t07_smithchart

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TUTORIAL #7

Using the Smith Chart

1. [2.19 P expanded] Use the Smith chart to find the following quantities for the transmission-line

circuit in the figure below.

inZ

0.4L

0 50Z 60 50LZ j

(a)The SWR on the line.(b)The reflection coefficient at the load.(c)The load admittance.(d)The input impedance of the line.(e)The distance from the load to the first voltage minimum.(f) The distance from the load to the first voltage maximum.(a)

0

60 501.2 1.0

50

LL

Z jz j

Z

; plot SWR circle on Smith chart read out SWR from

the resistance value on the right, SWR = 2.46.

(b)Read out || either from the nomograph or as the ratio /R.| | 34 / 79 0.43 .

Read out using the degree scale on the outer rim of the chart. If a degree scale is not

available, you can use a protractor or the scale of distance-in-wavelengths (toward load,

counter-clockwise), which is always available on the outer rim of the Smith chart.

(0.325 0.25)0.94 rad 54

0.25

(c)Find diametrically opposite point ofzL on Smith chart and read out values.0.49 0.41 9.8 8.2 mS

50

LL L

yy j Y j

(d)Follow SWR circle in direction toward generator (clockwise) moving a distance of 0.4.As the loadzL corresponds to relative-to-short position of 0.176, the relative-to-short

position ofzin would be at

0.176 0.4 0.576 .

Since the chart goes only to 0.5, you will have to go beyond 0.0 and continue to 0.076.This corresponds to input impedance of


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0.49 0.41 50 24.5 20.3in in inz j Z z j .

(e)The location of a voltage minimum (which is also the location of a current maximum)corresponds to a minimum resistance value. This location is found on the Smith chart as

the intersection point of the SWR circle and the abscissa on the left where the scale for theresistance circles resides. Conveniently, this location coincides with the Charts zero-

reference location. Move in the direction toward load (counter-clockwise) until youreach the location of the load; read out this location from the toward load scale:

min 0.325l .

(f) The location of a voltage maximum (which is also the location of a current minimum)corresponds to a maximum resistance value. This location is found on the Smith chart asthe intersection point of the SWR circle and the abscissa on the right where the scale for

the resistance circles resides. It corresponds to a quarter-wavelength-reference location on

the Chart. Move from this location toward the load:

max (0.325 0.25) 0.075l .


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2. [Impedance match with a shunt stub] A transmission line of characteristic

impedance 0 50Z is terminated with a load impedance of 60 50LZ j as shown in the

figure.

(a)What is the admittance 0Y of the transmission line?(b)Find the length L (in terms of) of the piece of this transmission line such that at L L

the input admittance inY

satisfies 10 0Re inY Y Z

. What is Im inY at this position?

(c)Use a short-circuited or an open-circuited piece of a 50- transmission line (whichever isshorter) to cancel Im inY by connecting it in parallel at the position L as shown in the

figure. This piece of transmission line is called ashunt stub. What is the length of the stubstubL ?

(d)What is now the input impedance inZ at L L ?(e)What is the input impedance at 2L L ?

?inZ

?L

0 50Z 60 50LZ j

stub ?L

stub

0

50

Z

?

SOLUTION:

(a) 0 01/ 1/ 50 0.02Y Z S

(b) 60 50LZ j

0

1.2 1L

L

Zz j

Z (same as in Problem 1)

Find Ly from impedance Smith chart. Same as in Problem 1: 0.49 0.41Ly j

( 9.836 8.197LY j mS)

We now start using the Smith chart as an admittance chart. Go toward the generator on theSmith chart until intersecting the unity admittance circle.

0.074

(0.5 0.426) 0.16 0.234 0.234L

L

1 0.93iny j


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0Im 0.93 0.93 0.02 0.0186inY Y S

(c) We now need to connect to inY a purely reactive load of susceptance

stub 0Im Im 0.93 0.0186inY Y Y S (inductive susceptance)

A short-circuited stub would achieve the above susceptance with a shorter line than the open-

circuited stub.

stub (0.3815 0.25) 0.1315L

(d) At L L , stub Re 1 0.02 0.02in in inY Y Y Y S

1/ 50in inZ Y

(e) At 2L L , the input impedance is still 50 because the line is matched at the positionL L .


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3. [Impedance match with a series stub] A transmission line of characteristic

impedance 0 50Z is terminated with a load impedance of 60 50LZ j as shown in the

figure.

?inZ

?L

0 50Z 60 50LZ j

stub

?L

stub0 50Z

?

(a)Find the length L (in terms of) of the piece of this transmission line such that at L L the real part of the input impedance inZ satisfies 0Re inZ Z . What is Im inZ at this

position?

(b)Use a short-circuited or an open-circuited piece of a 50- transmission line (whichever isshorter) to cancel Im inZ by connecting it in series at the position L as shown in the

figure. This piece of transmission line is called aseries stub. What is the length of the stubstubL ?

(c)What is now the input impedance inZ at L L ?SOLUTION:

(a)

0

1.2 1L

L

Zz j

Z

(same as in Problems 1 and 2)

Trace the SWR circle toward generator (clockwise) until it crosses the unity-resistance circle.

The normalized impedance at this position is ( ) 1 0.93z L j . The length L is

(0.34 0.176) 0.164L

Im 0.93 50 46.5inZ

(b) The stub must have purely reactive impedance of value

stub stubIm ( ) 0.93 ( 46.5 )z j z L Z j - inductive reactance

A short-circuit stub would be shorter than an open-circuit one. The length of the short-circuit stub

must be stub 0.12L .

(c) 0 50inZ Z


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