Chapter 6 Metallic Waveguide and Cavity Resonators
6-1 General Metallic Waveguides
How to study the theory of metallic waveguides (by
L. J. Chu, 朱蘭成):
1. Specify a proper coordinate system, and derive
waveguide’s equations to express the transverse
components of the E- and H-fields in terms of the
longitudinal components by Maxwell’s equations.
2. Caculate the eigenmodes (TM mode, TE mode, TEM mode or other types of
modes) of the waveguide, and obtain the eigenvalues and the longitudinal
field-components of the corresponding eigenmodes by solving the wave equations.
Substituting the longitudinal field-components into the longitudinal components,
we can obtain the other components. If the eigenmode is injected into a waveguide,
it can propagate along an infinitely-long straight waveguide without any
deformation. However, in case the input EM wave is not an eigenmode, some
power loss occurs and then it becomes the eigenmode gradually. All the
eigenmodal functions in an infinitely-long straight metallic waveguide are
orthogonal to each other. Moreover, these eigenmodes form a complete set (a
basis in a vector space), such that any electromagnetic fields within the waveguide
can be uniquely expressed by the eigenmodal functions.
3. Obtain the quantities of the physical characteristics for a given eigenmode, such as
the cutoff frequency (fc), the propagation constant (γ=α+jβ), the phase velocity
(vp=ω/β), the group velocity ( v g   /  ), the impedance Z, etc.
Waveguide’s equations: According to Ampere’s law and Faraday’s law, we obtain
H x0  
E x0  
H z0
E z0
H z0
E z0
1
1
0
,
(


j

)
H


(


j

)
y
x
y
y
x
h2
h2
E z0
H z0
E z0
H z0
1
1
2
2
2
0
,
(


j

)
E


(


j

) , where h =γ +k ,
y
2
2
x
y
y
x
h
h





2
2
2
2
2
2
2


 E  k E  0
 t E  (  k ) E  0  [ t  h ]E
 2 
and  2 



2
2
2
2
2

 H  k H  0 
 t H  (  k ) H  0  [ t  h ]H
Case 1 TEM mode: Ez=Hz=0
h 2  0   TEM  k 2   TEM  jk  j  , v p 
2
Z TEM 
E x0
j  TEM



0

j
Hy
TEM
1



1

zˆ  E
  and H 
Z TEM

Note: All frequencies make γTEM is pure imaginary  TEM wave can propagate at
any frequency, no cutoff
Case 2 TM mode: Hz=0, Ez≠0 and ▽t2Ez+h2Ez=0
j E z0
 E z0
 E z0
j E z0
0
0
0
,
,
,
H


E


E


y
x
y
h 2 x
h 2 x
h 2 y
h 2 y

E y0
1
E x0

j and 
H
( zˆ  E )
 Z TM  0   0 
(
)
Z TM
j

Hy
Hx
H x0 
  h2  k 2  h 1  (
h
f 2
) , where f c 
fc
2 
If f>fc,   j  j  1  ( f c ) 2  Z TM   1  ( f c ) 2 , v p 
f
f
1
 1  ( f c / f ) 2
Case 3 TE mode: Ez=0, Hz≠0 and ▽t2Hz+h2Hz=0
H x0  
 Z TE 
 H z0
h 2 x
, H y0  
 H z0
h 2 y
, E x0  
j H z0
j H z0
0
,
E

y
h 2 x
h 2 y


E y0
E x0
j

ˆ  H)
and
E


Z
(
z



(

)
TE

j
H y0
H x0
If f>fc,   j  jk 1  ( f c ) 2  j  1  ( f c ) 2
f
 Z TE 

1  ( fc / f )2
, vp 
f
1
 1  ( f c / f ) 2
A case of longitudinal vp>0 but longitudinal vg=0 in barber’s pole.
6-2 Parallel–Plate Waveguides
Case 1 TMn mode: Hz=0, E z ( x, y)  E z0 ( y)e z
dE z0 ( y )
 h 2 E z0 ( y )  0 , E z0 ( y)  0 at y=0 and b
2
dy
 Eigenvalues: h=
n
ny
) , n=0, 1, 2, 3, …
, E z0 ( y )  An sin(
b
b
 0
j dE z0 ( y ) j
ny
H
(
y
)


An cos(
)
 x
2
dy
h
b
h
n

and   h 2   2   ( ) 2   2 

0
b
 E 0 ( y )    dE z ( y )   h A cos( ny )
y
n
2

dy

b
h
Cutoff frequency: fc=
n
2b 
fulfills γ=0. (Note: n=0 is the TEM mode)
Case 2 TEn modes: Ez=0, H z ( y, z )  H z0 ( y)e z
d 2 H z0 ( y )
dH z0 ( y )
ny
2
0
) and
,

h
H
(
y
)

0
 0 at y=0 and b  H z0 ( y )  Bn cos(
z
2
b
dy
dy
n
fc=
2b 
 0
j dH z0 
ny
 Bn sin(
)
H y ( y)   2
dy
h
b
h
n

and γ= ( ) 2   2  , n=1, 2, 3, …

0
b
 E 0 ( y )    dH z  j B sin( ny )
x
n

h
b
h 2 dy
Eg. (a)Write the instantaneous field
expression for TM1 mode in a
parallel-plate waveguide. (b) Sketch the
E- & H- field lines in the yz-plane.
(Sol.) (a) For n=1, E y ( y, z, t ) 
H z ( y, z , t )  
(b)
b
y
A1 cos( ) sin( t  z )

b
b
y

A1 cos( ) sin( t  z ) ,    2   ( ) 2

b
b
dy dz
dy E y ( y, z,0)
b
y
y



  cot( ) tan z  cos( ) cos z =constant
b
E y Ez
dz E z ( y, z,0)

b
Eg. (a) Write the instantaneous field expression for TE1 mode in a parallel-plate
waveguide. (b) Sketch the electric and magnetic field lines in the yz-plane.
(Sol.) (a) For n=1, H z ( y, z; t )  B1 cos(
H y  ( y, z; t )  
E x ( y , z; t )  
y
b
) cos(t  z ) ,
b
y
B1 sin( ) sin( t  z )

b
b
y
B1 sin( ) sin( t  z ) , where β is the same as that of the TM1

b
mode.
(b) At t=0,
b
y
B1 sin( ) sin z ,

b
dy H y ( y, z;0) b
y


tan( ) tan z
dz H z ( y, z;0) 
b
E x ( y, z;0)  
Eg. Find the electric and the magnetic fields of the propagating wave in a
parallel waveguide b=5cm, filled with a dielectric (4ε0,μ0) and excited by

H  yˆ cos 40x sin 8 10 9 t .
(Sol.) f=4×109Hz, cos40πx= cos(
( f c )TM 2 
2
2b  0 4 0
    0 4 0 1  (
2x
) : TM2 mode
0.05
 3 109 , ( f c )TM 2  f , ∴ TM2 mode can propagate!
( f c ) TM 2
f

9
) 2  110.82  H  yˆ cos(40x) sin( 8 10 t  110.82 z )

E  xˆ124.67 cos(40x) sin( 8 109 t  110.82 z )  zˆ141.37 sin( 40x) cos(8 109 t  110.82 z )
Eg. Find the electric field of the propagating wave in an air parallel waveguide

b=5cm excited by E  yˆ10(sin 20x  0.5 sin 60x) sin 1010 t .
(Sol.) f=5×109Hz, sin20πx= sin(
( f c )TE1 
1
2b  0 0
x
0.05
 3 109 , ( f c )TE3 
) : TE1 mode, sin60πx= sin(
3
2b  0 0
 9 109 , ( f c )TE1  f  ( f c )TE3
∴ Only TE1 mode can propagate!     0  0 1  (

80z
E  yˆ10 sin( 20x) sin( 1010 t 
)
3
3x
) : TE3 mode
0.05
( f c ) TE1
f
)2 
80
3
Energy-transport velocity and attenuation in parallel-plate waveguides:
(P )
Energy velocity: ven  z' and (ven ) TE  (ven ) TM
Wav
Energy velocity of TM mode:


 
1
1
( Pz ) av   Pav  dS   Re[ E  H *]  zˆdy   Re(  zˆE y0 H x0  yˆE z0 H x0 )  zˆdy
2
2
s
s
s

b
4h
w'av 
2
An
2

 
 

A 2
ny
2
ny
  2 2
ny
2
Re( E  E*)  Re( H  H *)  n [sin 2 (
)  2 cos 2 (
)]  ( 2 ) An cos(
)
4
4
4
b
b
4 h
b
h
b
W ' av   w' av dy 
0
b
4h
k 2 An  ven 
2
2

k
2

1

1 (
fc 2
)
f
Attenuation constant: α=αd +αc
α of TEM mode:  d 
G

R0 
2
2
R
1 f
 
 ' '

 
 , c 
2R0 b  c

2
2
α of TM mode:
  j[ 2  (1 
 j   1  (
j

)(
n 2 1 / 2
) ]  j  2   ( n ) 2  {1  j [ 2   ( n ) 2 ]1}
b
b
2
b
fc 2
j
)  {1 

f
2
     1  (

1
,
}   d  j   d 
fc 2
fc 2
1 ( )
2 1 ( )
f
f
fc 2
)
f
b
bA
PL ( z )
1
c 
, where P( z )  w  ( E y0 )( H x0 ) * dy  wb( n ) 2 ,
2 P( z )
2
2n
0
J sz0  H x0 ( y  0) 
Rs 
2
bAn
bAn 2
1
 PL ( z )  2w( J sz0 Rs )  w(
) Rs ,
n
2
n
f c
c f c
2
1
 c 


c
b
c
( f c / f )[1  ( f c / f ) 2 ]
α of TE mode: αd is the same as the expression in TM mode
P( z )  w
b
0
bB
1 0
w bBn 2 b 2 ny
( E x )( H y0 ) * dy 
(
)  sin (
)dy  wb( n ) 2 ,
0
2
2
n
b
2n
     1  (
PL ( z )  2w(
fc 2
)
f
2
1 0 2
J sx Rs )  w H z0 ( y  0) Rs  wB 2 Rs
2
2 Rs n 2
2Rs f c2
( fc / f )
PL ( z )
2 c f c
c 

( ) 

2
2
2 P( z ) b b
b  c
1  ( fc / f )2
bf 1  ( f c / f )
 
 
Note: ∵  c  of the higher-order modes >  c  of the lower-order modes, ∴
 d 
 d 
the lowest-order mode is often utilized in communication systems. Otherwise, the
signal decays very soon.
Eg. A waveguide is formed by two parallel copper sheets, which is separated by a
5cm thick lossy dielectric εr=2.25, μr=1, σ=10-10(S/m). For an operating frequency
of 10GHz, find αd, αc, β, vp, vg and λg for (a) the TEM mode, (b) the TM1 mode.
(Sol.) σc=5.8×107(S/m)
(a) TEM:      314.16rad / m ,  d 
1 f
 2.078  10 3 N p / m , v p 
b c
c 
g 
vp
f
vp 

2
 1.257  10 8 N p / m ,
 2  10 8 m / s , v g  v p ,
 0.02m
(b) TM1: f c 
d 
1

1
2b 

2 1  ( fc / f )
2
 2 109 Hz ,      1  (
 1.282  10 8 N p / m ,  c 
fc 2
)  307.88rad / m
f
1
1

 4.237  10 3 N p / m
2
b ( f c / f )[1  ( f c / f ) ]
v
(3 108 ) 2

 1.96 108 m / s , g  p  0.0204m
 2.041108 m / s , v g 
2.25v p

f
Eg. A parallel-plate waveguide made of two perfectly conducting infinite planes
spaced 3cm apart in air operates at a frequency 10GHz. Find the maximum
time-average power that can be propagated per unit width of the without a
voltage breakdown for (a) the TEM mode, (b) the TM1 mode, (c) the TE1 mode.
1
(Sol.) Without breakdown: Emax=3×106V/m, b=3×10-2m, f c 
 5 109 ( Hz ) ,
2b 
     1  (
fc 2
)
f
2
bE
1
TEM: P  Re( E y H x* )  b  max  3.581108
2
2
bA
b
3 10 6 
, Pav  b( 1 ) 2  2.067 108 W
A1  3 106  A1 
2

b
6
bB
b
3 10 
TE1:
B1  3 106  B1 
 Pav  b( 1 )  1.55 108 W

b
2
TM1:
6-3 Rectangular Waveguides
Case 1 TMmn modes: Hz=0, E z ( x, y, z )  E z0 ( x, y)e z
( t2  h 2 ) E z0 ( x, y )  0  (
2
2

 h 2 ) E z0 ( x, y)  0 ,
2
2
x
y
E z0 ( x, y)  0 at x=0, a, and y=0, b
Eigenvalues: h 2  (
m 2
n
mx
ny
)  ( ) 2 , E z0 ( x, y )  E 0 sin(
) sin(
)
a
b
a
b
Waveguide’s equations 
 m
m
n
 n
m
n
 0
0
E x ( x, y )   h 2 ( a ) E0 cos( a x) sin( b y ), E y ( x, y )   h 2 ( b ) E0 sin( a x) cos( b y )

H 0 ( x, y)  j ( n ) E sin( m x) cos( n y ), H 0 ( x, y)   j ( m ) E cos( m x) sin( n y )
x
0
y
0
a
b
a
b
h2 b
h2 a

  j  j  2   (
m 2 n 2
)  ( ) . Note: TMmn mode, neither m nor n can be zero.
a
b
Cutoff frequency: ( f c )TM mn 
1
2 
m
n
( )2  ( )2
a
b
In case of f>fc: waves can propagate, else if f<fc: evanescent waves (cutoff).
Case 2 TEmn modes: Ez=0, H z ( x, y, z )  H z0 ( x, y)e z
 H z0
 0 (or E y  0)

x  0, a
2
2
( 2  2  h 2 ) H z0 ( x, y )  0 ,  x 0
at
x
y
y  0, b
 H z  0 (or E  0)
x
 y
 H z0 ( x, y )  H 0 cos( mx ) cos( ny )
a
b
Waveguide’s equations 
j n
m
n
j m
m
n
 0
0
E
(
x
,
y
)

(
)
H
cos(
x
)
sin(
y
),
E
(
x
,
y
)


(
)
H
sin(
x
)
cos(
y)
x
0
y
0
2
2

b
a
b
a
a
b
h
h

H 0 ( x, y )   ( m ) H sin( m x) cos( n y ), H 0 ( x, y )   ( n ) H cos( m x) sin( n y )
0
y
0
 x
a
b
a
b
h2 a
h2 b
where   j  2   (
1
m 2 n 2
)  ( ) and ( f c ) TEmn 
a
b
2 
m
n
( )2  ( )2
a
b
In case of 2b>a>b, the fundamental mode of the rectangular waveguide is TE10 mode.
1
v
It has the lowest cutoff frequency ( f c ) TE10 

( Hz )
2a  2a
Note: We prefer a waveguide that is single-mode because the fundamental mode has
the lowest attenuation and avoid modal dispersion. The bandwidth of the
single-mode waveguide is (fc)1<f<(fc)2.
Dispersion: Waves have different velocities
Modal Dispersion: Different modes have distinct phase velocities.
Magic-T-junction: (made of rectangular metallic waveguide)
Eg. Calculate and compare the values of β, vp, vg, λg and Z TE1 0 for a 2.5cm×1.5cm
rectangular waveguide operating at 7.5GHz. if the waveguide is hollow.
(Sol.) μ=μ0,ε=ε0, ( f c )TE10 
     1  (
1 1
 6 109 Hz ,
2a 
fc 2
) =94.25rad/m, v p 
f
λg=vp/f=0.067m, Z TE1 0 
120
1 ( fc f )
2
1

1 (

fc 2
) =0.6
f
1
1 ( fc f )
=628.3Ω, v g 
2
= 5×108 m/s
f
1
 1  ( c ) 2 =1.8×108 m/s
f

Eg. Calculate and list in ascending order the cutoff frequencies of an a×b
rectangular waveguide for the following modes: TE01, TE10, TE11, and TE02 if
a=2b.
(Sol.) f c 
1
1
m
n
, ( f c ) TE01 
( ) 2  ( ) 2 , a=2b  ( f c ) TE10 
a
b
2b 
4b 
1
2 
( f c ) TE11  ( f c ) TM11 
5
4b 
, ( f c ) TE02 
1
b 
Eg. An air-filled a×b (b<a<2b) rectangular waveguide is to be constructed to
operate at 3GHz in the dominant mode. We desire the operating frequency to be
at least 20% higher than the cutoff frequency of the dominant mode and also at
least 20% below the cutoff frequency of the next higher-order mode. (a) Give a
typical design for the dimensions a and b. (b) Calculate for your design β, vp, λg
and the wave impedance at the operating frequency. [台大電研]
(Sol.) (a) f c 
1
m
n
 ( ) 2  ( ) 2 . b<a<2b, the dominant mode: TE10, the next
a
b
2 
mode: TE01
( f c ) TE10 
1
2a 
, ( f c ) TE01 
(1 2b  )  3  10 9
(1 2b  )
1
2b 
,
3  10 9  (1 2a  )
(1 2a  )
 20%  a  0.06m , b  0.04m , and a<2b
(b) Choose a=0.065m, b=0.035m, ( f c )TE10  2.3 109 ( Hz ) ,
     1  (
g 
vp
f
 20% ,
fc 2
)  40.15rad / m , v p 
f
 0.157m , Z TE10  0
1


1 (
1
1 ( fc f )
2
fc 2
)  0.679 ,
f
 4.7  108 m / s ,
1  ( f c f ) 2  120 / 0.639  590
Eg. A 3cm×1.5cm rectangular waveguide operating at 6GHz has a dielectric
discontinuity between medium 1 (μ0,ε0) and medium 2 (μ0,4ε0). (a) Find the SWR
in the free-space region. (b) Find the length and the permittivity of a
quarter-wave section to achieve a match between two media.
1
1
1
1
(Sol.) For TE10 mode, f c1 
 5  10 9 Hz , f c 2 
 2.5  10 9 Hz
2a  0  0
2a  0 4 0
Z1   0  0
1  ( f c1 f ) 2  682 , Z 2  0 4 0
1  ( f c 2 f ) 2  207
1 
Z  Z1
 2
 0.5337  SWR= 1  =3.289
Z 2  Z1
f c3 
1
1
and Z 3   0  r  0
2a  0  r  0
1  ( f c3 f ) 2  Z1 Z 2  εr=1.6995
d=λ3/4=vp3/4f=1.24685×10-2m
Eg. (a) Write the instantaneous field expressions for the TM11 mode in a
rectangular waveguide of sides a and b. (b) Sketch the electric and magnetic field
lines in a typical xy-plane and in a typical yz-plane.
(Sol.) (a) m=1, n=1
E x ( x, y , z ; t ) 
 
E z ( x, y, z; t )  E 0 sin(
H y ( x, y , z ; t ) 


 


( ) E 0 cos( x) sin( y ) sin( t  z ) , E y ( x, y, z; t )  2 ( ) E 0 sin( x) cos( y ) sin( t  z ) ,
a
b
a
b
h b
h a
2

a
 
x) sin(


b
y ) cos(t  z ) , H x ( x, y, z; t )    ( ) E 0 sin(  x) cos(  y) sin( t  z) ,
2
h

( ) E 0 cos( x) sin( y ) sin( t  z ) , where  
a
b
h a
2
(b) In a typical xy-plane,
(
b
a
b


k 2  h 2   2   ( ) 2  ( ) 2
a
b
dy
a


dy
b


) E  tan( x) cot( y ) , ( ) H   cot( x) tan( y )
dx
b
a
b
dx
a
a
b
Eg. For an a×b rectangular waveguide operating at the TM11 mode, derive the
expressions for the surface current densities on the conducting walls.
(Sol.)


j 
y
j 
y

J s ( x  0)  xˆ  [ yˆ ( 2 ( ) E 0 sin( )]   zˆ 2 ( ) E 0 sin( ), J s ( x  b)  J s ( x  0)


a
b
a
b
h
h



j

j



x


x
 J ( y  0)  yˆ  [ xˆ
( ) E 0 sin( )]   zˆ 2 ( ) E 0 sin( ), J s ( y  b)  J s ( y  0)
s
2

b
a
b
a
h
h

Eg. (a) Write the instantaneous field expression for the TE10 mode in a
rectangular waveguide having sides a and b. (b) Sketch the electric and magnetic
field lines in typical xy-, yz-, and xz-planes. (c) Sketch the surface currents on the
guide walls.
(Sol.) (a) m=1, n=0,
,
,
E x ( x, y, z; t )  0
E z ( x, y, z; t )  0
E y ( x, y , z ; t ) 
H x ( x, y , z ; t )  
 
 

( ) H 0 sin( x) sin( t  z ) ,
a
h a
2

( ) H 0 sin( x) cos(t  z ) , H y ( x, y, z; t )  0
a
h a
2
dx
 


where   k 2  h 2   2   ( ) 2 (b) ( ) H  2 ( ) tan( x) tan z
dz
a
h a
a



(c) J s  aˆ n  H . At t=0, J s ( x  0)   yˆH z (0, y, z;0)   yˆH 0 cos z


J s ( x  a)  yˆH z (a, y, z;0)  J s ( x  a)


 

J s ( y  0)  xˆH z ( x,0, z;0)  zˆH x ( x,0, z;0)  xˆH 0 cos( x) cos( z )  zˆ 2 ( ) H 0 sin( x) sin z
a
a
h a


J s ( y  b)   J s ( y  0)
Eg. The eigenmodes in the rectangular waveguide.
Bending and T-Branches of Waveguides:
Attenuation in the rectangular waveguide:
α=αd +αc, where  d 

2 1  ( fc f )2
and  c 
PL ( z )
.
2 P( z )
Consider TE10 mode:
P( z )  
b
0

a
0
b
1
1
a
 ( E y0 )( H x0 ) * dxdy   ( ) 2 H 02 
0
2
2

a
 sin
0
2
aH

( x)dxdy  ab( 0 ) 2
a
2
and


 J s0 ( x  0)  J s0 ( x  a)   yˆ y H z0 ( x  0)   yˆH 0

0
 0

a

0
0
H 0 sin( x)
 J s ( y  0)   J s ( y  b)  xˆH z ( y  0)  zˆH x ( y  0)  xˆH 0 cos( x)  zˆ
a

a

PL ( z )  2[ PL ( z )] x 0  2[ PL ( z )] y 0 , [ PL ( z )] x 0  
b
0
and [ PL ( z )] y 0  
a
0
2
1 0
b
J s ( x  0) Rs dy  H 02 Rs
2
2
2
2
1 0
a
a
[ J sx ( y  0)  J sx0 ( y  0) ]Rs dx  [1  ( ) 2 ]H 02 Rs
2
4

a
a 
a f

 PL ( z )  b  [1  ( ) 2 ] H 02 Rs  [b  ( c ) 2 ]H 02 Rs
2

2 f


( c ) TE10 
Rs [1  (2b a)( f c f ) 2 ]
b 1  ( f c f ) 2

Similar approach  ( c ) TM 
11
f c
1
2b f
f c
[1  ( c ) 2 ] , Rs 
2
c
b  c [1  ( f c f ) ]
a f
2 R s [(b a 2 )  (a b 2 )]
ab 1  ( f c f ) 2 [(1 a 2 )  (1 b 2 )]




2 Rs
f
b f
1

( c ) TE mn 
 1  ( )( c ) 2  [  ( c ) 2 ] 
2
a f
2
f

b 1  ( f c f ) 

General Cases: 


b

m2 ( )3  n2

2 Rs
a

( c ) TM mn 
2
b 1  ( f c f ) m 2 ( b ) 2  n 2


a


b b 2
( m  n2 )

a a

2
b
2
2 
m n

a2

Eg. A TE10 wave at 10GHz propagates in a brass σc=1.57×107(S/m) rectangular
waveguide with inner dimensions a=1.5cm and b=0.6cm, which is filled with
εr=2.25, μr=1, loss tangent=4×10-4. Determine (a) the phase constant, (b) the
guide wavelength, (c) the phase velocity, (d) the wave impedance, (e) the
attenuation constant due to loss in the dielectric, and (f) the attenuation constant
due to loss in the guide walls.
v
3 108
2 108


 0.02m
f
1010
2.25 1010
v
2  10 8
For TE10 mode, f c 

 0.667  1010 Hz
2
2a 2  (1.5  10 )
(Sol.) f=1010Hz,  


vp 
v
1 (
fc 2

)  234rad / m ,.  g 
 0.0268m
f
1  ( fc f )2
v
1  ( fc f )2
 2.68  108 m / s , Z TE 
10
  4  10 4   5  10 4 S / m ,  d 
Rs 
c 

2
 
1  ( fc f )2
 337.4()
Z TE10  0.084 Np / m  0.73dB / m
f c
 0.05101() ,
c
Rs [1  (2b a)( f c f ) 2 ]
b 1  ( f c f ) 2
 0.0526 Np / m  0.457dB / m .
Eg. (a) Determine the value of fc/f at which the attenuation constant due to
conductor losses in an a×b rectangular waveguide for the TE10 mode is a
minimum. What is the minimum obtainable αc in a 2cm  1cm guide? At what
frequency? (b) Determine the value of ( f f c ) TM 1 1 at which this attenuation
constant is a minimum.
(Sol.) (a) ( c )TE10 
f c
1
2b f c 2 d ( c ) TE10
0


[
1

( ) ],
df
b  c [1  ( f c f ) 2 ]
a f
(6b  3a)  (6b  3a) 2  8ab 1 / 2
 f  fc [
]
2a
(b) ( c )TM 11 
2 Rs (b a  a b 2 )
2
ab 1  ( f c f )  (1 / a  1 / b )
2
2
2
,
d ( c ) TM
df
11
 0  f  3 fc
Eg. An air-filled rectangular waveguide made of copper and having transverse
dimensions a=7.20cm and b=3.40cm operates at a frequency 3GHz in the
dominant mode. Find (a) fc, (b) λg, (c) αc, and (d) the distance over which the field
intensities of the propagating wave will be attenuated by 50%.
1
(Sol.) σc=5.8×107S/m (a) ( f c )TE10 
 2.083GHz  3GHz
2a 

(b)  g 
 0.109m
1 ( fc f )2
(c) ( c )TE 
10
f c
1
2b f
 [1  ( c ) 2 ]  2.26  10 3 N p / m
2
b  c [1  ( f c f )
a f
(d) 0.5  e  c d  d  307.25m
Eg. An average power of 1kW at 10GHz is to be delivered to an antenna at the
TE10 mode by an air-filled rectangular copper waveguide 1m long and having
sides a=2.25cm and b=1.00cm. Find (a) the attenuation constant due to conductor
losses, (b) the maximum values of the electric and magnetic field intensities
within the waveguide, (c) the maximum value of the surface current density on
the conducting walls, (d) the total amount of average power dissipated in the
waveguide. [台大電研]
1
f
(Sol.) (a) ( f c ) TE10 
 6  10 9 ( Hz ) , f  10  10 9 Hz , 1  ( c ) 2  0.7454
f
2a 
( c ) TE10 
f c
1
2b f

 [1  ( c ) 2 ]  0.13Np / m
2
b  c [1  ( f c f ) ]
a f
(b) P  ab(
aH 0 2

) ,      1  ( f c f ) 2  156.1rad / m , h 2  ( ) 2
2
a
 1000  3.56  10  2 H 02  H 0  167 A / m  E max  E y0 ( x, y )
max

 
 H 0  94800V / m
h2 a
a
H 0  187.4 A / m , H z0 ( x, y)
 H 0  167 A / m .
max
max

If at input end, all factor×e0.13 (=1.138)



(c) J s ( x  0)   yˆH 0 cos z  J s ( x  0)  H 0 , J s ( y  0)  yˆ  ( xˆH x0  zˆH z0 ) y 0
H x0 ( x, y )

1/ 2
E
f
E  f
f
x 
 J s ( y  0)  0  ( c ) 2  [1  2( c ) 2  sin( )] , ∴ Max of J s ( y  0)  0  1  ( c ) 2
0
f
0  f
f
 
At the input end, this factor×1.138
(d) P  1000  [e 2c 1  1]  26.2W (∵ P 
E
2
 e 2  )
Eg. Find the maximum amount of 10GHz average power that can be transmitted
through an air-filled rectangular waveguide a=2.25cm, b=1.00cm at the TE10
mode without a breakdown. (Without breakdown: E max  3 10 6 V / m )
(Sol.) H 0  5305 A / m , P( z )  ab(aH 0 2 ) 2  1MW
6-4 Circular Waveguides
Circular waveguide’s equations:
E z  H z
H z
j
j  E z

 E r   h 2 [  r  r  ], H r  h 2 [ r    r ]


 E  j [  E z   H z ], H   j [ E z   H z ]

  h 2 r 
r
r
r 
h2
Case 1 TMnp modes: Hz=0, and ▽t2Ez+h2Ez=0

1  E z
1  2 Ez
(r
) 2
 h 2 Ez  0
r r r
r  2
E z0 (r , )  Cn J n (hr ) cos n and E z (r , , z )  E z0 (r , )e z
j
jn
 0
Er  
C n J ' n (hr ) cos n , E0  2 C n J n (hr ) sin n


h
h r

j

n
j
H 0  
C n J n (hr ) sin n , H 0  
C n J ' n (hr ) cos n
r
2

h
h r

E z0  H z0  0 at r=a and h2=γ2+k2  h fulfills Jn(ha)=0 (the first root of J0(x) is
2.405)
For TM01 mode, cutoff frequency: ( f c ) TM 0 1 
hTM 0 1
2 

0.383
a 
( hTM 01 
2.405
)
a
Note: h fulfills the pth root of Jn(ha)=0 for TMnp mode, and h fulfills the pth root of
J’n(ha)=0 for TEnp mode.
Case 2 TEnp modes: Ez=0, ▽t2Hz+h2Hz=0  H z0 (r ,  )  C ' n J n (hr ) cos n
Waveguide’s equations
j
j n
 0
0
 H r   h C ' n J ' n (hr ) cos n , H   h 2 r C ' n J n (hr ) sin n

 E 0  jn C ' J (hr ) sin n , E 0   j C ' J ' (hr ) cos n
r
n
n

n
n
h
h2r

H z0
 0 at r=a  h fulfills J’n(ha)=0 (the first root of J’1(x) is 1.841)
E  0,
r
0
z
Cutoff frequency of TE11 mode: ( f c ) TE1 1 
hTE1 1
2 

0.293
a 
( hTE11 
1.841
)
a
Note: TE11 mode is the fundament (dominant) mode of a circular waveguide.
Eg. (a) A 10GHz signal is to be transmitted inside a hollow circular conducting
pipe. Determine the inside diameter of the pipe such that its lowest cutoff
frequency is 20% below this signal frequency. (b) If the pipe is to operate at
15GHz, what waveguide modes can propagate in the pipe?
0.293
0.879
(Sol.) (a) ( f c )TE11 

108 ( Hz ) , 10  (1-20%)=8, 2a=0.022m.
a
a  0 0
(b) fc of waveguide with a=0.011(m) is ( f c ) TE =8GHz<15GHz.
11
( f c ) TM 01 
2.405
2a  0 0
 10.45(GHz ) , ( f c ) TE21 
3.054
2a  0  0
 13.27(GHz)
6-5 Rectangular Cavity Resonators
Case 1 TMmnp mode: Hz=0, neither m nor n =0, p can be 0.
m
n
p

 E z ( x, y, z )  E 0 sin( a x) sin( b y ) cos( d z )

 E ( x, y, z )   1 ( m )( p ) E cos( m x) sin( n y ) sin( p z )
0
 x
d
a
b
d
h2 a

1 n p
m
n
p

) E 0 sin(
x) cos(
y ) sin(
z) ,
 E y ( x, y, z )   2 ( )(
d
a
b
d
h b

j n
m
n
p

 H x ( x, y, z )  h 2 ( b ) E 0 sin( a x) cos( b y ) cos( d z )

 H ( x, y, z )   j ( m ) E cos( m x) sin( n y ) cos( p z )
0
 y
a
a
b
d
h2
where h 2  (
m 2
n
p 2
)  ( )2  (
)
a
b
d
Case 2 TEmnp mode: Ez=0, p≠0. Either m or n =0, but not both.
m
n
p

 H z ( x, y, z )  H 0 cos( a x) cos( b y ) sin( d z )

 E ( x, y, z )  j ( n ) H cos( m x) sin( n y ) sin( p z )
0
 x
b
a
b
d
h2

j m
m
n
p

) H 0 sin(
x) cos(
y ) sin(
z)
 E y ( x, y , z )   2 (
a
a
b
d
h

1 m p
m
n
p

 H x ( x, y, z )   h 2 ( a )( d ) H 0 sin( a x) cos( b y ) cos( d z )

 H ( x, y, z )   1 ( n )( p ) H cos( m x) sin( n y ) cos( p z )
0
 y
d
a
b
d
h2 b
Both have the same resonant frequency (degenerate modes): f mnp 
1
m n
p
 ( )2  ( )2  ( )2
b
d
2  a
Note: TE101 mode is the dominant mode of the rectangular resonator in case of
a>b<d.
Eg. Given an air-filled lossless rectangular cavity resonator with dimensions
8cm  6cm  5cm, find the first twelve lowest-order modes and their resonant
frequencies.
c
m 2
n 2
p 2
m
n
p
(
) (
) (
)  1.5  1010  ( ) 2  ( ) 2  ( ) 2
2 0.08
0.06
0.05
8
6
5
 f r  3.125GHz , TE101  f r  3.54GHz ,
(Sol.) f r 
TM 110
TE111, TM111  f r  4.33GHz , ……
Quality Factor of Rectangular Cavity Resonators: Q=
W
, W=We+Wm
PL
Q=2π×(Time-average energy stored at a resonant frequency)/(Energy dissipated in
one period)
Consider TE101 mode:
We 
0
4

2
E y dv 
 0  2  02 2
4h 4 a 2
H 02 
d
0
b
a
0
0
 
sin 2 (

a
x) sin 2 (

d
z )dxdydz 
2
 0 101
 02 2 2 a d
H 0 ( )b( )
2 2
4 2
1
2
  0  02 a 3 bdf 101
H 02
4
2
0
0 2 d b a  4




2
{
H

H
}
dv

H 0    { 4 2 2 sin 2 ( x) cos 2 ( z )  cos 2 ( x) sin 2 ( z )}dxdydz
x
y

0
0
0
4
4
had
a
d
a
d
 2 a2 a d
a d
0
a2
 H 0 { 2 ( )b( )  ( )b( )} 
abd ( 2  1) H 02
4
d 2 2
2 2
16
d
Wm 


1
1
1
where h2= ( ) 2  ( ) 2 and the resonant frequency: ( f r ) TE101 
 2 2
a
d
2  0 0 a d
0
a2
2
At resonance, W=2We=2Wm=  H 0 abd ( 2  1)
8
d
Power loss per unit area: Pav=
PL   Pav ds  Rs


b
0

a
0
1
1 2
2
J s Rs  H Rs
2
2
H x ( z  0) dxdy  
2
d
0

b
0
H z ( x  0) dydz  
2
d
0

a
0
H x dxdz  
2
d
0

a
0
2
f101 0 abd (a 2  d 2 )
Rs H 02  a 2 b 1
b 1 

Q

(

)

d
(

)


TE101
2 d d 2
a 2 
Rs [2b(a 3  d 3 )  ad (a 2  d 2 )]
Similarly, the expression for the Q of an air-filled a×b×d rectangular resonator for the
TM110 mode is (Q) TM110 
 ( f r ) TM  0 abd (a 2  b 2 )
110
Rs [2d (a  d 3 )  ab(a 2  b 2 )]
3
.

H z dxdz
Eg. An air-filled rectangular cavity with brass walls σ=1.57×107(S/m) has the
following dimensions: a=4cm, b=3cm, and d=5cm. (a) Determine the dominant
mode and its resonant frequency for this cavity. (b) Find the Q and the
time-average stored electric and magnetic energies at the resonant frequency,
assuming H0 to be 0.1A/m.
(Sol.) (a) f r 
(b) (Q) TE101 
c m 2
n 2
p 2
(
) (
) (
) , dominant mode: TE101, ( f r )TE1 01  4.8GHz
2 0.04
0.03
0.05
 ( f r ) TE  0 abd (a 2  d 2 )
101
Rs [2b(a 3  b 3 )  ad (a 2  d 2 )]
 6869 , Rs 
 ( f r ) TE  0
101
c
0
2
 2 a 3bd ( f r )TE
H 02  7.73  10 14 ( J )
101
4
Eg. For an air-filled rectangular copper cavity resonator, determine how much b
should be increased in order to make Q 20% higher.
At the resonant frequency, We  Wm 
(Sol.)
Q2
b
 1.2  2  b2  1.44b1
Q1
b1
Eg. (a) What should be the size of a hollow cubic cavity made of copper in order
for it to have a dominant resonant frequency of 10GHz? (b) Find the Q at that
frequency.
(Sol.) (a) For a cubic cavity, a=b=d, TM110, TE011, and TE101 are degenerate dominant
modes. f101 
3  108
3  108
 1010 ( Hz ) , a 
 2.12  10 2 m .
2a
f  a a
(b) Q101  101 0 
f101 0
3R s
3
2  10
For copper,   5.80  10 7 ( S / m) , Q101  (
2.12
10  2 )  1010 (4 10 7 )(5.80 10 7 )  10700.
3
10
6-6 Circular Cavity Resonators
For an air-filled circular cylindrical cavity resonator of
radius a and length d. The resonant frequencies are
( f r ) TM mnp 
( f r )TEmnp 
1
(
2 
1
2 
(
X mn 2 p 2
)  ( ) , where Jm(Xmn)=0
a
d
X 'mn 2 p 2
)  ( ) , where J’m(X’mn)=0
a
d
In case of 2d>2a>d, the dominant mode of the circular cylindrical cavity is TM010
mode:
E z  C 0 J 0 (hr )  C 0 J 0 (
jC
jC0
2.405
2.405
r ) , H    0 J ' 0 (hr ) 
J1 (
r)
a
0
0
a
Quality factor of the TM010 mode: QTM 010 
W  2We 
0
2 
E z dv 
2
v

 0 C0
2
a
(2d )  J 02 (
0

W
PL
.
a
2.405
r )rdr  (0 d )C 02 [ J 12 (2.405)]
2
a

a
a
2
2
Rs
2
2
2  J r 2rdr  (2ad ) J z  Rs 2 H  rdr  (ad ) H  (r  a )
0
0
2
2
R C 2  a
2.405
 aRs C 0
 s 2 0 2 J 12 (
r )dr  (ad ) J 12 (2.405) 
(a  d ) J 12 (2.405)
2
0
a
0 
0

PL 
 QTM 010  (
0
Rs
)

2.405
2.405
0.115
, ( f r )TM 010 

109 Hz
a
2(1  a / d )
2a  0 0
Eg. A hollow circular cylindrical cavity resonator is to be constructed of copper
such that its length d equals its diameter 2a. (a) Determine a and d for a resonant
frequency of 10GHz at the TM010 mode. (b) Find the Q of the cavity at resonance.
(Sol.) (a) ( f r )TM 010 
(b) Rs 
0.115
10 9  10 10 9 , a  1.15  10 2 m , d=2a=2.30cm.
a
f 0
 1010  (4 10 7 )
120
2.405
2
Q

(
)
 11,580.
,


2
.
61

10
(

)

2
2.61  10 2(1  1 / 2)

5.80 10 7
Eg. In some microwave applications, ring-shaped cavity resonators with a very
narrow center part are used. A cross section of such a resonator is shown in the
figure, in which d is very small in comparison with the resonant wavelength.
Assuming that this the narrow center part and the inductance of the rest of the
structure, find (a) the approximate resonant frequency. (b) the approximate
resonant wavelength.
a 2
h b
ln( )
2
a
d
1
1
(a) f r 
(b) r 
2 LC
f r 
(Sol.) C 
, L
6-7 Excitations of Waveguides
6-8 Directional couplers