Slide of computer networks introduction to computer networks

Prentation of computer networks

1. 1
Prof. David R. Jackson
Dept. of ECE
Notes 4
ECE 5317-6351
Microwave Engineering
Fall 2011
Waveguides Part 1:
General Theory

2. 2
In general terms, a waveguide is a devise that confines electromagnetic
energy and channels it from one point to another.
Examples
–Coax
–Twin lead (twisted pair)
–Printed circuit lines
(e.g. microstrip)
–Optical fiber
–Parallel plate waveguide
–Rectangular waveguide
–Circular waveguide
Waveguide Introduction
Note: In microwave engineering, the term “waveguide” is often used to
mean rectangular or circular waveguide (i.e., a hollow pipe of metal).

3. 3
General Solutions for TEM, TE and TM Waves
Assume e
jt
time dependence and homogeneous source-free materials.
Assume wave propagation in the zdirectionz
jk zz
ee

    ˆ, , , ,
z
jk z
tz
E x y z e x y z e x y e
    , , , ˆ,
z
jk z
tz
H x y z h x y h x y ez
 ,
zz
jk k j     z
y
x,,
PEC
transverse
componentsJE

4. 4
Helmholtz EquationE j H
H j E J


  
    E j H j j E J         0
v
E
H


 
 
Vector Laplacian definition : 
2
E E E       
2 2 2 2
ˆˆ ˆ
x y z
E x E y E z E      
where

5. 5
Helmholtz Equation  
2
22
22
22
v
v
v
E E j j E J
E E j J
E E j J
E E j E
 

  


  


  

      

    



     



     

  E j H j j E J        
Assume Ohm’s
law holds:JE

6. 622 v
E E j E

  


    

 22
22
22
v
v
c
v
E j E
EE
E k E

  


 




     
 
 

    



    


Next, we examine the term on the right-hand side.
Helmholtz Equation (cont.)22
c
k 

7. 7  
  
 0
0
0
v
H j E J
H j E J
H j E E
Ej
E


 
 

  
      
      
   
  

To do this, start with Ampere’s law:
In the time-harmonic (sinusoidal steady
state, there can never be any volume
charge density inside of a linear,
homogeneous, isotropic, source-free
region that obeys Ohm’s law.
Helmholtz Equation (cont.)22 v
E k E



   



8. 8
Helmholtz equation22
0E k E  
Hence, we have
Helmholtz Equation (cont.)

9. 9 
 
  
  
   
    
2
c
c
c
H j E J
H j E E
H j E
H j E
H j j H
H H j j H

 
 

 
 
      
      
     
    
    
     
Similarly, for the magnetic field, we haveH j E J  
Helmholtz Equation (cont.)

10. 1022
0H k H  
Hence, we have
Helmholtz equation
Helmholtz Equation (cont.)

11. 1122
0H k H  
Summary
Helmholtz equations22
0E k E  
Helmholtz Equation (cont.)
These equations are valid for a source-free homogeneous isotropic
linear material.

12. 12
Assume a guided wave with a field variation in the zdirection
of the formz
jk z
e
Field Representation
Then all six of the field components can be expressed in terms
of these two fundamental ones: ,
zz
EH

13. 13
Types of guided waves:
Field Representation (cont.)
TEM
z:E
z= 0, H
z= 0
TM
z:E
z0, H
z= 0
TE
z:E
z= 0, H
z0
Hybrid:E
z0, H
z0
Microstrip
h
w

r
TEM
z
TM
z,TE
z
Hybrid

14. 14
Assume a source-free region with a variationz
jk z
e E j H   1)
z
z y x
E
jk E j H
y


  
 2)
z
z x y
E
jk E j H
x


  
 3)
y x
z
E E
jH
xy

 
  
 4)
z
z y c x
H
jk H j E
y



 5)
z
z x c y
H
jk H j E
x



 6)
y x
z
H H
jE
xy

 


Field Representation: Proof c
H j E 

15. 15
Combining 1)and 5)2
2
22
2
1
()
c
z z z z
x
cc
z
zz
z z x x
c
zz
x c z
c
k
z
c z z x
E k H k
jH
y x j
EH
j jk k k H
EH
jk jk H j H
y j x
EHj
Hk
yx
k y x




 


 
      
 

  



  





  
1/2
22
cz
k k k
Cutoff wave
number
Field Representation: Proof (cont.)
A similar derivation holdsfor the other three transverse field components.

16. 162
2
2
2
zz
x c z
c
zz
y c z
c
zz
xz
c
zz
yz
c
EHj
Hk
k y x
EHj
Hk
k x y
EHj
Ek
k x y
EHj
Ek
k y x













  






Summary
These equations give the
transverse field
components in terms of
longitudinalcomponents,
E
zand H
z.
Field Representation (cont.)22
c
k   
1/2
22
cz
k k k

17. 17
Therefore, we only need to solve the Helmholtz equations for the
longitudinal field components (E
zand H
z).
Field Representation (cont.)22
0
zz
H k H   22
0
zz
E k E  

18. 18
Transverse Electric (TE
z) Waves0
z
E
In general, E
x,E
y,H
x,H
y,H
z 0
To find the TE
zfield solutions (away from any sources), solve22
( ) 0
z
kH   2 2 2
2
2 2 2
22
) 0(0
zz
kH
x y z
kH  
  
    
  
The electric field is “transverse”
(perpendicular) to z.

19. 19
Recall that the field solutions we seek are assumed to
vary as z
jk z
e ( , , ) ( , )
jk z
z
zz
H x y z h x y e 
2
22
22
22
,0
z
c
z
k
k k h x y
xy


    



 2 2 2
cz
k k k 
22
2
22
,0
cz
k h x y
xy

   

Solve subject to the appropriate
boundary conditions.2 2 2
2
2 2 2
0
z
kH
x y z
  
   
  
Transverse Electric (TE
z) Waves (cont.)  
22
2
22
,,
z c z
h x y k h x y
xy

   

(This is an eigenvalueproblem.)2
.
c
kThe eigenvalue is alwaysreal

20. 20
Once the solution for H
zis obtained, 22
22
z z z
xx
cc
z z z
yy
cc
jk H H j
HE
k x k y
jk H H j
HE
k y k x


 
  




TE wave impedanceTE
z
Z
k

 yx
y x z
EE
H H k

 
For a wave propagating in the positive zdirection (top sign):yx
y x z
EE
H H k

  
Transverse Electric (TE
z) Waves (cont.)
For a wave propagating in the negative zdirection (bottom sign):

21. 21
Also, for a wave propagating in the positive zdirection, 
ˆˆˆ
ˆˆˆ
1
ˆ()
t x y
t TE x y
TE t
tt
x TE y
y TE
TE
x
z e ye xe
z e Z
e
xh yh
Zh
hz
h
e Z h
e
Z
Z


  
   

  
   ˆˆ, , ,
t x y
e x y xe x y ye x y
Similarly, for a wave propagating in the negative zdirection, 1
ˆ()
tt
TE
h z e
Z
  
Transverse Electric (TE
z) Waves (cont.)   
1
ˆ,,
tt
TE
h x y z e x y
Z
 

22. 220
z
H
Transverse Magnetic (TM
z) Waves
In general, E
x,E
y,E
z,H
x,H
y0
To find the TE
zfield solutions (away from any sources), solve22
( ) 0
z
kE   2 2 2
2
2 2 2
22
) 0(0
zz
kE
x y z
kE  
  
    
  

23. 23
22
2
22
,0
cz
k e x y
xy

   

solve subject to the appropriate
boundary conditions
2
22
22
22
,0
z
c
z
k
k k e x y
xy


    



 2 2 2
cz
k k k
Transverse Magnetic (TM
z) Waves (cont.)  
22
2
22
,,
z c z
e x y k e x y
xy

   

(Eigenvalue problem)

24. 2422
22
c z z z
xx
cc
c z z z
yy
cc
j E jk E
HE
k y k x
j E jk E
HE
k x k y



  


  

TM wave impedancez
TM
c
k
Z

 yx z
y x c
EE k
HH 
  yx z
y x c
EE k
HH 
  
Once the solution for E
zis obtained,
For a wave propagating in the positive zdirection (top sign):
For a wave propagating in the negative zdirection (bottom sign):
Transverse Magnetic (TM
z) Waves (cont.)

25. 25
Also, for a wave propagating in the positive zdirection, 
ˆˆˆ
ˆˆˆ
1
ˆ()
t x y
t TM x y
TM t
tt
x TM y
y TM
TM
x
z e ye xe
z e Z
e
xh yh
Zh
hz
h
e Z h
e
Z
Z


  
   

  
   ˆˆ, , ,
t x y
e x y xe x y ye x y
Similarly, for a wave propagating in the negative zdirection, 1
ˆ()
tt
TM
h z e
Z
     
1
ˆ,,
tt
TM
h x y z e x y
Z
 
Transverse Magnetic (TM
z) Waves (cont.)

26. 26
Transverse ElectroMagnetic(TEM) Waves0, 0
zz
EH  
From the previous equations for the transverse field components, all of
them are equal to zero if E
zand H
z are both zero.
Unless2
0
c
k
For TEM waves 2 2 2
0
cz
k k k   zc
kk  
In general, E
x,E
y,H
x,H
y0
(see slide 16)
Hence, we have

27. 27
In a linear, isotropic, homogeneous source-free region, 0E   
 
 
0
,0
,0
,0
z
z
t
jk z
tt
jk z
tt
tt
E
e x y e
e e x y
e x y
  
   
   
    ˆˆ
t
xy
xy

  

In rectangular coordinates, we have
Notation:0
yx z
EE E
x y z
 
  
    ,0
tt
e x y  
Transverse ElectroMagnetic(TEM) Waves (cont.)

28. 28
Also, for the TEM
zmode we have from Faraday’s law (taking the z
component):  ˆˆ 0
z
z E z j H j H        0
y x
EE
xy


  ,0
tt
e x y   0
y x
ee
xy


 ,0
tt
e x y   ˆˆ
t
xy
xy

  

Notation:
or
Transverse ElectroMagnetic(TEM) Waves (cont.)
Taking the zcomponent of the curl, we have
Hence

29. 29 ,0
tt
e x y    ,,
tt
e x y x y    ,0
tt
e x y    ,0
tt
xy     
2
,0
t
xy  
Hence
Transverse ElectroMagnetic(TEM) Waves (cont.)

30. 30
2
,0xy  
Since the potential function that describes the electric field in
the cross-sectional plane is two dimensional, we can drop the
“t” subscript if we wish:
Transverse ElectroMagnetic(TEM) Waves (cont.)
Boundary Conditions: 

,
,
a
b
a
b
x y V
x y V


conductor " "
conductor " "
This is enough to make the potential function unique. Hence,
the potential function is the same for DC as it is for a high-
frequency microwave signal. 
2
,0xy  
a b

31. 31
Notes:
A TEM
zmode has an electric field that has exactly the same shape as
a static (DC) field. (A similar proof holds for the magnetic field.)
This implies that the Cand Lfor the TEM
zmode on a transmission
line are independent of frequency.
This also implies that the voltage drop between the two conductors of
a transmission line carrying a TEM
zmode is path independent.
A TEM
zmode requires two or more conductors (a static field cannot
be supported by a single conductor such as a hollow metal pipe.
Transverse ElectroMagnetic(TEM) Waves (cont.)

32. 32
TEM Solution Process
A) Solve Laplace’s equation subject to appropriate
B.C.s.:
B) Find the transverse electric field:
C) Find the total electric field:
D) Find the magnetic field:
2
,0xy    
1
ˆ;
TEM
H z E z
Z
    propagating TEM
z
Z
kk
   

 
      ,,
t
e x y x y   , , , ,
z
jk z
tz
E x y z e x y e k k

