Smith Chart Problems

Home , Slotted line, Smith Chart

1. The 0:1 length line shown has acharacteristic imp edance of 50 and is

terminated with a load imp edance of Z =5+j 25 .

=0:1+j 0:5on the Smith chart. (a) Lo cate z =

See the point plotted on the Smith chart.

(b) What is the imp edance at ` =0:1?

Since we want to move away from the load (i.e., toward the

generator), read read 0:074 on the wavelengths toward

generator scale and add ` =0:1 to obtain 0:174 on the

wavelengths toward generator scale. A radial line

from the center of the chart intersects the constant re ec-

tion co ecient magnitude circle at z = 0:38 + j 1:88: Hence

Z = zZ =50(0:38 + j 1:88) = 19 + 94 .

Find VSWR = z =13on the horizontal line to the right

max

of the chart's center. Or use the SWR scale on the chart.

(d) What is ?

From the reflection coefficient scale below the chart,

nd j j = 0:855. From the angle of reflection coef-

ficient scale on the p erimeter of the chart, nd the angle of

j 126:5

= 126:5 . Hence =0:855e .

L L

(e) What is at ` =0:1 from the load?

Note that jj = j j =0:855. Read the angle of the re ection

co ecient from the angle of reflection coefficient

j 55:0

scale as 55:0 . Hence =0:855e . Problem 1 l = 0.1λ 01 Z50=Ω =+ Ω 0 ZL Z525[]L j

l = 0.1λ 01

0.174λ .TG . . 0.074λ .TG . .

AngΓ= 55.00 Γ= 0 AngL 126.5 =+ z0.10.5L j z=+ 0.38j 1.88

Γ=0.855 55.00 Γ= 0 L 0.855126.5

== VSWRzmax 13

VSWR= 13

Γ=Γ= L 0.855

2. A transmission line has Z =1:0, Z = z =0:2 j 0:2 :

0 L L

(a) What is z at ` = =0:25?

From the chart, read 0:467 from the wavelengths to-

ward generator scale. Add 0:25 to obtain 0:717 on

the wavelengths toward generator scale. This is not

on the chart, but since it rep eats every half wavelength, it

is the same as 0:717 0:500 = 0:217. Drawing a ra-

dial line from the center of the chart, we nd an intersec-

tion with the constant re ection co ecient magnitude circle

at z = Z =2:5+j 2:5.

(b) What is the VSWR on the line?

From the intersection of the constant re ection co ecientcir-

cle with the right hand side of the horizontal axis, read VSWR

= z =5:3:

max

minimum?

The voltage minimum o ccurs at z which is at a distance of

min

0:500 0:467 =0:033 from the load. Or read this distance

directly on the wavelengths toward load scale.

The current minimum o ccurs at z which is a quarter of a

max

wavelength farther down the line or at 0:033 +0:25 =0:283

from the load. Problem 2 l = 0.25λ

Z=Ω 1.0 == − ΩΩ 0 ZL ZLL z 0.2j 0.2 [ / ]

l = 0.25λ

0.217λ .TG . .

zZ2.52.5== +j

zmin

λ == 0.033 VSWRzmax 5.3

=− z0.20.2L j

0.467λ .TG . .

VSWR= 5.3

3. The air- lled two-wire line has a characteristic imp edance of 50 and is

op erated at f =3 GHz. The load is Z = 100 + j 40 .

(a) For the line ab ove, nd z on the chart.

100 + j 40 Z

= The normalized load is z = = 2:0+ j 0:8.

Z 50

See the Smith chart for lo cation of p oint.

(b) What is the line imp edance 2:5cm from the load?

3 10 cm/sec c

= = 10 cm: Since we are Note that =

f 3 10 Hz

going to move toward the generator (away from the load), at

the normalized load p osition, rst read 0:217 on the wave-

lengths toward generator scale. Then add 2.5 cm/10

cm = 0:25 to this value to obtain 0:467 on the wave-

lengths toward generator scale. A radial line from

the center at this p oint intersects the constant re ection co-

ecient magnitude circle at z =0:43 j 0:17, so Z = zZ =

50(0:43 + j 0:17) = 21:5 j 8:5 .

From the intersection of the re ection co ecient circle and

the horizontal axis on the right hand side of the chart, read

VSWR = 2.4. Or use the SWR scale belowthe chart. Problem 3 l = 2.5 [cm]

Z=Z z → Z=Ω= 50 ,f 3GHz =+ Ω 0 0 ZL ZL 100j 40 [ ]

0.217λ .TG . .

=+ z2.00.8L j

== VSWRzmax 2.4

z=− 0.43j 0.17

0.467λ .TG . .

l = 0.25λ

4. The line shown is 3=8 long and its normalized input imp edance is z =

j 2:5:

(a) What is the normalized receiving or load end imp edance, z ?

Since wemust movetoward the load to nd the load imp edance,

lo cate z = j 2:5 on the chart and nd the intersection on the

wavelengths toward load scale at 0:189. Add the dis-

tance 3=8 = 0:375 to get 0:064 on the wavelengths

toward load scale. At the intersection of the line from

the center of the chart and the constant re ection co ecient

circle, read z = j 0:425.

(b) What is the distance from the load to the rst voltage minimum?

The voltage minimum o ccurs at the imp edance minimum on

the horizontal line to the left of the chart's center. This dis-

tance can be read directly o the wavelengths toward

load scale (why?) as 0:064. Note that the line imp edance

at the voltage minimum would b e that of a short circuit.

From the p osition of z ,wewanttomovetoward the new load

(i.e., toward the end of the op en-circuited line) until wereach

the op en circuit condition z = 1. This o ccurs at 0:25 on the

wavelengths toward load scale and the total distance

moved is hence 0:25 0:064 =0:186, which is the length

of op en-circuited line needed to replace z . L Problem 4 λ l = 3 8

z2.5=−j → zL

""" """! open circuited line section replacing original load zL

λ l = 3 8

Vmin z =∞

0.064 λ

z2.5=−j =− zL j 0.425

0.064λ .T.L. 0.189λ .T.L.

0.186 λ

5. An air- lled 50 coaxial line has a loaded VSWR of 3.3 at a frequency

of 3GHz. Replacing the load with a short causes the voltage minimum

to move 1.0 cm towards the generator. What is the normalized load

imp edance?

For an air- lled line, the wavelength is

c 3 10 cm/sec

= = = 10 cm:

f 3 10 Hz

With the VSWR lo cated at z =3:3, draw the circle of con-

max

stant re ection co ecient magnitude. With the load in place, our

only information is the lo cation of a voltage minimum (equivalent

to an imp edance minimum). But thinking of the Smith chart as

a crank diagram, we know that the voltage minimum must be

at the intersection of the constant re ection co ecient circle and

the horizontal axis on the left hand side of the chart between

z =0 and z =1:0, (i.e., at z =0:3). With this information,

min

we have thus established a corresp ondence between a lo cation

on the line and a point on the chart. Now, move from this

1:0cm

minimum voltage point on the chart =0:1 toward the

10:0cm

generator to the lo cation where the voltage minimum was mea-

sured in the shorted line measurement. Since it was obtained

from a shorted line measurement, the lo cation of this p ointmust

be an integral number of half-wavelengths from the load (and

hence must have the same line imp edance as at the load). The

normalized imp edance here is therefore z =0:44 + j 0:63. For a

line with a 50 characteristic imp edance, the load imp edance is

thus Z = 50(0:44 + j 0:63) = 22 + j 31:5: L Problem 5

0.10 λ

=+ zL 0.44j 0.63

Vmin

VSWR= 3.3

6. A slotted line measurement on an air- lled TEM line yields a VSWR of

1.6 at a frequency of 1 GHz. When the load is replaced by a short, the

voltage minimum moves 3 cm towards the load. Find the normalized load

imp edance.

Lo cate the re ection co ecient circle by its intersection with

z = VSWR = 1:6. The wavelength for an air- lled TEM

max

line is

3 10 cm/s

= c=f = =30cm:

10 Hz

Hence the shift in the voltage minimum is

3cm

=0:1:

30 cm=

Lo cate the voltage minimum at the imp edance minimum, z =

min

0:625 and move 0:1 toward the load to a p osition an integral

numb er of half-wavelengths from the load. The imp edance there,

z =0:77 j 0:39,isthe same as that of the load. L Problem 6

Vmin

VSWR= 1.6

=− zL 0.77j 0.39

0.10 λ

7. A slotted line measurement yields the following parameter values:

(a) Voltage minima at 9.2 cm and 12.4 cm measured away from the load

with the line terminated in a short.

(b) VSWR = 5.1 with the line terminated in the unknown load; a voltage

minimum is lo cated 11.6 cm measured away from the load.

What is the normalized line imp edance?

Note that this data could have come from either a waveguide

or a TEM line measurement. If the transmission system is a

waveguide, then the wavelength used is actually the guide wave-

length. From the voltage minima on the shorted line, the (guide)

wavelength may b e determined:

= 12:4cm 9:2cm = 3:2cm

= 6:4cm:

Hence the shift in the voltage minimum when the load is replaced

by ashort is

12:4cm 11:6cm

= 0:125

6:4cm=

toward the generator. Lo cate the re ection co ecient magni-

tude circle by its intersection with z =VSWR=5:1 on the

max

horizontal axis. Then from the voltage minimum opp osite z ,

max

move0:125 toward the generator to nd a p osition an integral

numb er of half-wavelengths from the load. The imp edance there

is the same as that of the load, z =0:38 + j 0:93. Alternatively,

move 0:5 0:125 = 0:375 toward the load to lo cate the

g g g

same value. Problem 7

0.125 λ =+ zL 0.38j 0.93

Vmin

VSWR= 5.1

8. Use a single parallel stub tuner to match the ab ove line to its load. Use

ashorted stub and nd its distance from the load, ` , and its length ` .

1 2

First, lo cate the load imp edance z = 0:4 j 0:4 on the chart

and re ect it to obtain the load admittance, y = 1:24 + j 1:24.

Tomatch the line, the parallel combination of the load and stub

line sections must present a unit (normalized) admittance at the

junction point, i.e. y + y =1:0. Since the stub section is loss-

1 2

less, its input admittance must be purely susceptive, i.e. of the

form y = jb, and hence the input admittance of the loaded sec-

tion must have the form y = 1:0 jb. The lo cus of y is thus

1 1

the unit conductance circle.

To determine the unknown susceptive part of y , we move from

the p oint on the chart corresp onding to y toward the gener-

ator until we reach the intersection with the unit conductance

circle, yielding y = 1:0 j 1:15 and thus y = j 1:15. The

1 2

electrical distance moved between the points y and y is ` =

L 1 1

0:334 0:181 =0:153 measured on the wavelengths to-

ward generator scale.

To nd ` , lo cate y = j 1:15 and movetoward the load to the in-

2 2

nite admittance p ointonthechart, corresp onding to a shorted

stub. The shorted stub length is therefore ` =0:386, the sum

of 0:25 measured on the wavelengths toward load scale

and 0:136 measured on the wavelengths toward genera-

tor scale (Why two di erent scales?).

Note that a second solution may b e obtained by rotating past the

rst intersection to the second intersection with the unit conduc-

tance circle at y =1:0+j 1:15. This will yield a longer section

of line ` . Also, wewillhave y = j 1:15 and the corresp onding

1 2

length of the shorted stub section ` will b e shorter than in the

previous solution. In this case it is not clear which solution is

preferable. In general, however, if one solution results in both `

and ` being shorter, then that con guration normally has the

greater bandwidth. Problem 8

0.136λ .T.G.

= 0.181λ .T.G. yj2 1.15

l = λ 2 0.386 l = λ 1 0.153

∞ 2 y= 1

=− locus of yjb1 1

3 =− zL 0.40j 0.40

=− yj1 11.15

Reflecting across chart 1 converts impedance to admittance

Rotating about chart center 2 changes line locations 0.334λ .T.G. Moving along a constant conductance (resistance) circle 3 is equivalent to adding susceptance (reactance) in parallel (series).

9. Use a single series stub tuner to match the ab ove line to its normalized

load. Use a shorted stub and nd its distance from the load, ` , and its

length ` .

This problem is similar to the parallel stub tuner except that,

since the stub is in series with the line, constructions are done

using imp edances rather than admittances. First, lo cate the load

imp edance z = 0:4 j 0:4 on the chart. To match the line,

the series combination of the load and stub line sections must

present a unit (normalized) imp edance at the junction point,

i.e. z + z = 1:0. Since the stub section is lossless, its input

1 2

imp edance must b e purely reactive, i.e. of the form z = jx,and

hence the input admittance of the loaded section must have the

form z = 1:0 jx. The lo cus of z is thus the unit resistance

1 1

circle.

The unknown reactive part of z is determined by moving from

the p ointonthechart corresp onding to z toward the generator

until the intersection with the unit resistance circle is reached,

yielding z = 1:0+ j 1:13 and thus z = j 1:13. The electrical

1 2

distance moved b etween the p oints z and z is ` =0:235, the

L 1 1

sum of 0:069 measured on the wavelengths toward load

scale and 0:166 measured on the wavelengths toward gen-

erator scale (Why di erent scales?).

To nd ` ,locatez = j 1:13 and movetoward the load until the

2 2

imp edance is zero, corresp onding to a shorted stub. The shorted

stub length is therefore ` =0:365, which can be measured di-

rectly on the wavelengths toward generator scale.

A second solution maybefound by rotating past the rst inter-

section to the second intersection with the unit resistance circle,

yielding z = 1:0 j 1:13 and a longer section of line ` . Then

1 1

z = j 1:13 and the corresp onding length of the shorted stub sec-

tion ` will b e shorter than in the previous solution. In this case

it is not clear which solution is preferable. In general, however,

if one solution results in both ` and ` being shorter, then that

1 2

con guration normally has the greater bandwidth. Problem 9

0.166λ .T.G.

=+ zj1 11.13

3 2 z=0

=− locus of z1 1 jx l = λ 1 0.235 =− z0.40.4L j

l = λ 2 0.365

=− zj2 1.13

0.069λ .T.L.

0.365λ .T.G.

10. Use a single parallel stub tuner to matchtheabove line to its normalized

load. Use a shorted stub and nd its distance from the load, ` , and its

length ` .

First, lo cate the load imp edance z = 0:3+j 0:5 on the chart

and re ect it to obtain the load admittance, y = 0:89 j 1:48.

Tomatch the line, the parallel combination of the load and stub

line sections must present a unit (normalized) admittance at the

junction point, i.e. y + y =1:0. Since the stub section is loss-

1 2

less, its input admittance must be purely susceptive, i.e. of the

form y = jb, and hence the input admittance of the loaded sec-

tion must have the form y = 1:0 jb. The lo cus of y is thus

1 1

the unit conductance circle.

To determine the unknown susceptive part of y , we move from

the p oint on the chart corresp onding to y toward the generator

until we reach the intersection with the unit conductance circle,

yielding y = 1:0+j 1:56 and thus y = j 1:56. The electrical

1 2

distance moved b etween the p oints y and y is ` =0:350, the

L 1 1

sum of 0:172 measured on the wavelengths toward load

scale and 0:178 measured on the wavelengths toward gen-

erator scale (Why di erent scales?).

To nd ` , lo cate y = j 1:56 and move toward the load to the

2 2

in nite admittance p ointonthechart, corresp onding to a shorted

stub. The shorted stub length is therefore ` =0:250 0:159 =

0:091, where b oth distances are measured on the wavelengths

toward load scale.

As usual, a second solution may b e obtained by rotating past the

rst intersection to the second intersection with the unit conduc-

tance circle at y =1:0 j 1:56. This will yield a longer section

of line ` . Also, we will have y = j 1:56 with the corresp ond-

1 2

ing length of the shorted stub section ` also longer than in the

previous solution. In this case one would generally cho ose the

solution with b oth ` and ` shorter, that con guration normally

1 2

having the greater bandwidth. Problem 10

0.178λ .T.G. l = λ 1 0.350

=+ yj1 11.56 =+ z0.30.5L j

1 y=∞

locus of yjb=− 1 2 1

yL l = λ 2 0.091

=− yj2 1.56

0.172λ .T.L.

0.159λ .T.L.