Smith Chart Examples

Smith Chart

Smith Chart:

Graphical Chart by P. H. Smith

0

0

jL

L

Z Re

Z R

In a lossless transmission line, the voltage reflection coefficient is

defined as

The normalized load impedance can be written as

0 0

L L LL

Z R jXz r jx

R R

where r and x are normalized resistance and normalized reactance

1

1

Lr i

L

zj

z

where i and r and x are normalized resistance and normalized

reactance

11

1 1

j

L j

ez

e

1

1

r i

r i

jr jx

j

1 . 1

1 1

r i r i

r i r i

j jr jx

j j

1 . 1

1 1

r i r i

r i r i

j jr jx

j j

2 2

2 2

1

1

r i

r i

r

2 2

2

1

i

r i

x

2 2

2 2

1

1

r i

r i

r

2 2

2 1

1 1r i

r

r r

2 2

2

1

i

r i

x

2 2

2 1 11r i

x x

2 2

2 1

1 1r i

r

r r

; 01

r i

r

r

Centre Radius 1

1 r

Resistive Circles

•The centers of all r-circles lie on the r axis.

•The r = 0 circle, unity radius, centre at origin

•The r-circles becomes progressively smaller as r increases from 0 toward

ending at (r = 1, i = 0) point for open circuit

•All r – circles pass through the (r = 1, i = 0) point

2 2

2 1

1 1r i

r

r r

2 2

2 1 11r i

x x

Reactance Circles

11;r i

x

Centre Radius 1

x

•The centers of all x-circles lie on the r = 1 lines;

for x > 0 (inductive reactance) lie above r axis.

for x < 0 (capacitive reactance) lie below r axis.

• The x = 0 circle becomes the r axis

•The x-circles becomes progressively smaller as x from 0 toward

ending at (r = 1, i = 0) point for open circuit

•All x–circles pass through the (r = 1, i = 0) point

2 2

2 1 11r i

x x

•Smith Chart: a chart of r and x circle in r and I plane for 1

•R and x circle are everywhere orthogonal to one another

•The intersection of r and x circle defines a points that represents a

normalized load impedance

•Actual Impedance is ZL R0 (r + jx)

Lz r jx

0.52.0

3.0-3.0

-2.0

-1.0

-0.5

0.5 1.0 2.0 5.0Гr

Гx

X/ZO=0.2

X/ZO=-0.2

-1

C

A

1

R/Z

O=

0.2

1.0

•r = -1 and i = 0 corresponds to r = 0

and x = 0: short circuit

•r = 1 and i = 0 corresponds to

infinite impedance: open circuit

i

Smith chart can also be marked as polar coordinates:

Magnitude of

Phase angle of

1

1

SC

OC

Pm PM

Each circle intersect the real axis at two points:

PM on positive real axis

Pm on negative real axis

Since x = 0 along the real axis: PM and Pm both represents a purely

resistive load

RL > R0 (r > 1) : PM

RL > R0 (r < 1) : Pm

0Lif R R0

LRS r

R

The value of the r-circle passing through the point PM is numerically

equal to the standing wave ratio

0Lif R R0 1

L

RS

R r

The value of the r-circle passing through the point Pm on negative real

axis is numerically equal to the 1/S

1r

S

1. All the circles are centered at the origin and their radii vary from 0 to

1.

2. The angle measured from the positive real axis, of the line drawn from

the origin through the point representing zL equal .

3. The value of the r-circle passing through the intersection of the circle

and the positive real axis equals the standing wave ratio

Constant Circle

The input impedance looking towards the load end at a distance z’

from the load is

' 2 '

0

' 2 '

0

0

1' 2''

12

z zLL

iz zL

L

IZ Z e eV z

Z zII z

Z Z e eZ

2 '

0 2 '

1'

1

j z

i j z

eZ z Z

e

Normalized input impedance

2 '

2 '

0

' 1

1

j zi

i j z

Z z ez

Z e

2 '

2 '

0

' 1

1

j z

i

i j z

Z z ez

Z e

2 '

2 '

0

' 1

1

j z

i

i j z

Z z ez

Z e

0

' 1

1

i

i

Z z ez

Z e

2 'j z

At z’ = 0; 2 'j z

1

1

j

i L j

ez z

e

1

1S

The magnitude of remains constant, therefore VSWR

are not changed by additional length

Keeping constant, subtract (rotate clockwise direction) from

an angle to .

4 '2 '

zz

This will locate the point for , which determine je iz

2 'j z

The outer scale on smith chart is marked “wavelength toward

generator” in clockwise direction (increasing z’)

The inner scale is marked “wavelength toward load” in counter

clockwise direction (decreasing z’)

4 '2 '

zz

4 'f ' / 2; 2 ' 2

zi z z

Therefore complete revolution gives the z’ of /2

Example:1 Find L if the load impedance ZL is 25+j100 and characteristic

impedance of transmission line is 50

0.5 2.0Lz j

Example:2: ZL = 25 + j 100; z’ = length of transmission line d = 0.18;

Find Zin and (d) 0.8246

50.906

L

L

0.1793

0.3593

0.5 2.0Lz j

Example:3: ZL = 25 + j 100; Find the location of first voltage maximum (dmax)

and first voltage minimum from load end (dmin)

0.5 2.0Lz j

Example:4: ZL = 25 - j 100; Find the location of first voltage maximum (dmax)

and first voltage minimum from load end (dmin)

0.3207

0.5 2.0Lz j

Example:5: Find the VSWR on transmission line

(i) if ZL1 = 25 + j 100 and Z0 = 50 ; (ii) if ZL1 = 25 - j 100 and Z0 = 50

Circle of Constant

resistance r = 10.4

Example:5:

Given: R0 = 50 , S = 3.0, = 0.4 m, First voltage minima zm’ = 0.05 m:

Find (i) , (ii) ZL

' 0.050.125

0.4mz

0.6 0.8Lz j

50 0.6 0.8LZ j

30 40LZ j

jd e

2 / 4/ 4

j j j

j

d e e e

e d

1 1;

1 1n n

d dz d y d

d d

114

4 11

4

n n

dd

z d y dd

d

4

n nz d y d

0

0

.4 4

.

n

n

Z d Z z d

Y d Y y d

Actual Impedance

Actual Admittance where 0 01/Y Z

00

0

tan

tan

Lin

L

Z jZ lZ Z

Z jZ l

200 0in

L

jZZ Z Z

jZ

For l = /4

0

0

in

L

Z Z

Z Z

1in L

L

z yz

Example:5: ZL = 25 + j 100 and Z0 = 50 ; Find YL

1 1;

1 1n n

d dz d y d

d d

Impedance = Resistance + j Reactance

Z = R + j X

Open Circuit

Short Circuit

Admittance Chart

Transmission Line Impedance Matching:

Quarter Wave Transformer

;4 2

l l

tan l

00

0

tan

tan

Lin

L

Z jR lZ R

R jZ l

2

0i

L

RZ

Z 0 i LR Z Z

0 50.100 70.7R

50 100

/ 4

70.7

Single Stub Matching: d

R0

ZL

R0

yi

ys

yB yL

l

B

B’

i B SY Y Y

0

0

1iY Y

R

0 0 0

i SBY YY

Y Y Y

For matching

i B Sy y y

1 B Sy y

d

R0

ZL

R0

yi

ys

yB yL

l

B

B’

The input admittance of short circuited stub is

purely susceptive ys

1 B Sy y

1B By jb

s By jb

Example: A 50 transmission line is connected to a load impedance ZL

= 35 – j 47.5 . Find the position and length of a short-circuited stub

required to match the line.

Solution:

0

0

50

35 47.5

0.70 0.95

L

LL

R

Z j

Zz j

R

P1: zL = 0.70 – j 0.95

P2: yL = 0.50 + j 0.68

P2’: 0.109

Move Constant || circle from

P2 to P3 or P2 to P4 and

reach g = 1 circle

P3: yB1= 1 + j1.2 = 1 + jbB1

P4: yB2= 1 - j1.2 = 1 + jbB2

P3’: 0.168

P4’:0.332

Solution for location of stub

For P3: (from P2’ to P3’) d1 = 0.168 - 0.109 = 0.059

For P4: (from P2’ to P4’) d2 = 0.332 - 0.109 = 0.223

Solution for length of stub

For P3: (PSC on the extreme right of chart to P3’’ which represents – jbB1

= -j 1.2

l1 = 0.361 - 0.250 = 0.111

For P4: (from PSC to P4’’: + j 1.2) l2 = 0.139 + 0.250 = 0.389

yS = - jbB

First Solution: d1 = 0.059 and l1 = 0.111

Second Solution: d2 = 0.223 and l2 = 0.389

Double Stub Matching:

d0 is constant and can be arbitrary chosen as /8 or 3/8

d0

R0

ZL

R0

yi

ySB

yB

yA

lB

B

B’

ySA

A’

A

yA = ySA + yL

yi = ySB + yB

lA

R0

i B SBY Y Y

0

0

1iY Y

R

0 0 0

i SBBY YY

Y Y Y

For matching

i B SBy y y

1 B SBy y

d0

R0

ZL

R0

yi

ySB

yB

yA

lB

B

B’

ySA

A’

A

yA = ySA + yL

yi = ySB + yB

lA

R0

The input admittance of short circuited stub is purely susceptive ysB

1 B SBy y 1B By jb

sB By jb

Example: A 50 transmission line is connected to a load impedance ZL

= 60 + j 80 . A double stub tuner spaced /8 apart is used to matched

the load as shown below. Find the required lengths of short circuited

stubs.

Solution:

01 500.30 0.40

60 80L

L L

Ry j

z Z j

d0

R0

ZL

R0

yi

ySB

yB

yA

lB

B

B’

ySA

A’

A

yA = ySA + yL

yi = ySB + yB

lA

R0

Draw g0 = 1 circle

Rotate this circle by /8 towards

load

yL = 0.30 – j 0.40 as PL

Move on constant g circle

(g = 0.3) which intersects

rotated g circle at PA1 and

PA2.

PA1: yA1 = 0.30 + j 0.29

PA2: yA2 = 0.30 + j 1.75

Move /8 on Constant ||

circle from PA1 or PA2

and reaches PB1 or PB2

respectively

PB1: yB1 = 1 + j 1.38

PB2: yB2 = 1 – j 3.5

First stub length

(ySA)1 = yA1 – yL = j 0.69

(ySA)2 = yA2 – yL = j 2.15

A1: j 0.69

lA1 = (0.096 + 0.25) = 0.346

A2: j 2.15

lA2 = (0.181 + 0.25) = 0.431

PA1: yA1 = 0.30 + j 0.29

PA2: yA2 = 0.30 + j 1.75

yL = 0.30 – j 0.40 as PL

Second stub length

(ySB)1 = - j 1.38

(ySB)2 = + j 3.5

B1: - j 1.38

lB1 = (0.35 - 0.25) = 0.10

B2: + j 3.5

lB2 = (0.206 + 0.25) = 0.456

PB1: yB1 = 1 + j 1.38

PB2: yB2 = 1 – j 3.5

d0

R0

ZL

R0

yi

ySB

yB yA

lB

B

B’

ySA

A’

A

yA = ySA + yL

yi = ySB + yBlA

dL

The solution for this problem is that if first stub is connected at some

distance from load end such that it comes out of forbidden region

If yL lies inside the g0 = 1 circle; no value of stub susceptance b1 could

bring the load point to intersect the rotated 1 + jb circle.

Then the region inside g0 = 1 circle is called forbidden range of load

admittance, which can not be matched with this particular double stub tuning

arrangement.