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Smith chart
The Smith Chart: allows to compute the input impedance to a transmission line
( )( )
( )0 0
ˆ ˆ1 0 1ˆ ˆ0
ˆ ˆ1 0 1
in
in in
in
Z z Z Z Z + Γ + Γ = = = = − Γ − Γ
The load reflection coefficient and the input coefficient are related as4
2ˆ ˆ ˆj
j
in L Le eπ
β λ−
−Γ = Γ = Γ
We write the normalized input impedance0
ˆ1ˆˆ
ˆ1
inin
in
in
Zz r jx
Z
+ Γ = = = + − Γ
And the reflection coefficient as4
ˆ ˆj
in L e p jqπ
λ−
Γ = Γ = +
Combining both equations1
ˆ1
in
p jqz r jx
p jq
+ += + =
− −
Solving the real and imaginary parts
( )
2
2
2
1
1 1
rp q
r r
− + =
+ +
( )2
2
2
1 11p q
x x
− + − =
Circles of radius( )
1
1r +centered at 0
1
rp q
r= =
+
Circles of radius1
xcentered at
11p and q
x= =
EE 342—Spring 2010 #115
( )
2
2
2
1
1 1
rp q
r r
− + =
+ +Circles of radius( )
1
1r +centered at 0
1
rp and q
r= =
+
( )2
2
2
1 11p q
x x
− + − =
Circles of radius
1
xcentered at
11p and q
x= =
EE 342—Spring 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 342—Spring 2010 #117
Smith chart
Relation between the normalized input impedance to the line and the reflection coefficient
We plot the normalized input impedance
ˆinz r jx= +
The point defines the magnitude and
the angle of the reflection coefficient
2 2ˆin p qΓ = +
( )ˆ2 4
inθ β π
λΓ= ∠ − = ∠ −
EE 342—Spring 2010 #118
Smith chart
1ˆ
1in
p jqz r jx
p jq
+ += + =
− −
4
ˆ ˆj
in L e p jqπ
λ−
Γ = Γ = +
How determine the input
impedance
We plot the normalized load impedance
0
ˆˆ L
L L L
Zz r j x
Z= = +
Rotate (with a compass) an angle
( )ˆ2 4
inθ β π
λΓ= ∠ − = ∠ −
Clockwise TG (towards generator)
If we know the input impedance we calculate
0
ˆˆ
inin in in
Zz r jx
Z= = +
Rotate (with a compass) an angle
( )ˆ2 4
inθ β π
λΓ= ∠ = ∠
Counterclockwise TL (towards load)
EE 342—Spring 2010 #119
Smith chart
Example: a coaxial cable (εr = 2.25), length 10m. Frequency source 34 MHz. The
characteristic impedance is Z0 = 50 Ω. The line is terminated with a load ZL = (50+j100) Ω.
Determine the input impedance
Propagation velocity80
2 10
r
v mvsε
= =
Wavelength5.882 1.7
vm
fλ λ= = ⇒ =
Normalized load impedance
50 100ˆ 1 2
50L
jz j
+= = +
Rotate 1.7 λ TG (3 turns plus 0.2 λ)
ˆ 0.29 0.82inz j= −
And unnormalizing
0
ˆ ˆ 14.5 41in inZ z Z j= = −
EE 342—Spring 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)
viii) Matched load
min
in
C
Z
Z
max
in
C
Z
Z
( )0Γ =
i) D
ii) A
iii) E
iv) C
v) B
vi) D
vii) B
viii) A
EE 342—Spring 2010 #121
Smith chart
Example:
( )ˆ 20 40inZ j= − Ω
( )ˆ 20 40LZ j= + Ω
0
ˆ 100Z = Ω
( )ˆ 0.2 0.4inz j= −
We calculate( )ˆ 0.2 0.4Lz j= +
Determine the length
of the line in
wavelengths
0.062 TG
0.436 TG
0.436 0.062 0.374λ λ λ= − =
0.438 TL
0.064 TL
0.438 0.064 0.374λ λ λ= − =
EE 342—Spring 2010 #122
Smith chart
Exercise: Determine ZL attached to
a line with Z0 = 100 Ω. Removing
the load yields an input impedance
Zin= -j80Ω.With the unknown
impedance attached the input
impedance is (30 + j 40) Ω.
Determine ZL
With open circuit80
ˆ 0.8100L
in Z
jz j
=∞
− Ω= = −
Ω
0.107 TL
0.393 TG
ˆLz = ∞
0.25 TL
0.25 TG
( )0.393 0.25 0.143λ λ= − =
( )0.25 0.107 0.143λ λ= − =
With the load attached
( )30 40ˆ 0.30 0.40
100Lin Z
jz j
+ Ω= = +
Ω
0.065 TG
0.435 TL
Rotate TL 0.143 λ
0.435 λ + 0.143 λ = 0.578 λ = 0.078 λ
ˆ 0.32 0.49 32 49L Lz j Z j= − = −
EE 342—Spring 2010 #123
Smith chart
Exercise: Determine the load
impedance, VSWR and load
reflection coefficient for :
( )ˆ 50 100inZ j= − Ω
0
ˆ 50Z = Ω 0.4 λ=
ˆ 1 2inz j= −
0.187λ TL
450
Rotate TL (CCW)
0.187 0.4 0.587 0.087λ λ λ λ+ = =
ˆ 0.22 0.58Lz j= −
( )ˆ 11 29LZ j= − Ω
-1180
00.73 118LΓ = ∠ −
7VSWR =
EE 342—Spring 2010 #124
Smith chart