prof. d. r. wilton note 3 transmission lines (bounce diagram) ece 3317
TRANSCRIPT
Prof. D. R. Wilton
Note 3 Note 3 Transmission LinesTransmission Lines
(Bounce Diagram)(Bounce Diagram)
ECE 3317
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
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)
Bounce Diagram
z = 0
RL
z = L
V0 [V]
t = 0
+
-
Rg
gV t Z0
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
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 usedthe geometric series formula
Steady-State Solution
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:
Steady-State Solution
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
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
0( , ) L
L g
RV z V
R R
Hence we finally have:
This is just the voltage divider equation!
Continuing with the simplification:
Note: the steady-state solution does not depend on the transmission line length or characteristic impedance.
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
Example (cont.)
The bounce diagram can be used to get an “oscilloscope trace” at any point on the line.
[ns]t1 2 3 4 5
1 [V]
0.5 [V]0.375 [V] 0.4375 [V]
0.25 [V]
34( , ) ( )V L t oscilloscope trace
steady state voltage: 0( , ) 0.400 [V]L
L g
RV z V
R R
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
4z L
0.75 [ns]
1.25 [ns]
2.75 [ns]
3.25 [ns]
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]z4
L
0.375 [V]0.25 [V]
( , 3.75 [ns]) ( )V z snapshot
2
L 3
4
LL
To obtain current bounce diagram from 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.)
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
[m]z
Note: We can also just change the signs of the reflection coefficients, as shown.
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
Note: These diagrams are for the normalized current, defined as Z0 I (z,t).
I
current current
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
oscilloscope trace of current
Example (cont.)
steady state current: 0( , ) 0.016 [A]L g
VI z
R R
0 ( , ) 0.016 75 1.20[A]Z I z
1 2 3 4 5
1
1.5
1.125 1.1875
1.25
30 4( , )Z I L t
[ ]t ns
2.75 [ns]
3.25 [ns]
0.75 [ns]
1.25 [ns]
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
snapshot of current
[m]z
4
L
1.1251.25
0 ( , 3.75 [ns])Z I z
2
L 3
4
LL
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)
Example (cont.)
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]
150 75 1
225 32
13
J
J JT
41
375 150 1
225 3
J 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]
[m]z
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+-
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
Example: Pulse
-
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]zW
0.25
1.00 [ns]
1.50 [ns]
3.00[ns]
3.50[ns]
5.00 [ns]
5.50 [ns]
z = 0.75 L
W = 0.25 [ns]
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
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
L / 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
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
Capacitive Load
C
z = 0 z = L
V0 [V]
t = 0
+
-
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.
Capacitive Load
CL
z = 0 z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
0t
T
2T
3T
L t0g
V
L t V
0V
1 L t V
t
z0
00 0
0 / 2
ZV V
Z Z
V
Capacitive Load (cont.)
CL
z = 0 z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
At t = 0: capacitor acts as a short circuit. 0 1L
At t = : capacitor acts as an open circuit. 1L
Between t = 0 and t = , there is an exponential time-constant behavior.
/1 1 1 ,t TL t e t T
0 LZ C
/0 tF t F F F e
Time-constant formula: Hence we have:
(valid for t < T)
Capacitive Load (cont.)
CL
z = 0 z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
0t
T
2T
3T
L t0g
V
L t V
0V
1 L t V
t
z
t
V(0,t)
T 2T
V0 / 2
V0 steady-state
Inductive Load
At t = 0: inductor as a open circuit. 0 1L
At t = : inductor acts as a short circuit. 1L
Between t = 0 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
+
-
Z0
gV t Z0
(valid for t < T)
Inductive Load (cont.)
0t
T
2T
3T
L t0g
V
L t V
0V
1 L t V
t
z
t
V(0,t)
T 2T
V0 / 2
V0
steady-state
LL
z = 0 z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
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.
z = 0
load
z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
(matched source)
t
V (0, t)
resistive load, RL > Z0
t
V (0, t)
resistive load, RL < Z0
Time-Domain Reflectometer (cont.)
z = 0
load
z = L
V0 [V]
t = 0
+
-
Z0
gV t Z0
(matched source)
capacitive load inductive load
t
V (0, t)
t
V (0, 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.)