antennas and microwaves engineering …...voltage standing wave ratio (vswr): it is the ratio of the...
TRANSCRIPT
Philadelphia University
Faculty of Engineering Communication and Electronics Engineering
Part 2 Dr. Omar R Daoud 1
ANTENNAS and MICROWAVES
ENGINEERING
(650427)
Transmission Lines
General Considerations It is a two-port network connecting a generator circuit to
a load.
The impact of a transmission line on the current and
voltage in the circuit depends on the length of line, l and
the frequency, f of the signal provided by generator.
At low frequency, the impact is negligible
At high frequency, the impact is very significant
2/27/2018 Dr. Omar R Daoud 2
Transmission Lines
General Considerations It may be classified into two
types:
Transverse electromagnetic
(TEM) transmission lines:
waves propagating along these
lines having electric and
magnetic field that are entirely
transverse to the direction of
propagation
Higher order transmission
lines:
waves propagating along these
lines have at least one
significant field component in
the direction of propagation
2/27/2018 Dr. Omar R Daoud 3
Transmission Lines
General Considerations It is represented by a parallel-wire
configuration regardless of the specific
shape of the line, i.e coaxial line, two-
wire line or any TEM line.
It can be modelled as a Lumped
element:
It consists of four basic elements called
‘the transmission line parameters’ :
R’ : combined resistance of both
conductors per unit length, in Ω/m
L’ : the combined inductance of both
conductors per unit length, in H/m
G’ : the conductance of the insulation
medium per unit length, in S/m
C’ : the capacitance of the two conductors
per unit length, in F/m
2/27/2018 Dr. Omar R Daoud 4
The following parameters µ, σ, ε
refer to the insulating material
between conductors
Transmission Lines
General Considerations In the TEM transmission lines:
The followings should be fulfilled
The propagation constant is
The line characteristics impedance is
The wave propagation phase velocity is
2/27/2018 Dr. Omar R Daoud 5
The following parameters µ, σ, ε
refer to the insulating material
between conductors
''CL
'
'
C
G
j
CjG'LjR'
''
''
''0
CjG
LjRZ
fu p
Transmission Lines
Example For an air line with characteristic impedance of 50Ω and phase
constant of 20 rad/m at 700MHz, find the
Capacitance per meter and
the inductance per meter of the line.
2/27/2018 Dr. Omar R Daoud 6
An air line (a transmission
line for which air is the
dielectric material present
between the two conductors,
which renders G’ = 0. In
addition, the conductors are
made of a material with high
conductivity so that R’ ≈0 ).
'
'
'
' and ''''Im 0
C
L
Cj
LjZCLCjLj
pF/m 9.90501072
20'
8
0
ZC
nH/m 227109.9050''' 122
0 LCLZ
Transmission Lines
Lossless Transmission Lines In the lossless transmission lines, R’
and G’ are set to zero (very small
values) : The propagation constant is
The line characteristics impedance is
2/27/2018 Dr. Omar R Daoud 7
The following parameters µ, σ, ε
refer to the insulating material
between conductors
line) (lossless '
'
0,G' and 0R' since
''
''
0
0
C
LZ
CjG
LjRZ
line) (lossless ''
line) (lossless 0
CL
Transmission Lines
Lossless Transmission Lines In the lossless transmission lines, R’
and G’ are set to zero (very small
values) :
The wave propagation phase velocity is
The wavelength is
2/27/2018 Dr. Omar R Daoud 8
The following parameters µ, σ, ε
refer to the insulating material
between conductors
(m/s) 1
(rad/m)
pu
rr
p
f
c
f
u
01
Transmission Lines
Reflection Coefficient
Voltage Reflection Coefficient (Γ) :
It is the ratio of the amplitude of the reflected voltage wave, V0- to the
amplitude of the incident voltage wave, V0+ at the load.
Z0 for lossless line is a real number while ZL in general is a complex
number. Hence,
A load is matched to the line if ZL = Z0 because there will be no
reflection by the load (Γ = 0 and V0−= 0).
When the load is an open circuit, (ZL=∞), Γ = 1 and V0- = V0
+.
When the load is a short circuit (ZL=0), Γ = -1 and V0- = V0
+.
2/27/2018 Dr. Omar R Daoud 9
less)(dimension 1
1
0
0
0
0
0
0
ZZ
ZZ
ZZ
ZZ
V
V
L
L
L
L
rje
Transmission Lines
Example A 100-Ω transmission line is connected
to a load consisting of a 50-Ω resistor in
series with a 10pF capacitor. Find the
reflection coefficient at the load for a
100-MHz signal.
2/27/2018 Dr. Omar R Daoud 10
Hz10MHz100 ,100 F,10 ,50 80
11LL fZCR
15950
10102
150
/
118
LLL
jj
CjRZ
7.6076.0
159.15.0
159.15.0
1/
1/
0L
0L
j
j
ZZ
ZZ
Transmission Lines Standing Waves
The Interference between the reflected
wave and the incident wave along a
transmission line creates a standing
wave.
Constructive interference gives maximum
value, and occurs at
Destructive interference gives minimum
value, and occurs at
The repetition period for standing wave
pattern is λ/2 (it is λ for incident and
reflected wave individually).
2/27/2018 Dr. Omar R Daoud 11
For a matched line, ZL = Z0, Γ = 0 and
= |V0+| for all values of z. zV
~
For a short-circuited load, (ZL=0), Γ = -1.
For an open-circuited load, (ZL=∞), Γ = 1.
The wave is shifted by λ/4 from short-circuit case.
0 if ...... 2, 1, 0, n
0 if ...... 3, 2, 1, n
0 where24
r
r
rmax
n
nl
4/ if 4/
4/ if 4/
maxmax
maxmax
min
ll
lll
Transmission Lines
Standing Waves Voltage Standing Wave Ratio
(VSWR):
It is the ratio of the maximum voltage
amplitude to the minimum voltage
amplitude and provides a measure of
the mismatching between the load and
the transmission line.
For a matched load, if
Γ = 0, VSWR = 1
|Γ| - 1, VSWR = ∞
2/27/2018 Dr. Omar R Daoud 12
For a matched line, ZL = Z0, Γ = 0 and
= |V0+| for all values of z. zV
~
For a short-circuited load, (ZL=0), Γ = -1.
For an open-circuited load, (ZL=∞), Γ = 1.
The wave is shifted by λ/4 from short-circuit case.
less)(dimension 1
1~
~
min
max
V
VVSWR
Transmission Lines
Example A 50- transmission line is terminated in a load with ZL = (100 + j50)Ω . Find
the voltage reflection coefficient and
the voltage standing-wave ratio (VSWR).
2/27/2018 Dr. Omar R Daoud 13
6.26
0L
0L 45.05050100
5050100
1/
1/ jej
j
ZZ
ZZ
6.245.01
45.01
1
1
VSWR
Transmission Lines
Lossless TL: input impedance (Zin) It is the ratio of the total voltage (incident and reflected voltages)
to the total current at any point z on the line.
For a line terminated in a short-circuit, ZL = 0:
For a line terminated in an open circuit, ZL = ∞:
Then:
2/27/2018 Dr. Omar R Daoud 14
ljZ
lI
lVZ tan~
~
0
sc
scsc
in
ljZlI
lVZ cot~ 0
oc
ococin
ocin
scin ZZZo oc
in
scintan
Z
Zl
0
0
0 ZeeV
eeV
lI
lVZ
ljlj
ljlj
in
02
2
1
1Z
e
elj
lj
ljZZ
ljZZZ
L
L
tan
tan
0
00
Transmission Lines
Example A source with 50 source
impedance drives a 50
transmission line that is 1/8 of
wavelength long, terminated in a
load ZL = 50 – j25 . Calculate:
the voltage reflection
coefficient
the voltage standing-wave
ratio (VSWR)
The input impedance seen
by the source.
2/27/2018 Dr. Omar R Daoud 15
Transmission Lines
Example
2/27/2018 Dr. Omar R Daoud 16
076
0
0
242.0502550
502550 j
L
LL
ej
j
ZZ
ZZ
64.11
1
L
LVSWR
48
2
14
tan
8.38.30
255050
50255050
tan
tan
0
00
j
j
jj
jZZ
jZZZZ
L
Lin
Transmission Lines
Lossless TL: length of line At the line length of where n is an integer, then
If the transmission line is a quarter wavelength, with
then
2/27/2018 Dr. Omar R Daoud 17
2/nl
0tan
2//2tantan
n
nl
2/for ZZ Lin nl
2/4/ nl
24
2
l
2/4/for Z
ZZ
L
2
0in nl
Transmission Lines Example
A 50-Ω lossless transmission line is to be
matched to a resistive load impedance
with ZL=100Ω via a quarter-wave section
as shown, thereby eliminating reflections
along the feedline. Find the characteristic
impedance of the quarter-wave
transformer.
2/27/2018 Dr. Omar R Daoud 18
To eliminate reflections at terminal AA’, the input impedance Zin looking into the quarter-wave line should be equal to Z01 (the characteristic impedance of the feedline). Thus, Zin = 50Ω .
Since the lines are lossless, all the incident power will end up getting transferred into the load ZL.
7.701005002
L
202
in
Z
Z
ZZ
Transmission Lines
Lossless TL: Power Flow Average power for incident wave
Average power for reflected wave
The net average power delivered to the
load:
2/27/2018 Dr. Omar R Daoud 19
(W) 2 0
2
0iav
Z
VP
iav
2
0
2
02rav
2P
Z
VP
(W) 12
2
0
2
0rav
iavav
Z
VPPP
Transmission Lines
Lossless TL: Return Loss (RL)
When the load is mismatched, not all the of the available
power from the generator is delivered to the load.
This “loss” is called return loss, and is defined (in dB) as:
At matching load case (|Γ|=0),
RL is dB (no reflected power).
At total reflection case (|Γ|=1),
RL is 0 dB (all incident power is reflected).
So return loss will have only values between 0 < RL <
2/27/2018 Dr. Omar R Daoud 20
log20RL
Transmission Lines Summary
2/27/2018 Dr. Omar R Daoud 21
Transmission Lines
Smith Chart
It is the TL calculator and
used to analyze and design
TL circuits.
Smith Chart parameters are:
The reflection coefficient:
Γ = |Γ|ejθ, (|Γ| ≤ 1), (-180°≤ θ ≤180°)
The normalized impedance:
z = Z / Z0
The normalized admittance: y = 1 / z
SWR and RL scale
2/27/2018 Dr. Omar R Daoud 22
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
1
1
L
L
z
z
LL jxrz
1
1L
1
11
L
Lz
y
Transmission Lines Smith Chart
2/28/2018 Dr. Omar R Daoud 23
The families of circle for rL and xL.
The complex Γ plane. Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
For Z0 = 50Ω ,
a ZL = 0 (S.C.)
b ZL = ∞ (O.C.)
c ZL = 100 + j100 Ω
d ZL = 100 - j100 Ω
e ZL = 50 Ω
yL=0.25 - j0.6
zL=0.6 + j1.4
2/27/2018 Dr. Omar R Daoud 24
Transmission Lines
Smith Chart
The Smith Chart is a plot of
normalized impedance. For example, if a Z0 = 50 Ω transmission line is
terminated in a load ZL = 50 + j100 Ω. To
locate this point on Smith Chart, Normalize the load impedance, ZNL = ZL/ZN to obtain
ZNL = 1 + j2 Ω
The normalized load impedance is located at the
intersection of the r =1 circle and the x =+2 circle.
2/28/2018 Dr. Omar R Daoud 25
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
(c)
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
(c)
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
(c)
Transmission Lines
Smith Chart: Input Impedance The input impedance, Zin:
Γ is the voltage reflection coefficient at the
load. The phase angle of Γ is shifted by 2βl
to get ΓL ( from zin to zL; the same |Γ|, but the
phase is changed by 2βl = 2(2π/λ)l= 2π). On
the Smith chart, this means rotating in a
clockwise direction (WTG).
For one complete rotation corresponds to l =
λ/2.
The objective of shifting Γ to ΓL is to find Zin at
an any distance l on the transmission line.
2/28/2018 Dr. Omar R Daoud 26
lj
lj
ine
eZZ
2
2
01
1
A 50-Ω transmission line is
terminated with ZL=(100-j50)Ω.
Find Zin at a distance l =0.1λ from
the load.
Transmission Lines
Smith Chart: Reflection Coefficient
The reflection coefficient has a magnitude
and an angle : The magnitude can be measured using a scale for
magnitude of reflection coefficient provided below
the Smith Chart,
The angle is indicated on the angle of reflection coefficient scale shown outside the circle
on chart.
2/28/2018 Dr. Omar R Daoud 27
L
jL e
1L
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Scale for magnitude of reflection coefficient
0457.0 j
jL
e
e
Transmission Lines
Smith Chart: VSWR Point A is the normalized load impedance
with zL=2+j1.
VSWR = 2.6 (at Pmax).
The distance between the load and the first
voltage maximum is lmax=(0.25-0.213)λ.
The distance between the load and the first
voltage minimum is lmin=(0.037+0.25)λ.
Move along the constant circle is
akin to moving along the transmission line.
2/28/2018 Dr. Omar R Daoud 28
j
L e
Transmission Lines Example
Given that the voltage standing-wave
ratio, VSWR = 3. On a 50-Ω line, the
first voltage minimum occurs at 5 cm
from the load, and that the distance
between successive minima is 20 cm,
find the load impedance.
The distance between successive minima
is equal to λ/2. Hence, λ = 40 cm.
First voltage minimum (in wavelength unit)
is at
on the WTL scale from point B.
Intersect the line with constant SWR circle
= 3.
The normalized load impedance at point
C is:
De-normalize (multiplying by Z0) to get ZL:
2/28/2018 Dr. Omar R Daoud 29
125.040
5min l
8.06.0L jz
40308.06.050L jjZ
Transmission Lines Smith Chart
The Smith Chart is a plot of normalized
impedance. For example, if a Z0 = 50 Ω transmission line is terminated in
a load ZL = 50 + j100 Ω and has a line length of 0.3λ.
Normalize the load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2 Ω
The normalized load impedance is located at the intersection of
the r =1 circle and the x =+2 circle.
Moving away from the load (towards generator) corresponds
to moving in the clockwise direction on the Smith Chart.
Moving towards the load corresponds to moving in the
anti-clockwise direction on the Smith Chart.
To find ZIN, move towards the generator by:
Drawing a line from the center of chart to outside
Wavelengths Toward Generator (WTG) scale, to get
starting point a at 0.188λ
Adding 0.3λ moves along the constant circle
to 0.488λ on the WTG scale.
Read the corresponding normalized input
impedance point c, ZNIN = 0.175 - j0.08Ω
Denormalizing, to find an input impedance,
VSWR is at point b;
2/28/2018 Dr. Omar R Daoud 30
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
(c)
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
(c)
j
L e
475.8
0
jZ
ZZZ
IN
NININ
9.5VSWR
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
(c)
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Transmission Lines Smith Chart ZL= 50 - j25 and Z0=50 Ohm. Find Zin, VSWR and ΓL
using the Smith Chart.
Locate the normalized load, and label it as point a,
where it corresponds to
Draw constant circle.
It can be seen that
Move from point a (at 0.356λ) on the WTG scale,
clockwise toward generator a distance λ/8 or
0.125λ to point b, which is at 0.481λ.
We could find that at this point, it corresponds to
2/28/2018 Dr. Omar R Daoud 31
5.01 jZNL
j
L e
076245.0 j
L e
66.1VSWR
07.062.0 jZNIN
5.331 jZIN
Transmission Lines Smith Chart The input impedance for a 100 Ω lossless transmission
line of length 1.162 λ is measured as 12 + j42Ω.
Determine the load impedance.
Locate th Normalize the input impedance:
Locate the normalized input impedance and label it
as point a
Take note the value of wavelength for point a at
WTL scale.
At point a, WTL = 0.436λ Move a distance 1.162λ
towards the load to point b (WTL = 0.436λ +
1.162λ = 1.598λ ). But, to plot point b, 1.598λ –
1.500λ = 0.098λ.
Read the point b:
Denormalized it:
2/28/2018 Dr. Omar R Daoud 32
42.012.0100
4212
0
jj
Z
Zz in
in
7.015.0 jZNL
7015
0
j
ZZZ NLL
Transmission Lines Smith Chart On a 50 lossless transmission line, the VSWR is
measured as 3.4. A voltage maximum is located
0.079λ away from the load (towards generator).
Determine the load.
Use the given VSWR to draw a constant circle.
Then move from maximum voltage at WTG =
0.250λ (towards the load) to point a at WTG =
0.250λ - 0.079λ = 0.171λ.
At this point we have
ZNL = 1 + j1.3 Ω,
or ZL = 50 + j65 Ω.
2/28/2018 Dr. Omar R Daoud 33
j
L e
Transmission Lines
Smith Chart: Impedance Matching
Transmission line is matched to the load when Z0 =
ZL; which no reflection occurs.
This is usually not possible since ZL is used to serve
other application.
Alternatively, we can place an impedance-
matching network between load and transmission
line (The load impedance will be transformed to be
equal to Z0 of the transmission line.
The important of impedance matching or tuning:
Maximum power is delivered when the load is matched
to the line.
The power loss in the feed line is minimized. (increase
power handling capability by optimizing VSWR)
Improved the signal-to-noise ratio (SNR). (e.g with
controlled mismatch, an amplifier can operate with
minimum noise generation)
Reduced the amplitude and phase errors.
2/28/2018 Dr. Omar R Daoud 34
Matching
Network
Load
ZL
Z0
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Adding an impedance-matching network ensures that all power will make it to the
load.Techniques of impedance matching :
• Quarter-wave transformer
• Single / double stub tuner
• Lumped element tuner
• Multi-section transformer
Transmission Lines
Smith Chart: Impedance Matching
Quarter Wave Transformer It can only be constructed if the load impedance is all
real (no reactive component)
Designed to achieve matching at a single
frequency/narrow bandwidth
Easy to build and use
The input impedance, Zin (looking into the quarter
wave long section of lossless ZS impedance line
terminated in a resistive load RL):
3/1/2018 Dr. Omar R Daoud 35
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Quarter-wave transformer.
ljRZ
ljZRZZ
LS
SLSin
tan
tan
, 24
2
l
l tan
LS RZZ 0
Transmission Lines
Smith Chart: Impedance Matching
Quarter Wave Transformer
Example
Calculate the position and characteristic impedance of
a quarter wave transformer that will match a load
impedance, RL = 15Ω; to a 50 Ω line.
The transformer’s impedance
the position of quarter-wave transformer from the
load:
d = 0.25 λ
3/1/2018 Dr. Omar R Daoud 36
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Quarter-wave transformer.
39.27)15)(50(0 LS RZZ
Transmission Lines
Smith Chart: Impedance Matching
Quarter Wave Transformer
Example
A transistor has an input impedance of ZL = 25 Ω, at
an operating frequency of 500 MHz.
Assume:
Thickness of the dielectric is d = 1 mm,
The relative dielectric constant εr is 4.
The surface resistance, R and shunt conductance, G, are
be neglected.
Find:
The length, l
The width, w
The characteristic impedance of the quarter-wave
parallel plate line transformer for which matching is
achieved.
3/1/2018 Dr. Omar R Daoud 37
Transmission Lines Smith Chart: Impedance Matching
Quarter Wave Transformer
Example
ZL = 25 Ω, f = 500 MHz, d= 1mm, εr = 4, and R=G=0
The characteristic of the line is
From the table,
The line length l follows from the condition
The input impedance of the combined transmission line and
the load is:
where d = l = λ/4 and the reflection coefficient is
3/1/2018 Dr. Omar R Daoud 38
355.35)25)(50(0 LS RZZ
w
d
C
LZ
p
Line mmZ
dw
rline
p329.5
0
0
mnHw
dL
p/8.235
mpFd
wC
p
/6.188
o = 8.85 x 10-12 F/m
µo = 4 x 10-7 H/m
mmLCf
l 967.744
1
4
)(1
)(1
tan
tan
d
dZ
djZZ
djZZZZ line
Lline
lineLlinein
d
v
fj
ZZ
ZZed
plineL
lineLdj 22exp)( 2
0
Transmission Lines
Smith Chart: Impedance Matching
Stub Matching
Shunt stub matching network
The matching network has to
transform the real part of load
impedance, RL to Z0 and reactive
part, XL to zero Use two adjustable
parameters – e.g. shunt-stub. Thus,
The main idea of shunt stub matching
network is to:
Find length d and l in order to get yd
and yl .
Ensure total admittance ytot = yd + yl =
1 for complete matching network.
3/1/2018 Dr. Omar R Daoud 39
YLY0
d
lY=1/Z
Y0
Open
or
shorted
stub(a)
YLZ0
d
lZ=1/Y
Z0
Open
or
shorted
stub(b)
Z0
Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.
Transmission Lines
Smith Chart: Impedance Matching
Stub Matching
Shunt stub matching network using
Smith Chart
Locate the normalized load impedance ZNL.
Draw constant SWR circle and locate YNL.
Move clockwise (WTG) along circle to
intersect with 1 ± jB value of yd.
The length moved from YNL towards yd is the
through line length, d.
Locate yl at the point jB .
Depends on the shorted/open stub, move
along the periphery of the chart towards yl
(WTG).
The distance traveled is the length of stub, l .
3/1/2018 Dr. Omar R Daoud 40
YLY0
d
lY=1/Z
Y0
Open
or
shorted
stub(a)
YLZ0
d
lZ=1/Y
Z0
Open
or
shorted
stub(b)
Z0
Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.
j
L e
Transmission Lines Smith Chart: Impedance Matching
Shunt stub matching network using Smith Chart
Example
Construct the shorted shunt stub matching network for a
50Ω line terminated in a load ZL = 20 – j55Ω
Locate the normalized load impedance,
ZNL = ZL/Z0 = 0.4 – j1.1Ω
Draw constant circle.
Locate YNL. (0.112λ at WTG)
Moving to the first intersection with the ± jB circle, which
is at 1 + j2.0 yd
Get the value of through line length, d from 0.112λ to
0.187λ on the WTG scale, so d = 0.075λ
Locate the location of short on the Smith Chart (note: when
short circuit, ZL = 0, hence YL = ∞) on the right side of the
chart with WTG=0.25λ.
Move clockwise (WTG) until point jB, which is at 0 - j2.0,
located at WTG= 0.324λ yl
Determine the stub length, l 0.324λ – 0.25λ = 0.074 λ
3/1/2018 Dr. Omar R Daoud 41
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Figure 6-28bc (p. 302) (b) Adding shunt admittances. (c) Using the Smith Chart to find through line and
stub lengths. Values on the chart apply to Example 6.7.
j
L e
Thus, the values are:
d = 0.075 λ
l = 0.074 λ
yd = 1 + j2.0 Ω
yl = -j2.0 Ω
Where
YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1
Transmission Lines Smith Chart: Impedance Matching
Shunt stub matching network using Smith Chart
Example
Construct the open ended shunt stub matching network
for a 50Ω line terminated in a load ZL = 150 + j100Ω
Locate the normalized load impedance,
ZNL = ZL/Z0 = 3.0 + j2.0Ω
Draw constant circle.
Locate YNL. (0.474λ at WTG)
Moving to the first intersection with the ± jB circle, which
is at 1 + j1.6 yd
Get the value of through line length, d from 0.474λ to
0.178λ on the WTG scale, so d = 0.204λ
Locate the location of open end on the Smith Chart (note:
when open circuit, ZL = ∞, hence YL = 0) on the left side of
the chart with WTG=0.00λ.
Move clockwise (WTG) until point jB, which is at 0 – j1.6,
located at WTG= 0.339λ yl
Determine the stub length, l 0.339λ – 0.00λ = 0.339 λ
3/1/2018 Dr. Omar R Daoud 42
j
L e
Thus, the values are:
d = 0.204 λ
l = 0.339 λ
yd = 1 + j1.6 Ω
yl = -j1.6 Ω
Where
YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1
Transmission Lines Smith Chart: Impedance Matching
Shunt stub matching network using Smith Chart
Example
50-Ω transmission line is connected to an antenna with load
impedance ZL = (25 − j50)Ω. Find the position and length of
the short-circuited stub required to match the line.
The normalized load impedance is
Value of yL at B is which locates at position 0.115λ on the WTG scale.
Draw constant SWR circle that goes through points A and B.
There are two possible matching points, C and D where the
constant SWR circle intersects with circle rL=1 (now gL =1 circle).
First matching points, C.
At C, is at 0.178λ on WTG scale.
Distance B and C is d= (0.178λ- 0.115λ)= 0.063λ
Normalized input admittance
at the juncture is:
E is the admittance of short-circuit stub, yL=-j∞.
Normalized admittance of −j 1.58 at F and position 0.34λ on the
WTG scale gives: l1= (0.34λ- 0.25λ)= 0.09λ
3/1/2018 Dr. Omar R Daoud 43
jj
Z
Zz
5.0
50
5025
0
LL
(located at A).
8.04.0L jy
58.11d jy
58.1
58.1101
s
s
sin
jy
jyj
yyy d
Transmission Lines Smith Chart: Impedance Matching
Shunt stub matching network using Smith Chart
Example
50-Ω transmission line is connected to an antenna with load
impedance ZL = (25 − j50)Ω. Find the position and length of
the short-circuited stub required to match the line.
The normalized load impedance is
Value of yL at B is which locates at position 0.115λ on the WTG scale.
Draw constant SWR circle that goes through points A and B.
There are two possible matching points, C and D where the
constant SWR circle intersects with circle rL=1 (now gL =1 circle).
Second matching points, D.
At D, is at 0.322λ on WTG scale.
Distance B and D is d= (0.322- 0.115λ)= 0.207λ
Normalized input admittance at the juncture is: at G
Rotating from point E to point G, we get
3/1/2018 Dr. Omar R Daoud 44
jj
Z
Zz
5.0
50
5025
0
LL
(located at A).
8.04.0L jy
58.11 jyd
58.1s jy
41.016.025.02 l