Berkeley
The Smith Chart
Prof. Ali M. Niknejad
U.C. Berkeley
c 2016 by Ali M. Niknejad
Copyright February 2, 2016
1 / 27
The Smith Chart
The Smith Chart is simply a graphical calculator for
computing impedance as a function of reflection coefficient
z = f (ρ)
More importantly, many problems can be easily visualized with
the Smith Chart
This visualization leads to a insight about the behavior of
transmission lines
All the knowledge is coherently and compactly represented by
the Smith Chart
Why else study the Smith Chart? It’s beautiful!
There are deep mathematical connections in the Smith Chart.
It’s the tip of the iceberg! Study complex analysis to learn
more.
2 / 27
0.1
0.39
0.4
0.38
0.11
0.12
0.13
0.36
5
0.9
1.2
0.7
1.4
0.8
-55
1.0
0
-5
-4
0.15
0.37
0.14
0.35
0.41
0.09
-4
0
0.34
0.6
0.4
3
0.0
7
1.6
-30
0.08
-35
CAP
AC
ITI
VE
-65
1.8
0.2
0.42
0.16
-60
0.5
2.0
1
0.3
0.17
-70
(-j
0.4
RE
A
CT
AN
CE
CO
M
PO
N
EN
T
6
0.6
0.0
0.8
4
1.0
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
0.4
0.6
0.8
0.29
9
0.1
0.33
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.1
0.4
85
IND
UCT
IVE
0.8
0.6
80
0.2
20
10
0.04
0.46
1.0
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
0.2
90
0.3
-85
4.0
5.0
0.28
0.4
0.3
)
/Yo
(-jB
CE
AN
PT
CE
-75
US
ES
IV
CT
DU
IN
R
O
),
Zo
X/
1.0
0.22
0.3
8
-80
0.0
5
(+
jX
/Z
0.4
20
10
20
50
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.23
0.04
0.6
0.28
5
0.0
5
70
0.4
4
6
65
0.5
0.0
0.6 60
2.0
1.8
0.3
0.22
5
25
0.21
0.4
1.6
30
0.2
2
0.2
0.3
55
1.2
45
50
1.0
0.9
0.8
1.4
0.7
0.16
0.34
0.21
-20
0.4
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
0.8
0.29
0.2
35
-25
0.35
0.3
0.49
0.48
D <—
RD LOA
TOWA
THS
0.47
ENG
VEL
WA
<—
-90
1
0.1
40
0.1
Yo)
jB/
E (+
NC
TA
EP
SC
SU
0.36
0.3
0.2
R
,O
o)
VE
TI
CI
PA
CA
0.37
9
0.4
1.0
3
0.4
0.13
0.38
3.0
7
0.12
4.0
0.0
0.4
-15
0.42
0.39
5.0
0.11
-10
0.1
50
0.08
0.09
0.41
0.1
0.46
An Impedance Smith Chart
Without further ado, here it is!
0.14
0.15
0.33
0.17
2
0.1
8
0.4
3.0
15
10
3 / 27
Generalized Reflection Coefficient
In sinusoidal steady-state, the voltage on the line is a T-line
v (z) = v + (z) + v − (z) = V + (e −γz + ρL e γz )
Recall that we can define the reflection coefficient anywhere
by taking the ratio of the reflected wave to the forward wave
ρ(z) =
v − (z)
ρL e γz
=
= ρL e 2γz
v + (z)
e −γz
Therefore the impedance on the line ...
Z (z) =
v + e −γz (1 + ρL e 2γz )
v + −γz
(1 − ρL e 2γz )
Z0 e
4 / 27
Normalized Impedance
...can be expressed in terms of ρ(z)
Z (z) = Z0
1 + ρ(z)
1 − ρ(z)
It is extremely fruitful to work with normalized impedance
values z = Z /Z0
z(z) =
Z (z)
1 + ρ(z)
=
Z0
1 − ρ(z)
Let the normalized impedance be written as z = r + jx (note
small case)
The reflection coefficient is “normalized” by default since for
passive loads |ρ| ≤ 1. Let ρ = u + jv
5 / 27
Dissection of the Transformation
Now simply equate the real (<) and imaginary (=)
components in the above equaiton
r + jx =
(1 + u + jv )(1 − u + jv )
(1 + u) + jv
=
(1 − u) − jv
(1 − u)2 + v 2
To obtain the relationship between the (r , x) plane and the
(u, v ) plane
1 − u2 − v 2
r=
(1 − u)2 + v 2
x=
v (1 − u) + v (1 + u)
(1 − u)2 + v 2
The above equations can be simplified and put into a nice
form
6 / 27
Completing Your Squares...
If you remember your high school algebra, you can derive the
following equivalent equations
r
u−
1+r
2
+ v2 =
1
(u − 1) + v −
x
2
1
(1 + r )2
2
=
1
x2
These are circles in the (u, v ) plane! Circles are good!
We see that vertical and horizontal lines in the (r , x) plane
(complex impedance plane) are transformed to circles in the
(u, v ) plane (complex reflection coefficient)
7 / 27
Resistance Transformations
r=2
r=1
r=0
.5
r=
0
v
u
r
r=0 r=.5 r=1
r=2
r = 0 maps to u 2 + v 2 = 1 (unit circle)
r = 1 maps to (u − 1/2)2 + v 2 = (1/2)2 (matched real part)
r = .5 maps to (u − 1/3)2 + v 2 = (2/3)2 (load R less than Z0 )
r = 2 maps to (u − 2/3)2 + v 2 = (1/3)2 (load R greater than
Z0 )
8 / 27
Reactance Transformations
x
v
x=2
x=1
x
=
x=2
.
5
x=1
x=0
u
r
x = -1
-.5
x = -2
-2
x=
x = -1
x
=
x = ±1 maps to (u − 1)2 + (v ∓ 1)2 = 1
x = ±2 maps to (u − 1)2 + (v ∓ 1/2)2 = (1/2)2
x = ±1/2 maps to (u − 1)2 + (v ∓ 2)2 = 22
Inductive reactance maps to upper half of unit circle
Capacitive reactance maps to lower half of unit circle
9 / 27
Complete Smith Chart
r>1
v
match
x=2
x=1
x
=
r=
0
.5
r=2
r=1
r=0
.5
open
inductive
u
-.5
capacitive
-2
x=
=
x = -1
x
short
10 / 27
Reading the Smith Chart
11 / 27
Reading the Smith Chart
load
First map zL on the Smith Chart as ρL
To read off the impedance on the T-line at any point on a
lossless line, simply move on a circle of constant radius since
ρ(z) = ρL e 2jβz
12 / 27
Motion Towards Generator
wards generato
nt to
r
me
ve
mo
L
CIRC E
R
W
S
load
Moving towards generator means ρ(−`) = ρL e −2jβ` , or
clockwise motion
For a lossy line, this corresponds to a spiral motion
We’re back to where we started when 2β` = 2π, or ` = λ/2
Thus the impedance is periodic (as we know)
13 / 27
SWR Circle
wards generato
nt to
r
me
ve
o
m
load
Since SWR is a function of
|ρ|, a circle at origin in (u,v)
plane is called an SWR circle
Recall the voltage max
occurs when the reflected
wave is in phase with the
forward wave, so
ρ(zmin ) = |ρL |
This corresponds to the intersection of the SWR circle with
the positive real axis
Likewise, the intersection with the negative real axis is the
location of the voltage min
14 / 27
Example of Smith Chart Visualization
load
Prove that if ZL has an inductance reactance, then the
position of the first voltage maximum occurs before the
voltage minimum as we move towards the generator
Proof: On the Smith Chart start at any point in the upper half
of the unit circle. Moving towards the generator corresponds
to clockwise motion on a circle. Therefore we will always cross
the positive real axis first and then the negative real axis.
15 / 27
Admittance Chart
65
06
0.
0.
44
70
0
45
50
55
60
N
EN
4
PO
0.3
0.28
In fact everything is switched
around a bit and you can
construct a combined
admittance/impedance Smith
Chart. You can also use an
impedance chart for admittance
if you simply map x → b and
r →g
Be careful ... the caps are now on the top of the chart and
the inductors on the bottom
0.2
10
0.
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
COEFFICIENT IN
0.27
REFLECTION
DEGR
L E OF
EES
A NG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EE S
ANG
8
0.6
0.1
0.4
0.2
0.1
0.2
0.3
0.4
0.6
0.5
0.7
0.8
1.0
0.9
1.2
1.4
1.6
2.0
1.8
3.0
4.0
20
10
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
50
20
0.4
10
0.15
0.35
0
-4
0.14
-80
0.36
5
-70
-90
0.12
0.13
0.38
0.37
0.11
-100
-110
0.1
CO
M
N
(-j
-1
EN
T
-70
PO
-65
0.07
30
-1
0.43
-120
8
0.0
0.4
9
0.0
19
)
/ Yo
(-jB
-85
E
NC
TA
EP
SC
EA
CT
AN
CE
99
9
1
0.4
0.4
0.389999
4
6
SU
VE
R
0.0
VE
TI
R
0
-35
6
-4
4
-5
0.1
0.3
-55
-60
CIT
I
-60
7
0.1
3
0.3
1.0
0.9
1.2
0.8
1.4
0.7
CAP
A
-150 -80
C
DU
IN
0.2
1.6
-30
5.0
4.0
-75
O
-25
31
0.6
0.18
0
-5
40
),
Zo
X/
0.
0.5
1.8
0.32
3.0
06
0.4
19
0.
-20
0.
2.0
44
0.
0.3
0
0.
4
0.2
0.05
0.8
0.6
0.4
1.0
-15
0.2
9
0.2
1
-30
0.3
0.28
0.22
99
-20
0.2
-4
99
0.1
0.6
0.
8
1.0
0.44
20
50
0.22
0.4
85
6
15
0
80
CTI
VE
RE
AC
TA
NC
EC
OM
9
IND
U
30
0.2
10
20
1
0.2
15
-10
99
14
(+
j
4
0.
T
0.3
20
0.6
0.8
1.0
1.0
5.0
0.2
40
0.05
31
99
0.32
50
25
0.
X/
Z
30
19
0.44
0.18
0.
75
0.2
7
0.5
0.0
0.1
0.3
3
60
0.4
4.0
5.0
0.0 Ð > WAVELE
0.49
NGTH
S TOW
AR D
0.0
0.479999
D <Ð
0.49
GEN
R D L OA
0.479999
ERA
TOWA
± 180
TO
THS
0.47
170
RÐ
-170
ENG
VEL
0.47
>
WA
160
<Ð
-90
90
-160
4
0.6
C
AN
PT
6
0.3
35
0.7
CE
US
ES
IV
0.1
70
40
0.8
T
CI
PA
CA
0.9
R
0.35
80
1.0
2.0
,O
o)
1.2
1.8
0.43
0
13
1.6
0.07
Yo)
jB/
E (+
Since y = 1/z = 1−ρ
1+ρ , you can
imagine that an Admittance
Smith Chart looks very similar
0.15
0.36
90
1.4
0.4
3.0
110
0.14
0.37
0.38
0.389999
100
0.4
1
0.4
9
99
19
120
0.13
0.12
0.11
0.1
9
0.0
8
0.0
The short and open likewise swap positions
16 / 27
Admittance on Smith Chart
Sometimes you may need to work with both impedances and
admittances.
This is easy on the Smith Chart due to the impedance
inversion property of a λ/4 line
Z0 =
Z02
Z
If we normalize Z 0 we get y
Z0
Z0
1
=
= =y
Z0
Z
z
17 / 27
Admittance Conversion
Thus if we simply
rotate π degrees on
the Smith Chart
and read off the
impedance, we’re
actually reading off
the admittance!
z
y
Rotating π degrees
is easy. Simply
draw a line through
origin and zL and
read off the second
point of intersection
on the SWR circle
18 / 27
Example Calculation
19 / 27
Remember This?
ZS
Z0
ZL
V0
z
z = -d
1
z=0
Consider the transmission line circuit shown below. A voltage
source generating 10V amplitude of sinewave at 10 GHz is
driving a transmission line terminated with load ZL = 80 - j40
ohm. The transmission line has a characteristic impedance of
Z0 (= 100Ω), effective dielectric constant of 4, and length
d =22.5 mm.
1
2
3
Find the reflection coefficient at the load (z = 0) and at the
source (z = −d). [ Note this is 1.5λ ]
Find the input impedance at the source (z = -d) and at z =
18.75 mm. [Note this is 1.25λ ]
Plot the magnitude of the voltage along the line. Find voltage
maximum, voltage minimum, and standing wave ratio.
20 / 27
0.4
0.39
0.38
0.1
0.11
0.12
0.13
0.37
0.14
5
0.9
1.2
1.6
0.5
0.0
6
-70
0.6
-65
1.8
2.0
R
O
),
Zo
X/
E
IV
CT
DU
IN
-75
0.8
3
7
0.0
0.4
-60
0.7
1.4
0.8
-55
1.0
0
-5
-4
0.15
0.34
0.36
0.41
0.09
0
0.16
0.35
2
0.4
-35
-4
8
0.0
CAP
AC
ITI
VE
0.2
-30
7
0.1
(-j
RE
AC
TA
NC
EC
OM
PO
N
EN
T
1
0.3
3
0.3
4
0.6
0.4
9
9
8
o)
jB/Y
E (NC
TA
EP
SC
SU
1.0
1.0
0.0
4
0.4
6
1.0
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
80
4.0
j ρ
15
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
85
IND
UCT
IVE
90
0.2
-85
1.0
0.2
0.4
0.6
0.8
0.2
0.1
|ρ|e
0.3
0.4
-80
0.6
4
0.8
0.1
0.4
0.2
10
6
0.8
0.28
0.0
(+
jX
/Z
5
0.0
5
0.4
20
0.22
0.4
0.6
10
20
50
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.23
1
5
0.0
70
4
0.4
65
0.5
6
0.0
2.0
1.8
1.6
55
1.2
45
50
1.0
0.9
0.8
1.4
0.7
0.6 60
2
0.2
5
0.4
0.3
0.2
0.1
0.3
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
25
5.0
0.28
0.49
0.3
3
2
0.2
30
0.3
0.16
0.34
-25
35
0.22
0.3
0.35
1
0.2
9
0.2
0.2
40
0.3
0.1
0.14
0.2
0.48
D <—
RD LOA
TOWA
THS
ENG
VEL
WA
<—
-90
1
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
3.0
Yo)
jB/
E (+
NC
TA
EP
SC
SU
0.36
0.3
0.4
R
,O
o)
VE
TI
CI
PA
CA
0.37
9
0.4
-20
3
0.4
2
0.13
0.38
4.0
0
0.4
0.12
-15
0
0.39
5.0
0.11
-10
.07
0.4
20
0.1
50
.08
0.09
0.41
0.1
0.47
Homework Problem with Aid from Mr. Smith
0.15
0.1
7
0.1
8
0.4
3.0
10
|ρ| = .25
= −.056 − .24j
ρ = 257◦
21 / 27
Impedance Matching with Smith Chart
22 / 27
Impedance Matching Example
45
50
1.0
0.9
55
1.4
0.8
1.6
1.8
2.0
6
0.0
4
0.4
70
5
0.4
20
3.0
0.6
r=1 CIRCLE
0.3
0.8
0.21
4.0
1.0
5.0
0.8
0.6
10
0.4
0.2
SW
RC
10
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
IND
UCT
IVE
LE
IRC
15
0.28
1.0
0.22
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
0.4
5
(+
jX
/Z
0.0
0.4
0.29
0.46
8
2
0.3
0.04
R
,O
o)
0.1
0.3
25
0.2
80
0.17
0.33
30
0.2
1
85
0.16
0.34
35
0.3
90
0.15
0.35
40
9
20
0.1
10
20
50
5.0
4.0
3.0
2.0
1.8
0.4
10
)
/Yo
(-jB
CE
AN
PT
CE
-75
US
ES
IV
CT
DU
IN
R
O
LOAD
0.6
0.8
1.0
-80
1.0
4.0
0.8
0.6
1.2
0.5
2.0
0
1.0
0.9
1.4
0.14
0.36
5
-4
0.12
0.13
0.38
0.37
0.11
0.39
0.1
0.4
0.41
0.09
-70
The match is at the
center of the circle. Grab
a reactance in series or
shunt to move you there!
0.35
-4
-5
0
0.15
(-j
6
Simply find the
intersection of the SWR
circle with the r = 1
circle
0.34
-35
0.8
0.16
-65
1.8
1.6
-30
-55
0.17
CO
M
PO
N
EN
T
0.0
Single stub impedance
matching is easy to do
with the Smith Chart
0.33
0.7
2
CAP
AC
ITI
VE
RE
AC
TA
NC
E
0.2
0.6
0.3
8
-60
1
0.3
4
-25
9
0.4
0.4
0.1
0.1
0.0
7
0.4
0.08
0.42
5
0.0
),
Zo
X/
3.0
-20
0.3
5
0.4
-15
0.29
3
0.04
5.0
0.28
0.2
0.4
0.21
0.3
0.22
0.2
-10
-85
0.1
0.46
1.6
1.4
1.2
1.0
0.8
0.9
0.7
0.6
0.5
0.4
0.3
0.1
0.49
0.48
D <—
RD LOA
TOWA
THS
0.47
ENG
VEL
WA
<—
-90
50
0.2
20
0.2
0.2
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
50
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
0.14
0.36
0.1
)
/Yo
(+jB
CE
AN
PT
CE
US
ES
IV
IT
C
PA
CA
0.6 60
3
0.37
0.5
0.4
0.13
0.38
65
7
0.0
0.42
0.12
0.7
0.08
0.39
0.4
1.2
0.11
0.1
0.09
0.41
23 / 27
0.4
0.38
0.39
0.12
0.13
0.14
0.37
0.11
0.1
0.35
5
-4
0.9
1.2
7
1.6
0.5
6
-70
-65
1.8
2.0
R
O
),
Zo
X/
-85
1.0
E
NC
TA
EP
SC
SU
-80
E
IV
CT
DU
IN
-75
4
)
/Yo
(-jB
0.1
SW
RC
80
0.0
4
0.4
6
1.0
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
1.0
4.0
15
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
85
IND
UCT
IVE
90
0.2
6
0.8
3
0.4
-60
0.7
1.4
0.8
-55
1.0
0
-5
0.34
0.15
-4
0.16
3
0.36
0.41
0.09
0
.42
0.0
0.2
8
-30
0.1
0
8
0.0
CAP
AC
ITI
VE
0.6
1
0.3
-35
0.0
9
7
0.1
(-j
RE
A
CT
AN
CE
CO
M
PO
N
EN
T
0.1
0.3
4
0.6
0.4
9
0.3
0.4
0.0
1.0
0.4
(+
jX
/Z
5
0.0
5
0.4
20
0.2
0.4
0.6
0.8
0.2
5
r=1 CIRCLE
0.28
1
0.0
70
4
0.4
65
0.5
6
0.0
2.0
1.8
1.6
55
1.2
45
50
1.0
0.9
0.8
1.4
0.7
0.6 60
2
0.2
5
0.4
0.2
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
25
0.2
-20
3.0
0.49
0.3
2
0.3
3
0.22
4.0
0.3
0.8
30
0.3
0.6
-25
0.2
-15
0.16
0.34
10
20
50
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.23
0.3
35
5.0
0.28
LOAD
0.35
0.22
0.1
40
1
0.2
9
0.2
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.14
0.3
LE
IRC
5.0
Yo)
jB/
E (+
NC
TA
EP
SC
SU
0.36
0.2
0.48
D <—
RD LOA
TOWA
THS
ENG
VEL
WA
<—
-90
1
0.4
0.37
0.3
0.4
R
,O
o)
VE
TI
CI
PA
CA
10
3
0.13
0.38
-10
0.4
0.12
9
0.2
7
2
0.39
20
0.11
50
0.0
0.4
0.4
8
0.4
0.6
0.1
0.8
0.0
0.09
0.41
0.1
0.47
Series Stub Match
0.15
0.1
7
0.1
8
0.4
3.0
10
24 / 27
Shunt Stub Match
45
50
1.0
0.9
55
1.4
0.8
1.6
1.8
2.0
6
0.0
4
0.4
70
5
0.4
20
3.0
0.6
r=1 CIRCLE
0.3
0.8
0.21
4.0
1.0
5.0
0.8
0.6
10
0.4
0.2
SW
RC
10
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
IND
UCT
IVE
LE
IRC
15
0.28
1.0
0.22
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
0.4
5
(+
jX
/Z
0.0
0.4
0.29
0.46
8
2
0.3
0.04
R
,O
o)
0.1
0.3
25
0.2
80
0.17
0.33
30
0.2
1
85
0.16
0.34
35
0.3
90
0.15
0.35
40
9
20
0.1
10
20
50
5.0
4.0
3.0
2.0
1.8
0.4
10
)
/Yo
(-jB
CE
AN
PT
CE
-75
US
ES
IV
CT
DU
IN
R
O
LOAD
0.6
0.8
1.0
-80
1.0
4.0
0.8
0.6
1.2
0.9
1.0
5
-4
0.12
0.13
0.38
0.37
0.11
0.39
0.1
0.4
0.41
0.09
-70
(-j
0.5
2.0
1.8
1.6
0.14
0.36
0.8
1.4
0.35
-4
0
0
0.15
-5
0.34
-35
-55
0.16
-65
0.6
-30
0.7
0.17
0.33
CO
M
PO
N
EN
T
6
To find the shunt stub
value, simply convert the
value of z = 1 + jx to
y = 1 + jb and place a
reactance of −jb in shunt
2
CAP
AC
ITI
VE
RE
AC
TA
NC
E
0.2
0.0
Let’s now solve the same
matching problem with a
shunt stub.
0.3
8
-60
1
0.3
4
-25
9
0.4
0.4
0.1
0.1
0.0
7
0.4
0.08
0.42
5
0.0
),
Zo
X/
3.0
-20
0.3
5
0.4
-15
0.29
3
0.04
5.0
0.28
0.2
0.4
0.21
0.3
0.22
0.2
-10
-85
0.1
0.46
1.6
1.4
1.2
1.0
0.8
0.9
0.7
0.6
0.5
0.4
0.3
0.1
0.49
0.48
D <—
RD LOA
TOWA
THS
0.47
ENG
VEL
WA
<—
-90
50
0.2
20
0.2
0.2
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
50
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
0.14
0.36
0.1
)
/Yo
(+jB
CE
AN
PT
CE
US
ES
IV
IT
C
PA
CA
0.6 60
3
0.37
0.5
0.4
0.13
0.38
65
7
0.0
0.42
0.12
0.7
0.08
0.39
0.4
1.2
0.11
0.1
0.09
0.41
25 / 27
0.4
0.39
0.38
0.1
0.11
0.12
0.13
0.14
0.37
0.36
0.41
5
-4
0.9
1.2
1.0
0
-5
0.08
0.4
7
3
1.6
-60
0.7
1.4
0.8
-55
-4
0
0.34
0.15
0.35
0.42
0.09
-35
0.16
0.6
-65
0.0
-30
0.5
1.8
6
0.0
-70
2.0
0.6
4
0.8
0.4
-20
(-j
0.4
CO
M
PO
N
EN
T
0.3
1
0.17
1.0
0.2
80
0.04
0.46
1.0
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
1.0
4.0
15
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
85
IND
UCT
IVE
90
0.6
-85
)
/Yo
(-jB
CE
AN
PT
CE
-75
US
ES
IV
CT
DU
IN
R
O
),
Zo
X/
0.29
0.1
9
0.33
-80
0.8
0.04
0.0
5
(+
jX
/Z
0.4
20
0.28
CAP
AC
ITI
VE
RE
AC
TA
NC
E
5
0.0
r=1 CIRCLE
0.22
0.3
0.2
5
0.4
5
70
0.4
4
0.5
6
0.0
65
0.6 60
2.0
1.8
2
0.2
-25
0.2
0.1
10
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
25
0.21
2
0.3
0.3
1.6
30
LOAD (z)
10
20
0.2
50
0.4
0.6
0.4
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.23
1.0
0.3
0.8
8
0.6
0.1
0.2
5.0
0.28
0.3
55
1.2
45
50
1.0
0.9
0.8
1.4
0.7
0.16
0.34
0.22
0.2
35
0.21
LOAD (y)
0.35
0.29
0.1
40
0.3
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.14
0.2
0.49
0.48
D <—
RD LOA
TOWA
THS
0.47
ENG
VEL
WA
<—
-90
1
g=1 CIRCLE
3.0
Yo)
jB/
E (+
NC
TA
EP
SC
SU
0.36
0.3
0.4
0.37
9
0.4
R
,O
o)
VE
TI
CI
PA
CA
0.13
0.38
4.0
3
0.4
0.12
-15
7
0.39
5.0
0.11
0.8
-10
0.0
0.42
0.4
20
0.1
50
0.08
0.09
0.41
0.1
0.46
Matching with Lumped Components (I)
0.15
0.33
0.17
0.1
8
0.4
3.0
10
Suppose the load is inside the 1 + jx circle.
26 / 27
0.4
0.39
0.38
0.1
0.11
0.12
0.13
0.14
0.37
0.36
0.41
5
-4
0.9
1.2
1.0
0
-5
0.08
0.4
7
3
1.6
-60
0.7
1.4
0.8
-55
-4
0
0.34
0.15
0.35
0.42
0.09
-35
0.16
0.6
-65
0.0
-30
0.5
1.8
6
0.0
-70
2.0
0.6
4
0.8
0.4
-20
(-j
0.4
CO
M
PO
N
EN
T
0.3
1
0.17
1.0
0.2
80
0.04
0.46
1.0
RE
AC
TA
75
NC
EC
OM
PO
N
EN
T
1.0
4.0
15
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
85
IND
UCT
IVE
90
0.6
-85
)
/Yo
(-jB
CE
AN
PT
CE
-75
US
ES
IV
CT
DU
IN
R
O
),
Zo
X/
0.29
0.1
9
0.33
-80
0.8
0.04
0.0
5
(+
jX
/Z
0.4
20
0.28
0.3
CAP
AC
ITI
VE
RE
AC
TA
NC
E
5
0.0
r=1 CIRCLE
0.22
0.2
0.2
5
0.4
5
70
0.4
4
0.5
6
0.0
65
0.6 60
2.0
1.8
2
0.1
10
20
0.2
50
0.4
0.6
0.4
10
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
0.27
REFLECTION COEFFICIENT IN DEG
REES
LE OF
ANG
ISSION COEFFICIENT IN
TRANSM
DEGR
LE OF
EES
ANG
0.23
0.2
0.0 —> WAVELE
NGTH
S TOW
ARD
0.0
0.49
GEN
ERA
0.48
TO
R—
0.47
>
25
0.21
2
0.3
0.3
1.6
30
5.0
0.28
1.0
0.3
0.8
-25
0.6
8
0.2
0.1
55
1.2
45
50
1.0
0.9
0.8
1.4
0.7
0.16
0.34
0.22
0.3
35
0.21
0.2
0.35
0.29
0.1
40
0.3
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.14
0.2
0.49
0.48
D <—
RD LOA
TOWA
THS
0.47
ENG
VEL
WA
<—
-90
1
g=1 CIRCLE
3.0
LOAD (z)
0.36
0.3
0.4
0.37
9
0.4
R
,O
o)
VE
TI
CI
PA
CA
0.13
0.38
4.0
3
0.4
Yo)
jB/
E (+
NC
TA
EP
SC
SU
0.12
-15
7
0.39
5.0
0.11
0.8
-10
0.0
0.42
0.4
20
0.1
50
0.08
0.09
0.41
0.1
0.46
Matching with Lumped Components (II)
0.15
0.33
0.17
0.1
8
0.4
3.0
10
Suppose the load is outside the 1 + jx circle.
27 / 27