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ECE 4380/5390 Spring 2013 Instructor: Dr. Raymond Rumpf Office: A-337 E-Mail: [email protected]
Topic #6 Smith Charts
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Outline
• Construction of the Smith Chart • Admittance and impedance • Circuit theory • Determining VSWR and • Impedance transformation • Impedance matching
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Construction of the Smith Chart
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Polar Plot of Reflection Coefficient
The Smith chart is based on a polar plot of the voltage reflection coefficient . The outer boundary corresponds to || = 1. The reflection coefficient in any passive system must be|| ≤ 1.
j e
radius on Smith chart angle measured CCW from right side of chart
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Normalized Impedance
All impedances are normalized. This is usually done with respect to
the characteristic impedance of the transmission line Z0.
Z z Z0
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Reflection Coefficient form Normalized Impedance
We can write the reflection coefficient in terms of normalized impedances.
Z Z L 0 ZZ Z Z z 1 L 000 L Z Z ZZLL0 L 0 z1 ZZ00
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Derivation of Smith Chart: Solve for Load Impedance
Solving the previous equation for load impedance, we get
z 1 L zL 1 zz11 LL 1 zzLL 1 zL zzLL1 1
zL 11 1 z L 1
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Derivation of Smith Chart: Real and imaginary parts
The load impedance and reflection coefficient can be written in terms of real and imaginary parts.
zrjxLL L r j i
Substituting these into the load impedance equation yields 1 z L 1
1rij rjxLL 1rij
1ri j rjxLL 1ri j
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Derivation of Smith Chart:
Solve for rL and xL
We solve or previous equation for rL and xL by setting the real and imaginary parts equal.
1 j rjx ri LL 22 1ri j 1ri 11rijj ri r L 2 2 11 jj ri ri 1ri 11 jj 1 1 2 rr riiri 2 2 1ri 122jj jj ririirii 2 2 2 1ri i xL 2 1222 j 2 ri i 2 1 2 ri 1ri 1222 rij i 2222 11ri ri Smith Charts 10
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Derivation of Smith Chart:
Rearrange equation for rL
We rearrange the equation for rL so that it has the form of a circle.
122 r ri L 2 2 1ri 22 22 2 2 1ri rrLL2 r L1 1ri ri 0 rL rrLL11 r L 1 2 2 22 2 2 1 r i rrrLLL2 1 10ri ri rrLL r L rrrLLL111 2 2 2 22r i 1 2 210rr i rrLL2 rrLL11 rrrLLL ri 22 rL 1 rr11 210rr 22 r 22 r LL Lr Lr r Li i L 2 22 22 rrr1 21110rr r r LLL2 Lr L r L i L ri 22 r 1 L rrLL11 22rLr rL 1 2 0 2 rir 1 r 1 r 1 L L L 2 ri 2 can be factored r 1 L rL 1
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Derivation of Smith Chart:
Rearrange equation for xL
We rearrange the equation for xL so that it has the form of a circle.
2 x i L 2 2 1ri
2 2 2i 1ri xL
2 2 10 2 ri i xL swap terms can be factored 2 2 11 ri10 2 xxLL
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Derivation of Smith Chart: Two families of circles Constant Resistance Circles Constant Reactance Circles
22 22 rL 2 1 2 11 ri ri1 rrLL11 xLLx These have centers at These have centers at
rL 1 ri 0 ri1 rL 1 xL Radii Radii 1 1
1 rL xL
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Derivation of Smith Chart: Putting it all together
Lines of constant Lines of constant Lines of constant resistance inductive reactance reflection coefficient Superposition
++=
Lines of constant capacitive reactance
We ignore what is outside the || = 1 circle.
We don’t draw the constant || circles.
This is the Smith chart!
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Alternate Way of Visualizing the Smith Chart
Lines of constant resistance Lines of constant reactance Reactance Regions
open L circuit
short circuit C
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3D Smith Chart
The 3D Smith Chart unifies passive and active circuit design.
2D 3D
EE3321 ‐‐ Final Lecture
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Summary of Smith Chart
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Impedance and Admittance on the Smith Chart
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Admittance Coordinates
We could have derived the Smith chart in terms of admittance.
You can make an admittance Smith chart by rotating the standard Smith chart by 180.
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Impedance/Admittance Conversion The Smith chart is just a plot of complex numbers. These could be admittance as well as impedance. To determine admittance from impedance (or the other way around)… 1. Plot the impedance point on the Smith chart. 2. Draw a circle centered on the Smith chart that passes through the point (i.e. constant VSWR). 3. Draw a line from the impedance point, through the center, and to the other side of the circle. 4. The intersection at the other side is the admittance.
impedance admittance
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Visualizing Impedance/Admittance Conversion
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Example #1 – Step 1 Plot the impedance on the chart
zj0.2 0.4
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Example #1 – Step 2 Draw a constant VSWR circle
zj0.2 0.4
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Example #1 – Step 3 Draw line through center of chart
zj0.2 0.4
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Example #1 – Step 4 Read off admittance
zj0.2 0.4
yj1.0 2.0
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Example #2 – Step 1 Plot the impedance on the chart
zj0.5 0.3
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Example #1 – Step 2 Draw a constant VSWR circle
zj0.5 0.3
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Example #2 – Step 3 Draw line through center of chart
zj0.5 0.3
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Example #2 – Step 4 Read off admittance
zj0.5 0.3 yj1.5 0.9
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Circuit Theory
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Adding a Series Capacitor
Suppose we have an initial impedance of
z = 0.5 + j0.7
And we add a series capacitor
of zC = -j1.0.
Since we do not change the resistance, we walk toward capacitance (CCW) around the constant resistance circle.
The “angular” distance covers X=-j1.0 around the constant R circle.
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Adding a Series Inductor
Suppose we have an initial impedance of
z = 0.8 - j1.0
And we add a series inductor
of zL = j1.8.
Since we do not change the resistance, we walk toward Inductance (CW) around the constant resistance circle.
The “angular” distance covers X=j1.8 around the constant R circle.
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Adding a Shunt Capacitor
Here we use the Smith chart rotated by 180° for admittance.
In terms of admittance, capacitance is a positive quantity, but the positive direction is now downward on the Smith chart.
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Adding a Shunt Inductor
We again use the Smith chart rotated by 180° for admittance.
In terms of admittance, inductance is a negative quantity, but the negative direction is now upward on the Smith chart.
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Summary
Series C Shunt C
Series L Shunt L
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Example #1 – Circuit Analysis
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Example #2 – Circuit Analysis
3.3157 nH
Z ? 50 1 in 1.9894 pF ZRjL in jC || 1 RjL jC 1 RjL f 2.4 GHz jC RjL Z0 50 12 LC j RC 50j 1.5080 1010 3.3157 10 9 2 1 1.5080 1010 3.3157 10 9 1.9894 10 12j 1.5080 10 10 50 1.9894 10 12 20j 40 50 20j 40 0.0769 j 0.6154 0.62 82.9 50 20j 40 1 0.0769j 0.6154 VSWR 4.2654 1 0.0769j 0.6154
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Example #2 – Circuit Analysis Normalize Impedances
j1.0
1 zin ? 1 j1.5 50 r 1.0 50 j2 2.4 1099 3.3157 10 Lj1.0 50 1 912 j2 2.4 10 1.9894 10 1 C 50 j 1.5
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Example #2 – Circuit Analysis Plot load impedance j1.0
1 1 zin ? j1.5
z 1
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Example #2 – Circuit Analysis Add series inductor j1.0
1 1 zin ? j1.5
zj1
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Example #2 – Circuit Analysis Convert to admittance 1 j1.0
1 j1.5 yin ? yj0.5 0.5
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Example #2 – Circuit Analysis Add shunt capacitance 1 j1.0
1 j1.5 yin ?
yj0.5 1.0
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Example #2 – Circuit Analysis Convert to impedance j1.0
1 1 zin ? j1.5
zj0.4 0.8
We now know zin
zin z 0.4 j 0.8
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Example #2 – Circuit Analysis Denormalize the impedance j1.0
1 1 zin ? j1.5
zj0.4 0.8
ZZzin 0 in 20 j 40
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Example #2 – Circuit Analysis All together 3.3157 nH
50 1.9894 pF
Zjin 20 40
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Determining VSWR and
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The Horizontal Bar on the Smith Chart
VSWR
Reflectance 2
Reflection Coefficient
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Determining VSWR
1. Plot the normalized load impedance on the Smith chart. 2. Draw a circuit centered on the Smith chart that intersections this point. 3. The VSWR is read where the circle crosses the real axis on right side.
Example: 50 line connected to 75+j10 load impedance. Z 75 j 10 zjL 1.5 0.2 Z0 50
impedance 1 VSWR
VSWR = 1.55
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Example #1 – Circuit Analysis What is the VSWR? 3.3157 nH
50 1.9894 pF
Zjin 20 40
VSWR 4.6
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Example #1 – Circuit Analysis What is the reflection coefficient?
0.62
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Impedance Transformation
Normalized Impedance Transformation Formula Our impedance transformation formula was
ZjZL 0 tan zjL tan ZZin 0 zin ZjZ0 L tan 1tan jzL We can write this in terms of the reflection coefficient.
jj jj ZjZcos sin 0.5ZeL e 0.5 Ze0 e ZZL 0 Z in 0 0 jj jj ZjZ0 cos L sin 0.5Ze0 e 0.5 ZeZ e jjjj jj ZeLL Ze Ze00 Ze ZZeLL00 ZZe ZZ00jjjj j j Ze00 Ze ZeLL Ze Z L Z 0 e Z LZe0 ZZe j 1 L 0 j j 2 ZZeL 0 1e ZZ00 j j 2 ZZeL 0 1e 1 j ZZeL 0
We normalize by dividing by Z0.
1e j2 z in 1e j2
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Normalized Admittance Transformation Formula Our impedance transformation formula was
ZjZLL00tan YjY tan ZZin 0 YY in 0 ZjZ00LLtan YjY tan
zjLLtan yj tan zyin in 1jzLL tan 1 jy tan
We can write this in terms of the reflection coefficient.
11eejj22 zy in11eejj22 in
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Interpreting the Formula
The normalized impedance transformation formula was
1e j2 z in 1e j2
Recognizing that = ||ej, this equation can be written as
11eejj 2 ej2 z in 1eejj 2 1e j2
Thus we see that traversing along the transmission line simply changes the phase of the reflection coefficient.
As we move away from the load and toward the source, we subtract phase from . On the Smith chart, we rotate clockwise (CW) around the constant VSWR circle by an amount 2l. A complete rotation corresponds to /2.
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Impedance Transformation on the Smith chart
1. Plot the normalized load impedance on the Smith chart. 2. Move clockwise around the middle of the Smith chart as we move away from the load (toward generator). One rotation is /2 in the transmission line. 3. The final point is the input impedance of the line.
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Example #1 – Impedance Trans. Normalized the parameters
0.67
Z0 50 ZjL 50 25
0.67
zjL 10.5
zjL tan 1jj 0.5 tan 2 0.67 zjin 1.299 0.485 1tan110.5tan20.67jzL j j
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Example #1 – Impedance Trans. Plot load impedance
0.67
zjL 10.5
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Example #1 – Impedance Trans.
Walk away from load 0.67 0.145
0.67
zjL 10.5
Since the Smith chart repeats every 0.5, traversing 0.67 is the same as traversing 0.17.
Here we start at 0.145 on the Smith chart.
We traverse around the chart to 0.145 + 0.17 = 0.315.
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Example #1 – Impedance Trans. Determine input impedance
0.67
Zin zjL 10.5
Reflection at the load will be the same regardless of the length of line.
Therefore the VSWR will the same.
The input impedance must lie on the same VSWR plane.
zjin 1.3 0.5
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Example #1 – Impedance Trans. Denormalize
0.67
Zin zjL 10.5
To determine the actual input impedance, we denormalize.
ZZzin 0 in 50 1.3 j 0.5 65 j 25
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Example #2 – Calculate Line Length
3.5 cm 30
L 200 nH m Zin 5 nH C 163 pF m
f 1.5 GHz
Free Space Wavelength c 310 8 0 20 cm 0 f 1.5 109 Wavelength on the Line 21 2 f LC f LC 1 11.68 cm 1.5 109912 200 10 163 10
Note: ≠ 0 so these are NOT the same. Use instead of 0 for Smith chart analysis. Smith Charts 61
Example #2 – Calculate Z’s
3.5 cm 30
L 200 nH m Zin 5 nH C 163 pF m
f 1.5 GHz
Line Impedance L 200 109 Z 35.0 0 C 163 1012 Load Impedance
ZRjLL 30j 2 1.5 1099 5 10 30j 47.1
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Example #2 – Normalize
3.5 cm 30
L 200 nH m Zin 5 nH C 163 pF m
f 1.5 GHz
Normalize Load Impedance L 200 109 Z 35.0 0 C 163 1012 Load Impedance
ZjL 30 47.1 zjL 0.86 1.34 Z0 35.0
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Example #2 – Transform
0.168 3.5 cm 30
L 200 nH m Zin 5 nH C 163 pF m
f 1.5 GHz
Line Length in Wavelengths
3.5 cm 0.3 11.68 cm
Azimuthal Distance Start: 0.168 0.468 End: 0.168 0.30 0.46 8 Transformed Impedance
zjin 0.26 0.19
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Example #2 – Denormalize
0.168 3.5 cm 30
L 200 nH m Zin 5 nH C 163 pF m
f 1.5 GHz
Denormalize Input Impedance
ZZzin 0 in 35.0 0.26 j 0.19 9.21j 6.65
0.468 Exact Answer
Zjin 9.42 7.03
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Example #2 – Component Values
0.168 3.5 cm 30
L 200 nH m Zin 5 nH C 163 pF m
f 1.5 GHz
Component Values 1 ZR 9.21 j 6.65 in jC R 9.21 1 C 15.9 pF 2 1.5 109 6.65 0.468
9.21
Zin 15.9 pF
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Impedance Matching
Two‐Element Matching L Network
The most common impedance matching circuit is the L network.
Circuit Q is fixed and cannot be controlled.
Considerations: 1. Elimination of stray reactance. 2. Need for filtering. 3. Need to pass or block DC.
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Three‐Element Matching Pi and T Networks
Three‐element matching networks are advantageous when it is desired to control the circuit Q.
Q must be greater than the Q possible with an L network.
Pi Network T Network
fc minZZsL , f maxZZsL , Q 1 Q c 1 fR fR
Virtual resistance R is used only during design. It is not a physical component in the network.
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Two‐Element Impedance Matching with a Smith Chart
Zjsrc 25 15
ZjL 100 25
Solution 1. Matching network must be low‐pass to conduct DC. 2. This dictates series L and shunt C. 3. We walk along constant X circles until the input impedance is the source impedance.
Zj25 15 src 159 nH
38.8 pF
ZjL 100 25
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What is Stub Tuning? (1 of 6)
Zsrc Z , 0 ZL
Power is reflected due to an impedance mismatch. Z Z L 0 ZL Z0
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What is Stub Tuning? (2 of 6)
d A
Zsrc Z , 0 ZL
A
We want to add a short circuit stub to match the impedance. 0
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Stub Tuning Concept (3 of 6)
dA
Zsrc Z , 0 ZL
11 YZAA Z 0 jB A
We back off from the load until the real part of the
input admittance is 1/Z0.
At this point, the real part of admittance is matched to the transmission line.
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Stub Tuning Concept (4 of 6)
d A
zsrc Z , 0 YSA zL
1 YZin 0
We can match perfectly to this admittance by introducing a shunt element with the conjugate susceptance. YjBSA A
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Stub Tuning Concept (5 of 6)
A
YjBSA A Z0 ,
To realize this shunt susceptance with a short‐circuit
stub, we back off some distance from a short A circuit load until the input admittance is –jBA.
This is our stub.
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Stub Tuning Concept (6 of 6)
d A
zsrc Z , 0 zL
A
Last, we add the stub at position dA from the load to cancel the susceptance of the load. 1 YZin 0
The load is matched and we have zero reflection!
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Voltage on Transmission Line Before and After Stub Tuning
zsrc Z , 0 zL
V(z) without stub tuner
V(z) with stub tuner
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Single‐Stub Tuning Procedure
Step 1 – You will be given Z0, ZL, n and f. Normalize the impedances and calculate .
ZL c0 zL Z0 nf
Step 2 – Plot zL and then invert to find the corresponding admittance yL. Step 3 – Walk CW around the constant VSWR circle until the R=G=1 circle intersects it. There will be two intersections (A=closest, B=farthest). Step 4 – Pick A or B. They will be complex conjugates. Assume A for the following steps.
Step 5 – How far in the CW direction did you travel to get to A and B? These are dA and dB. Step 6 – yA is the admittance where the stub is about to be placed. This is probably point A. We need to cancel the reactive component A of this.
ygjA AA y SAA 1 j
Step 7 – Find the –jA circle on the chart and follow it to the outside of the chart. Step 8 – Start at the far right side of the chart and move CW to the point above.
Step 9 – Determine the distance (in wavelengths) this represents. This is lA.
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Single‐Stub Tuning Example – Step 1
Problem: A 50 transmission line with an air‐core operates at 100 MHz and is connected to a
load impedance of ZL = 27.5 + j35 . Design a single‐stub tuner.
Step 1 – Normalize impedance and calculate .
ZL 27.5j 35 zjL 0.55 0.7 Z0 50
c 310 8 ms 0 3 m nf 1.0 100 106 Hz
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Single‐Stub Tuning Example – Step 2
Plot impedance and invert to find admittance.
We read
yjL 0.72 0.87
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Single‐Stub Tuning Example – Step 3
Walk CW around the constant VSWR circle until the R=G=1 circle intersects it.
We read A A 11.1j B 11.1j
B
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Single‐Stub Tuning Example – Step 4
Pick A or B.
We will pick A because it leads to the shortest stub. A
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Single‐Stub Tuning Example – Step 5 0.165 How far CW did we traverse to get to A?
First part: 0.5 – 0.364 = 0.136 A Second part: 0.165
Total: 0.136 + 0.165 = 0.301 0 or 0.5 d A 0.301
d A 90.3 cm
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Single‐Stub Tuning Example – Step 6
yA is the admittance where the stub is about to be placed. We chose point A.
A 11.1j A
We need to cancel the reactive
component A of this.
yjSA 1.1
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Single‐Stub Tuning Example – Step 7
Find the –jA circle on the chart and follow it to the outside of the chart.
yjSA 1.1
We are setting up to do an admittance transformation in the stub to realize a –j1.1 input admittance.
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Single‐Stub Tuning Example – Step 8
Start at the far right side of the chart and move CW to the point above (move away from short).
Here we are doing an admittance transformation to realize –j1.1.
The far right side of the Smith chart is a short circuit for admittances.
Admittance transformation
A Smith Charts 86
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Single‐Stub Tuning Example – Step 9
Determine the distance (in wavelengths) this represents.
This is lA.
Line starts at 0.25.
Line ends at 0.367.
Total length: 0.367 – 0.25 = 0.117
A 0.117
A 0.117 3 m 0.351 m 35.1 cm
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