Lecture 10 Waveguides
1
Microstrip
 Parallel plate
 Two-wire TL
 Coaxial cable


Waveguide refers to the structure that does not support
TEM mode, bring up “the cutoff frequency”

The wave characteristics are examined along straight
guiding structures with a uniform cross section such as
rectangular waveguides
We can write E in the instantaneous form as
E ( z , t )  E 0 cos(t   z )
or
S
E  E 0 e  z
We begin with Helmholz’s equations:
assume WG is filled in with a charge-free
lossless dielectric
2 E  k 2 E  0
2 H  k 2 H  0
where
k  u
2

 2 E  ( 2xy   2z ) E  ( 2xy  2 ) E   2xy E   2 E
z

 2xy E  ( 2   u2 ) E  0
and
 2xy H  ( 2   u2 ) H  0.
We can write E
and H in phasor forms as
E ( x, y , z )  ( E x a x  E y a y  E z a z ) e   z
and
H ( x, y , z )  ( H x a x  H y a y  H z a z )e   z
E
From
we have
  E   j H
Ez
  E y   j H x
y
Ez
 Ex 
  j H y
x
E y
Ex

  j H z
x
y
H z
  H y  j E x
y
H z
 H x 
 j E y
x
H y H x

 j Ez
x
y
H
E
H
We can express Ex, Ey, Hx, and Hy in terms of z-component
by substitution so we get for lossless media  = j,
Hy 
 j  H z
j Ez

u2   2 y  u2   2 x
j Ez
j  H z
Hx  2
 2
2
 u   y  u   2 x
 j H z
j  Ez
Ex  2
 2
2
 u   y  u   2 x
Ey 
j H z
j  Ez

 u2   2 x  u2   2 y

Transverse ElectroMagnetic (TEM) waves, no Ez or
Hz

Transverse Magnetic (TM), non-zero Ez but Hz = 0

Transverse Electric (TE), non-zero Hz but Ez = 0

Since Ez and Hz are 0, TEM wave exists only when
u2   2  0
u     
u p ,TEM 
ZTEM

1

rad / m
m/s
Ex
j



Hy
j


A single conductor cannot support TEM

From
2xy Ez  ( 2  u2 ) Ez  0
We can solve for Ez and then solve for other components from
(1)-(4) by setting Hz = 0, then we have
ZTM
Ex  E y




Hy
Hx
j

Notice that  or j for TM is not equal to that for TEM .
We define
h 2   2   u2
Solutions for several WG problems will exist only for real
numbers of h or “eigen values” of the boundary value problems,
each eigen value determines the characteristic of
the particular TM mode.
From
  h2  u2
The cutoff frequency exists when  = 0 or
h
or
fc 
We can write
f 2
  h 1 ( ) .
fc
2 
Hz.
h 2  u2  c2 
2
 f 
 f  1
 c
f  fc
or
and  is imaginary
  2   h2
Then
  j   j u 1  (
h
u
)  j u
2
fc 2
1 ( ) .
f
This is a propagating mode with a phase constant :
  u 1  (
fc 2
)
f
rad / m
Wavelength in the guide,
g 
2


u
 fc 
1  
 f 
2
m.
where u is the wavelength of a plane wave with a frequency
f in an unbounded dielectric medium (, )
2
up
u 


u f 
f
1
m
The phase velocity of the propagating wave in the guide is

up  

uu
 f 
1  c 
 f 
2
g
 u

The wave impedance is then
ZTM
 fc 
  1  
 f 
2

m/s
2
 f 
 f  1
 c
or
f  fc
  2   h2
Then
    u
f 2
1  ( )  real
fc
Wave diminishes rapidly with distance z.
ZTM is imaginary, purely reactive so there is no power flow
From
2xy H z  ( 2  u2 ) H z  0
We can solve for Hz and then solve for other components from
(1)-(4) by setting Ez = 0, then we have
ZTE
Ex  E y j



Hy
Hx


Notice that  or j for TE is not equal to that for TEM .
 Cutoff frequency fc,, g, and up are similar to those in TM mode.
 But
ZTE 

 fc 
1  
 f 
 Propagating mode f > fc
 Evanescent mode f < fc
2


Finding E and H components in terms of z, WG geometry,
and modes.
From
2xy Ez  ( 2  u2 ) Ez  0,
Expanding for z-propagating field gets
 2 Ez  2 Ez
2


(

u   ) Ez  0
2
2
x
y
where
E z  E z ( x , y )e   z
Assume
Ez ( x, y)  XY
where X = f(x) and Y = f(y).
Substituting XY gives
d2X
d 2Y
2
Y

X

(

u   ) XY  0
2
2
dx
dy
and we can show that
2
2
1
d
Y
1
d
X
2
2
u   

.
2
2
Y dy
X dx
Let
and
then we can write
2
1
d
X
2
x  
X dx 2
2
1
d
Y
 y2  
Y dy 2
  u2   x2   y2 .
We obtain two separate ordinary differential equations:
d2X
2


x X 0
2
dx
d 2Y
2


yY  0
2
dy
X ( x)  c1 cos   x x   c2 sin   x x 
Y ( y )  c3 cos   y y   c4 sin   y y 
Appropriate forms must be chosen to satisfy boundary
conditions
1. in the x-direction
Et at the wall = 0 Ez(0,y) and Ez(a,y) = 0
and X(x) must equal zero at x = 0, and x = a.
Apply x = 0, we found that C1 = 0 and X(x) = c2sin(xx).
Therefore, at x = a, c2sin(xa) = 0.
  x a  m
m
x 
.
a
(m  0,1, 2,3,...)
2. in the y-direction
Et at the wall = 0 Ez(x,0) and Ez(x,b) = 0
and Y(y) must equal zero at y = 0, and y = b.
Apply y = 0, we found that C3 = 0 and Y(y) = c4sin(yy).
Therefore, at y = a, c4sin(yb) = 0.
  y b  n
n
y  .
b
(n  0,1, 2,3,...)
 m   n 
    
  b 
a

 

2
2
u
 m   n 
h



 a   b 
2
and
2
rad / m
2
therefore we can write
Ez  E0 sin
m x
n y  j z
sin
e
V /m
a
b
 Every combination of integers m and n defines possible
mode for TMmn mode.
 m = number of half-cycle variations of the fields in the xdirection
 n = number of half-cycle variations of the fields in the ydirection
 For TM mode, neither m and n can be zero otherwise Ez
and all other components will vanish therefore TM11 is the
lowest cutoff mode.
fc ,mn 
h
2 
c ,mn 

2
1
2 
m n
 a   b 
   
2
2
m n
 a   b 
   
2
m
2
Hz
a)
Find h, fc, and c for TM11 mode
b)
If the operating frequency is 15% higher than the cutoff
frequency, find (Z)TM11, ()TM11, and (g)TM11. Assume the
wg to be lossless for propagating modes.