EECS 117
Lecture 6: Lossy Transmission Lines and the Smith
Chart
Prof. Niknejad
University of California, Berkeley
University of California, Berkeley
EECS 117 Lecture 6 – p. 1/33
Dispersionless Line
To find the conditions for the transmission line to be
dispersionless in terms of the R, L, C , G, expand
p
(jωL0 + R0 )(jωC 0 + G0 )
γ =
s
RG
G
R
2
)
+
+
(jω) LC(1 +
=
2
jωL jωC (jω) LC
q
√
(jω)2 LC =
Suppose that R/L = G/C and simplify the term
R2
2R
+
=1+
jωL (jω)2 L2
University of California, Berkeley
EECS 117 Lecture 6 – p. 2/33
Dispersionless Line (II)
For R/L = G/C the propagation constant simplifies
=
R
1+
jωL
2
√
R
γ = −jω LC 1 +
jωL
Breaking γ into real and imaginary components
r
√
C
− jω LC = α + jβ
γ=R
L
The attenauation constant α is independent of
frequency. For low loss lines, α ≈ − ZR0 X
The propagation constant β is a linear function of
frequency X
University of California, Berkeley
EECS 117 Lecture 6 – p. 3/33
Lossy Transmission Line Attenuation
The power delivered into the line at a point z is now
non-constant and decaying exponentially
+ |2
|v
1
−2αz
e
< (Z0 )
Pav (z) = < (v(z)i(z)∗ ) =
2
2|Z0 |
2
For instance, if α = .01m− 1, then a transmission line of
length ` = 10m will attenuate the signal by 10 log(e2α` ) or
2 dB. At ` = 100m will attenuate the signal by 10 log(e2α` )
or 20 dB.
University of California, Berkeley
EECS 117 Lecture 6 – p. 4/33
Lossy Transmission Line Impedance
Using the same methods to calculate the impedance for
the low-loss line, we arrive at the following line
voltage/current
v(z) = v + e−γz (1 + ρL e2γz ) = v + e−γz (1 + ρL (z))
v + −γz
e (1 − ρL (z))
i(z) =
Z0
Where ρL (z) is the complex reflection coefficient at
position z and the load reflection coefficient is unaltered
from before
The input impedance is therefore
e−γz + ρL eγz
Zin (z) = Z0 −γz
e
− ρL eγz
University of California, Berkeley
EECS 117 Lecture 6 – p. 5/33
Lossy T-Line Impedance (cont)
Substituting the value of ρL we arrive at a similar
equation (now a hyperbolic tangent)
ZL + Z0 tanh(γ`)
Zin (−`) = Z0
Z0 + ZL tanh(γ`)
For a short line, if γδ` 1, we may safely assume that
Zin (−δ`) = Z0 tanh(γδ`) ≈ Z0 γδ`
√
p
Recall that Z0 γ = Z 0 /Y 0 Z 0 Y 0
As expected, input impedance is therefore the series
impedance of the line (where R = R0 δ` and L = L0 δ`)
Zin (−δ`) = Z 0 δ` = R + jωL
University of California, Berkeley
EECS 117 Lecture 6 – p. 6/33
Review of Resonance (I)
We’d like to find the impedance of a series resonator
1
+R
near resonance Z(ω) = jωL + jωC
Recall the definition of the circuit Q
time average energy stored
Q = ω0
energy lost per cycle
For a series resonator, Q = ω0 L/R. For a small
frequency shift from resonance δω ω0
!
1
1
+R
Z(ω0 + δω) = jω0 L + jδωL +
δω
jω0 C 1 + ω
0
University of California, Berkeley
EECS 117 Lecture 6 – p. 7/33
Review of Resonance (II)
Which can be simplified using the fact that ω0 L =
1
ω0 C
Z(ω0 + δω) = j2δωL + R
Using the definition of Q
δω
Z(ω0 + δω) = R 1 + j2Q
ω0
For a parallel line, the same formula applies to the
admittance
δω
Y (ω0 + δω) = G 1 + j2Q
ω0
Where Q = ω0 C/G
University of California, Berkeley
EECS 117 Lecture 6 – p. 8/33
λ/2 T-Line Resonators (Series)
A shorted transmission line of length ` has input
impedance of Zin = Z0 tanh(γ`)
For a low-loss line, Z0 is almost real
Expanding the tanh term into real and imaginary parts
j sin(2β`)
sinh(2α`)
+
tanh(α`+jβ`) =
cos(2β`) + cosh(2α`) cos(2β`) + cosh(2α`)
Since λ0 f0 = c and ` = λ0 /2 (near the resonant
frequency), we have
β` = 2π`/λ = 2π`f /c = π + 2πδf `/c = π + πδω/ω0
If the lines are low loss, then α` 1
University of California, Berkeley
EECS 117 Lecture 6 – p. 9/33
λ/2 Series Resonance
Simplifying the above relation we come to
πδω
Zi n = Z0 α` + j
ω0
The above form for the input impedance of the series
resonant T-line has the same form as that of the series
LRC circuit
We can define equivalent elements
Req = Z0 α` = Z0 αλ/2
Leq
πZ0
=
2ω0
University of California, Berkeley
Ceq
2
=
Z0 πω0
EECS 117 Lecture 6 – p. 10/33
λ/2 Series Resonance Q
The equivalent Q factor is given by
β0
π
1
=
=
Q=
2α
αλ0
ω0 Req Ceq
For a low-loss line, this Q factor can be made very large.
A good T-line might have a Q of 1000 or 10,000 or more
It’s difficult to build a lumped circuit resonator with such
a high Q factor
University of California, Berkeley
EECS 117 Lecture 6 – p. 11/33
λ/4 T-Line Resonators (Parallel)
For a short-circuited λ/4 line
Zin
tanh α` + j tan β`
= Z0 tanh(α + jβ)` = Z0
1 + j tan β` tanh α`
Multiply numerator and denominator by −j cot β`
Zin
1 − j tanh α` cot β`
= Z0
tanh α` − j cot β`
For ` = λ/4 at ω = ω0 and ω = ω0 + δω
π πδω
ω0 ` δω`
= +
+
β` =
2ω0
2
v
v
University of California, Berkeley
EECS 117 Lecture 6 – p. 12/33
λ/4 T-Line Resonators (Parallel)
So cot β` = −tan πδω
2ω0 ≈
Zin
−πδω
2ω0
and tanh α` ≈ α`
Z0
1 + jα`πδω/2ω0
≈
= Z0
α` + jπδω/2ω0
α` + jπδω/2ω0
This has the same form for a parallel resonant RLC
circuit
1
Zin =
1/R + 2jδωC
The equivalent circuit elements are
Req
Z0
=
α`
Ceq
π
=
4ω0 Z0
University of California, Berkeley
Leq
1
= 2
ω0 Ceq
EECS 117 Lecture 6 – p. 13/33
λ/4 T-Line Resonators Q Factor
The quality factor is thus
β
π
=
Q = ω0 RC =
2α
4α`
University of California, Berkeley
EECS 117 Lecture 6 – p. 14/33
The Smith Chart
The Smith Chart is simply a graphical calculator for
computing impedance as a function of reflection
coefficient z = f (ρ)
More importantly, many problems can be easily
visualized with the Smith Chart
This visualization leads to a insight about the behavior
of transmission lines
All the knowledge is coherently and compactly
represented by the Smith Chart
Why else study the Smith Chart? It’s beautiful!
There are deep mathematical connections in the Smith
Chart. It’s the tip of the iceberg! Study complex analysis
to learn more.
University of California, Berkeley
EECS 117 Lecture 6 – p. 15/33
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University of California, Berkeley
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An Impedance Smith Chart
Without further ado, here it is!
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EECS 117 Lecture 6 – p. 16/33
Generalized Reflection Coefficient
In sinusoidal steady-state, the voltage on the line is a
T-line
v(z) = v + (z) + v − (z) = V + (e−γz + ρL eγz )
Recall that we can define the reflection coefficient
anywhere by taking the ratio of the reflected wave to the
forward wave
ρL eγz
v − (z)
= −γz = ρL e2γz
ρ(z) = +
e
v (z)
Therefore the impedance on the line ...
v + e−γz (1 + ρL e2γz )
Z(z) = v+
−γz (1 − ρ e2γz )
e
L
Z0
University of California, Berkeley
EECS 117 Lecture 6 – p. 17/33
Normalized Impedance
...can be expressed in terms of ρ(z)
1 + ρ(z)
Z(z) = Z0
1 − ρ(z)
It is extremely fruitful to work with normalized
impedance values z = Z/Z0
1 + ρ(z)
Z(z)
=
z(z) =
1 − ρ(z)
Z0
Let the normalized impedance be written as z = r + jx
(note small case)
The reflection coefficient is “normalized” by default
since for passive loads |ρ| ≤ 1. Let ρ = u + jv
University of California, Berkeley
EECS 117 Lecture 6 – p. 18/33
Dissection of the Transformation
Now simply equate the < and = components in the
above equaiton
((1 + u + jv)(1 − u + jv)
(1 + u) + jv
=
r + jx =
(1 − u)2 + v 2
(1 − u) − jv
To obtain the relationship between the (r,x) plane and
the (u,v) plane
1 − u2 − v 2
r=
(1 − u)2 + v 2
v(1 − u) + v(1 + u)
x=
(1 − u)2 + v 2
The above equations can be simplified and put into a
nice form
University of California, Berkeley
EECS 117 Lecture 6 – p. 19/33
Completing Your Squares...
If you remember your high school algebra, you can
derive the following equivalent equations
r
u−
1+r
2
1
+v =
(1 + r)2
2
1
(u − 1) + v −
x
2
2
1
= 2
x
These are circles in the (u,v) plane! Circles are good!
We see that vertical and horizontal lines in the (r,x)
plane (complex impedance plane) are transformed to
circles in the (u,v) plane (complex reflection coefficient)
University of California, Berkeley
EECS 117 Lecture 6 – p. 20/33
Resistance Transformations
r=2
r=1
r=0
.5
r=
0
v
u
r
r=0 r=.5 r=1
r=2
r = 0 maps to u2 + v 2 = 1 (unit circle)
r = 1 maps to (u − 1/2)2 + v 2 = (1/2)2 (matched real
part)
r = .5 maps to (u − 1/3)2 + v 2 = (2/3)2 (load R less than
Z0 )
r = 2 maps to (u − 2/3)2 + v 2 = (1/3)2 (load R greater
than Z0 )
University of California, Berkeley
EECS 117 Lecture 6 – p. 21/33
Reactance Transformations
x
v
x=2
x=1
x
=
x=2
.5
x=1
x=0
u
r
-2
x=
x = -1
x
x = -2
=.5
x = -1
x = ±1 maps to (u − 1)2 + (v ∓ 1)2 = 1
x = ±2 maps to (u − 1)2 + (v ∓ 1/2)2 = (1/2)2
x = ±1/2 maps to (u − 1)2 + (v ∓ 2)2 = 22
Inductive reactance maps to upper half of unit circle
Capacitive reactance maps to lower half of unit circle
University of California, Berkeley
EECS 117 Lecture 6 – p. 22/33
Complete Smith Chart
r>1
v
match
x=2
x=1
x
=
r=
0
.5
r=2
r=0
.5
inductive
r=1
open
u
=.5
capacitive
University of California, Berkeley
-2
x=
x = -1
x
short
EECS 117 Lecture 6 – p. 23/33
Load on Smith Chart
load
First map zL on the Smith Chart as ρL
To read off the impedance on the T-line at any point on
a lossless line, simply move on a circle of constant
radius since ρ(z) = ρL e2jβ
University of California, Berkeley
EECS 117 Lecture 6 – p. 24/33
Motion Towards Generator
Moving towards
generator means
ρ(−`) = ρL e−2jβ` , or
clockwise motion
towards generator
ent
m
ve
mo
For a lossy line, this
corresponds to a
spiral motion
load
We’re back to where
we started when
2β` = 2π , or ` = λ/2
Thus the impedance
is periodic (as we
know)
University of California, Berkeley
EECS 117 Lecture 6 – p. 25/33
SWR Circle
Since SWR is a function of |ρ|, a circle at origin in
(u,v) plane is called an SWR circle
Recall the voltage max
occurs when the reflected
wave is in phase with the
forward wave, so
ρ(zmin ) = |ρL |
S
voltage min
W
R
CIRC
LE
ρL = |ρL |ejθ
voltage max
ρ = |ρL |ej(θ−2β`)
This corresponds to the
intersection of the SWR
circle with the positive real
axis
Likewise, the intersection
with the negative real axis
is the location of the voltge
min
University of California, Berkeley
EECS 117 Lecture 6 – p. 26/33
Example of Smith Chart Visualization
Prove that if ZL has an inductance reactance, then the
position of the first voltage maximum occurs before the
voltage minimum as we move towards the generator
A visual proof is easy using Smith Chart
On the Smith Chart start at any point in the upper half
of the unit circle. Moving towards the generator
corresponds to clockwise motion on a circle. Therefore
we will always cross the positive real axis first and then
the negative real axis.
University of California, Berkeley
EECS 117 Lecture 6 – p. 27/33
Impedance Matching Example
Single stub impedance matching is easy to do with the
Smith Chart
Simply find the intersection of the SWR circle with the
r = 1 circle
The match is at the center of the circle. Grab a
reactance in series or shunt to move you there!
University of California, Berkeley
EECS 117 Lecture 6 – p. 28/33
Series Stub Match
University of California, Berkeley
EECS 117 Lecture 6 – p. 29/33
Admittance Chart
1−ρ
, you can imagine that an
Since y = 1/z = 1+ρ
Admittance Smith Chart looks very similar
In fact everything is switched around a bit and you can
buy or construct a combined admittance/impedance
smith chart. You can also use an impedance chart for
admittance if you simply map x → b and r → g
Be careful ... the caps are now on the top of the chart
and the inductors on the bottom
The short and open likewise swap positions
University of California, Berkeley
EECS 117 Lecture 6 – p. 30/33
Admittance on Smith Chart
Sometimes you may need to work with both
impedances and admittances.
This is easy on the Smith Chart due to the impedance
inversion property of a λ/4 line
2
Z
Z0 = 0
Z
If we normalize Z 0 we get y
1
Z0
Z0
= =y
=
z
Z
Z0
University of California, Berkeley
EECS 117 Lecture 6 – p. 31/33
Admittance Conversion
Thus if we simply rotate π degrees on the Smith Chart
and read off the impedance, we’re actually reading off
the admittance!
Rotating π degrees is easy. Simply draw a line through
origin and zL and read off the second point of
intersection on the SWR circle
University of California, Berkeley
EECS 117 Lecture 6 – p. 32/33
Shunt Stub Match
Let’s now solve the same matching problem with a
shunt stub.
To find the shunt stub value, simply convert the value of
z = 1 + jx to y = 1 + jb and place a reactance of −jb in
shunt
University of California, Berkeley
EECS 117 Lecture 6 – p. 33/33