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8/2/2019 Lecture 2 Transmission Line Discontinuities
1/22
EECS 117
Lecture 2: Transmission Line Discontinuities
Prof. Niknejad
University of California, Berkeley
University of California, Berkeley EECS 117 Lecture 2 p. 1/
8/2/2019 Lecture 2 Transmission Line Discontinuities
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Energy to Charge Transmission Line
Rs
Z0
i+ =v+
Z0
+Vs
+
v
+
The power flow into the line is given by
P+line = i+(0, t)v+(0, t) =
v+(0, t)
2Z0
Or in terms of the source voltage
P
+
line = Z0
Z0 + Rs2
V2s
Z0 =
Z0
(Z0 + Rs)2V
2
s
University of California, Berkeley EECS 117 Lecture 2 p. 2/
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Energy Stored in Inds and Caps (I)
But where is the power going? The line is lossless!
Energy stored by a cap/ind is 12
CV2/12
LI2
At time td, a length of = vtd has been charged:
1
2
CV2 =1
2
CZ0
Z0 + Rs2
V2s
1
2LI2 =
1
2L
Vs
Z0 + Rs
2
The total energy is thus
1
2 LI2
+
1
2 CV2
=
1
2
V2s(Z0 + Rs)2
L
+ C
Z2
0
University of California, Berkeley EECS 117 Lecture 2 p. 3/
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Energy Stored (II)
Recall that Z0 =
L/C. The total energy stored on the
line at time td = /v:
Eline(/v) = L V2s(Z0 + Rs)2
And the power delivered onto the line in timetd
:
Pline
v=
lv
Z0V2s
(Z0 + Rs)2=
L
C
LC
V2s(Z0 + Rs)2
As expected, the results match (conservation ofenergy).
University of California, Berkeley EECS 117 Lecture 2 p. 4/
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Transmission Line Termination
Rs
i+ =v+
Z0
+Vs
+v+
Z0, td i =vL
RL
Consider a finite transmission line with a termination
resistance
At the load we know that Ohms law is valid: IL = VL/RL
So at time t = /v, our pulse reaches the load. Sincethe current on the T-line is i+ = v+/Z0 = Vs/(Z0 + Rs)and the current at the load is VL/RL, a discontinuity isproduced at the load.
University of California, Berkeley EECS 117 Lecture 2 p. 5/
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Reflections
Thus a reflected wave is created at discontinuity
VL(t) = v+(, t) + v(, t)
IL(t) =1
Z0v+(, t) 1
Z0v(, t) = VL(t)/RL
Solving for the forward and reflected waves
2v+(, t) = VL(t)(1 + Z0/RL)
2v
(, t) = VL(t)(1 Z0/RL)
University of California, Berkeley EECS 117 Lecture 2 p. 6/
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Reflection Coefficient
And therefore the reflection from the load is given by
L =V(, t)
V+(, t)=
RL Z0RL + Z0
Reflection coefficient is a very important concept fortransmisslin lines:
1
L
1
L = 1 for RL = 0 (short)L = +1 for RL = (open)
L = 0 for RL = Z0 (match)Impedance match is the proper termination if we dontwant any reflections
University of California, Berkeley EECS 117 Lecture 2 p. 7/
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Propagation of Reflected Wave (I)
If L = 0, a new reflected wave travels toward thesource and unless Rs = Z0, another reflection alsooccurs at source!
To see this consider the wave arriving at the source.Recall that since the wave PDE is linear, asuperposition of any number of solutins is also a
solution.At the source end the boundary condition is as follows
Vs IsRs = v+
1 + v
1 + v
+
2
The new term v+2
is used to satisfy the boundarycondition
University of California, Berkeley EECS 117 Lecture 2 p. 8/
8/2/2019 Lecture 2 Transmission Line Discontinuities
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Propagation of Reflected Wave (II)
The current continuity requires Is = i+1
+ i1
+ i+2
Vs = (v+1
v1
+ v+2
)Rs
Z0+ v+
1+ v
1+ v+
2
Solve for v+2
in terms of known terms
Vs =
1 + Rs
Z0
(v+
1+ v+
2) +
1 Rs
Z0
v1
+
But v+
1 =Z0
Rs+Z0 Vs
Vs =Rs + Z0
Z0
Z0Rs + Z0
Vs +
1 Rs
Z0
v1
+
1 +
RsZ0
v+2
University of California, Berkeley EECS 117 Lecture 2 p. 9/
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Propagation of Reflected Wave (III)
So the source terms cancel out and
v+2
=Rs Z0Z0 + Rs
v1
= sv
1
The reflected wave bounces off the source impedancewith a reflection coefficient given by the same equation
as before(R) =
R Z0R + Z0
The source appears as a short for the incoming waveInvoke superposition! The term v+
1took care of the
source boundary condition so our new v+2
only needed
to compensate for the v
1 wave ... the reflected wave isonly a function of v
1University of California, Berkeley EECS 117 Lecture 2 p. 10/
8/2/2019 Lecture 2 Transmission Line Discontinuities
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Bounce Diagram
We can track the multiple reflections with a bouncediagram
Time
Space
td
2td
3td
4td
5td
6td
/2/4 3/4
v+1
v
1= Lv
+
1
v+2 =
sv
1 = sLv+1
v
2= Lv
+
2= s
2Lv+
1
v+3 = sv
2 = 2s
2Lv
+
1
v
3= Lv
+
3=
2s3Lv+
1
v+4 = sv
3 = 3s
3Lv
+1
University of California, Berkeley EECS 117 Lecture 2 p. 11/
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Freeze time
If we freeze time and look at the line, using the bouncediagram we can figure out how many reflections haveoccurred
For instance, at time 2.5td = 2.5/v three waves have
been excited (v+1
,v1
, v+2
), but v+2
has only travelled a
distance of /2
To the left of /2, the voltage is a summation of three
components: v = v+1
+ v1
+ v+2
= v+1
(1 + L + Ls).
To the right of /2, the voltage has only two
components: v = v+1
+ v1
= v+1
(1 + L).
University of California, Berkeley EECS 117 Lecture 2 p. 12/
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Freeze Space
We can also pick at arbitrary point on the line and plotthe evolution of voltage as a function of time
For instance, at the load, assuming RL > Z0 andRS > Z0, so that s,L > 0, the voltage at the load will will
increase with each new arrival of a reflection
v+1 = .4
v1 = .2 v+2 = .04 v2 = .02 v+3 = .004 v3 = .002
Rs = 75
RL = 150
s = 0.2
L = 0.5
vss = 2/3.6.64 .66 .664 .666
td 2td 3td 4td 5td 6td
vL(t)
t
University of California, Berkeley EECS 117 Lecture 2 p. 13/
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Steady-State Voltage on Line (I)
To find steady-state voltage on the line, we sum over allreflected waves:
vss = v+
1 + v
1 + v+
2 + v
2 + v+
3 + v
3 + v+
4 + v
4 + Or in terms of the first wave on the line
vss = v+
1 ( 1 + L + Ls + 2
Ls + 2
L2
s + 3
L2
s + 3
L3
s + Notice geometric sums of terms like kL
ks and
k+1L
ks .
Let x = Ls:
vss = v+1
(1 + x + x2 + + L(1 + x + x2 + ))
University of California, Berkeley EECS 117 Lecture 2 p. 14/
S d S V l Li (II)
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Steady-State Voltage on Line (II)
The sums converge since x < 1
vss = v+1
1
1 Ls+
L
1 LsOr more compactly
vss = v+1
1 + L
1 Ls
Substituting for L and s gives
vss = VsRL
RL + Rs
University of California, Berkeley EECS 117 Lecture 2 p. 15/
Wh H d h T Li ?
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What Happend to the T-Line?
For steady state, the equivalent circuit shows that thetransmission line has disappeared.
This happens because if we wait long enough, the
effects of propagation delay do not matter
Conversly, if the propagation speed were infinite, thenthe T-line would not matter
But the presence of the T-line will be felt if wedisconnect the source or load!
Thats because the T-line stores reactive energy in thecapaciance and inductance
Every real circuit behaves this way! Circuit theory is anabstraction
University of California, Berkeley EECS 117 Lecture 2 p. 16/
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PCB Interconnect
Suppose = 3cm, v = 3 108m/s, so thattp = /v = 10
10s = 100ps
On a time scalet < 100ps
, the voltages on interconnectact like transmission lines!
Fast digital circuits need to consider T-line effects
conductor
ground
dielectric
logic gate
PCB substrate
University of California, Berkeley EECS 117 Lecture 2 p. 17/
E l O Li (I)
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Example: Open Line (I)
Source impedance is Z0/4, so s = 0.6, load is openso L = 1
As before a positive going wave is launched v+1
Upon reaching the load, a reflected wave of of equalamplitude is generated and the load voltage overshoots
vL = v+1
+ v1
= 1.6V
Note that the current reflection is negative of the voltage
i =
i
i+ = v
v+ = vThis means that the sum of the currents at load is zero(open)
University of California, Berkeley EECS 117 Lecture 2 p. 18/
E l O Li (II)
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Example: Open Line (II)
At source a new reflection is created v+2
= Lsv+1
, and
note s < 0, so v+2
= .6 0.8 = 0.48.
At a time 3tp, the line charged initially to v+
1 + v
1 dropsin value
vL = v+1
+ v1
+ v+2
+ v2
= 1.6
2
.48 = .64
So the voltage on the line undershoots < 1
And on the next cycle 5tp the load voltage again
overshootsWe observe ringing with frequency 2tp
University of California, Berkeley EECS 117 Lecture 2 p. 19/
E l O Li Ri i
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Example: Open Line Ringing
Observed waveform as a function of time.
University of California, Berkeley EECS 117 Lecture 2 p. 20/
Ph sical Int ition: Shorted Line (I)
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Physical Intuition: Shorted Line (I)
The intitial step charges the first capacitor through thefirst inductor since the line is uncharged
There is a delay since on the rising edge of the step, the
inductor is an open
Each successive capacitor is charged by its inductor
in a uniform fashion ... this is the forward wave v+1
L L L L
i+
v+ v+ v+ v+
i+ i+ i+
University of California, Berkeley EECS 117 Lecture 2 p. 21/
Physical Intuition: Shorted Line (II)
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Physical Intuition: Shorted Line (II)
The volage on the line goes up from left to right due tothe delay in charging each inductor through theinductors
The last inductor, though, does not have a capacitor tocharge
Thus the last inductor is discharged ... the extra charge
comes by discharging the last capacitor
As this capacitor discharges, so does its neighboringcapacitor to the left
Again there is a delay in discharging the caps due tothe inductors
This discharging represents the backward wave v1
University of California, Berkeley EECS 117 Lecture 2 p. 22/