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PART 1: Wave propagation
1
Introduction
PROPAGATION & REFLECTION OF PLANE WAVES
ELECTRIC AND MAGNETIC FIELDS FOR PLANE WAVE
PLANE WAVE IN LOSSY DIELECTRICS – IMPERFECT DIELECTRICS
PLANE WAVE IN LOSSLESS (PERFECT) DIELECTRICS
PLANE WAVE IN FREE SPACE
 PLANE WAVE IN CONDUCTORS
POWER AND THE POYNTING VECTOR
2
PROPAGATION & REFLECTION OF PLANE
WAVES
Will discuss the effect of propagation of EM wave in four medium : Free space
; Lossy dielectric ; Lossless dielectric (perfect dielectric) and Conducting media
.
Also will be discussed the phenomena of reflections at interface between differe
nt media.
Ex : EM wave is radio wave, TV signal, radar radiation and optical wave in opti
cal fiber.
Three basics characteristics of EM wave :
- travel at high velocity
- travel following EM wave characteristics
- travel outward from the source
These propagation phenomena for a type traveling wave called pl
ane wave can be explained or derived by Maxwell’s equations.
3
ELECTRIC AND MAGNETIC FIELDS FOR PLAN
E WAVE
From Maxwell’s equations :
∂
B
∂
H
∇ E   -
∂
t
∂
t
∂
D
∂
E
∇ H  J 
 J 
∂
t
∂
t
∇ D   v
∇ B  0
4
Assume the medium is free of charge :
∂
H
∂
t
∂
E
∇ H  
∂
t
∇ D  0
∇ E  - 
∇ B  0
From vector identity and taking the curl
of (1)and substituting (1) and (2)
∇ (∇ E )  ∇(∇ E ) -∇2 E
where ∇(∇ E )  0
 v  0, J  0
(1)
( 2)
(3)
( 4)

∂
H
  -∇2 E
 ∇  - 
∂
t 

∂
→ ∇2 E   (∇ H )
∂
t
∂2 E
2
∴ ∇ E  
∂
t2
5
ie Helmholtz ' s equation for electric field
∂2 E
2
3
∇ E  
(
Vm
)
2
∂
t
Similarly in the same way, from vect
or identity and taking the curl of (2)a
nd substituting (1) and (2)
In Cartesian coordinates :
2
2
2
2E ∂
E ∂
E
∂
E


 
2
2
2
x
∂
y
∂
z
∂
t2
∂2 H
3
∇ H  
A
m
∂
t2
2
Assume that :
(i) Electric field only has x component
(ii) Propagate in the z direction
2
2
∂
Ex
∂
Ex
 
2
∂
z
∂
t2
6
2
2
∂
Ex
∂
Ex
 
2
∂
z
∂
t2
The solution for this equation :
E x  E x cos(t - z )  E x- cos(t  z )
Incidence wave propagate in
+z direction
To find H field :
Reflected wave propagate i
n -z direction
∂
H
∇ E  -
∂
t
∂
Ex
∂
Ex
∇ E 
yˆ zˆ
∂
z
∂
y


 E x sin( t - z ) - E x- sin( t  z ) yˆ
7
∂
Hy
∂
∂
H
Hx
∂
Hz
-
 - 
xˆ 
yˆ 
∂
t
t
∂
t
∂
t
 ∂
On the right side equ
ation :

zˆ 

Equating components on both side = y component
-
∂
Hy
∂
t


  E x sin( t -  z ) -  E x- sin( t  z )
E x
E x- H y  ∫ sin( t -  z )dt - 
sin( t   z ) dt


 
 E x cos(t -  z ) 
E x cos(t   z )


 

Hy 
E x cos(t -  z ) E x - cos(t   z )


 H y cos(t -  z ) - H y- cos(t   z )
8
Hence :
E x  E x cos(t - z )  E x- cos(t  z )
H y  H y cos(t - z) - H y- cos(t  z)
These equations of EM wave are called PLANE WAVE.
Main characteristics of EM wave :
(i) Electric field and magnetic field always perpendicular.
(ii) NO electric or magnetic fields component in the direction of propagation.
(iii)
E  H will provides information on the direction of propagation.
9
PLANE WAVE IN LOSSY DIELECTRICS – IMPER
FECT DIELECTRICS
  0 ;   0 r ;    0 r
Assume a media is charged free , ρv =0
∂
D
∇ H  J 
   j E
∂
t
∂
B
(2)
∇ E   - jH
∂
t
(1)
Taking the curl of (2) :
∇∇ E  - j ∇ H 
10
From vector identity
:
∇∇ A
 ∇∇ A -∇ A
2
∇∇ E  -∇2 E  - j   j E
∇2 E - j   j E  0
∇2 E -  2 E  0
Where :
 2 -  2  - 2  (Re) (6)
  j   j 
 -   j
  propagatio n constant
Define :
∇∇ E  - j ∇ H 
Equating (4) and (5) for Re and
Im parts :
2
2
∂
D
   j E (1)
∂
t
∂
B
∇ E   - jH
(2)
∂
t
∇ H  J 
(4)
2   (Im) (7)
    j
 2   2 -  2   2 j
(5)
11
Magnitude for (5) ;
   
2
2
2
Add (10) and (6) :
(8)
Hence :
2 2    2   2 -  2 
Magnitude for (4) ;
 
2
- 
2
    
2
   2   2
2
(9)
Equate (8) and (9) :
 2   2    2   2
 2 -  2  - 2  (Re) (6)
(10)
2

  2  1  2 2 -  2 

2


2 
2


2
 1  2 2 - 1




 
2


2
 1  2 2 - 1 Np / m (11)



 is known as attenuation constant as a measure
of the wave is attenuated while traveling in a m
edium.
12
Substract (10) and (6) :
2 2    2   2   2 
 
 

2
 1  2 2  1
2 
 

rad / m (12)
 is phase constant
If the electric field propagate in +z direction and has component x, the equation o
f the wave is given by :
E ( z , t )  E0 e
-z
cost -  z xˆ
(13)
And the magnetic field :
H ( z, t )  H 0e-z cost - z -  yˆ
(14)
13
where ;
H0 
E0
E ( z , t )  E0 e -z cost - z xˆ
H ( z, t )  H 0e-z cost - z -  yˆ (15)
(15)

(14)
Intrinsic impedance :
j
j

  ∠   e , ()
  j
where ;
 
 /
2 1/ 4
    
 
1  
    
, tan 2 
(16)

, 0 ≤  ≤450

(17)
Conclusions that can be made for the wave propagating in lossy dielectrics mater
ial :
(i) E and H fields amplitude will be attenuated by
e -z
(ii) E leading H by

14
Wave velocity ;
 
 /
   2 
 
1  
    
1/ 4
Jd
E



 tan 

jE

, 0 ≤  ≤450 (17)

From (17) and (18)
Loss tangent ;
J
, tan 2 
  2
(18)
Loss tangent values will determine types of media :
tan θ small (σ / ωε < 0.1) – good dielectric – low loss
tan θ large (σ /ωε > 10 ) - good conductor – high loss
Another factor that determined the characteristic of the media is operating frequency. A m
edium can be regarded as a good conductor at low frequency might be a good dielectric at
higher frequency.
15
E ( z, t )  E0 e -z cost - z xˆ
 E0 e -z e jt - z  xˆ
H0 
(14)
E0

H ( z , t )  H 0 e -z cost - z -  yˆ

E0

e -z e
jt - z -

(15)
yˆ
Graphical representation of E field in lossy dielectric
E0 x
E ( z , t )  E0 e -z cost -  z xˆ
e z
z
y
16
PLANE WAVE IN LOSSLESS (PERFECT) DIELECTRICS
Characteristics:
  0,    0 r ,   0 r
Substitute in (11) and (12) :
  0,    
u

1
2

, 



 o

0

(20)
(21)
(19)

 
2


2
1

1

 Np / m (11)
2 2




 
2


2
 1  2 2  1 rad / m (12)



j
j

  ∠   e  , ()
  j
(22)
The zero angle means that E and H fields are in phase at each fixed location
.
17
PLANE WAVE IN FREE SPACE
Free space is nothing more than the perfect dielectric media :
Characteristics:
  0,    0 ,   0
(23)
Substitute in (20) and (21) :
  0 ,    0 0   / c

1
2
u 
c ,  


0 0
where
u
(24)

(25)
u  c  3 108 m / s
  0 
0
 120 
0
  0,    
(20)

1
2

, 
(21)




 0o

(22)
  0  4 10 7 H / m
(26)
   0  8.854 10 12 
1
10 9 F / m
36
18
The field equations for E and H obtained :
E  E0 cos(t - z ) xˆ
H 
E0
0
cos(t - z ) yˆ
E ( z , t )  E0 e -z cost -  z xˆ
(27)
(14)
H ( z, t )  H 0e-z cost - z -  yˆ
(28)
E and H fields and the direction of propagation :
x
Ex+
kos(-z)
Generally :
kˆ  zˆ
Eˆ  Hˆ  kˆ
z
(at t = 0)
y
Hy+ kos(-z)
19
(15)
PLANE WAVE IN CONDUCTORS

In conductors :    or
→∞

 ~  ∞,    0 ,   0 r
With the characteristics :

 
2
 
1

1

 Np / m (11)
2 2
2 


Substitute in (11 and (12) :
 


2
 f
(29)
(30)

 45o E leads H by 450


 
2
(31)



2
 1  2 2  1 rad / m (12)



j
j
  ∠   e  , ()
  j
The field equations for E and H obtained :
E  E0e-z cos(t - z ) xˆ
H 
E0
0
(32)
e-z cos(t - z - 45o ) yˆ
(33)
20
It is seen that in conductors E and
H waves are attenuated by e
-z
From the diagram  is referred to as the skin depth. It refers to the amplitude of th
e wave propagate to a conducting media is reduced to e-1 or 37% from its initial va
lue.
E0 e -  E0 e -1
In a distance :
∴  1 /  
1
f
It can be seen that at higher freque
ncies
 is decreasing.
(34)
x
E0
0.368E0
z

21
Ex : A lossy dielectric has an intrinsic impedance of 200∠30  at the particular frequency
. If at that particular frequency a plane wave that propagate in a medium has a magnetic field
given by :
o
H  10e-x cos(t - x/2 ) yˆ A / m. Find E and .
Solution :
Eˆ  Hˆ  kˆ
→ Eˆ  yˆ  xˆ
∴E  - zˆ
From intrinsic impedance, the magnitude of E field :

E0
 200∠30o
H0
→ E0  2000∠30o
It is seen that E field leads H field :
  300   / 6
E  -2000e-x cos(t - x / 2   / 6) zˆ (V / m)
22
E  -2000e-x cos(t - x / 2   / 6) zˆ (V / m)
To find
:




1






 1 




 - 1 

  1
2
1/ 2
2
 tan 2  tan 600  3
  2 -1 
1
; and we know
∴ 


  2  1
3

→ 
 0.2887 Np / m
1/ 2
  1/ 2
3
Hence:
E  -2000e-0.2887x cos(t - x / 2   / 6) zˆ (V / m)23
POWER AND THE POYNTING VECTOR
∂
H
∇ E  -
∂
t
(35)
∂
E
∇ H  E  
∂
t
Dot product (36) with
(36)
:E
∂
E
E  ∇ H   E  E  
∂
t
2
(37)
From vector identity:
∇ A  B   B  ∇ A - A  ∇ B 
Change
(38)
A  H , B  Ein (37) and use (38) , equation (37) becomes :
∂
E
2
H  ∇ E   ∇ H  E   E  E  
(39)
∂
t
24
∂
E
H  ∇ E   ∇ H  E   E  E  
∂
t
2
(39)
H
∂
∇ E  -
t
∂
(35)
And from (35):
 ∂
H
 ∂
  H  ∇ E   H   - 
H H
∂
t 
2∂
t

(40)
Therefore (39) becomes:
-
 ∂H 2
2 ∂
t
-∇ E  H   E
2
∂
E
 E 
∂
t
(41)
where:
∇ H  E   -∇ E  H 
Integration (41) throughout volume v :
∂ 1 2 1
2
2


∇

E

H
dv


E


H
dv

E
dv
∫
∫
∫


∂
t v 2
2

v
v
(42)
25
∂ 1 2 1
2
2


∇

E

H
dv


E


H
dv

E
dv
∫
∫
∫


∂
t v 2
2

v
v
(42)
Using divergence theorem to (42):
∂ 1 2 1
2
2


E

H

d
S


E


H
dv

E
dv
∫
∫
∫


∂
t v 2
2

s
v
Total energy flow leavi
ng the volume
The decrease of the energy densi
ties of energy stored in the electr
ic and magnetic fields
(43)
Dissipated ohmi
c power
Equation (43) shows Poynting Theorem and can be written a
s:
 E  H W / m
2
26
Poynting theorem states that the total power flow leaving the volume is equal to t
he decrease of the energy densities of energy stored in the electric and magnetic fi
elds and the dissipated ohmic power.
The theorem can be explained as shown in the diagram below :
Output power
σ
Ohmic losses
E
J
H
Stored electric fi Stored magnetic
field
eld
Input power
27
Given for lossless dielectric, the electric and magnetic fields are :
E  E0 cos(t - z ) xˆ
H 
E0

cos(t - z ) yˆ
The Poynting vector becomes:
  E  H W / m2

E 20

cos 2 (t - z ) yˆ
28
To find average power density :
Integrate Poynting vector and divide with interval T = 1/f :
Pave
1

T
T
E02
∫

cos 2 (t   z ) dt
0
2 T
0
1 E

2T 
1  cos( 2t - 2 z )dt
∫
0
T


1 E
1

sin( 2t - 2  z ) 
t 
2T  
2
0 


1 E02
 P ave 
W / m2
2 
2
0
Average power through area S :
1 E02
P ave 
S (W )
2 
29
Given for lossy dielectric, the electric and magnetic fields are :
E  E0e-z cos(t - z ) xˆ
H 
E0
0
e -z cos(t -  z -  ) yˆ
The Poynting vector becomes:

E 20

e  2z cos(t - z ) cos(t - z   )
Average power :
1 E02 2z
P ave 
e
cos
2 
30
Ex. 2: A uniform plane wave propagate in a lossless dielectric in the +z direction. T
he electric field is given by :
E ( z, t )  377 cost  4 / 3z   / 6xˆ (V / m)
The average power density measured was 377 W / m 2
Find:
(i) Dielectric constant of the material if
  0
(ii) Wave frequency
(iii) Magnetic field equation
Solution:
(i) Average power :
1 E2
Pave 
 377
2 
1 (377) 2
 377
2 
   377 / 2  188.5
31
For lossless dielectric :
0



 r 0

r 
1

0
 1.9986
0
  r  4.0
E ( z, t )  377 cost  4 / 3z   / 6xˆ (V / m)
(ii) Wave frequency :
  4 / 3    0

4
3  0
2f  3.9946 1016
 f  99.93 106  (100 MHz )
32
E ( z, t )  377 cost  4 / 3z   / 6xˆ (V / m)
(iii) Magnetic field equation :
H ( z, t ) 
377
cos(t  (4 / 3) z   / 6) yˆ

 2 cos(t  (4 / 3) z   / 6) yˆ ( A / m)
33
Chapter 2.
Transmission Line Theory
34
2.1 Transmission Lines
•
A transmission line is a distributed-parameter network, where voltages and currents
can vary in magnitude and phase over the length of the line.
Lumped Element Model for a Transmission Line
• Transmission lines usually consist of 2 parallel conductors.
• A short segment Δz of transmission line can be modeled as a lumped-element
circuit.
Figure 2.1 Voltage and current definitions and equivalent circuit for an incremental
length of transmission line. (a) Voltage and current definitions. (b) Lumped-element
equivalent circuit.
35
• R = series resistance per unit length for both
conductors
• L = series inductance per unit length for both
conductors
• G = shunt conductance per unit length
• C = shunt capacitance per unit length
• Applying KVL and KCL,
i ( z, t )
v( z , t )  Rzi ( z, t )  Lz
 v( z  z , t )  0 (2.1a)
t
v( z  z , t )
i ( z, t )  Gzv( z  z , t )  C z
 i ( z  z , t )  0 (2.1b)
t
36
• Dividing (2.1) by Δz and Δz  0,
v( z , t )
i ( z , t )
  Ri ( z, t )  L
(2.2a)
z
t
i ( z, t )
v( z , t )
 Gv( z , t )  C
(2.2b)
z
t
 Time-domain form of the transmission line,
or telegrapher, equation.
• For the sinusoidal steady-state condition with
cosine-based phasors,
dV ( z )
 ( R  j L) I ( z ) (2.3a)
dz
dI ( z )
 (G  jC )V ( z ) (2.3b)
dz
37
Wave Propagation on a Transmission Line
• By eliminating either I(z) or V(z):
2
d 2V ( z )
d
I ( z)
2
2



V
(
z
)
(2.4
a
)



I ( z ) (2.4b)
2
2
dz
dz
where     j  ( R  jL)(G  jC) the
complex propagation constant. (α = attenuation
constant, β = phase constant)
• Traveling wave solutions to (2.4):
V ( z )  V0 e  z +V0 e  z , I ( z )  I 0 e  z  I 0e  z (2.6)
Wave
propagation in
+z directon
Wave
propagation in z directon
38
• Applying (2.3a) to the voltage of (2.6),
I ( z) 

V0 e z +V0 e z 
R  j L
• If a characteristic impedance, Z0, is defined as
Z0 
R  j L

R  j L

, (2.7)
G  jC
V0
V0
 Z0   

I0
I0
• (2.6) can be rewritten
V0  z V0  z
I ( z) 
e 
e
(2.8)
Z0
Z0
39
• Converting the phasor voltage of (2.6) to the
time domain:
v( z, t )  V0 cos(t   z   + )e  z  V0 cos(t   z    )e z (2.9)
• The wavelength of the traveling waves:
=
2

(2.10)
• The phase velocity of the wave is defined as
the speed at which a constant phase point
travels down the line,
vp =
dz 
= = f
dt 
(2.11)
40
Lossless Transmission Lines
• R = G = 0 gives     j   j LC
or
   LC ,   0 (2.12)
L
Z0 =
C
(2.13)
• The general solutions for voltage and current
on a lossless transmission line:
V ( z )  V0 e j z +V0 e j  z ,
I 0  j z
I ( z) 
e
 I 0 e j  z (2.14)
Z0
41
• The wavelength on the line:
=
• The phase velocity on the line:
2

vp =
=
2
 LC

1
=

LC
(2.15)
(2.16)
42
2.2 Field Analysis of Transmission Lines
• Transmission Line Parameters
Figure 2.2
Field lines on an arbitrary TEM transmission
line.
43
• The time-average stored magnetic energy for 1
m section of line:
Wm 

4

S
H  H ds
• The circuit theory gives
 L   2 S H  H ds
Wm  L | I 0 |2 / 4
| I0 |

• Similarly, We  S E  E ds,
4



C
2
| V0 |

S
We  C | V0 |2 / 4
E  E ds
44
• Power loss per unit length due to the finite
conductivity (from (1.130))
Pc 
Rs
2

C1  C2
H  H dl
• Circuit theory  Pc  R | I 0 |2 / 2
Rs

R
H

H
dl (H || S)
2 C  C
| I0 |
1
2
• Time-average power dissipated per unit length
in a lossy dielectric (from (1.92))
Pd 
 
2

S
E  E  ds
45
The Telegrapher Equations Derived form Field
Analysis of a Coaxial Line
• Eq. (2.3) can also be obtained from ME.
• A TEM wave on the coaxial line: Ez = Hz = 0.
• Due to the azimuthal symmetry, no φ-variation
 ə/əφ = 0
• The fields inside the coaxial line will satisfy
ME.
 E   j H
 H  j E
where      j 
46
E

E
ˆ 
1 
 ˆ

 zˆ
(  E )   j ( ˆ H   ˆH )
z
z
 
H ˆ H 
1 
ˆ


 zˆ
(  H )  j ( ˆ E  ˆE )
z
z
 
Since the z-components must vanish,
E 
f ( z)

, H 
g ( z)

From the B.C., Eφ = 0 at ρ = a, b  Eφ = 0 everywhere
H  0
E
z
  j H ,
H
z
 j E
E 
h( z )

47
h( z )
g ( z )
  j g ( z ),
  j h( z ),
z
z
The voltage between 2 conductors
V ( z)  
b
 a
E (  , z ) d   h ( z ) 
b
 a
d
b
 h( z ) ln

a
The total current on the inner conductor at ρ = a
I ( z)  
2
 0
H (a, z )ad  2 g ( z )
V ( z )
 ln b / a
I ( z )
2 V ( z )
j
I ( z ),
  j (   j )
z
2
z
ln b / a
V ( z )
I ( z )
  j LI ( z ),
 (G  jC )V ( z )
z
z
48
Propagation Constant, Impedance, and Power Flow
for the Lossless Coaxial Line
• From Eq. (2.24)
 2 E
z
2
 2   2 
   E  0
2
• For lossless media,       LC
• The wave impedance

Zw 

  / 
H

E
Ex 2.1
• The characteristic impedance of the coaxial
line
V0 E ln b / a  ln b / a
 ln b / a
Z0 
I0

2 H

2


2
49
• Power flow ( in the z direction) on the coaxial
line may be computed from the Poynting
vector as

2 b
V
I
1
1
1


0 0
P   E  H  ds   

d

d


V
I
0 0
2 s
2  0  a 2 2 ln b / a
2
• The flow of power in a transmission line takes
place entirely via the E & H fields between
the 2 conductors; power is not transmitted
through the conductors themselves.
50
2.3 The Terminated Lossless Transmission Lines
The total voltage and current on the line
V ( z )  V0 e j z +V0 e j  z ,
V0  j z V0  j  z
I ( z) 
e

e
(2.34)
Z0
Z0
51
• The total voltage and current at the load are
related by the load impedance, so at z = 0
V (0) V0  V0
ZL =
= 
Z0

I (0) V0  V0
Z L  Z0 
V 
V0
Z L  Z0

0
• The voltage reflection coefficient:
V0 Z L  Z 0
  
V0
Z L  Z0
(2.35)
• The total voltage and current on the line:
V ( z )  V0 e j  z +e j  z  ,
V0  j  z
e
I ( z) 
 e j  z  (2.36)
Z0
52
• It is seen that the voltage and current on the
line consist of a superposition of an incident
and reflected wave.  standing waves
• When Γ= 0  matched.
• For the time-average power flow along the line
at the point z:
Pavg
 2
0

1
1V
2

 2 j  z
2 j z
 Re V ( z ) I ( z ) 
Re 1   e
 e

2
2 Z0

 2
0
1V

2 Z0


1   
2
53
• When the load is mismatched, not all of the
available power from the generator is delivered
to the load. This “loss” is return loss (RL):
RL = -20 log|Γ| dB
• If the load is matched to the line, Γ= 0 and
|V(z)| = |V0+| (constant)  “flat”.
• When the load is mismatched,
V ( z )  V0 1  e 2 j  z  V0 1  e 2 j  l  V0 1   e j (  2  l ) (2.39)
Vmax  V0 1    , Vmin  V0 1    (2.40)
54
• A measure of the mismatch of a line, called the
voltage standing wave ratio (VSWR)
1 
(1< VSWR<∞)
SWR 
1 
• From (2.39), the distance between 2 successive
voltage maxima (or minima) is l = 2π/2β = λ/2
(2βl = 2π), while the distance between a
maximum and a minimum is l = π/2β = λ/4.
• From (2.34) with z = -l,
V0 e j l
(l )   j l  (0)e2 j l (2.42)
V0 e
55
• At a distance l = -z,
 e j  l  e  j  l 
1  e 2 j  l
 Z0
 j l
 j l 
2 j  l
e


e
1


e


( Z L  Z 0 )e j  l  ( Z L  Z 0 )e  j  l
 Z0
( Z L  Z 0 )e j  l  ( Z L  Z 0 )e  j  l
V0
V (l )
Z in 
 Z0 
I (l )
V0
(2.43)
Z L cos  l  jZ 0 sin  l
 Z0
Z 0 cos  l  jZ L sin  l
 Z0
Z L  jZ 0 tan  l
Z 0  jZ L tan  l
(2.44)
 Transmission line impedance equation
56
Special Cases of Terminated Transmission Lines
• Short-circuited line
ZL = 0  Γ= -1
V ( z )  V0 e  j  z  e j  z   2 jV0 sin  z ,

V0  j z
V
e
I ( z) 
 e j  z   2 0 cos  z
Z0
Z0
Zin  jZ0 tan  l
(2.45)
57
Figure 2.6
(a) Voltage, (b) current,
and (c) impedance (Rin
= 0 or ) variation
along a short-circuited
transmission line.
58
• Open-circuited line
ZL = ∞  Γ= 1
V ( z )  V0 e  j  z  e j  z   2V0 cos  z,

V0  j  z

2
jV
0
e
I ( z) 
 e j  z  
sin  z
Z0
Z0
(2.46)
Zin   jZ0 cot  l
59
Figure 2.8
(a) Voltage, (b) current, and
(c) impedance (Rin = 0 or )
variation along an opencircuited transmission line.
60
• Terminated transmission lines with special
lengths.
• If l = λ/2, Zin = ZL.
• If the line is a quarter-wavelength long, or, l =
λ/4+ nλ/2 (n = 1,2,3…), Zin = Z02/ZL. 
quarter-wave transformer
Figure 2.9
Reflection and transmission at the junction of two
transmission lines with different characteristic impedances.
61
2.4 The Smith Chart
• A graphical aid that is very useful for solving
transmission line problems.
Derivation of the Smith Chart
• Essentially a polar plot of the Γ(= |Γ|ejθ).
• This can be used to convert from Γto
normalized impedances (or admittances), and
vice versa, using the impedance (or admittance)
circles printed on the chart.
62
Figure 2.10
The Smith chart.
63
• If a lossless line of Z0 is terminated with ZL, zL
= ZL/Z0 (normalized load impedance),
zL  1

  e j
zL  1
zL 
1   e j
1   e j
• Let Γ= Γr +jΓi, and zL = rL + jxL.
1   2r  i2
rL 
(1   r ) 2  i2
(1   r )  ji
rL  jxL 
(1   r )  ji
2
xL 
2i
(1   r ) 2  i2
2

 1 
rL 
2
 r 
  i  
 ,
1  rL 

 1  rL 
2

1   1 
  r  1   i     
xL   xL 

2
2
64
• The Smith chart can also be used to
graphically solve the transmission line
impedance equation of (2.44).
1  e2 j  l
Zin  Z 0
1  e2 j  l
(2.57)
• If we have plotted |Γ|ejθ at the load, Zin seen
looking into a length l of transmission line
terminates with zL can be found by rotating the
point clockwise an amount of 2βl around the
center of the chart.
65
• Smith chart has scales around its periphery
calibrated in electrical lengths, toward and
away from the “generator”.
• The scales over a range of 0 to 0.5 λ.
66
Ex 2.2 ZL = 40+j70, l = 0.3λ, find Γl, Γin and Zin
Figure 2.11
Smith chart for Example 2.2.
67
The Combined Impedance-Admittance Smith Chart
• Since a complete revolution around the Smith
chart corresponds to a line length of λ/2, a λ/4
transformation is equivalent to rotating the
chart by 180°.
• Imaging a give impedance (or admittance)
point across the center of the chart to obtain
the corresponding admittance (or impedance)
point.
68
Ex 2.3 ZL = 100+j50, YL, Yin ? when l = 0.15λ
Figure 2.12
ZY Smith chart with solution for Example 2.3.
69
The Slotted Line
• A transmission line allowing the sampling of E
field amplitude of a standing wave on a
terminated line.
• With this device the SWR and the distance of
the first voltage minimum from the load can be
measured, from this data ZL can be determined.
• ZL is complex  2 distinct quantities must be
measured.
• Replaced by vector network analyzer.
70
Figure 2.13
An X-band waveguide slotted line.
71
• Assume for a certain terminated line, we have
measured the SWR on the line and lmin , the
distance from the load to the first voltage
minimum on the line.
SWR  1
|  |
SWR  1
• Minimum occurs when e j ( 2  l )  1
• The phase of Γ =     2 lmin
• Load impedance Z L  Z 0 1  
1 
72
Ex 2.4
• With a short circuit load, voltage minima at z =
0.2, 2.2, 4.2 cm
• With unknown load, voltage minima at z =
0.72, 2.72, 4.72 cm
• λ = 4 cm,
• If the load is at 4.2 cm, lmin = 4.2 – 2.72 = 1.48
cm = 0.37 λ
  ?,  ?, Z L  ?
73
Figure 2.14
Voltage standing wave patterns for Example
2.4. (a) Standing wave for short-circuit load. (b) Standing
wave for unknown load.
74
Figure 2.15
Smith chart for Example 2.4.
75
2.5 The Quarterwave Transformer
Impedance Viewpoint
RL  jZ1 tan  l
Z in  Z1
Z1  jRL tan  l
• For βl = (2π/λ)(λ/4) = π/2
Z12
Z in 
RL
• In order for Γ = 0, Zin = Z0
Z1  Z 0 RL
76
Figure 2.16
The quarter-wave matching transformer.
77
Ex 2.5 Frequency Response of a Quarter-Wave
Transformer
• RL = 100, Z0 = 50
Z1  Z0 RL  70.71
Zin  Z0
|  |
Zin  Z0
 2   0      0      f 
l            
    4   2      2   f0 
78
Figure 2.17
Reflection coefficient versus normalized
frequency for the quarter-wave transformer of Example 2.5.
79
The Multiple Reflection Viewpoint
Figure 2.18 (p. 75)
Multiple reflection
analysis of the quarterwave transformer.
80
Z1  Z 0
Z 0  Z1
RL  Z1
1 
, 2 
 1 , 3 
Z1  Z 0
Z 0  Z1
RL  Z1
2Z 0
2Z1
T1 
, T2 
Z1  Z 0
Z1  Z 0
  1  T1T23  T1T232  T1T2232 

 1  T1T23  ( 23 ) n
n 0
T1T2 3
1  1 23  T1T23
 1 

1   23
1   23
81
• Numerator
1  1 23  T1T23  1  3 (12  T1T2 )
( Z1  Z 0 )( RL  Z1 )  ( RL  Z1 )( Z1  Z 0 )
 1  3
( Z1  Z 0 )( RL  Z1 )
2( Z12  Z 0 RL )

( Z1  Z 0 )( RL  Z1 )
82
2.6 Generator and Load Mismatches
• Because both the generator and load are
mismatched, multiple reflections can occur on
the line.
• In the steady state, the net result is a single
wave traveling toward the load, and a single
reflected wave traveling toward the generator.
• In Fig. 2.19, where z = -l,
1  l e2 j l
Zl  jZ 0 tan  l
Zin  Z 0
 Z0
2 j  l
1  l e
Z 0  jZl tan  l
l 
Zl  Z0
Zl  Z0
(2.67)
(2.68)
83
Figure 2.19
Transmission line circuit for mismatched load
and generator.
84
• The voltage on the line:
V (l )  Vg
Zin
 V0 (e j  l  l e  j  l )
Z in  Z g
Zin
1
V  Vg
Z in  Z g e j  l  l e  j  l

0
Z0
e j l
V  Vg
Z 0  Z g 1  l  g e 2 j  l

0


(2.70)
(2.71)
• Power delivered to the load:
SWR 
By (2.67)
&
g 

Z g  Z0
1  l
1  l
Z in
1
1 
1 


2 1 
Pl  Re Vin I in  Re | Vin |    Re | Vg |2
2
2 
Zin  2 
Z in  Z g

Rin
1
2
 | Vg |
(2.39)
2
2
2
( Rin  Rg )  ( X in  X g )

Z g  Z0
2
1 

 
Z in 

85
• Case 1: the load is matched to the line, Zl = Z0,
Γl = 0, SWR = 1, Zin = Z0,
Z0
1
2
Pl  | Vg |
2
( Z0  Rg )2  X g2
(2.40)
• Case 2: the generator is matched to the input
impedance of a mismatched line, Zin = Zg
Rg
1
2
Pl  | Vg |
2
4 Rg2  X g2


(2.41)
• If Zg is fixed, to maximize Pl,
2 Rin ( Rin  Rg )
Pl
1
0

0
2
2
2
Rin
( Rin  Rg )  ( X in  X g )
( Rin  Rg ) 2  ( X in  X g ) 2 
86
or
Rg2  Rin2  ( X in  X g )2  0
2 X in ( X in  X g )
Pl
0
0
2
2
2
X in
( Rin  Rg )  ( X in  X g ) 
or X in ( X in  X g )  0
• Therefore, Rin = Rg and Xin = -Xg, or Zin = Zg*
• Under these conditions
Pl 
1
1
| Vg |2
2
4Rg
(2.44)
• Finally, note that neither matching for zero reflection
(Zl = Z0), nor conjugate matching (Zin = Zg*),
necessary yields a system with the best efficiency.
87