lecture 3 transmission line junctions time harmonic exitation
TRANSCRIPT
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EECS 117
Lecture 3: Transmission Line Junctions / Time
Harmonic Excitation
Prof. Niknejad
University of California, Berkeley
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Transmission Line Menagerie
striplinemicrostripline coplanar
rectangular
waveguide
coaxialtwo wires
T-Lines come in many shapes and sizes
Coaxial usually 75 or 50 (cable TV, Internet)Microstrip lines are common on printed circuit boards(PCB) and integrated circuit (ICs)
Coplanar also common on PCB and ICsTwisted pairs is almost a T-line, ubiquitous forphones/Ethernet
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Waveguides and Transmission Lines
The transmission lines weve been considering havebeen propagating the TEM mode or TransverseElectro-Magnetic. Later well see that they can also
propagation other modesWaveguides cannot propagate TEM but propagationTM (Transverse Magnetic) and TE (Transverse Electric)
In general, any set of more than one losslessconductors with uniform cross-section can transmitTEM waves. Low loss conductors are commonly
approximated as lossless.
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Cascade of T-Lines (I)
Z01 Z02
z = 0
i1 i2
v1 v2
Consider the junction between two transmission linesZ01 and Z02
At the interface z = 0, the boundary conditions are thatthe voltage/current has to be continuous
v+1
+ v1
= v+2
(v
+
1 v
1 )/Z01 = v
+
2 /Z02
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Cascade of T-Lines (II)
Solve these equations in terms of v+1
The reflection coefficient has the same form (easy to
remember) =
v1
v+1
=Z02 Z01Z01 + Z02
The second line looks like a load impedance of valueZ02
Z01 Z02
z = 0
i1
+v1
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Transmission Coefficient
The wave launched on the new transmission line at theinterface is given by
v+
2 = v+
1 + v
1 = v+
1 (1 + ) = v+
1
This transmitted wave has a coefficient
= 1 + = 2Z02Z01 + Z02
Note the incoming wave carries a power
Pin =|v+1|2
2Z01
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Conservation of Energy
The reflected and transmitted waves likewise carry apower of
Pref = |v
1 |22Z01
= ||2 |v+1 |22Z01
Ptran = |v+2 |22Z02
= ||2 |v+1 |22Z02
By conservation of energy, it follows that
Pin = Pref + Ptran
1
Z02 2
+
1
Z012
=
1
Z01
You can verify that this relation holds!
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Bounce Diagram
Consider the bouncediagram for the follow-ing arrangement
Z01 Z02
Rs
RL
1 2
Time
Space
td
2td
3td
4td
5td
6td
v+1
1v+1
L1v+
1
L12v
+
1
sL12v+1
jv+1
jsv+1
s2jv
+1
1 1 + 2
td1
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Junction of Parallel T-Lines
Z01Z02
z = 0
Z03
Again invoke voltage/current continuity at the interface
v+1
+ v1
= v+2
= v+3
v+1 v
1
Z01=
v+2
Z02+
v+3
Z02
But v+2
= v+3
, so the interface just looks like the case oftwo transmission lines Z01 and a new line with char.impedance Z01
||Z02.
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Reactive Terminations (I)
Rs
Vs
Z0, td L
Lets analyze the problem intuitively first
When a pulse first sees the inductance at the load, itlooks like an open so 0 = +1
As time progresses, the inductor looks more and more
like a short! So = 1
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Reactive Terminations (II)
So intuitively we might expect the reflection coefficientto look like this:
1 2 3 4 5
-1
-0.5
0.5
1
t/
The graph starts at +1 and ends at 1. In between wellsee that it goes through exponential decay (1st order
ODE)
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Reactive Terminations (III)
Do equations confirm our intuition?
vL = Ldi
dt= L
d
dtv
+
Z0 v
Z0
And the voltage at the load is given by v+ + v
v + LZ0
dv
dt= L
Z0dv+
dt v+
The right hand side is known, its the incomingwaveform
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Solution for Reactive Term
For the step response, the derivative term on the RHSis zero at the load
v+ =Z0
Z0 + RsVs
So we have a simpler case dv+
dt = 0
We must solve the following equation
v +L
Z0
dv
dt=
v+
For simplicity, assume at t = 0 the wave v+ arrives atload
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Laplace Domain Solution I
In the Laplace domain
V(s) +sL
Z0V(s)
L
Z0v(0) =
v+/s
Solve for reflection V(s)
V(s) = v
(0)L/Z01 + sL/Z0
v+
s(1 + sL/Z0)
Break this into basic terms using partial fraction
expansion
1
s(1 + sL/Z0)
=1
1 + sL/Z0+
L/Z0
1 + sL/Z0
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L l D i S l i (II)
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Laplace Domain Solution (II)
Invert the equations to get back to time domain t > 0
v(t) = (v(0) + v+)et/ v+
Note that v(0) = v+ since initially the inductor is anopen
So the reflection coefficient is(t) = 2et/ 1
The reflection coefficient decays with time constantL/Z0
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Ti H i S d S
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Time Harmonic Steady-State
Compared with general transient case, sinusoidal case
is very easy t j
Sinusoidal steady state has many importantapplications for RF/microwave circuits
At high frequency, T-lines are like interconnect fordistances on the order of
Shorted or open T-lines are good resonators
T-lines are useful for impedance matching
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Wh Si id l St d St t ?
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Why Sinusoidal Steady-State?
Typical RF system modulates a sinusoidal carrier(either frequency or phase)
If the modulation bandwidth is much smaller than the
carrier, the system looks like its excited by a puresinusoid
Cell phones are a good example. The carrier frequency
is about 1 GHz and the voice digital modulation is about200 kHz(GSM) or 1.25 MHz(CDMA), less than a 0.1% ofthe bandwidth/carrier
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G li d Di ib d Ci i M d l
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Generalized Distributed Circuit Model
Z Z Z Z Z
Y Y Y Y Y
Z: impedance per unit length (e.g. Z = jL + R)
Y
: admittance per unit length (e.g. Y
= jC
+ G
)A lossy T-line might have the following form (but wellanalyze the general case)
L
C
R
G
L
C
R
G
L
C
R
G
L
C
R
G
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Ti H i T l h E ti
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Time Harmonic Telegraphers Equations
Applying KCL and KVL to a infinitesimal section
v(z + z) v(z) = Zzi(z)
i(z + z) i(z) = Yzv(z)Taking the limit as before (z 0)
dv
dz= Zi(z)
didz = Y v(z)
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Si St d St t (SSS) V lt /C t
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Sin. Steady-State (SSS) Voltage/Current
Taking derivatives (notice z is the only variable) wearrive at
d
2
vdz2 = Zdidz = Y Zv(z) = 2v(z)
d2i
dz2 = Ydv
dz = Y Zi(z) =
2
i(z)Where the propagation constant is a complex function
= + j =
(R
+ jL
)(G
+ jC
)
The general solution to D2G 2G = 0 is ez
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L l Li f SSS
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Lossless Line for SSS
The voltage and current are related (just as before, butnow easier to derive)
v(z) = V+
ez
+ V
ez
i(z) =V+
Z0ez V
Z0ez
Where Z0 =
Z
Y is the characteristic impedance of the
line (function of frequency with loss)
For a lossless line we discussed before, Z = jL andY = jC
Propagation constant is imaginary
=
jLjC = j
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B k t Ti D i
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Back to Time-Domain
Recall that the real voltages and currents are the and parts of
v(z, t) = ezejt = ejtz
Thus the voltage/current waveforms are sinusoidal inspace and time
Sinusoidal source voltage is transmitted unaltered ontoT-line (with delay)
If there is loss, then has a real part , and the wavedecays or grows on the T-line
ez = ezejz
The first term represents amplitude response of theT-line
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Passive T Line/Wave Speed
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Passive T-Line/Wave Speed
For a passive line, we expect the amplitude to decaydue to loss on the line
The speed of the wave is derived as before. In order to
follow a constant point on the wavefront, you have tomove with velocity
d
dt (t z = constant)
Or, v = dzdt =
=
1
LC
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