ECE 391
supplemental notes - #11
Adding a Lumped Series Element
Consider the following T-line circuit:
Z0,1!
R
Z0,2!
ZL
z in,1 = rin,1 + jxin,1
Z in,1 = z in,1Z 0,1
z in,2
z L,1 = rin,2
è
Z
zL = L
Z
= rin,2 + jxin,2 0,2
Z 0,2
Z
R
+
+ jxin,2 0,2
Z 0,1 Z 0,1
Z 0,1
renormalization
All calculations can be done on Smith chart
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ECE391– Transmission Lines
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1
Smith Chart as Admittance Chart
For circuits involving parallel (shunt) connections of circuit
elements, it is more convenient to work with admittances.
How can the Smith chart be used with admittances?
è How is Y=1/Z = G + jB related to reflection coefficient Γ ?
Consider normalized admittance y
y=
Y G + jB
Z 1
1
=
= g + jb = Y Z 0 = 0 = =
Y0
Y0
Z z r + jx
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Smith Chart as Admittance Chart
Reflection coefficient Γ
Γ=
z −1 1 y −1
y −1
=
=−
z +1 1 y +1
y +1
Smith chart can directly be used with y=g+jb after
rotating reflection coefficient by 180°.
y −1
= −Γ = Γ e jπ
y +1
g = const
b = const
circles
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Impedance vs. Admittance
Coordinates
Impedance Chart
Admittance Chart
Im(Γ)
inductive
r + jx
0
SC
capacitive
r - jx
Im(-Γ)
capacitive
Re(Γ)
OC
g + jb
0
OC
inductive
SC
Re(-Γ)
g - jb
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Example 4
Given the normalized load admittance
yL = 0.5 + j2.0
Determine the normalized admittance at distance
d = λ/16 = 0.0625λ
Solution: use Smith chart with admittance
coordinates è see web applet (example 4)
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Example 5
Given a 50Ω T-line that is terminated in ZL = ZT= 25Ω and has a
shunt resistance Rsh = 50Ω connected at distance d = 1m from
the termination. The wavelength on the line is λ = 3m.
Determine the input impedance at the location of the
shunt resistance.
Solution: see web applet, example 5
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Additional Example
Z L = (25 + j 25)Ω
Z 0 = 50Ω
z L = 0.5 + j 0.5
l = 0.1025λ
Results
SWR ≈ 2.6
ΓL ≈ 0.45 θ L ≈ 116.5
yL =1 − j
zin ≈ 1.5 + j1.1
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Impedance Matching
Implementation Issues
§ Types of Networks
- lumped elements
- distributed elements
- mixed
- reactive vs. resistive
§ Design Criteria
- bandwidth
- lossless/lossy
- complexity
- parasitics
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Impedance Matching
è Matching network needs to have at least
two degrees of freedom (two knobs)
to match a complex load impedance
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Transmission-Line Matching
Idea: Use impedance transformation property
of transmission line
2βz’
load zL (or yL)
transformation circle
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Transmission-Line Matching
Transformation to real z or y
2βz’
load zL (or yL)
real z (or y)
real z (or y)
transformation circle
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Transmission-Line Matching
Transformation to r=1 (or g=1) circle
2βz’
z=1+jx
(y=1+jb)
load zL (or yL)
r=1 or g=1
transformation circle
z=1-jx
(y=1-jb)
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Single Stub Matching
Stub tuners are useful for matching complex load
impedances. The two design parameters in single stub
matching networks are
§ distance from load at which stub is connected
§ length of stub
Other design considerations are
§ open or shorted stub
§ series or shunt connected stub
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Basic Design Principle
Yin = Y0
Objective: Match load admittance by
1. transforming along T-line to Yin = Y0+jB
2. tuning out remaining jB with a parallel shunt element of
admittance –jB
3. shunt element can be realized with a T-line stub
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Design Steps
1. Determine distance d/λ from the load YL (or ZL) at
which the
§ input admittance is Y0 +jB (for shunt stub matching)
§ input impedance is Z0+jX (for series stub matching)
2. Realize the stub with a length l/λ of open- or shortcircuited transmission line (stub).
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Example 6
Using a short-circuited stub, match the load
impedance ZL=(80 + j70)Ω to a transmission line
with Z0 = 50Ω.
ZL
Z0
d
Solution: use Smith chart with admittance coordinates
è see web applet (example 6)
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Additional Example
RL =125Ω
f 0 = 1GHz
CL = 2.54pF Z 0 = 50Ω
z L = 0.5 − j
y in_1,2 ≈ 1 ± j1.58
Need shunt element with
Ysh_1,2 =  j1.58 / Z 0 =  j 31.6mS
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9
Quarter-Wave Transformer
λ / 4 at design frequency f0
Z1 =
Z0Z L
increased mismatch
NOTE: for complex load, insert λ/4 transformer where Zin=R
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Example: Quarter-Wave Transformer
RL =125Ω
f 0 = 1GHz
CL = 2.54pF Z 0 = 50Ω
z L = 0.5 − j
z in,1 ≈ 0.24
Zin,1 ≈ 12Ω
From Zin,2 = Z 0 =
Z 02,T
Z in,1
Z0,T ≈ 24.5Ω
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Bandwidth Performance
at f 0 = 1GHz
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Lumped Element Matching
Reactive L-Section Matching Networks
RL > Z 0
B=
RL < Z 0
X L ± RL Z 0 RL2 + X L2 − RL Z 0
RL2 + X L2
X =
Z
1 X L Z0
+
− 0
B
RL
B RL
different solutions
B=±
(Z 0 − RL )
RL
Z0
X = ± RL (Z 0 − RL ) − X L
may result in different bandwidths
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Example
Design criteria
§ bandwidth
§ realizability of components
§ parameter sensitivity
§ … cost … ?
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Lumped-Distributed and
Distributed Matching Networks
Series Reactance Matching
Shunt Reactance Matching
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Examples
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Double-Stub Tuner
• Double-stub (and multi-stub) tuners use fixed
stub locations (typical spacing: λ/8)
• Advantages:
- easier to accommodate different loads
- better bandwidth performance
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Smith Chart with Lossy T-Lines
Recall for a lossless T-line
jφ ( z')
Z in (z ') 1+ Γ L e
z in (z ') =
=
Z0
1− Γ L e jφ( z')
φ (z ') = θ L − 2 β z '
For low-loss T-lines, Z0 ≈ real and γ = α + jβ
Then
−2 α z' jφ ( z')
Z in (z') 1+ Γ L e −2γ z' 1+ Γ L e e
z in (z') =
=
=
Z0
1− Γ L e −2γ z' 1− Γ L e −2α z'e jφ ( z')
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Smith Chart with Lossy T-Lines
Γ ( z') = Γ L e −2α z'
Γim
As z ' → ∞
z in → 1
(Z in → Z 0 )
Γre
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Approach
Γim
1. Assume lossless line
and rotate by -2βz’
zL
2. Change |Γ| by factor
e −2α z'
Γre
Γ lossy
= Γ lossless
e−2α z'
in
in
Γ inlossless = Γ L e jφ
Oregon State University
ECE391– Transmission Lines
195
Spring Term 2014
Example 7
Given a lossy T-line with characteristic impedance
Z0 = 100Ω, length l = 1.5m (l < λ/2), and
ZSC(z’=l) = 40 – j280 Ω.
(a) Determine α and β.
(b) Determine Zin for ZL = 50 + j50 Ω an l = 1.5m.
Solution using Smith chart
è see web applet (example 7)
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