Transcript 6341 Notes 10 Dielectric Rod and Wirex

ECE 6341
Spring 2016
Prof. David R. Jackson
ECE Dept.
Notes 10
1
Dielectric Rod
z
a
 r , r
This serves as a model for a fiber-optic guide.
2
Fiber Optic Guides
Two types of fiber-optic guides:
1) Single-mode fiber
This fiber carries a single mode (HE11). This requires the fiber
diameter to be on the order of a wavelength. It has less loss,
dispersion, and signal distortion. It is often used for long-distances
(e.g., greater than 1 km). The diameter is typically 8 m. It requires
more expensive electro-optic equipment.
2) Multi-mode fiber
This fiber has a diameter that is large relative to a wavelength
(e.g., 10 wavelengths). It operates on the principle of total internal
reflection (critical-angle effect). It can handle more power than the
single-mode fiber, but has more dispersion. The diameter is
typically 50 m. It is often used for LANs, etc.
3
Single-mode Fiber
It usually has a yellow jacket.
4
Multi-mode Fiber
A 1.25 Gbit/s multi-mode fiber
http://en.wikipedia.org/wiki/Multi-mode_fiber
5
Fiber-optic Cable
Fibers (single-mode or multi-mode) may be bundled together into
a “fiber optic cable” that has one or more fibers.
Simplex: single fiber in a cable
Duplex: two fibers in a cable
Cable Types: (L to R): Zipcord, Distribution, Loose Tube, Breakout
http://www.thefoa.org/tech/ref/basic/cable.html
6
Single-mode Fiber: Operation
Fields decay away from the rod
HE11 mode
A single mode (HE11 mode) propagates on the dielectric rod. The mode is
similar to the TM0 mode on a grounded slab. It has a zero cutoff frequency.
7
Multi-mode Fiber: Operation
Higher index core region
A multimode fiber can be
explained
using down an
A laser bouncing
acrylic rod, illustrating
the
geometrical
optics and
total
internal
reflectionThe
of light
internal
reflection.
in a multi-mode
fiber
“ray”
of light is optical
actually
a
superposition of many
waveguide modes (hence
the name “multimode”).
http://en.wikipedia.org/wiki/Optical_fiber
8
Single-mode vs. Multi-mode Fibers
Graded index
9
Dielectric Rod
z
0
1
 r , r
a
Modes are hybrid unless

0

For example, assume TMz:
(n  0)
Note:
We can have
TE0p, TM0p modes
  Az
1 
 
 jk z 
1

E 

j z j 
H 
10
Dielectric Rod (cont.)
At  = a :
1H 1  0 H  0
E 1  E 0
so
 1  0



1  1
1  0

11   0 0 
Hence, for n > 0 we have
11  0 0
(Not true in general!)
11
Dielectric Rod (cont.)
Representation of potentials inside the rod:
Az1  A J n  k 1  sin  n  e  jkz z
Fz1  B J n  k 1  cos  n  e
where
k21  k12  kz2
 jk z z
 < a:
(kz is unknown)
12
Dielectric Rod (cont.)
To see choice of sin/cos, examine the field components (for example E):
1 Fz k z Az
E  

   
The field E is assumed (arbitrarily) to vary as sin(n) for our choice of potentials.
Fz  cos  n 
Az  sin  n 
We then have for the field components:
E  sin  n 
H   cos  n 
E  cos  n 
H  sin  n 
Ez  sin  n 
H z  cos  n 
13
Dielectric Rod (cont.)
 > a:
Representation of potentials outside the rod:
Use
H n 2 (k 0  )  H n 2 ( j  0  )
where
k 0   k  k
2
0

2 1/2
z
  j  0
 0  k z2  k02
Note: 0 is interpreted as a positive real number in order to have
decay radially in the air region, for a bound (non-leaky) mode.
14
Dielectric Rod (cont.)
Useful identity:
Another useful
identity:
H n ( jx)   1
 2
K n ( x) 
n 1

2
j
H n1 ( jx)
n 1
1
H n ( jx)
Kn (x) = Modified Bessel function of the second kind.
15
Dielectric Rod (cont.)
Modified Bessel function of the second kind
1
1
0.8
K1  x 
0.6
K0( x)
K1( x)
0.4
K0  x 
0.2
3
3.69110
0
0
1
3
510
2
3
xx
4
5
5
16
Dielectric Rod (cont.)
The modified Bessel function of the first kind grows exponentially, so
we don’t want this one.
I n  x    i  J n  ix 
n
I0
I1
I2
x
17
Dielectric Rod (cont.)
Hence, we choose
Az 0  CKn (  0  )sin(n ) e  jkz z
Fz 0  DKn (  0  ) cos(n ) e  jkz z
18
Dielectric Rod (cont.)
Summary of potentials:
Az1  A J n  k 1  sin  n  e  jkz z
Fz1  B J n  k 1  cos  n  e  jkz z
Az 0  CKn (  0  )sin(n ) e  jkz z
Fz 0  DKn (  0  ) cos(n ) e
 jk z z
19
Dielectric Rod (cont.)
Match Ez , Hz , E , H at  = a:
 M 11
M
 21
 M 31

 M 41
M 12
M 22
M 32
M 13
M 23
M 33
M 42
M 43
Ez1  Ez 0
Example:
1
j11
M 14   A  0
M 24   B  0
  
M 34   C  0
   
M 44   D  0
 2
2
Ez 

k

 Az
j  z 2

1
 k12  kz2  AJ n  k1a  
1
j0 0
k
2
0
 k z2  CK n   0 a 
2
0
 k z2  K n   0 a 
so
M 11 
1
j11
k
2
1
 k z2  J n  k 1a 
M 13 
1
j0 0
k
M 12  M 14  0
20
Dielectric Rod (cont.)
 M 11
M
 21
 M 31

 M 41
M 12
M 22
M 32
M 13
M 23
M 33
M 42
M 43
M 14   A  0
M 24   B  0
  
M 34   C  0
   
M 44   D  0
To have a non-trivial solution, we require that
det  M (k z ,  )   0
kz = unknown (for a given frequency )
21
Dielectric Rod (cont.)
Cutoff frequency:
Note:
This is an open structure, so
cutoff means the boundary
between proper and improper
behavior (kz = k0).
k z  k0  c 0 0
Set
Then
det  M (c 0 0 , c )   0
The unknown is now c.
22
Dielectric Rod (cont.)
Dominant mode (lowest cutoff frequency): HE11 (fc = 0)
E
Note: The notation HE means that the mode is hybrid, and has both Ez
and Hz, although Hz is stronger. (For an EH mode, Ez would be stronger.)
The field shape is somewhat similar to the TE11 circular waveguide mode.
The physical properties of the fields are similar to those of the TM0 surface wave on
a slab (For example, at low frequency the field is loosely bound to the rod.)
23
Dielectric Rod (cont.)
When will the next mode be at cutoff?
This determines the upper frequency limit for the single-mode fiber.
The next mode (i.e., with the next lowest cutoff frequency) is the TM01 mode.
(Recall: For n = 0, the modes are TMz and TEz.)
k z  k0
Cutoff:
k   k  0  0 (in air region)
 2
E 
j z
1
  Az
 2
1
2
Ez 

k
k2 
 2
 
j  z
j

1
The tangential field in the air
region at the boundary becomes
zero at the cutoff frequency: the
rod acts like a circular
waveguide with a PEC wall.
24
Dielectric Rod (cont.)
TM01 mode at cutoff:
k 1a  x01  2.405
k12  k02 a  2.405
or
k0 a n12  1  2.405
Hence, to have only the HE11 mode, we have the following frequency restriction:
k0 a 
2.405
n 1
2
1
n1 = index of refraction of rod
25
Impedance of Wire
A round wire made of conducting material is examined.
Goal: Determine the impedance per unit length (in the z direction).
a

z
The wire has a conductivity of .
We neglect the z variation of the fields inside the wire.
(The wire is short compared with a wavelength.)
26
Impedance of Wire (cont.)
Inside the wire:
Ez  AJ 0  k  
(The field must be finite on the z axis.)
k    c
 

   1  j




 

     j




a
  j
z

   e j /4 
 2

2
e
 j /4

27
Impedance of Wire (cont.)
Hence, we have
  j /4 

Ez  AJ 0  2 e




a
where

2

z

(skin depth)
We can also write the field as
 j 3 /4 
 j 3 /4 


Ez  AJ 0   2 e
  AJ 0  2 e







28
Impedance of Wire (cont.)
 j 3 /4 

Ez  AJ 0  2 e




Definition of Kelvin functions:


 Im  J  xe

Ber  x   Re J  xe
j 3 /4
Bei  x 
j 3 /4

a
z

Therefore, we can write


 


Ez  A  Ber0  2   jBei0  2  

 



29
Impedance of Wire (cont.)
The current flowing in the wire is
I   J z dS
S
2 a

 J
z
 d  d



Ez  AJ 0  2 e j 3 /4 



0 0
a
 2  J z  d 
a
0
a
 2  Ez  d 
z

0



I  2 A J 0  2 e j 3 /4   d 



0
a
Hence
30
Impedance of Wire (cont.)
The internal impedance per unit length is defined as:
Ez  a 
Zl 
I
a
Hence,


AJ 0  2 e j 3 /4 



Zl 
a



2 A J 0  2 e j 3 /4   d 



0
a
z

Note: The internal impedance
accounts for the internal stored
energy and power dissipation.
31
Impedance of Wire (cont.)
We also have the following helpful
integration identity:
 J  x  xdx  xJ  x 
0
1
a
z
Hence
a

0

 j 3 /4 
   j 3 /4 

J0  2 e
e
 d   


2




 xJ1  x  0
L
 LJ1  L 
where
L 2
a

e j 3 /4
2 L
 J  x  xdx
0
0
Use :
x 2
 j 3 /4
e

 2  j 3 /4
dx  d  
 e
  
32
Impedance of Wire (cont.)
Hence, we have
a j 3 /4 

J0  2 e




Zl 
2
a j 3 /4  
a j 3 /4 
   j 3 /4  
2 
e
2
e
J
2
e
 1

 


 

 2
 
where
a
a

a
a
J 0  e j 3 /4   Ber0    jBei0  


 
 
a

a
a
J 1  e j 3 /4   Ber1    jBei1  


 
 
z

33
Impedance of Wire (cont.)
At low frequency (a << ):
Zl 
1
 0 r 

j



  a 2 
 8 
a
At high frequency (a >> ):
Zs
Zl 
2 a
z

where
Z s  Rs 1  j 

Rs 


2
1
(surface resistance of metal)
34
Impedance of Wire (cont.)
Transmission Line Model
An extra impedance per unit length Zi is added to the TL model in
order to account for the internal impedance of the conductors.
Note: The form of Zi will depend on the shape of the conductors:
we have only analyzed the case of a round wire.
L0 Dz
R Dz
Zi Dz
C Dz
G Dz
The external inductance per unit length L0 is calculated assuming perfectly
conducting wires.
35
Impedance of Wire (cont.)
Transmission Line Model
For a twin lead (two wires running parallel) we have:
Zi  2Z l
We assume that the wires are far enough apart so that the
current is uniform inside each one.
i
+
-
v
z
36