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### Transcript of Smith Chart Examples

• Smith chart

The Smith Chart: allows to compute the input impedance to a transmission line

( ) ( )( )0 0 1 0 1

0 1 0 1

inin in

in

Z z Z Z Z + +

= = = =

The load reflection coefficient and the input coefficient are related as42

jjin L Le e

pi = =

We write the normalized input impedance0

1

1inin

inin

Zz r jx

Z

+ = = = +

And the reflection coefficient as4

jin L e p jq

pi = = +

Combining both equations1

1inp jq

z r jxp jq

+ += + =

Solving the real and imaginary parts

( )2

22

11 1

rp qr r

+ = + +

( )2

22

1 11p qx x

+ =

1r +centered at 0

1rp q

r= =

+

centered at11p and qx

= =

EE 342Spring 2010 #115

• ( )2

22

11 1

rp qr r

+ = + +

1r + centered at 01rp and q

r= =

+

( )2

22

1 11p qx x

+ =

1x

centered at11p and qx

= =

EE 342Spring 2010 #116

Smith chart

• The graphs relate the real and imaginary part of the reflection coefficient at a point (p,q)

with the real and imaginary part of the normalized input impedance (r,x)

EE 342Spring 2010 #117

Smith chart

• Relation between the normalized input impedance to the line and the reflection coefficient

We plot the normalized input impedance

inz r jx= +

The point defines the magnitude and

the angle of the reflection coefficient

2 2

in p q = +

( ) 2 4in pi = =

EE 342Spring 2010 #118

Smith chart

1

1inp jq

z r jxp jq

+ += + =

4

jin L e p jq

pi = = +

• How determine the input

impedance

We plot the normalized load impedance

0

L

L L LZ

z r j xZ

= = +

Rotate (with a compass) an angle

( ) 2 4in pi = =

Clockwise TG (towards generator)

If we know the input impedance we calculate

0

in

in in inZ

z r jxZ

= = +

Rotate (with a compass) an angle

( ) 2 4in pi = = Counterclockwise TL (towards load)

EE 342Spring 2010 #119

Smith chart

• Example: a coaxial cable (r = 2.25), length 10m. Frequency source 34 MHz. The

characteristic impedance is Z0 = 50 . The line is terminated with a load ZL = (50+j100) .

Determine the input impedance

Propagation velocity80 2 10

r

v mvs

= =

Wavelength5.882 1.7v mf = = =

50 100 1 2

50Lj

z j+= = +

Rotate 1.7 TG (3 turns plus 0.2 )

0.29 0.82inz j=

And unnormalizing

0

14.5 41in inZ z Z j= =

EE 342Spring 2010 #120

Smith chart

• A B

C

D

E

Exercise:

Match the following

normalized impedances

with points A,B,C,D and E

on the Smith chart

i) 0+j0

ii) 1+j0

iii) 0-j1

iv) 0+j1

v) +j

vi)

vii)

min

in

C

ZZ

max

in

C

ZZ

( )0 =

i) D

ii) A

iii) E

iv) C

v) B

vi) D

vii) B

viii) A

EE 342Spring 2010 #121

Smith chart

• Example:

( ) 20 40inZ j= ( ) 20 40LZ j= +

0 100Z =

( ) 0.2 0.4inz j= We calculate( ) 0.2 0.4Lz j= +

Determine the length

of the line in

wavelengths

0.062 TG

0.436 TG

0.436 0.062 0.374 = =

0.438 TL

0.064 TL

0.438 0.064 0.374 = =

EE 342Spring 2010 #122

Smith chart

• Exercise: Determine ZL attached to

a line with Z0 = 100 . Removing

the load yields an input impedance

Zin= -j80.With the unknown

impedance attached the input

impedance is (30 + j 40) .

Determine ZL

With open circuit80

0.8100Lin Z

jz j

=

= =

0.107 TL

0.393 TG

Lz =

0.25 TL

0.25 TG

( )0.393 0.25 0.143 = =( )0.25 0.107 0.143 = =

( )30 40 0.30 0.40

100Lin Zj

z j+ = = +

0.065 TG

0.435 TL

Rotate TL 0.143

0.435 + 0.143 = 0.578 = 0.078

0.32 0.49 32 49L Lz j Z j= =

EE 342Spring 2010 #123

Smith chart

reflection coefficient for :

( ) 50 100inZ j= 0

50Z = 0.4 =

1 2inz j=

0.187 TL

450

Rotate TL (CCW)

0.187 0.4 0.587 0.087 + = =

0.22 0.58Lz j= ( ) 11 29LZ j=

-1180

00.73 118L =

7VSWR =

EE 342Spring 2010 #124

Smith chart