V]
t = 0
+
-
Rg = 225 []
gV t Z0 = 75 [] T = 1 [ns]
0
1
2
3
4
5
6
1
2L 1
2g
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 V
R Z
0
0
1
2g
gg
R Z
R Z
0
0
1
2L
LL
R Z
R Z
9
Example (cont.)The bounce diagram can be used to get an “oscilloscope trace” at any point on the line.
Steady state voltage: 0( , ) 0.400 [V]L
L g
RV z V
R R
[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]
3
4z L
0.75 [ns]
1.25 [ns]
2.75 [ns]
3.25 [ns]
0
1
2
3
4
5
6
1
2L 1
2g
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
10
The bounce diagram can also be used to get a “snapshot” of the line voltage at any point in time.
Example (cont.)
0
1
2
3
4
5
6
1
2L 1
2g
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
1
2L 1
2g
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
3.75 [ns]t
L/4
[m]z
4
L
0.375 [V]0.25 [V]
( , 3.75 [ns]) ( )V z snapshot
2
L 3
4
LL
Wavefront is moving to the left
[V]
11
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
1
2
1
4
1
16
1
32
0
1
1.5
1.25
1.125
1.1875
1
64
1.203125
[ns]t
1
8
1.21875
0
1
2
3
4
5
6
1
2L 1
2g
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
1
2L 1
2g
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 Current
Example (cont.)
12
Note: We can also just change the signs of the reflection coefficients, as shown.
Note: These diagrams are for the normalized current, defined as Z0 I (z,t).
I
0
1
2
3
4
5
6
1
1
2
1
4
1
16
1
32
0
1
1.5
1.25
1.125
1.1875
1
64
1.203125
[ns]t
1
8
1.21875
1
2Ig 1
2IL
[m]z
Current
Example (cont.)
0
0
,
,
, 1/
, 1/
,
,
IL
L
I L t
I L t
V L t Z
V L t Z
V L t
V L t
13
0
1
2
3
4
5
6
1
1
2
1
4
1
16
1
32
0
1
1.5
1.25
1.125
1.1875
1
64
1.203125
[ns]t
1
8
1.21875
1
2Ig 1
2IL
[m]z
Current
Example (cont.)
Steady state current: 0( , ) 0.016 [A]L g
VI z
R R
0 ( , ) 0.016 75 1.20Z I z
1 2 3 4 5
1
1.5
1.125 1.1875
1.25
30 4( , )
(
Z I L t
oscilloscope trace of current)
[ ]t ns
2.75 [ns]
3.25 [ns]
0.75 [ns]
1.25 [ns]
3
4z L
(units are volts)
14
Example (cont.)
0
1
2
3
4
5
6
1
1
2
1
4
1
16
1
32
0
1
1.5
1.25
1.125
1.1875
1
64
1.203125
[ns]t
1
8
1.21875
1
2Ig 1
2IL
[m]z
Current
3.75 [ns]t
L/4
[m]z
4
L
1.1251.25
0 ( , 3.75 [ns])Z I z
(snapshot of current)
2
L 3
4
LL
Wavefront is moving to the left
(units are volts)
15
Example
Reflection and Transmission Coefficient at Junction Between Two Lines
z = 0
RL = 50 []
z = L
V0 = 4 [V]
t = 0
+
-
Rg = 225 []
gV t Z0 = 75 [] Z0 = 150 []
T = 1 [ns]T = 1 [ns]
150 75 1
225 34
13
J
J JT
75 150 1
225 32
13
J
J JT
Junction
KVL: TJ = 1 + J(since voltage must be continuous across the junction)
J
JT
JT
J
16
Example (cont.)
1
32
3
J
JT
4
31
3
J
J
T
Bounce Diagram for Cascaded Lines
z = 0
RL = 50 []
z = L
V0 = 4 [V]
t = 0
+
-
Rg = 225 []
gV t Z0 = 75 [] Z0 = 150 []
T = 1 [ns]T = 1 [ns]
0
1
2
3
4
1
2g
1 [V]
0.3333 [V]
0.1667[V]
0[V]
1 [V]
1.3333 [V]
1.5000 [V]
[ns]t
1
2L
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
17
Pulse Response
Superposition can be used to get the response due to a pulse.
0gV t V u t u t W
t
gV t 0V
W
We thus subtract two bounce diagrams, with the second one being a shifted version of the first one.
RL
z = 0 z = L
Vg (t)+
-
Rg
gV t Z0+-
18
Example: Pulse
RL = 25 []
z = 0.75 L
z = 0 z = L
Rg = 225 []
Z0 = 75 [] T = 1 [ns]Vg (t) +-
W = 0.25 [ns]
V0 = 4 [V]
t
gV t 0V
W
1 [V]V
Oscilloscope trace
19
Example: Pulse (cont.)
Subtract 0gV t V u t u t W
0
1
2
3
4
5
6
1
2L 1
2g
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
1
2L 1
2g
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]z
W0.25
1.00 [ns]
1.50 [ns]
3.00[ns]
3.50[ns]
5.00 [ns]
5.50 [ns]
W = 0.25 [ns]3
4z L 3
4z L
20
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]
RL = 25 []
z = 0.75 L
z = 0 z = L
Rg = 225 []
Z0 = 75 [] T = 1 [ns]Vg (t) +-
W = 0.25 [ns]
V0 = 4 [V]
t
gV t 0V
W
21
Example: Pulse (cont.)
t = 1.5 [ns]
1
2L 1
2g
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
1
2L 1
2g
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
subtract
W = 0.25 [ns]
Snapshot 0gV t V u t u t W
22
Example: Pulse (cont.)
Snapshot of voltage
[m]z
0.5 [V]
( , 1.5 [ns])V z
L0.5L 0.75L0.25L
t = 1.5 [ns]
RL = 25 []
z = 0 z = L
Rg = 225 []
Z0 = 75 [] T = 1 [ns]Vg (t) +-
W = 0.25 [ns]
V0 = 4 [V]
t
gV t 0V
W
Pulse is moving to the left
23
Capacitive Load
C
z = 0 z = L
V0 [V]
t = 0
+
-
Rg = Z0
gV t Z0
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).
0g Hence
The reflection coefficient is now a function of time.
24
Capacitive Load (cont.)
0t
T
2T
3T
L t0g
V
L dt t V
0V
1 L dt t V
t
z0
00 0
0 / 2
ZV V
Z Z
V
CL
z = 0 z = L
V0 [V]
t = 0
+
- gV t Z0
Rg = Z0
/d dt L z c dt
z
25
Capacitive Load (cont.)
At t = T: The capacitor acts as a short circuit: 1L T
At t = : The capacitor acts as an open circuit: 1L
Between t = T and t = , there is an exponential time-constant behavior.
/1 1 1 ,t TL t e t T
0 LZ C
/t TF t F F T F e
General time-constant formula: Hence we have:
CL
z = 0 z = L
V0 [V]
t = 0
+
-
Rg = Z0
gV t Z0
t T
26
Capacitive Load (cont.)
0t
T
2T
3T
L t0g
V
L t T V
0V
1 L t T V
t
z
t
V(0,t)
T 2T
V0 / 2
V0 steady-state
/ /1 2 , 1 2 1t T t TL Lt e t e
CL
z = 0 z = L
V0 [V]
t = 0
+
-
Rg = Z0
gV t Z0
Assume z = 0
V(0,t)+
-
Oscilloscope trace
0 / 2V V
2 /0 1 t TV e
27
Inductive Load
At t = T: inductor as a open circuit: 1L T
At t = : inductor acts as a short circuit: 1L
Between t = T and t = , there is an exponential time-constant behavior.
/1 1 1 ,t TL t e t T
0/LL Z
LL
z = 0 z = L
V0 [V]
t = 0
+
-
Rg = Z0
gV t Z0
28
Inductive Load (cont.)
0t
T
2T
3T
L t0g
V
L t T V
0V
1 L t T V
t
z
t
V(0,t)
T 2T
V0 / 2
V0
steady-state
LL
z = 0 z = L
V0 [V]
t = 0
+
-
Rg = Z0
gV t Z0
/ /1 2 , 1 2t T t TL Lt e t e
0 / 2V V
2 /0
t TV e
Assume z = 0
V(0,t)+
-
29
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, RL > Z0resistive load, RL < Z0
t
V (0, t)
t
V (0, t)
The time indicates where the break is.
dt t
z = 0
Load
z = L
V0 [V]
t = 0
+
-
Rg = Z0
gV t Z0 Fault
(dt round - trip time down to fault)
The fault is modeled as a load resistor at z = zF.
z = zF
2 /d F dt Z c
30
Time-Domain Reflectometer (cont.)
Capacitive load Inductive load
t
V (0, t)
t
V (0, t)
z = 0
Load
z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
(matched source)
The reflectometer can also tell us what kind of a load we have.
31
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.)
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