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ECE 4380/5390 Spring 2013 Instructor: Dr. Raymond Rumpf Office: A-337 E-Mail: [email protected]

Topic #6 Smith Charts

Smith Charts 1

Outline

• Construction of the Smith Chart • and impedance • Circuit theory • Determining VSWR and  • Impedance transformation •

Smith Charts 2

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Construction of the Smith Chart

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Polar Plot of

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   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 of the 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 ZZLL0 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  zzLL1 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

1rij rjxLL 1rij 

1ri j  rjxLL 1ri 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 1ri j  1ri  11rijj  ri r   L 2 2 11 jj     ri ri 1ri  11  jj 1      1  2  rr riiri 2 2 1ri  122jj jj  ririirii 2 2 2 1ri  i xL  2 1222  j  2  ri i 2 1  2 ri 1ri  1222   rij i 2222 11ri   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.

122  r  ri L 2 2 1ri  22 22 2 2 1ri  rrLL2 r L1 1ri   ri   0 rL rrLL11 r L  1 2 2 22 2 2 1 r i rrrLLL2 1 10ri      ri   rrLL r L rrrLLL111 2 2 2 22r i 1 2 210rr  i    rrLL2 rrLL11 rrrLLL ri 22  rL 1 rr11 210rr   22   r  22   r   LL Lr Lr r Li i L 2 22 22 rrr1 21110rr r r LLL2 Lr L r L i L ri 22  r 1 L rrLL11 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 1ri 

2 2 2i 1ri   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   rrLL11 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

zj0.2 0.4

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Example #1 – Step 2 Draw a constant VSWR circle

zj0.2 0.4

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Example #1 – Step 3 Draw line through center of chart

zj0.2 0.4

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Example #1 – Step 4 Read off admittance

zj0.2 0.4

yj1.0 2.0

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Example #2 – Step 1 Plot the impedance on the chart

zj0.5 0.3

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Example #1 – Step 2 Draw a constant VSWR circle

zj0.5 0.3

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Example #2 – Step 3 Draw line through center of chart

zj0.5 0.3

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Example #2 – Step 4 Read off admittance

zj0.5 0.3 yj1.5 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 (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 (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  12 LC j RC 50j  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 Lj1.0 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

zj1

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Example #2 – Circuit Analysis Convert to admittance 1 j1.0

1 j1.5 yin  ? yj0.5 0.5

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Example #2 – Circuit Analysis Add shunt capacitance 1 j1.0

1 j1.5 yin  ?

yj0.5 1.0

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Example #2 – Circuit Analysis Convert to impedance j1.0

1 1 zin  ? j1.5

zj0.4 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

zj0.4 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 zjL 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  ZZL 0 Z in 0 0 jj jj   ZjZ0 cos L sin 0.5Ze0  e 0.5 ZeZ  e jjjj jj  ZeLL Ze Ze00 Ze ZZeLL00 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 1e ZZ00 j  j 2 ZZeL  0 1e 1 j ZZeL  0

We normalize by dividing by Z0.

1e j2 z  in 1e j2

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Normalized Admittance Transformation Formula Our impedance transformation formula was

ZjZLL00tan YjY tan  ZZin 0 YY in 0 ZjZ00LLtan YjY tan 

zjLLtan yj tan  zyin in 1jzLL tan 1 jy tan 

We can write this in terms of the reflection coefficient.

11eejj22  zy in11eejj22 in 

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Interpreting the Formula

The normalized impedance transformation formula was

1e j2 z   in 1e j2

Recognizing that  = ||ej, this equation can be written as

11eejj 2   ej2  z   in 1eejj 2  1e j2 

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 2l. 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  1jj 0.5 tan 2  0.67  zjin   1.299 0.485 1tan110.5tan20.67jzL  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 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 109 Z  35.0  0 C 163 1012 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 109 Z  35.0  0 C 163 1012 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

 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 minZZsL , f maxZZsL , Q  1 Q c 1 fR fR

Virtual resistance R is used only during design. It is not a physical component in the network.

Smith Charts 69

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.

Zj25 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

Smith Charts 71

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.

Smith Charts 73

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 . 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.

Smith Charts 75

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

Smith Charts 77

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 –jA 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 27.5j 35 zjL  0.55  0.7 Z0 50 

c 310 8 ms  0 3 m nf 1.0 100 106 Hz

Smith Charts 79

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

Smith Charts 81

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 part: 0.165

Total: 0.136 + 0.165 = 0.301 0 or 0.5 d A  0.301

d A  90.3 cm

Smith Charts0.364 83

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 –jA 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.

Smith Charts 85

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 .

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

Smith Charts 87

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