bounce1
TRANSCRIPT
Prof. D. R. Wilton
Note 4 Note 4 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 = t2V +
00
0g
ZV VR Z
+⎛ ⎞
= ⎜ ⎟⎜ ⎟+⎝ ⎠
Generator circuit initially sees line characteristic impedance in series with generator impedance; hence by voltage divider
00
0
( / ) ( / )d dg
ZV t z c V u t z cR Z
+ ⎛ ⎞− = −⎜ ⎟⎜ ⎟+⎝ ⎠
or
Bounce Diagram
z = 0
RL
z = L
V0 [V]
t = 0+
-
Rg
( )gV t Z0
d
LTc
=
00
0g
ZV VR Z
+⎛ ⎞
= ⎜ ⎟⎜ ⎟+⎝ ⎠
0
0
gg
g
R ZR Z
⎛ ⎞−Γ = ⎜ ⎟⎜ ⎟+⎝ ⎠
0
0
LL
L
R ZR Z
⎛ ⎞−Γ = ⎜ ⎟+⎝ ⎠
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
t↓
Propagating voltages
Net line voltages
Steady-State Solution
2 2 3 3 2 2 3 3
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
−+
=
Sum of all right-traveling waves Sum of all left-traveling waves
( )( )( )( )
( )( )
( )( ) ( )( )
0 0
0 0
0 00
00 0
0 0
00 0
0 00
00 0 0 0
1
1
g L
g Lgg L
g L
Lg L
L
gg L g L
R Z R Z
R Z R Z
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, we have:
( )0
111
n
nz
zz
∞
=
=−
<
∑Note: We’ve usedthe geometric series formula:
Steady-State SolutionSimplifying, we have:
( )( )
( )( ) ( )( )
( )( )
( )( ) ( )( )
( )( ) ( )( )
00 0
0 00
00 0 0 0
0 00 0
000 0 0 0
0 0
0 0 0 0
0 02 20 0 0 0
1( , )
2
2
2
Lg L
L
gg L g L
Lg L
L
gg L g L
L
g L g L
L
g L L g 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 Z VR Z R Z R Z R Z
R Z VR R Z R Z R Z R R Z
⎛ ⎞−+ + +⎜ ⎟ ⎛ ⎞+⎝ ⎠∞ = ⎜ ⎟⎜ ⎟++ + − − − ⎝ ⎠
⎛ ⎞+ +⎜ ⎟ ⎛ ⎞+⎝ ⎠= ⎜ ⎟⎜ ⎟++ + − − − ⎝ ⎠
=+ + − − −
=+ + + − −
( )
( )
0 0
0 0
0 0
0
22
L g
L
L g
L
L g
R Z R Z
R Z VR Z R Z
R VR R
+ +
=+
=+
Steady-State Solution
0( , ) L
L g
RV z VR R
⎛ ⎞∞ = ⎜ ⎟⎜ ⎟+⎝ ⎠
Hence we finally have:
This is just the voltage divider equation, as if the transmission line were not present!
Note: the steady-state solution does not depend on the transmission line length or characteristic impedance.
Example
z = 0
RL = 25 [Ω]
z = L
V0 = 4 [V]
t = 0+
-
Rg = 225 [Ω]
( )gV t Z0 = 75 [Ω] T = 1 [ns]
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
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 = = −⎜ ⎟+⎝ ⎠
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 VR R
⎛ ⎞∞ = =⎜ ⎟⎜ ⎟+⎝ ⎠
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
34
z 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
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
3.75 [ns]t =
L/4
[m]z4L
0.375 [V]0.25 [V]
( , 3.75 [ns]) ( )V z snapshot
2L 3
4L L
To obtain a 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
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, which we define as Z0 I(z,t).
[m]z
voltage Z0 x current
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
[m]z
Note: We can also just change the signs of the reflection coefficients, as shown.
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
Note: These diagrams are for the normalized current, defined as Z0 I(z,t).
IΓ = −Γ
Z0 x current Z0 x current
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
Z0 x current
oscilloscope trace of current
Example (cont.)
steady state current: 0( , ) 0.016 [A] ;L g
VI zR R
⎛ ⎞∞ = =⎜ ⎟⎜ ⎟+⎝ ⎠
( )( )0 ( , ) 0.016 75 1.20[A]Z I z ∞ = =
1 2 3 4 5
1
1.5
1.125 1.18751.25
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
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
Z0 x current
3.75 [ns]t =
L/4
snapshot of current
[m]z4L
1.1251.25
0 ( , 3.75 [ns])Z I z
2L 3
4L L
ExampleReflection and Transmission Coefficients at a
Junction Between Two Lines of Differing Characteristic Impedance
RL = 50 [Ω]V0 = 4 [V]
150 75 1225 3
413
J
J JT
+
+ +
−Γ = =
= +Γ =
75 150 1225 3
213
J
J JT
−
− −
−Γ = = −
= +Γ =
KVL: TJ = 1 + ΓJ(since voltage and current must be continuous across the junction)
z = 0 z = L
t = 0+
-
Rg = 225 [Ω]
( )gV t Z0 = 75 [Ω] Z0 = 150 [Ω]
T = 1 [ns]T = 1 [ns]junction
Example (cont.)
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]
Bounce Diagram for Cascaded Lines
V0 = 4 [V]
[m]z
150 75 1225 3
213
J
J JT
+
− −
−Γ = =
= +Γ =
413
75 150 1225 3
J J
J
T + +
−
= +Γ =
−Γ = = −
z = 0
RL = 50 [Ω]
z = L
t = 0+
-
Rg = 225 [Ω]
( )gV t Z0 = 75 [Ω] Z0 = 150 [Ω]
T = 1 [ns]T = 1 [ns]
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 [Ω]
z = 0.75 L
z = 0 z = L
Rg = 225 [Ω]
Z0 = 75 [Ω] 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
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]zW0.25
1.00 [ns]
1.50 [ns]
3.00[ns]
3.50[ns]
5.00 [ns]
5.50 [ns]
z = 0.75 LW = 0.25 [ns]
Example: Pulse (cont.)
RL = 25 [Ω]
z = 0.75 L
z = 0 z = L
Rg = 225 [Ω]
Z0 = 75 [Ω] T = 1 [ns]Vg (t) +-
W = 0.25 [ns]
V0 = 4 [V]
t
( )gV t 0V
W
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]−
Example: Pulse (cont.)
-
t = 1.5 [ns]
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
L / 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
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 [Ω]
z = 0 z = L
Rg = 225 [Ω]
Z0 = 75 [Ω] T = 1 [ns]Vg (t) +-
W = 0.25 [ns]
V0 = 4 [V]
t
( )gV t 0V
W
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).
0gΓ =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 tΓ0gΓ =
V +
( )L t V +Γ
0V +
( )( )1 L t V ++Γ
t
z0
00 0
0 / 2
ZV VZ 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 1LΓ = −
At t = ∞: capacitor acts as an open circuit. ( ) 1LΓ ∞ =
Between t = 0 and t = ∞, there is an exponential time-constant behavior.
( ) ( ) ( )/1 1 1 ,t TL t e t Tτ− −⎡ ⎤Γ = + − − ≥⎣ ⎦
0 LZ Cτ =
( ) ( ) ( ) ( ) /0 tF t F F F e τ−⎡ ⎤= ∞ + − ∞⎣ ⎦
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 tΓ0gΓ =
V +
( )L t V +Γ
0V +
( )( )1 L t V ++Γ
t
z
t
V(0,t)
T 2T
V0 / 2
V0 steady-state
Inductive Load
At t = 0: inductor as a open circuit. ( )0 1LΓ =
At t = ∞: inductor acts as a short circuit. ( ) 1LΓ ∞ = −
Between t = 0 and t = ∞, there is an exponential time-constant behavior.
( ) ( ) ( )/1 1 1 ,t TL t e t Tτ− −⎡ ⎤Γ = − + − − ≥⎣ ⎦
0/LL Zτ =
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 tΓ0gΓ =
V +
( )L t V +Γ
0V +
( )( )1 L t V ++Γ
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.
t
V(0,t)
resistive load, RL > Z0
t
V(0,t)
resistive load, RL < Z0
1[V] 1[V]LΓ
LΓ
z = 0
load
z = L
V0 = 2[V]
t = 0+
-
Z0
( )gV t Z0
matched source inside the TDR
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)1[V] 1[V]
Example of a commercial product
The 20/20 Step Time Domain Reflectometer (TDR) was designed to provide a clear picture of coaxial or twisted pair cable lengths and to pin- point cable faults.
AEA Technology, Inc.
Time-Domain Reflectometer (cont.)