ch2
DESCRIPTION
Ch2TRANSCRIPT
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EE 330
1 transmission lines
Transmission Lines
V)10sin(50)( 5 ttvs π=
l
What if this is 5 km ???
+vC
−2.5 H
0.1 F
6 Ω
3 Ω)(tvs+−~
6 Ω
+vC
−2.5 H
0.1 F 3 Ω+−~)(tvs
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EE 330
2 transmission lines
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EE 330
3 transmission lines
Lumped Element Model
Δ ΔΔ
Δ Δ
Δ Δ Δ Δ Δ
Δ Δ Δ Δ Δ Δ
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EE 330
4 transmission lines
Transmission lines transmit energy and signals from a source (generator) to a load
Distinguishing characteristics of a transmission line:-
the devices to be connected are separated by distances comparable to or larger than the signal wavelength-
the parameters of the circuit are distributed and are evaluated on a per-unit length basis: R' –
resistance per unit length, Ω/mL' –
inductance per unit length, H/mG' –
conductance per unit length, S/mC' –
capacitance per unit length, F/m-
transmission lines are circuit elements that have complex impedances, which are functions of line length and signal frequency
TEM transmission lines-
two-wire line-
coaxial cable-
parallel-plate line
01.0≥λl+
−∼Vg
Rg
ZL
generator transmission line load
fu p=λ
l
01.0≥λl+
−∼Vg
Rg
ZL
generator transmission line load
fu p=λ 01.0≥
λl+
−∼+−∼Vg
Rg
ZL
generator transmission line load
fu p=λ
l
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EE 330
5 transmission lines
coaxial line
two wire line
parallel plate
microstrip
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EE 330
6 transmission lines
Notice: εμ=''CL
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EE 330
7 transmission lines
Power loss: due to non-ideal conductors and dielectrics
+−∼Vg
Rg
ZL
zL Δ' zL Δ' zL Δ' zL Δ'
zC Δ' zC Δ' zC Δ' zC Δ'
z – l 0
Lossless transmission line model (R' = 0, G' = 0)
Dispersion: due to frequency dependent properties of the materials
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EE 330
8 transmission lines
Examples:
1. Compute the line parameters for the following lossless transmission lines:
A. A coaxial cable with diameters of the concentric conductors 2a = 0.4 mm and 2b = 3 mm employs polyethylene as insulator between the conductors.
μH/m 4.04.0
3ln2
104ln2
'7
0 =⎟⎠⎞
⎜⎝⎛×
=⎟⎠⎞
⎜⎝⎛=
−
ππ
πμμ
abL r
pF/m 9.9
6.010ln
10854.8
ln'
120 =
⎟⎠⎞
⎜⎝⎛
××=
⎟⎠⎞
⎜⎝⎛
=−πεεπ
ad
C r
12
2
>>⎟⎠⎞
⎜⎝⎛
ad
B. The conducting wires of a two-wire line have 1.2-mm diameter and are a distance of 1 cm from one another. The two wires are in air.
Assume
pF/m 62
4.03ln
10854.825.22
ln
2'12
0 =⎟⎠⎞
⎜⎝⎛
×××=
⎟⎠⎞
⎜⎝⎛
=−πεεπ
ab
C r
μH/m 1.16.0
10ln104ln'7
0 =⎟⎠⎞
⎜⎝⎛×
=⎟⎠⎞
⎜⎝⎛=
−
ππ
πμμ
adL r
0'=R
0'=G
0'=R
0'=G
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EE 330
9 transmission lines
Examples:
2. Compute the line parameters at 1 MHz for a coaxial air line with
an inner conductor diameter of 0.6 cm and an outer conductor diameter of 1.2 cm. The conductors are made of copper. Assume the air has zero conductivity.
Ω×=×
×××== −
−4
7
760 106.2
108.510410 ππ
σμμπ
c
rcs
fR
/m1008.2006.01
003.01
2106.211
2' 2
4
Ω×=⎟⎠⎞
⎜⎝⎛ +
×=⎟
⎠⎞
⎜⎝⎛ += −
−
ππ baRR s
μH/m 14.06.02.1ln
2104ln
2'
70 =⎟
⎠⎞
⎜⎝⎛×
=⎟⎠⎞
⎜⎝⎛=
−
ππ
πμμ
abL r
pF/m 26.80
6.02.1ln
10854.82
ln
2'12
0 =⎟⎠⎞
⎜⎝⎛
××=
⎟⎠⎞
⎜⎝⎛
=−πεεπ
ab
C r
0 todue 0' == σG
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EE 330
10 transmission lines
Generalized Transmission Line
For AC currents and voltages, using phasor
notations and applying Kirchhoff’s voltage and current laws to a section of the line:
+
v(z,t)
–
zL Δ'
zC Δ'
Δz
+
v(z + Δz,t)
–
i(z,t) i(z + Δz,t)
Phasor
notations:
( ) ( ) ]~Re[, tjezVtzv ω=
( ) ( ) ]~Re[, tjezItzi ω=
( ) ( ) ( )
( ) ( ) ( )zVCjGtdzId
zILjRtdzVd
~''~
~''~
ω
ω
+=−
+=−
telegrapher’s equations
Transmission line equations:
KVL, KCL
zR Δ'
zG Δ'
Δ ΔΔ
Δ Δ
Δ Δ Δ Δ Δ
Δ Δ Δ Δ Δ Δ
Δ ΔΔ
Δ Δ
Δ Δ Δ Δ Δ
Δ Δ Δ Δ Δ Δ
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EE 330
11 transmission lines
Upon uncoupling the telegrapher’s equations, two second order differential equations are obtained: ( ) ( )
( ) ( ) 0~~
0~~
22
2
22
2
=−
=−
zItd
zId
zVtd
zVd
γ
γ
wave equations
( )( )'''' CjGLjRj ωωβαγ ++=+=
]Im[γβ =
– complex propagation constant
– phase constant, rad/m
The general solution to the wave equation is a sum of two waves,
one traveling from the generator toward the load and another one traveling from the load toward the generator:
( )( ) zz
zz
eIeIzI
eVeVzVγγ
γγ
−−+
−−+
+=
+=
00
00~
~
]Re[γα = – attenuation constant, Np/m
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EE 330
12 transmission lines
Traveling waves parameters:
amplitude:
frequency:
phase velocity on the line:
wavelength on the line:
zz eVeV αα −−+00 ,
πω2
=f
βω
=pu
fu p=λ
( ) ( ) ( )−−+−+ ++++−= φβωφβω αα zteVzteVtzv zz coscos, 00
( ) zjzjzjzj eeeVeeeVzV βαφβαφ −+ −−−+ += 00~
Phasor
notation :
incident wave,
reflected wave,
Instantaneous function (time domain):
0 z
Zg
+−∼ ZLgV~
TL
( )tv+ ( )tv−
( )tv+ ( )tv−
( ) zz eVeVzV γγ −−+ += 00~
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EE 330
13 transmission lines
Current in terms of voltage (using the solutions to the wave equations into the telegrapher’s equations):
( ) ( )
zz
zz
eZVe
ZV
eVeVLjR
zI
γγ
γγ
ωγ
0
0
0
0
00''~
−−
+
−−+
−=
=−+
=
where
0
0
0
0
0
''''
''
ZCjGLjR
LjRIV
IV
=++
=
=+
=−
= −
−
+
+
ωω
γω
– characteristic impedance of the line, Ω
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EE 330
14 transmission lines
A lossless coaxial cable with diameters of the concentric conductors 2a = 0.4 mm and 2b = 3 mm employs polyethylene as insulator between the conductors.
μH/m 4.0'=L
12
2
>>⎟⎠⎞
⎜⎝⎛
ad
The conducting wires of a two-wire lossless line have 1.2-mm diameter and are a distance of 1 cm from one another. The two wires are in air.
Assume
pF/m 62'=CFrom before: 0'=R 0'=G
Ω==++
= 3.80''
''''
0 CL
CjGLjRZ
ωω
μH/m 1.1'=L pF/m 9.9'=CFrom before: 0'=R 0'=G
Ω==++
= 333''
''''
0 CL
CjGLjRZ
ωω
Examples:
3. Compute the characteristic impedance of the transmission lines from examples 1 and 2.
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EE 330
15 transmission lines
Signal frequency of 1 MHz, a coaxial air line with an inner conductor diameter of 0.6 cm and an outer conductor diameter of 1.2 cm. The conductors
are made of copper. Assume the air has zero conductivity.
μH/m 14.0'=L pF/m 26.80'=CFrom before: Ω×= −21008.2'R 0'=G
Ω°−∠=×
=×
+×=
=×××
×××+×=
++
=
°−
°
−
−
−
−−
35.14210588.0
10588.01008.2
1026.801021014.01021008.2
''''
904
65.88
4
2
126
662
0
j
j
ee
jj
jj
CjGLjRZ
ππ
ωω
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EE 330
16 transmission lines
Lossless Transmission Line
''''2 CLjCL ωωγ =−=
εμωωγβ === '']Im[ CL
Complex propagation constant:
Phase constant:
Attenuation constant: 0]Re[ == γα
''
0 CLZ =Characteristic impedance:
εμβω 1
''1
===CL
u pPhase velocity:
εμωπ
ωπ
βπλ 2
''22
===CL
Wavelength:
0'=R 0'=G
(real quantity)
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EE 330
17 transmission lines
H/m]['L
F/m]['C
][''
0 Ω=CLZ
parameter
coaxial line two wire line parallel plate
( )ad 2>>
⎟⎠⎞
⎜⎝⎛
abln
2πμ
( )abln2 επ
wdη
wdμ
dwε
⎟⎠⎞
⎜⎝⎛
abln
2πη
⎟⎠⎞
⎜⎝⎛
adln
πμ
( )adlnεπ
⎟⎠⎞
⎜⎝⎛
adln
πη
εμη = – intrinsic impedance of the dielectric
material between the two conductors εμ=''CL
Lossless TEM Transmission Lines
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EE 330
18 transmission lines
Examples:
4. Design a parallel-plate transmission line according to the following requirements: (a) characteristic impedance of 50 Ω; (b) capacitance per unit length of 100 pF/m. You have at your disposal three dielectric materials, namely, teflon, polyethylene, and polystyrene, in the form of 1.9-mm thick strips. Assume the insulating materials are perfect dielectrics and the metal plates are perfect conductors.
μH/m 25.05010100'''' 2122
00 =××==⇒= −ZCLCLZ
wide.mm 9.6 bemust plates theand nepolyethyle bemust material dielectric The2.6 :epolystyren 2.25 : nepolyethyle 2.1 :teflon === rrr εεε
mm6.91025.0
0019.0104'
' 67
00 Ldw
wdL r =
××==⇒= −
−πμμμ
25.20096.010854.8
0019.010100'' 12
12
00 =
××××
==⇒= −
−
wdC
dwC rr ε
εεε
wd
CLZ η==
''
0 wdL μ='
dwC ε='
0εεε r=00 μμμμ == r
1material?
size?
?=rε
?=w
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EE 330
19 transmission lines
Generalized impedance on the line:
( ) ( )( ) ⎥
⎦
⎤⎢⎣
⎡−+
== −−+
−−+
zjzj
zjzj
eVeVeVeVZ
zIzVzZ ββ
ββ
00
000~
~
At z = 0, load impedance: ( ) 000
000 ZVVVVZZL ⎟⎟
⎠
⎞⎜⎜⎝
⎛−+
== −+
−+
At z = –
l, input impedance: ( )lZZin −=
+−⎟⎟⎠
⎞⎜⎜⎝
⎛+−
= 00
00 V
ZZZZV
L
L
Reflected wave amplitude in terms of incident wave amplitude:
Note: No reflected wave ( ), when !!!00 =−V 0ZZL =
( )
( ) zjzj
zjzj
eZVe
ZVzI
eVeVzV
ββ
ββ
0
0
0
0
00
~
~
−−
+
−−+
−=
+=
0 z
+−∼
z = –
l
( )zZ LZinZ
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EE 330
20 transmission lines
Relating the reflection coefficient to the load impedance through and using Euler’s identity and some trigonometry:
( )( )⎥⎦
⎤⎢⎣
⎡++
=lZjZlZjZZZ
L
Lin β
βtantan
0
00
+−⎟⎟⎠
⎞⎜⎜⎝
⎛+−
= 00
00 V
ZZZZV
L
L
( ) ( )( )⎥⎦
⎤⎢⎣
⎡−+−+
=zZjZzZjZZzZ
L
L
ββ
tantan
0
00
A distance from the load:zl −=
( ) ⎥⎦
⎤⎢⎣
⎡
−+
= −−+
−−+
zjzj
zjzj
eVeVeVeVZzZ ββ
ββ
00
000
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EE 330
21 transmission lines
–
reflection coefficient in terms of load impedance
0
0
ZZZZ
L
L
+−
=Γ
Generally, the load impedance is a complex number, therefore the reflection coefficient is also a complex number:
ΓΓ=Γ θje
Note: 1≤Γ
+−⎟⎟⎠
⎞⎜⎜⎝
⎛+−
= 00
00 V
ZZZZV
L
L
Voltage reflection coefficient at the load:
+
−
=−+
−
==Γ0
0
00
0
VV
eVeV
zzj
zj
β
β
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EE 330
22 transmission lines
( ) ( )( ) ( )zjzjzjzj
zjzjzjzj
eeZVe
ZVe
ZVzI
eeVeVeVzV
ββββ
ββββ
Γ−=−=
Γ+=+=
−+−
−+
−+−−+
0
0
0
0
0
0
000
~
~
( ) ( )( )
( )( ) ⎟⎟
⎠
⎞⎜⎜⎝
⎛Γ−Γ+
=⎥⎦
⎤⎢⎣
⎡Γ−Γ+
== −+
−+
zj
zj
zjzj
zjzj
eeZ
eeVeeVZ
zIzVzZ β
β
ββ
ββ
2
2
00
00 1
1~~
Voltage and current in terms of reflection coefficient:
Impedance in terms of reflection coefficient:
At z = – l, ( ) ( )( ) ⎥
⎦
⎤⎢⎣
⎡Γ−Γ+
=−−
=−= −
−
lj
lj
in eeZ
lIlVlZZ β
β
2
2
0 11
~~
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EE 330
23 transmission lines
Examples:
5. A 75-Ω
transmission line is terminated in two different loads: (a) ZL = 75 Ω, and (b) open circuit. Find the corresponding reflection coefficients
at the load.
line matched075757575a)
0
0
⇒=+−
=Γ
+−
=ΓZZZZ
L
L
111b)
0
0 ⎯⎯ →⎯+−
=Γ ∞→LZ
L
L
ZZZZ
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EE 330
24 transmission lines
Examples:
6. The reflection coefficient at the load is measured when a 75-Ω
transmission line is terminated in three different loads. The corresponding values of
Γ are as follows: (a) Γ
= –
j0.5, (b) Γ
= –
1, (c) Γ
= 0.5. Find the load impedances.
Ω=−+
×= 2255.015.0175c) LZ
Γ−Γ+
=11
0ZZL
Ω−==+−
×= °− 6045755.015.0175a) 13.53 je
jjZ j
L
circuitshort 0111175b) ⇒Ω=
+−
×=LZ
0
0
ZZZZ
L
L
+−
=Γ
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EE 330
25 transmission lines
Power flow:
( ) ( )( ) ( )zjzj
zjzj
eeZVzI
eeVzV
ββ
ββ
Γ−=
Γ+=
−+
−+
0
0
0
~
~
Instantaneous incident power at the load (z = 0):
( ) ( ) ( ) [ ] ( )+++
+++ +=⎥⎥⎦
⎤
⎢⎢⎣
⎡==
++
φωωφωφ tZ
Vee
ZV
eeVtitvtP tjjtjji 2
0
2
0
0
00 cosReRe
Instantaneous reflected power at the load (z = 0):
( ) ( ) ( ) [ ] =⎥⎥
⎦
⎤
⎢⎢
⎣
⎡Γ−Γ==
+Γ
+Γ
++−− tjjjtjjjr ee
Z
VeeeVetitvtP ωφθωφθ
0
00 ReRe
( )+Γ
+
++Γ−= φθωtZ
V2
0
202 cos
0 z
Zg
+−∼ ZLgV~ ( )tPi ( )tPr
– l
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EE 330
26 transmission lines
Time-average incident power at the load (z = 0):
( )0
20
0 21
Z
VtdtP
TP
Tii
av
+
== ∫
Time-average reflected power at the load (z = 0):
( ) iav
Trr
av PZ
VtdtP
TP 2
0
202
0 21
Γ−=Γ−==+
∫
The same result can be obtained easier using phasors. The average value of a product of two time-harmonic quantities is:
*~~Re21 IVPav ⋅=
Power delivered to the load: rav
iavav PPP +=
Percentage of the incident power delivered to the load:
211 Γ−=+=+
= iav
rav
iav
rav
iav
iav
av
PP
PPP
PP
10 ≤≤ iav
av
PP
Note:
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EE 330
27 transmission lines
Examples:
7. A load impedance ZL = 80 –
j100 Ω
terminates a 50-Ω
line. The operating frequency is 100 MHz and the wavelength on the line is 2 m. Find (a) the phase velocity of the wave on the line, and (b) the percentage of power absorbed by the load.
%4.59594.0637.011637.0
637.050100805010080b)
22
7.35
0
0
==−=Γ−=⇒=Γ
=+−−−
=+−
=Γ °−
iav
av
j
L
L
PP
ejj
ZZZZ
8. Consider the transmission line from the previous example and find what percentage of the incident power is delivered to the load when the line is terminated in a load impedance ZL = Z0
=
50 Ω.
%100110 2
0
0 ==Γ−=⇒=+−
=Γ iav
av
L
L
PP
ZZZZ
m/s 10221010022a) 86 ×=××==== λ
λππ
βω ffu p
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EE 330
28 transmission lines
Examples:
9. A voltage generator with internal impedance Zg = 100 Ω
is connected to a λ/8-long 50-Ω
transmission line. A load impedance ZL = 50 + j50 Ω
terminates the line. Find
(a)
the input impedance,
( ) ( )V 90120sin220 °−−= ttvg π
( )( )
( ) ( ) Ω−=⎥⎦
⎤⎢⎣
⎡++++
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ ×++
⎟⎠⎞
⎜⎝⎛ ×++
=
=⎥⎦
⎤⎢⎣
⎡++
=
5010050505050505050
82tan505050
82tan505050
50
tantan
0
00
jjjjj
jj
jj
lZjZlZjZZZ
L
Lin
λλπ
λλπ
ββ
Zin
0 z
Zg
+−∼ ZLgV~
– l
Z0
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(b)
the reflection coefficient at the load,
4.02.0505050505050
0
0 jjj
ZZZZ
L
L +=++−+
=+−
=Γ
(c)
the phasor
of the input voltage
( )
( ) V3.11950100100
50100220
~~~
5.12 °−=−+
−=
=+
=−=
j
ing
ingi
ej
j
ZZZV
lVV
Zin
0 z
Zg
+−∼ ZLgV~
– l
−
+
iV~
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(d)
the average power delivered to the load.
*~~Re21
LLav IVP ⋅= ( ) ( ) ( ) ( )Γ−==Γ+==+
+ 10~~,10~~
0
00 Z
VIIVVV LL
( ) ljljiljlj
i eeVVeeVV ββ
ββ−
+−+
Γ+=⇒Γ+=
~~00
2.
Plug in into the expressions about and +0V LI~LV~
W571.161Re215.14.107Re
21 457631 === °°°− jjj
av eeeP
Analysis:
Computation:
4.
Plug in and into the expression about the average powerLV~ ∗LI~
3.
Take the complex conjugate of LI~
1.
Compute +0V
( )V2.85
4.13.119
4.02.0
3.119~4.49
9.36
5.12
44
5.12
0°−
°
°−
−
°−
−+ ==
++
=Γ+
= jj
j
jj
j
ljlji e
ee
eje
eee
VV ππββ
( ) ( )
( ) ( ) A5.150
4.02.012.851~
V4.1074.02.012.851~
764.49
0
0
314.490
°−°−+
°−°−+
=−−
=Γ−=
=++=Γ+=
jj
L
jjL
ejeZVI
ejeVV
A5.1~ 76* °= jL eI
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Standing WavesThe interference of the incident and the reflected voltage waves
gives rise to a pattern called standing wave, that is, a wave-like spatial distribution of the voltage magnitude on the transmission line.
( )zV~
z
2λ
4λ
minl
maxl
2λ
( )max
~ zV
( )min
~ zV
( )zI~
z
2λ
4λ
maxl
minl
2λ
( )max
~ zI
( )min
~ zI
This “wave”
does not travel along the transmission line, it is “frozen”.Two neighboring maximums (or minimums) are spaced by a half-
wavelength.
A similar pattern is formed by the interfering current waves.
The two patterns are in phase opposition.
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disturbance amplitude
Standing wave on a string
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( ) ( ) ( ) ( )Γ+∗ +Γ+Γ+== θβ zVzVzVzV 2cos21~~~ 2
0
Mathematically, the voltage standing wave pattern is described by
when( ) ( )Γ+= + 1~0max
VzV ( ) 12cos =+ Γθβ zπθβ nz 22 −=+ Γ
⎩⎨⎧
<=≥=
+=+
==−Γ
ΓΓΓ
0if...2,10if...2,1,0
,242
2max θ
θλπθλ
βπθ
nnnnlz
when( ) ( )Γ−= + 1~0min
VzV ( ) 12cos −=+ Γθβ z( )πθβ 122 ±−=+ Γ nz
⎩⎨⎧
≥−<+
±=±+==− Γ
4if""4if""
,4424 max
maxmaxmin λ
λλλλπθλ
ll
lnlz
Voltage standing wave ratio (VSWR): ( )( ) Γ−
Γ+==
11
~
~
min
max
zV
zVS
The extrema
in the pattern are related to the magnitude of Γ, and their positions are related to the phase of Γ. Therefore, from the standing wave pattern, the reflection coefficient can be constructed and then the load impedance and the power delivered to the load can be found.
1≥S
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Computing Γ from the standing wave pattern:1.
Compute the VSWR by taking the ratio of the maximum and minimum values of the voltage2.
Compute the magnitude of Γ from VSWR.3.
Determine the distance from the load in wavelengths of the nearest maximum or minimum.4.
Find the phase of Γ, θΓ
, using the result from step 3.5.
Construct the reflection coefficient in polar form from the results from steps 2 and 4.
Plotting the standing wave pattern from Γ:1.
Compute VSWR from the magnitude of Γ. If the amplitude of the incident wave is given, compute the actual values of minimum and maximum voltage on the line.
2.
Determine the distance from the load in wavelengths of the nearest maximum or minimum from the phase of Γ. If the signal wavelength on the line is given or can be determined from the data, compute the actual physical distance from the load.
3.
Draw a coordinate system in which the load is placed at z = 0 and the transmission line occupies the negative values of z.
4.
Plot the standing wave pattern starting from the load and moving
to the left where
z < 0. In steps of λ/4 mark the positions of several minimums and maximums starting from the position determined in step 2 and altering maximums and minimums. Make sure your pattern goes up to the maximum voltage value at the positions of
the maximums and down to the minimum voltage value at the positions of the minimums. Note: If you do not know the amplitude of the incident voltage wave and the wavelength on the
line, draw a normalized pattern, i.e., in terms of incident voltage amplitude and wavelength.
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Examples:
10. Determine the reflection coefficient at the load from the measured voltage standing wave pattern.
–33.5
4
6
V [V]
z[cm]–83.5
( )( )
5.146
~
~
min
max ===zV
zVS
2.051
15.115.1
11
==+−
=+−
=ΓSS
m2m5.04
m835.0 m,335.0 minmax =⇒=⇒== λλll
°==⎟⎠⎞
⎜⎝⎛ −=⇒+= Γ
Γ 6.12067.02
424 maxmax πλ
λπθλ
πθλ nlnl
πθ 67.02.0 jj ee =Γ=Γ Γ
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Examples:11. Plot the normalized voltage standing wave pattern if a 50-Ω
line is terminated in a load ZL = 40 + j50 Ω.
rad 26.1,495.0495.0505040505040 26.1
0
0 ==Γ⇒=++−+
=+−
=Γ Γθj
L
L ejj
ZZZZ
( ) ( ) ++ =Γ+= 00max495.11~ VVzV ( ) ( ) ++ =Γ−= 00min
505.01~ VVzV
K
K
,2,1,0,2
35.025.0
,2,1,0,2
1.024
26.124
maxmin
max
=⎟⎠⎞
⎜⎝⎛ +=+=
=⎟⎠⎞
⎜⎝⎛ +=⎟
⎠⎞
⎜⎝⎛ +=⎟⎟
⎠
⎞⎜⎜⎝
⎛+= Γ
nnll
nnnnl
λλ
λλπ
λπ
θ
–0.1
0.505
1.495
–0.35
( )+
0
~
V
zV
λz–0.85 –0.6
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Examples:
12. When a transmission line is terminated in load A, the voltage standing wave ratio is S = 5.8. When the same line is terminated in load B, it is S = 1.5. What percentage of the incident power is delivered to load in these two cases?
22
1111 ⎟
⎠⎞
⎜⎝⎛
+−
−=Γ−=SS
PP
iav
av
%505.018.518.51 :A Load
2
==⎟⎠⎞
⎜⎝⎛
+−
−=iav
av
PP
%9696.015.115.11 :B Load
2
==⎟⎠⎞
⎜⎝⎛
+−
−=iav
av
PP
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A transmission line is used to deliver signals to a load.
The reflection coefficient at the load is
What is the voltage standing wave ratio on the line?
2
31 πj
e−=Γ
A.
2B.
1C.
1/2D.
0.8 + j0.6E.
0.8 –
j0.6
Probing Question
2311311
11
31
=−+
=Γ−Γ+
=⇒=Γ S
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Special Cases
Short-circuited line: 0=LZ0Z
z2λ
−4λ
−
+02V
( ) ( ) ,21~00max
++ =Γ+= VVzV
( ) ( ) ,01~0min
=Γ−= +VzV
24maxmaxλλ nlz +==−
2minminλnlz ==−
∞=Γ−Γ+
=11
S
10
0 −=+−
=ΓZZZZ
L
L πθ ==Γ Γ,1
01 2 =Γ−=iav
av
PP
All the power is reflected back toward the generator.
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( ) ( ) XjzZjzZ sc =−= βtan0
The impedance at any point on the line is pure imaginary,i.e., reactive (either inductive or capacitive):
X
X < 0, capacitive
X > 0, inductiveX = 0, short circuit
X = ∞, open circuit
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Open-circuited line: ∞=LZ0Z
( ) ( ) ,21~00max
++ =Γ+= VVzV
( ) ( ) ,01~0min
=Γ−= +VzV24minλλ nz +=−
2maxλnz =−
∞=Γ−Γ+
=11
S
111
0
0 =+−
=ΓL
L
ZZZZ 0,1 ==Γ Γθ
01 2 =Γ−=iav
av
PP
All the power is reflected back toward the generator.
z2λ
−4λ
−
+02V
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The impedance at any point on the line is pure imaginary,i.e., reactive (either inductive or capacitive):
X
X < 0, capacitive
X > 0, inductive
X = 0, short circuit
X = ∞, open circuit
( ) ( ) Xjz
ZjzZ oc =−
−=βtan
0
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Matched line: 0ZZL =0Z
111
=Γ−Γ+
=S
00
0 =+−
=ΓZZZZ
L
L 0,0 ==Γ Γθ
11 2 =Γ−=iav
av
PP
All the power is absorbed by the load.
( ) 0ZzZZin == –
the impedance at any point on the line is resistiveand equals the characteristic impedance.
( ) ( ) +Γ
+ =+Γ+Γ+= 02
0 2cos21~ VzVzV θβ
z
+0V
–
There is no reflected wave.
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Examples:
13. The input impedance of a 75-Ω
open circuited transmission line is Zin = j25 Ω. Determine the length of the line in terms of the signal wavelength.
( ) ⎟⎠⎞
⎜⎝⎛−=−=
λπβ lZjlZjZ oc
in 2cotcot 00
The line length can be any length that satisfies the above expression, the shortest being 0.3λ for n = 1.
K,2,1,0,2
2.023
1cot21
27525cot
21
2cot
21
cot2
11
0
1
0
1
=+−=+⎟⎠⎞
⎜⎝⎛−=+⎟
⎠⎞
⎜⎝⎛=+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+⎟⎟⎠
⎞⎜⎜⎝
⎛=
−−−
−
nnnnjjnZZjl
nZZjl
ocin
ocin
πππλ
πλ
π
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Examples:
14. The input impedance of a transmission line is j25 Ω when it is short circuited and –
j100 Ω
when it is open circuited. Determine the characteristic impedance line.
( )( )lZjZ
lZjZocin
scin
β
β
cot
tan
0
0
−=
=
( ) Ω=−== 50100250 jjZZZ ocin
scin
( )[ ] ( )[ ] 2000 cottan ZlZjlZjZZ oc
inscin =−= ββ
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Examples:
15. An air-filled 2-m long lossless coaxial cable is short circuited. Determine what
type of load (capacitive, inductive, resistive, short circuit, open circuit) this cable is to a source generating signals of frequency (a) 18.75 MHz, (b) 37.5 MHz, (c) 56.25 MHz, (d) 75 MHz.
( )
( ) inductive 4
tan103
1075.1822tanMHz75.18
2tan2tantan
008
6
0
000
⇒=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
××××
=
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛==
ZjZjZjZ
cflZjlZjlZjZ
scin
scin
ππ
πλπβ
(a)
(b)
(c)
(d)
( ) circuitopen 2
tan103
105.3722tanMHz5.37 08
6
0 ⇒∞=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
××××
=ππ ZjZjZ sc
in
( ) capacitive 4
3tan103
1025.5622tanMHz25.56 008
6
0 ⇒−=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
××××
= ZjZjZjZ scin
ππ
( ) ( ) circuitshort 0tan103
107522tanMHz75 08
6
0 ⇒==⎟⎟⎠
⎞⎜⎜⎝
⎛
××××
= ππ ZjZjZ scin
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Smith’s ChartWhat is Smith’s chart? A graphical tool.
Purpose of introducing Smith’s chart: To avoid heavy math and speed up the process of analyzing and designing transmission line circuits.
How is it constructed? On the basis of the reflection coefficient plane.
Γ=ΓΓ=ΓΓ+Γ=Γ=Γ Γ Im and Re where, irirj je θ
Γr
Γi
1–1
1
–1
Only a circle of radius 1 has a meaning since
Any point within the unit circle represents certain reflection coefficient.
Smith’s chart works with normalized impedances and admittances:
1≤Γ
( ) ( )0ZzZzz =
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LLL
L xjrZZz +=
Γ−Γ+
==11
0
The normalized load impedance has positive resistive part and reactive part that can be either positive or negative:
Equating the real and the imaginary parts on both sides of the upper equation, two parametric equations are obtained:
22
2
11
1 ⎟⎟⎠
⎞⎜⎜⎝
⎛+
=Γ+⎟⎟⎠
⎞⎜⎜⎝
⎛+
−ΓL
iL
Lr rr
r
( )22
2 111 ⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛−Γ+−Γ
LLir xx
These two parametric equations describe two families of circles:
The first family consists of circles centered on the Γr –
axis and represent the value of the normalized resistance, rL .
Γr
Γi
1–1
1
–1The second family consists of circles centered on vertical line at Γr = 1 and represent the value of the normalized reactance, xL .
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The load impedance can be represented on the Smith’s chart by a point, at which two circles intersect: the circle corresponding to rL and the circle corresponding to xL .
Examples:15. Determine the load impedance denoted by the red star on the chart if this load terminates a 50-Ω
transmission line.
Ω−== 40200 jZzZ LL
The point representszL = 0.4 –
j0.8 Then
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Probing Questi
Which one of the points on the Smith's chart represents the load impedance of an open-circuited transmission line?
A
B
C
D E
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|Γ|
If a generalized reflection coefficient at an arbitrary point on
the transmission line is defined
as
, then the normalized impedance at that point can be expressed through the generalized
reflection coefficient the same way the normalized load impedance is expressed in terms of the reflection coefficient at the load:
( ) ( )zj rez βθ 2+Γ=Γ
( ) ( ) ( )( )zz
ee
ZzZzz zj
zj
Γ−Γ+
=Γ−Γ+
==11
11
2
2
0β
β
This observation suggests that Smith’s chart can be used to represent the impedance at any point on the line.The reflection coefficient along a particular transmission line maintains its magnitude; only its phase changes. Therefore, the normalized impedance at any point on the transmission line must lie on a circle with radius
.Also, it is obvious that the impedance on the line repeats itself every λ/2 length of the line. Therefore, the Smith chart represents a λ/2-long section of a transmission line.
Γ
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How to find the impedance at a point on the line that is a certain distance Δl from the load?
1.
Normalize the load impedance.2.
Normalize the distance, i.e., express Δl in terms of λ: ln = Δl/λ 3.
Mark on the chart the point representing the load impedance.4.
Draw the |Γ|-circle through that point. 5.
Draw a line through the load impedance point and the center of the chart.6.
Read the relative position of the load on the outer scale of the
chart.7.
Move along the outer scale in clockwise direction (toward the generator) a distance of ln and draw a line through that point on the outer scale and the center of the chart.
8.
Mark the point at which the line intersects the |Γ|-circle.9.
Read the resistive and the reactive parts of the impedance using
the corresponding families of circles. The result is the normalized impedance a distance of Δl from the load.
10.
To obtain the actual impedance value, multiply by the characteristic impedance.
How to find the impedance at a point on the line that is a certain distance Δl from the line input?
Perform the same steps as above with the following differences:1.
Start from the input impedance instead of the load impedance.2.
Move along the second outer scale in counterclockwise direction (toward the load).
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Examples:16. Consider the load impedance from the previous example. Determine the impedance 20 cm away from the load if the signal wavelength on the line is 1 m.
|Γ|
lnzL
z(– ln )
relative position of the load
relative position of the point of interest
zL = 0.4 –
j0.8
2.012.0cm 20 ==
Δ=⇒=Δ
λlll n
Rel. pos. z = 0.385 + 0.2 = 0.585
( ) ( ) Ω+=−=Δ− 5.275.150 jlzZlZ n
z(–ln ) = 0.31 + j0.55
Rel. pos. load = 0.385
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How to find the admittance that corresponds to certain impedance?
( ) ( )n
n
lj
lj
lj
lj
n ee
ee
ZlZlz π
π
β
β
4
4
2
2
0 11
11
−
−
−
−
Γ−Γ+
=Γ−Γ+
=−
=−
Consider the normalized impedance and admittance a distance l from the load:
( ) ( )( )
( ) ⎟⎠⎞
⎜⎝⎛ +=
Γ−
Γ+=
Γ−Γ+
=Γ+Γ−
=−
=−⎟⎠⎞
⎜⎝⎛ +−
⎟⎠⎞
⎜⎝⎛ +−
+−
+−
−
−
41
1
111
111
414
414
2
2
2
2
nlj
lj
lj
lj
lj
lj
nn lz
e
eee
ee
lzly
n
n
π
π
πβ
πβ
β
β
Notice that the admittance at certain point on the line equals the impedance a quarter wavelength away. Then, admittances also can be represented by points on Smith’s chart by simply adding a quarter wavelength to the actual position of interest. That is, the impedance and the admittance at the same position on the transmission line are represented by two points that are diametrically opposite to each other on the |Γ|-circle.
To summarize, Smith’s chart can be used to display normalized admittances. In that case, the r-circles play the role of normalized conductance circles (g-circles), and the x-
circles are considered normalized susceptance
circles (b-circles).
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zL
yL
Examples:17. Consider a 50-Ω
transmission line. Determine the load admittance if the load impedance is denoted by the red star on the chart.
1
00
02.001.0505.0
15.0
−Ω+=+
=
===
+=
jjZ
yYyY
jy
LLL
L
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Single-Stub Impedance Matching
The power absorbed by the load is maximum (100% of the incident power) when the transmission line is matched, i.e., the load impedance equals the characteristic impedance, ZL = Z0
.Unfortunately, oftenFortunately, there are means to match the load to the transmission line via impedance-matching network. One of the possibilities is: Single-stub matching using a short-circuited section of a transmission line.Essentially, a short-circuited section of the same type transmission line as the feeding (main) line is connected in parallel to the main line at
a position that is relatively close to the load. The point of connection and the length of the short-
circuited stub can be chosen in such a way so that to ensure an input impedance Z(–
d) = Z0
of the matching network that contains the load.
0ZZL ≠
LZ0Z
d
l
( )dZ −
matching network
feeding line
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Single-stub matching network: Analysis
To use Smith’s chart, we must work with normalized quantities. Then, what is needed is z(–
d) = 1.
Because the stub is connected in parallel to the feeding line, it is easier to work with admittances instead of impedances. The required normalized input admittance is also equal to one:
y(– d) = 1
Note that the required admittance is entirely real and a short-circuited line has a pure imaginary admittance. Then, the shorted stub can be used to compensate for the non-zero susceptance
of the feeding network.
What must be done? –
Find a point on the feeding line where yd = 1 + jb. Then, at that point, connect in parallel a shorted piece of line
that has input admittance ystub = –
jb . Eventually, the total admittance at the point of connection becomes
y(–
d) = yd + ystub = 1
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Single-stub matching network: Analytical solution
Examples:18. Design a single-stub matching network for a 50-Ω
transmission line transmitting signals with λ
= 2 m on the line to a (20 –
j40)-Ω
load.
jjz
yjjzL
LL +=−
==⇒−=−
= 5.08.04.0
118.04.050
4020
( ) ( )( ) ( )
( )( )( ) ( )
( )( )[ ] ( ) ( )[ ]( ) ( )( ) ( ) ( )( ) bj
djddjddjddj
djddj
djj
djj
dyjdjydy
L
L
+=−−+−
−−++=
=+−++
=⎟⎠⎞
⎜⎝⎛ ×++
⎟⎠⎞
⎜⎝⎛ ×++
=+
+=−
1tan5.0tan1tan5.0tan1
tan5.0tan1tan15.0
tan5.0tan1tan15.0
22tan5.01
22tan5.0
tan1tan
πππππππ
πππ
π
π
ββ
( ) ( ) ( ) ( )
( ) ( ) cm 8.7m 087.0279.0tan1 cm, 37.4m 374.0387.2tan1279.0tan,387.2tan05.0tan2tan75.0
12
11
212
======
==⇒=+−
−−
ππ
ππππ
dd
dddd
( ) ( )( ) ( ) ( )( )( )( ) ( )
1tan5.0tan1
tan1tan5.0tan15.0Re222 =
+−++−
=−dd
ddddyππ
πππ
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59 transmission lines
∞=stubLy ,
( ) ( )( ) ( )
( )( )( ) ( )
( )( )[ ] ( ) ( )[ ]( ) ( )( ) ( ) ( )( ) bj
djddjddjddj
djddj
djj
djj
dyjdjydy
L
L
+=−−+−
−−++=
+−++
=
=⎟⎠⎞
⎜⎝⎛ ×++
⎟⎠⎞
⎜⎝⎛ ×++
=+
+=−
1tan5.0tan1tan5.0tan1
tan5.0tan1tan15.0tan5.0tan1
tan15.02
2tan5.01
22tan5.0
tan1tan
πππππππ
πππ
π
π
ββ
( ) ( ) ( )( ) ( )( )( )( ) ( )
bdd
ddddy =+−
−++−=−
πππππ
222
2
tan5.0tan1tan1tan1tan5.0Im
( ) ( ) ( ) 58.1,For ,58.1,For 22,211,1 jlydjlydbjly stubstubstub −=−=−⇒−=−
( ) ( )( ) ( ) ( )⎟⎟
⎠
⎞⎜⎜⎝
⎛−
−=⇒
−=
⎟⎠⎞
⎜⎝⎛ ×
=+
+=− −
lyjl
lj
ljlyjljy
lystubstubL
stubLstub
1
,
, tan1tan
22tan
1tan1tan
πππββ
cm 18m 18.0,For ,cm 82m 82.0,For 2211 ==== lblb
To summarize, there are two distinct solutions to the line-matching problem:(1)
A shorted stub of length 82 cm connected in parallel to the feeding line 37.4 cm away from the load.(2)
A shorted stub of length 18 cm connected in parallel to the feeding line 8.7 cm away from the load.
The above extremely involving math analysis can be avoided by an
approximate graphical solution using Smith’s chart.
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Single-stub matching network design: Step by step using Smith’s chart
1.
Mark the normalized load impedance, zL , on Smith’s chart2.
Draw the |Γ|-circle.3.
Find the relative position of the normalized load admittance, yL , on Smith’s chart.4.
Mark the two points where the |Γ|-circle intersects the (g = 1)-circle. 5.
Choose one of the above points to be your point of connection. Usually, the better choice is the point that is closer to the load.
6.
Find the relative position of the point of connection.7.
Compute its normalized distance from yL .8.
Compute the actual distance of the connection point from the load, d. 9.
Read the susceptance
at the point of connection, b. The stub must have an admittance (–
jb).10.
Find the circle corresponding to susceptance
(–
b).11.
Mark the point where the (–
b)-circle intersects the |Γ|-circle. This point is the input of the stub, ystub .
12.
Read the relative position of ystub .13.
The stub is short-circuited, therefore its load admittance is yL, stub = ∞. Find its relative position on the chart.
14.
Determine the normalized distance from yL, stub to ystub . This is the normalized length of the stub.
15.
Compute the actual length of the stub.
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zL
yL
Examples:18. Design a single-stub matching network for a 50-Ω
transmission line transmitting signals with λ
= 2 m on the line to a (20 –
j40)-Ω
load.
yL, stub
ystub
y(– d)
ln
dn
8.04.050
4020 jjzL −=−
=
Relative positions of the two possible connection points:0.1787 and 0.3213.
Choosing the point at 0.1787
25.0rel.pos. , ,, =∞= stubLstubL yy
cm 8.8m 088.02044.0044.01347.01787.0
==×===−=
λn
n
ddd
cm 8.17m 178.02089.0089.025.0339.0
==×===−=
λn
n
lll
339.0rel.pos. ,6.1 =−= stubstub yjy
1347.0rel.pos. ,5.0 =+= LL yjy