Smith Chart Examples
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Smith Chart
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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
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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
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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
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2 2
2 1
1 1r i
r
r r
; 01
r i
r
r
Centre Radius 1
1 r
Resistive Circles
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•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
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2 2
2 1 11r i
x x
Reactance Circles
11;r i
x
Centre Radius 1
x
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•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
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•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
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Smith chart can also be marked as polar coordinates:
Magnitude of
Phase angle of
1
1
SC
OC
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Pm PM
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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
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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
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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
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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
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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
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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’)
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4 '2 '
zz
4 'f ' / 2; 2 ' 2
zi z z
Therefore complete revolution gives the z’ of /2
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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
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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
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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
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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
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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
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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
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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
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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
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Example:5: ZL = 25 + j 100 and Z0 = 50 ; Find YL
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1 1;
1 1n n
d dz d y d
d d
Impedance = Resistance + j Reactance
Z = R + j X
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Open Circuit
Short Circuit
Admittance Chart
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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
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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
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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
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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
![Page 35: Smith Chart Examples](https://reader033.fdocuments.us/reader033/viewer/2022042520/577cc10e1a28aba711921b98/html5/thumbnails/35.jpg)
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
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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
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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
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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
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The input admittance of short circuited stub is purely susceptive ysB
1 B SBy y 1B By jb
sB By jb
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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
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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
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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
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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
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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.