Transcript Outline - 正修科技大學

dz 
1
vp 
 
is phase velocity, defined as a fixed phase point on
dt k
 the wave travels.
In free space, vp=c=2.998108 m/s.
2 2v p v p



k

f
is wavelength, defined as the distance between
two successive maximum (or minima) on the wave.
In wave equations, j k
 for following conditions.
 Plane wave in a general lossy medium

    j  j  1  j
 j  ( '  j " )  j  ' (1  j tan )

 : Complex propagation constant (m-1 )
 : Attenuation constant(Np/m;1Np/m=8.69dB/m),  : Phase constant(rad/m)
Good conductor
Condition: (1)  >>ω or (2) ’’>>’
is skin depth or penetration depth, defined as the
1
2
s  
amplitude of fields in the conductor decay by an amount

 1/e or 36.8%, after traveling a distance of one skin depth.
Example1.2 : The skin depth of several kinds of materials at a
frequency of 10GHz:
Aluminum
s (10-7 m)
8.14
Copper
6.60
Gold
7.86
Silver
6.40
Thin plating
Example1.3 : A plane wave propagating in a lossless dielectric
medium has the electric field intensity given as follow.
Determine the wavelength, phase velocity, wave impedance,
and dielectric constant?
Ex  E0cos(1.511010t  61.6zˆ)
V/m
Solution :
Analyzing Ex can find =1.511010 rad/s and k=61.6 m-1. Then
2
2
wavelength  

 0.102 m
k
61.6

1.51  1010
phase velocity v p  
 2.45  108 m/s
k
61.6
2
2
vp 
1
 c   3  108 

dielectric constant  r     
 1.5
 v p   2.45  108 
,and ’=r0

  
0
377

wave impedance  

 307.8 


r
1.5

Poynting’s Theorem
 Energy (power) conservation for EM fields and sources
1
*
*
(
E

J

H
 M s )dv  P0  Pl  2 j (Wm  We ) Power delivered by
s

the source Ps
2 v
1
1
*
P0   E  H  ds   S  ds , S  E  H * Power transmitted through
the surface S
2 s
2 s
Ps  
Pl 

2 v
We 

4
2
E dv 

2
 E  E dv
*
v
"
"
(

E


H )dv

2
2
v
Wm 

4
*
H

H
dv

v
where S is instantaneous Poynting vector
 Time-averaged Poynting power
entering good conductor
1
P  Re( E  H * )
2
Power loss to heat in the
volume v (Joule’s law)
Net reactive energy
stored in the volume v
Example1.4 : The electromagnetic fields of an antenna at a
large distance are given as follow. Find the power radiated by
120

this antenna?
ˆ
E ( r ,  )  j
cos( cos )e  jk r
V/m
r sin 
2
1

H ( r,  )  ˆj
cos( cos )e  jk r
A/m
r sin 
2
Solution :
1
60
*
2 
P  Re( E  H )  rˆ 2 2 cos ( cos  ) W/m 2
2
2
r sin 
0
0
Pradiated  
 2

 2
0
0
0
0
ˆ 2 sin  d d
P rr
60
2 
cos ( cos  )d d  1443.5 W [End]
sin 
2
 Surface resistivity of conductor
Rs  Re( )  Re[(1  j )
Pt 
Rs
2
s
2
J s ds 
Rs
2
s


1
]

2
2  s
2
2
H t ds 
2 E Rs
02
Wave Reflection
Ei  xˆE0e
 jk0 z
Er  xˆE0e
, H i  yˆ
 jk0 z
1

, H r  yˆ
Et  xˆTE0e  jrz , H t  yˆ
 - 0

  0
E0e  jk0 z ; incident field


T

E0e  jk0 z ; reflectedfield
E0e  jrz ; transmitted field
2
T  1  
  0
Oblique incidence
Total reflection
Surface waves
Chapter 2
Transmission Line Theory
Transmission-Line (TL) Theory
TL theory bridges the gap between field analysis and basic circuit theory.
l
Rs
c
, , 
ZL
 Lumped-element equivalent circuit
At DC or very low frequencies, the equivalent circuit can be simplified as
Rs
R
ZL
At medium and high frequencies, the equivalent circuit becomes
Rs
R
L
G
C
ZL
 Distributed equivalent circuit
At RF and microwave frequencies, a general two-conductor uniform line
divided into many sections can be used to describe the transmission-line
behavior.
l = NZ
Rs
N sections
ZL
Z
Rs
R Z L Z
G Z
R Z L Z
C Z
G Z
R Z
C Z
L,C,R,G are called distributed parameters.
L Z
G Z
C Z
ZL
where
R: Conductor resistance (Series resistance) per unitlength.
I2R/2: Time-average power dissipated due to conductor loss per unitlength.
L: Self inductance (Series inductance) per unitlength.
I2L/4: Time-average magnetic energy stored in a unitlength transmission line.
C: Self capacitance (Shunt capacitance) per unitlength.
V2C/4: Time-average electric energy stored in a unitlength transmission line.
G: Dielectric Conductance (Leakage conductance, Shunt conductance) per
unitlength.
V2G/2: Time-average power dissipated due to dielectric loss in a unitlength
transmission line.
At very low frequencies:
 0
Z L  j L  0
YC  j C 0  0
G ( )   ( )   0  0
( represents dielectric conductivity)
Thus, L,C,G can be ignored at very low frequencies. But at high frequencies,
effects due to L,C,G have to be considered.
 Solutions of L,C,G parameters
PDE: (Laplace’s Equation)
0
Et  tVt 0

2 0
0
0
t Vt (u, v)  0  Vt (u, v) 
 Et (u, v)
0
0
c
Ht 
aˆ z  Et
BCs:

Vt (u1, v1 )  V0
0
0
H t (u , v )
Vt0 (u2 , v2 )  0
 ,  ,
V0
S
Z=l
C
Vt0 (u , v)
Z=0

c    j

L,C,G Distributed parameters can be found as
I0 

0 
H t  dl (A)
C
1 2
LI 0 
4

1  0  0*
 H t  H t dS (Joule/m)
4
S
1
CV02 
4

1  0  0*
 Et  Et dS (Joule/m)
4
L
C
S
1
GV02 
2

1 0 0*
 Et  Et dS (Watt/m)
2
S
G

I 02

V02

V02
For distributed parameters of TEM transmission lines
LC   , C / G   / 

 0  0*
H t  H t dS (H/m)
S

S

S
 0  0*
Et  Et dS (F/m)
0 0*
Et  Et dS (S/m)
Example2.1: Find the TL parameters of coaxial Line?
Solution( another solution can refer to p.54 of the text book)
2 
1 

1

PDE: 
(r )  2 2  Vt0 (r , )  0
 r r r r  
Vt0 (r , )
BCs: Vt ( r  a, )  V0
0
b
Vt ( r  b, )  0
0

Due to symmetry, Vt (r , )  Vt (r ),
0

0
PDE becomes ODE:
d
d
(r ) Vt0 (r )  0
dr dr
BCs become
Vt0 (r  a)  V0 , Vt0 (r  b)  0
0
a
 ,  ,
V0
General solutions for electric potential at z=0
Vt0 (r )  C1 ln( r )  C2
Substitute BCs into general solutions to find the coefficients C1 and C2
 V0
V0
C1 
, C2 
ln(b)
ln(b / a)
ln(b / a)
Final solution
 V0
Vt (r ) 
ln(r / b) (V)
ln(b / a)
0
Electric and magnetic fields at z= 0
0
Et
0
Ht
0


V0 aˆr
0
0
Et (r )  tVt (r )  (aˆr  aˆ
) Vt (r ) 
r
r
ln( b / a) r
0
0
c
 c V0 aˆ
H t (r ) 
aˆ z  Et (r ) 
(A/m)

 ln(b / a) r
(V/m)
Current along the inner conductor at z=0
 0  2  c
I 0   H t  dl  0
V0 aˆ

V0
 aˆ rd  2 c
(A)
 ln(b / a) r
 ln(b / a)
c
Find distributed parameters L,C,G
 2 b 1

r
d
r
d


ln( b / a ) (H/m)
2
2 0 a 2
2
I0 S
(2 )
r
  0  0*

2
2 b 1
C  2  Et  Et dS 
r dr d  
(F/m)
2 0 a 2
L

V0
G

 0  0*
 H t  H t dS 
S
0 0*
 Et  Et dS 
V02 S
(ln b / a )

(ln b / a )
r

1
2
a r 2 rdrd  ln(b / a)
2 b
2 0
Check the following relations between LC and C/G
LC   , C / G   / 
ln( b / a )
(S/m)
⊕
Loss tangent of dielectric
tan  
Material
FR4
Ceramic
Teflon
GaAs
Silcon
 ( )
 0

=r0
r= 4.5
r= 9.9
r= 2.2
r= 12.9
r= 11.9
 Conductor resistance per unitlength
1
1
R

 c (2 at )  c (2 bt )
1 1
1

(  ) (/m)
 c t C1 C2
tanc
0.014
0.0001
0.0003
0.002
0.015
 c , c
t
C1
C2
t
Skin effect: At high frequencies, currents tend to concentrate on surface
of the conductor within a skin depth  or penetration depth
(Defined as amplitude of fields decay to 1/e)
If   t ,
R
1
1

 c (2 a )  c (2 b )
1
1
1
1
1

(  )  Rs (  ) (/m)
 c  C1 C2
C1 C2
 c , c


where   skin depth  1  f  c c (m)
Rs  surface resistance  1 ( c  )   f  c  c
()
Effective conductor thickness
tec
 t , f  fec
tec  
 ( f ), f  fec
(f)
t
f
fec