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Smith Chart

Smith Chart

djo ed β2)( −Γ=Γ

Reflection coefficientIn the previous section we observed that the reflection coefficient of a losslesstransmission line is given by:

)(dΓ

where

oL

oLo ZZ

ZZ+−

=Γ

These equations represents the basis of the Smith Chart

djo ed β2)( −Γ=Γ

Calculation of Reflection CoefficientExample: Calculate the reflection coefficients of a transmission line with Zo=50Ω and terminated with the following loads:(a) ZL=0 Ω(b) ZL→∞ (open circuits)(c) ZL=50 Ω(d) ZL=(16.67-j16.67) Ω(e) ZL=(50+j50) Ω

)(dΓ

oL

oLo ZZ

ZZ+−

=Γ

Solution:In general the ZL can be a complex number, consequently Гo will also be complex number

θjooior

oL

oLo ejZZ

ZZΓ=Γ+Γ=

+−

=Γ

Smith Chart

djo ed β2)( −Γ=Γ

(a) ZL=0 Ω)(dΓ

oL

oLo ZZ

ZZ+−

=Γ

050505050

=Γ⇒+−

=+−

=Γ ooL

oLo ZZ

ZZ

1500500

−=Γ⇒+−

=+−

=Γ ooL

oLo ZZ

ZZ

15050

=Γ⇒−∞−∞

=+−

=Γ ooL

oLo ZZ

ZZ(b) ZL→∞ (open circuit)

(c) ZL=50 Ω

(d) ZL= (16.67-j16.67) Ω

(e) ZL= (50+j50) Ω

oo

oL

oLo ZZ

ZZ 22154.050 j16.67)-(16.6750 j16.67)-(16.67

∠=Γ⇒+−

=+−

=Γ

oo

oL

oLo ZZ

ZZ 3483.050 j50)(5050 j50)(50

∠=Γ⇒++−+

=+−

=Γ

Smith Chart

(a) ZL=0 Ω

0=Γo

o

io e

1801

11

∠=

=−=Γ π

1=Γo(b) ZL→∞

(c) ZL=50 Ω

(d) ZL= (16.67-j16.67) Ω

(e) ZL= (50+j50) Ω

oo 22154.0 ∠=Γ

oo 3483.0 ∠=Γ

Graphical Representation

This relation can be understood as the transformation of an impedance into a reflection coefficient.

Smith Chart

Transformation of an impedance into a reflection coefficient

Consequently, any load impedance ZL can be “transformed” into a reflection coefficient.The Smith Chart is based on this observation.

R

X Z=R + j X

50Ω

50Ω50+j50Ω

16.66Ω

-16.66Ω 16.66-j16.66 Ω

0Ω

oL

oLo ZZ

ZZ+−

=Γ

Smith Chart

Normalized Impedances

Consider a transmission line with characteristic impedance Zoand load impedance ZL.

The normalized load impedances are defined as:)(dΓ

o

LL Z

Zz =

oZZz in

in =

Example: Calculate the normalized impedance zL. Consider that ZL=(50 + j 50) Ω and Zo=50 Ω.

11505050 jjzL +=

+=

Smith Chart

Creating the Smith ChartConsider the normalized load impedances are defined as:

)(dΓ

o

LL Z

Zz =oZ

Zz inin =

The reflection coefficient at the load is given by

11

+−

=+−

=+−

=ΓL

L

ooL

ooL

oL

oLo z

zZZzZZz

ZZZZ

1=o

o

ZZ

11

+−

=+−

=+−

=Γ zz

ZZzZZz

ZZZZ

oo

oo

o

o

A generic reflection coefficient is given by

jxrz +=

Note that:oZ

Zz =

Smith Chart

Creating the Smith Chart: Constant Resistance Lines

ir jjxrjxr

zz

Γ+Γ=++−+

=+−

=Γ 11

11

Smith Chart

Creating the Smith Chart: Constant Reactance Lines

ir jjxrjxr

zz

Γ+Γ=++−+

=+−

=Γ 11

11

capacitive

inductive

Smith Chart

Creating the Smith Chart: Examples

Smith Chart

The Smith

Chart

Smith Chart

Example 1: Impedance Transformation

ZL=(30 +j 60) Ω is connected to a 50 Ω transmission line of 2 cm and operated at 2 GHz. The phase velocity vp=0.5 c.

Find the input impedance Zin

)(dΓ

Zo=50Ω

ZL=(30+j60)Ω

=2 cm

Solution1st step: Normalized the impedance ZL

2.16.0506030 jj

ZZz

o

LL +=

+==

2nd step: Locate the normalize impedance in the Smith Chart

3rd step: Identify the corresponding load reflection coefficient (Гo) in the Smith Chart

oo 56.716325.0 ∠=Γ

Smith Chart

4th step: The reflection coefficient at the input (Гin) will have the same magnitude at the input but different phase.

θβ jo

djoin eecmld Γ=Γ===Γ=Γ − 2)2(

)99.19156.71(6325.0)2( ooin cmld −∠===Γ=Γ

In the Smith Chart rotate 191.99o counterclockwise

Find Гin.5th step: Find the normalize impedance zin from the Smith Chart

oin 43.1206325.0 −∠=Γ

53.03.0 jzin −=

6th step: Convert zin into the actual impedance Zin

⇒Ω−== 50)53.03.0( jZzZ oinin Ω−Ω= )5.2615( jZin

o

pp

radmsm

Hz

lvlvfll

99.191351.302.0/1035.0

1024

24222

8

9

−≡−=××

×−=

−=−=−=−=

πθ

ωπλπβθ π ≡ 180o

Smith Chart

Smith Chart for Example 1

Smith Chart

Example 2: Return lossThe return loss of a transmission line connected to load ZL is 10 dB.

(a) Find the region in the Smith Chart such that RL(dB)=10 dB.

(b) Calculate the maximum and minimum resistive load with RL=10 dB

)(dΓ

Zo=50Ω

Solution(a) The return loss is given by

(b) From the Smith Chart we obtain:

/20)(

10log20RL(dB)dBRL

oo−

=Γ⇒Γ−=

0.316210 2010

=Γ⇒=Γ−

oo

Ω=⇒Ω×=⇒=Ω=⇒Ω×=⇒=

265052.052.0965092.192.1

minminmin

maxmaxmax

RRrRRr

Smith Chart

Smith Chart for Example 2

Smith Chart

Example 3: Г, SWR, RLFour different load impedances:

(a) ZL=50Ω (b) ZL=48.5Ω (c) ZL=(75+j 25)Ω(d) ZL=(10-j5)Ω are sequentially connected to a 50 Ω transmission line. Find the reflection coefficients and the SWR circles and determine the return loss in dB.

Solution

5.05SWR dB,50.3RL(dB)66.066.01.02.050510(c)

77.1SWR dB,1.11RL(dB)15.023.05.05.1502575(c)

03.1SWR dB,3.36RL(dB)015.097.0505.48(b)

1SWR ,RL(dB)000.15050(a)

==⇒−=Γ⇒−=Ω−

=

==⇒+=Γ⇒+=Ω

+=

==⇒−=Γ⇒=Ω

=

=∞→⇒=Γ⇒=ΩΩ

=

jjjz

jjjz

z

z

L

L

L

L

Γ−Γ+

=Γ=+−

=Γ=11

SWR ; log-20RL(dB);;oL

oL

o

LL ZZ

ZZZZz

)(dΓ

Zo=50Ω

Smith Chart

Smith Chart for Example 3

RL=11.1 dB

RL=3.5 dB

RL=36.3dB

Smith Chart

Example 4: Open Circuit TransformationFor an open-ended 50Ω transmission line operated at 3 GHz, and with a phase velocity of 77% of the speed of light, find the line lengths to create a 2 pF capacitor and a 5.3 nH inductor. Use the Smith Chart to do the calculation.

Solution: (a) Capacitor

λλλ

ππω

172.025.0422.0ts). wavelenghof multiple(in length themeasureChart Smith In the:3 Step

ChartSmiththeinimpedancethisLocate

53.0505.26 :ImpedanceNomalized

26.510210322

1Capacitor

ChartSmith in theit locate andcapacitor theof impedance theCalculate :2 StepChartSmith in the 1 Locate :1 Step

1

1

129

=−=

→−=ΩΩ−

==

Ω−=××××

−=

−==⇒

=Γ

−

dd

jjZZz

jFHz

jCf

jCjZ

o

LL

L

=50Ω

θβ jo

djo ee −− Γ=Γ=Γ 2

Smith Chart

mm24.13mm77172.0172.0 Calculate:5 Step

mm77m6.81

222

m6.81/10377.0

10322

2 where λ.h wavelengt theCalculate:4 Step

111

1

1

18

9

2

=⇒×==

==⇒=⇒=

=××××

===

=Γ=Γ=Γ

−

−

−−

ddd

smHz

vf

v

dee

pp

jo

djo

λ

πλβπλλ

πβ

ππωβ

βθθβ

=50Ω

θβ jo

djo ee −− Γ=Γ=Γ 2

Solution: (b) Inductor

mm8.32mm77426.0426.025.0)176.5.0(ts). wavelenghof multiple(in length themeasureChart Smith In the:3 Step

ChartSmiththeinimpedancethisLocate

00.2509.99 :ImpedanceNomalized

99.9103.510322CapacitorChartSmith in theit locate andcapacitor theof impedance theCalculate :2 Step

ChartSmith in the 1 Locate :1 Step

222

1

99

=⇒×=⇒=−+=

→=Ω

==

Ω=××××===⇒

=Γ

−

dddd

jjZZz

jHHzjLfjLjZ

o

LL

L

λλλλ

ππω

Smith Chart

Smith Chart for Example 4

0.25

0.422

0.5

0.176

Smith Chart

Example 5: Short Circuit TransformationFor an short circuit-ended 50Ω transmission line operated at 3 GHz, and with a phase velocity of 77% of the speed of light, find the line lengths to create a 2 pF capacitor and a 5.3 nH inductor. Use the Smith Chart to do the calculation.

Solution: (a) Capacitor

λ

ππω

422.0ts). wavelenghof multiple(in length themeasureChart Smith In the:3 Step

ChartSmiththeinimpedancethisLocate

53.0505.26 :ImpedanceNomalized

26.510210322

1Capacitor

ChartSmith in theit locate andcapacitor theof impedance theCalculate :2 StepChartSmith in the 1 Locate :1 Step

1

1

129

=

→−=Ω

Ω−==

Ω−=××××

−=

−==⇒

−=Γ

−

dd

jjZZz

jFHz

jCf

jCjZ

o

LL

L

θβ jo

djo ee −− Γ=Γ=Γ 2

1−=Γo

Smith Chart

mm32.5mm77422.0422.0 Calculate:5 Step

mm7722m6.81 :4 Example From

λ.h wavelengt theCalculate:4 Step

11

1

1

=⇒×==

==⇒=

= −

ddd

λ

βπλλ

πβ

β

=50Ω

Solution: (b) Inductor

mm6.13mm77176.0176.0ts). wavelenghof multiple(in length themeasureChart Smith In the:3 Step

ChartSmiththeinimpedancethisLocate

00.2509.99 :ImpedanceNomalized

99.9103.510322CapacitorChartSmith in theit locate andcapacitor theof impedance theCalculate :2 Step

ChartSmith in the 1 Locate :1 Step

222

1

99

=⇒×=⇒=

→=Ω

==

Ω=××××===⇒

−=Γ

−

dddd

jjZZz

jHHzjLfjLjZ

o

LL

L

λ

ππω

θβ jo

djo ee −− Γ=Γ=Γ 2

1−=Γo

Smith Chart

Smith Chart for Example 5

0

0.176

0.422

Г=-1

Smith Chart

Website : http://education.tm.agilent.com/index.cgi?CONTENT_ID=5

ZL=R+jωL

ZL/Zo=(R+jωL)/Zo

Smith Chart

Example 7: Series Connection of R and C seriesPlot in the Smith Chart the input impedance (Zin) vs. frequency when the frequency changes from 500 MHz to 4 GHz. Analyze the following cases (a) r =0.3 (b) r =0.5 (c) r =0.7 (d) r =1. Assume Zo=50 Ω

Solution:

80.050F101Hz1042

1GHz4For

37.650F101Hz105002

1MHz500For

211

0.1,7.0,5.0,3.0

11

129

126

o

ooo

ininin

−=⇒Ω××××

−=⇒=

−=⇒Ω××××

−=⇒=

−=−=

=

+=−==⇒+=

−

−

xxf

xxf

ZCfZCx

r

jxrZCjZR

ZZzCjRZ

o

π

π

πω

ωω

Note that r is constant and xchanges with frequency

Smith Chart

Smith Chart for Example 7

Smith Chart

Admittance TransformationThe admittance is the inverse of the impedance:

The normalized admittance is given by:

The reflection coefficient is given by:

Note that a normalize impedance z’:

Then

oo

inin

1;1ZYZY ==

;11inin

oinin

in

o

o zZZZ

ZYYy ====

in

in

inin

in

inin 1

11-1

1Γ+Γ−

==⇒ΓΓ+

= zyz

inin ΓΓ'Assume'-1'1' −==Γ

ΓΓ+

= − πjez

π

π

j

j

eeyz −

−

Γ−Γ+

==Γ+Γ−

=in

inin

in

in

11

11'

Conclusion: The location of yin is identical to the location of z’. The reflection coefficient Г’ is 180o from Г.

Smith Chart

Example 8: Impedance/admittance conversion

1150)5050(

inin

in jzjZZz

o+=⇒

ΩΩ+

==

5.05.0in jy −=

π

π

j

j

ee

yzΓ−Γ+

==Γ+Γ−

=111

11'

in

Using the Smith Chart find the admittance of the impedance Zin=(50+j50)Ω. Assume Zo=50Ω.

Solution1st step: Normalize the impedance:

2nd step: Locate zin in the Smith Chart.

3rd step: Trace a circumference.

4th step: Locate yin=z’

5th step: Find yin

6th step: Find Yin= yin Yo ⇒Ω

−== 501)5.05.0(inin jYyY o SjY )01.001.0(in +=

Verification

Ω−

=Ω×

−=

Ω+−

=Ω+

−=

Ω+−

=Ω+

== 505.05.0

25011

)11(505050

)11(505050

)5050(5050

)5050(11

2222in

injjjjj

jZY

SjY )01.001.0(in +=

Smith Chart

Smith Chart for Example 8

Smith Chart

Y-Smith ChartInstead of rotating the reflection coefficient by 180o, we can rotate the Smith Chart itself. The Chart obtained by this transformation is known as the Admittance Smith Chart or Y-Smith Chart.

Z- Smith Chart Y- Smith Chart

-2/3

32

31 j+

z=0.6+j1.2

- +

+inductive

capacitive

inductive

capacitive

Smith Chart

Y-Smith ChartInstead of rotating the reflection coefficient by 180o, we can rotate the Smith Chart itself. The Chart obtained by this transformation is known as the Admittance Smith Chart or Y-Smith Chart.

+

Smith Chart

Z- Smith Chart Y- Smith Chart

-2/3

32

31 j−

z=0.6+j1.2

-+ -

+

ZY-Smith ChartIn many practical design application it is necessary to switch frequently from impedance to admittance representation and vice versa.

To deal with those situations a combined, or so called ZY-Smith Chart can be obtained by overlaying the Z- and Y- Smith Charts.

Smith Chart

ZY-Smith Chart

Smith Chart

Example 9: Use the ZY-Smith Chart

(a) Identify the normalized impedance z=0.5+j0.5

Using the ZY-Smith Chart find the corresponding normalized admittance y

(b) Identify the normalized admittance y=1+j2

Using the ZY-Smith Chart find the corresponding normalized impedance z

Solution: (See Smith Charts)

Smith Chart

Example 9:

Part (a)

Z=0.5+j0.5 y=1-j1

b=-1

g=1

x=0.5

x=0.5

Smith Chart

Example 9:

Part (b)

z=0.2+j0.4

y=1+j2

b=2

g=1

x=0.4

r =0.2

Smith Chart

Y-Smith Chart Interpretation

ir jjbgjbg

yy

Γ+Γ=++−−

=+−

=Γ 11

11

Smith Chart

Y-Smith Chart Interpretation

ir jjbgjbg

yy

Γ+Γ=++−−

=+−

=Γ 11

11

y=g+jb

Inductive

Capacitive

Smith Chart

Parallel and Series Combination

In the following examples several basic circuit element configuration are analyzed and their impedance responses are displayed in the Smith Chart as function of frequencies.

The aim is to develop insight into how the impedances/ admittance behaves over a range of frequencies for different combinations of lumped circuit parameters.

A practical understanding of these circuit responses is needed later in the design of matching networks and in the development of equivalent circuit models.

Smith Chart

Parallel Connection of R and L elements

The admittance is given byo

oin

in1;1ZYZY ==

⇒−=−=⎟⎠⎞

⎜⎝⎛ +=== L

ZjZRLZjR

ZZLjRZYYYy o

o

oooo

o ωωω /111

inin

in

The normalized admittance is given by

LZjgy o

ω−=in

Observations: (1) g is constant (2) the imaginary part (the normalized susceptance, b) varies with frequency.

Then, when we move in frequency g is constant and b=-Zo/ωL changes.

Smith Chart

Example 10: Parallel Connection of R and L Elements

Graph on the Smith Chart the admittance yin when the frequency varies from 500 MHz to 4 GHz. Assume that g=0.3, 0.5, 07 and 1. Assume Zo=50 Ω.

59.11010105002

5096 −=

××××Ω

=−= − HHzLZb o

πω

The normalized susceptance b=-Zo/ωL, changes with frequency:Solution

For f=500 MHz =>

2.010101042

5099 −=

××××Ω

=−= − HHzLZb o

πωFor f=4 GHz =>

Smith Chart

Smith Chart for Example 10.

Smith Chart

Parallel Connection of R and C elements

The admittance is given byo

oin

in1;1ZYZY ==

⇒+=+=⎟⎠⎞

⎜⎝⎛ +=== o

oo

ooo

oZCjZRZCjR

ZZCjRZYYYy ωωω /

11in

inin

The normalized admittance is given by

oZCjgy ω+=in

Observations: (1) g is constant (2) the imaginary part (the normalized susceptance, b) varies with frequency.

Then, when we move in frequency g is constant and b= ωC Zo changes.

Smith Chart

Example 11: Parallel Connection of R and C Elements

Graph on the Smith Chart the admittance yin when the frequency varies from 500 MHz to 4 GHz. Assume that g=0.3, 0.5, 07 and 1. Assume Zo=50 Ω.

16.050101105002 126 =Ω×××××== − FHzZCb o πω

The normalized susceptance b= ωC Zo , changes with frequency:Solution

For f=500 MHz =>

For f=4 GHz => 26.1501011042 129 =Ω×××××== − FHzZCb o πω

Smith Chart

Smith Chart for Example 11.

Smith Chart

Series Connection of R and L Elements

The impedance is given by ;in LjRZ ω+=

⇒+=+== xjrZLj

ZR

ZZz

ooo

ωinin

The normalized admittance is given by

xjrz +=in

Observations: (1) r is constant (2) the normalized reactance, x, varies with frequency.

Then, when we move in frequency r is constant and x= (ωL)/ Zo changes.

oo ZLxZ

Rr ω== andwhere

Smith Chart

Example 12: Series Connection of R and L Elements

Graph on the Smith Chart the impedance zin when the frequency varies from 500 MHz to 4 GHz. Assume that r=0.3, 0.5, 07 and 1. Assume Zo=50 Ω.

63.0501001105002 96

=Ω

××××==

− HHzZ

Lxo

πωThe normalized reactance x= (ωL)/ Zo , changes with frequency:

Solution

For f=500 MHz =>

For f=4 GHz => 03.55010011042 99

=Ω

××××==

− HHzZ

Lxo

πω

Smith Chart

Smith Chart for Example 12.

Smith Chart

Series Connection of R and C Elements

The impedance is given by ;1in CjRZ ω+=

⇒+=−=+== xjrZCjZR

ZCjZR

ZZz

ooooo ωω11in

in

The normalized admittance is given by

xjrz +=in

Observations: (1) r is constant (2) the normalized reactance, x, varies with frequency.

Then, when we move in frequency r is constant and x= -1/(ωCZo) changes.

⎪⎪⎩

⎪⎪⎨

⎧

−=

=

o

o

CZx

ZRr

ω1where

Smith Chart

Example 13: Parallel Connection of R and L Elements

Graph on the Smith Chart the impedance zin when the frequency varies from 500 MHz to 4 GHz. Assume that r=0.3, 0.5, 07 and 1. Assume Zo=50 Ω.

37.650101105002

11126 −=

Ω×××××−=−= − FHzZCb

o πω

The normalized reactance x= -1/(ωC Zo) , changes with frequency:Solution

For f=500 MHz =>

For f=4 GHz => 80.0501011042

11129 −=

Ω×××××−=−= − FHzZCb

o πω

Smith Chart

Smith Chart for Example 13.

Smith Chart

Example 14The inductor value changes from 10 nH to 80 nH. Assume that the frequency of operation is f=500 MHz. Graph the normalized impedance zin in the Smith Chart. Assume that Zo= 50 Ω.

Solution:.The normalized impedance for L=10 nH is given by

63.05.0501010105002

5025 96

inin jHHzjZ

LjZR

ZZz

ooo+=

Ω××××

+ΩΩ

=+==−πω

The normalized impedance for L=80 nH is given by

03.55.0501080105002

5025 96

inin jHHzjZ

LjZR

ZZz

ooo+=

Ω××××

+ΩΩ

=+==−πω

25Ω

L=10nH-80nH

f=500 MHz

Smith Chart

j0.63

j5.03

10 nH80 nH

0.5

f=500 HMzYZ Smith Chart for Example 14

25Ω

L=10nH-80nH

f=500 MHz

Smith Chart

Example 15The capacitor value changes from 2 pF to 8 pF. Assume that the frequency of operation is f=500 MHz. Graph the normalized impedance zin in the Smith Chart. Assume that Zo= 50 Ω.

Solution:.The normalized impedance for C=2 pF is given by

8.05.050108105002

150251

126in

in jFHz

jCZjZR

ZZz

ooo−=

Ω×××××−

ΩΩ

=+== −πω

18.35.050101105002

150251

126in

in jFHz

jCZjZR

ZZz

ooo−=

Ω×××××−

ΩΩ

=+== −πω

The normalized impedance for C=8 pF is given by

25 Ω

C=2pF- 8pF

Smith Chart

YZ Smith Chart for Example 15

2 pF

0.5

f=500 HMz

-j3.18

-j0.8

8 pF

25 Ω

C=2pF-8pF

Smith Chart

Example 16The capacitor value changes from 2 pF to 8 pF. Assume that the frequency of operation is f=500 MHz. Graph the normalized impedance zin in the Smith Chart. Assume that Zo= 50 Ω.

Solution:.The normalized admittance for C=2 pF is given by

31.05.05010110500210050 126in

in jFHzjCZjRZ

YYy o

o

o+=Ω×××××+

ΩΩ

=+== −πω

The normalized impedance for C=8 pF is given by

26.15.05010810500210050 126in

in jFHzjCZjRZ

YYy o

o

o+=Ω×××××+

ΩΩ

=+== −πω

100 Ω2pF- 8pF

f=500 MHz

Smith Chart

YZ Smith Chart for Example 16

2 pF

0.5

f=500 HMz

+j0.31

+j1.26

8 pF

100 Ω1pF-8pF

f=500 MHz

Smith Chart

Example 17The capacitor value changes from 10 nH to 80 nH. Assume that the frequency of operation is f=500 MHz. Graph the normalized impedance zin in the Smith Chart. Assume that Zo= 50 Ω.

Solution:.The normalized admittance for L=10 nH is given by

59.15.0501010105002

5010050

96in

in jHHz

jLjZ

RZ

YYy oo

o−=

Ω×××××Ω

−ΩΩ

=+== −πω

The normalized impedance for L=80 nH is given by

2.05.0501080105002

5010050

96in

in jHHz

jLjZ

RZ

YYy oo

o−=

Ω×××××Ω

−ΩΩ

=+== −πω

100Ω10nH- 80nH

f=500 MHz

Smith Chart

YZ Smith Chart for Example 17

80 nH

0.5

f=500 HMz

-j0.2

-j1.59

10 nH

100Ω

10nH-80nH

f=500 MHz

Smith Chart

T-Network Application.In previous examples only pure series or shunt configuration have been analyzed In reality, however, we often encounters combination both.

To show how easily the ZY Chart allows transitions between series and shunt connections, let us investigate by way of an example the behavior of a T-type network connected to the input of a bipolar transistor.

The input port of the transistor is modeled as a parallel RC network.

Smith Chart

Example 18: T-Network AnalysisAnalyze the behavior of the circuit shown below at a frequency f=2 GHz.

1st step: Normalize impedance/admittances

⇒=Ω

== 625.05025.31

o

LL Z

Rr

⇒Ω

××××==

−

501038.41022 99

11

nHHzjZLjjx

oL

πω

⇒−=Ω×××××

−=−= − 833.0

501091.110221

129 jFHz

jZCjjx

oLCL πω

6.11==

LL rg

2.11 jjxjbLC

CL==

1.11

jjxL =

⇒−=Ω×××××

−=−= − 666.0

501039.210221

129 jFHz

jZCjjx

oC πω 5.11 jjxjb

CC ==

⇒Ω

××××==

−

501098.31022 99

22

nHHzjZLjjxo

Lπω

0.12 jjxL =

Solution:

Smith Chart

The resulting (normalized) circuit is shown below.

gL=1.6jbCL=1.2

1.11

jjxL =

5.1jjbC =

0.12 jjxL =

2nd step: Locate the gL in the Y-Smith Chart Section. (Point A)

3rd step: Locate the y=1.6+j1.2. (Point B) The line between A and B represents the increment of CL from 0 pF to 1.91 pF. (R-C in parallel).

4th step: Find the (series) normalized impedance at point B.

zB=0.4 - j 0.3

Smith Chart

The resulting (normalized) circuit is:

5th step: Calculate the normalized impedance zC (impedance at point C)8.04.03.04.01.1

1jjjzjxz BLc +=−+=+=

6th step: Locate the impedance zC.

7th step: Find the admittance at point C (yc)

zB=0.4 - j 0.3

1.11

jjxL =

5.1jjbC =

0.12 jjxL =zc

15.0 jyc −=

Smith Chart

The resulting (normalized) circuit is:

8th step: Calculate the normalized admittance yD (admittance at point D)

5.05.015.05.1 jjjyjby CCD +=−+=+=

9th step: Locate the impedance yD.

10th step: Find the normalized impedance zD 11 jzD −=

5.1jjbC =

0.12 jjxL =

.15.0 jyC −=

yD

Smith Chart

The resulting (normalized) circuit is:

11th step: Calculate the normalized impedance zE (impedance at point C)

11112 =−+=+= jjzjxz DLE

12th step: Calculate Zin

Consequently, the input impedance is matched to 50 Ω. Normally amplifiers are connected (through an adapting network) to 50 Ω-generators.

Ω= 50inZ

zE

11 jzD −=

0.12 jjxL =

⇒Ω×== 501oEin ZzZ

Smith Chart

Example 18:

ZY Smith Chart

A

B

C

D

E

Smith Chart