Waveguide notes 2017
Electromagnetic waves in free space
We start with Maxwell’s equations for an LIH medum in the case that the
source terms  and  are both zero.
 ·
 =∇
 · 
 =0
∇
 ·
 =0
∇

 ×
 = − 
∇




 ×
 =
∇

Take the curl of Faraday’s law, and then use Ampere’s law:
³
´
 ×

 ×

∇
∇
 × ∇
 ×

∇
= −
= −


³
´




 ∇
 ·
 − ∇2 
 = −
∇
 
Use the first Maxwell equation (the "H" in LIH assures us that spatial derivatives
√
of  are zero1 ), and we obtain the wave equation with wave speed  = 1 

2
2


∇2 
= 

2
2


= 2 ∇2 
 Now let’s look at a plane
A similar derivation gives the same equation for 
wave solution:
³
´
 = 
 0 exp  ·  − 

³
´
 = 
 0 exp  ·  −  + 

where  =   the wave phase speed. By including the phase constant 
 we allow for a possible phase shift bewteen 
 and 

in the expression for 
Inserting these expressions into Maxwell’s equations, we have
 ·

∇
 ·

∇
0 = 0
= 0 ⇒  · 
0 = 0
= 0 ⇒  · 
(1)
 and 
 are perpendicular to the direction of propagation. From
Thus both 
Faraday’s law
³
´
³
´
 × 
 0 exp  ·  −  =  
 0 exp  ·  −  + 
1 Here
we also assume that  is independent of 
1

Since this relation must be true for all  and  and  is real, we have  = 0 (
 oscillate in phase) and
and 

0
0 =  × 
 0 = 1 ̂ × 



(2)
 is also perpendicular to 
 and its magnitude is  
Thus 
If the waves propagate in a vacuum, the derivation goes through in the same
√
way and the only difference is that the wave speed is  = 1 0 0  In an LIH
p
p
medium,  =  where the refractive index  = 0 0 ' 0 
Electromagnetic fields in a wave guide
A wave guide is a region with a conducting boundary inside which EM waves
are caused to propagate. In this confined region the boundary conditions create
constraints on the wave fields. We shall idealize, and assume that the walls are
perfect conductors. If they are not, currents flowing in the walls lead to energy
loss. See Jackson Ch 8 for a discussion of this case, especially sections 1 and 5.
The boundary conditions at the walls of our perfectly conducting guide are
(see notes 1 eqns 5,7,8 and 10):
 =Σ
̂ · 
(3)
where Σ is the free surface charge density on the wall,
 =0
̂ × 
(4)
 =0
̂ · 
(5)
2
and
 =

̂ × 
(6)
 is the free surface current density. (Note that we have assumed 
 =0
where 
inside the conducting material. This is true here because all our fields are


assumed to be time-dependent, and then non-zero  
implies non-zero 
 is not allowed inside a perfect conductor, and
by Faraday’s law. Non-zero 
 must be zero too. ) Since we do not know Σ or 
 equations (4) and (5)
so 
will be most useful.
Now we use cylindrical coordinates with ̂ along the guide in the direction
of wave propagation. The transverse coordinates will be chosen to match the
cross-sectional shape of the guide — Cartesian for a rectangular guide and polar
for a circular guide. Next we assume that all fields may be written in the form
 =
 0 () −

We are not making any special assumptions about the time variation, because
we can always Fourier transform the fields to get combinations of terms of this
form. Then Maxwell’s equations in the guide take the form:
 ×
 =  

∇
 ·
 =0
∇
 ·
 =0
∇
and
 ×
 = − 

∇
 ×
 from Ampere’s law, we
Taking the curl of Faraday’s law, and inserting ∇
get:
³
´
³
´
³
´
 × ∇
 ×

 ∇
 ·
 − ∇2 
 =  ∇
 ×
 =  −

∇
= ∇

∇2 

= − 2 
(7)
This equation is the same as we obtained for free space. Note that  is usually
a function of 
Next we look for solutions that take the form of waves propagating in the
−direction, that is:
 0 () = 
  ( ) 

The wave equation (7) then becomes:
 +  2 
 =0
 − 2 
∇2t 
(8)
where ∇2t is the Laplacian operator in the two transverse coordinates ( and 

or  and  for example.) Thus equation (8) is an equation for the function 
of the two transverse coordinates.

Since we were able to simplify the equations by separating the function 
into its dependence on the coordinates along and transverse to the guide, we
3
now try to do the same thing with the components. At first glance, and based
on equation (1), you might want to jump to the conclusion that there is no
−component of a wave propagating in the −direction, but in general there
is. The waves are propagating between conducting boundaries, and we have to
allow for the possibility that waves travel at an angle to the guide center-line,
 is perpendicular
and bounce back and forth off the walls as they travel. Since 

to the wave vector,  in such a bouncing wave has a −component. The total
electromagnetic disturbance in the guide is a sum of such waves. The sum is a
combination of waves that interfere constructively. Thus we take
  =  ̂ + 
t

and similarly
  =  ̂ + 
t

This decomposition simplifies the boundary conditions, since the normal ̂ on
the boundary has no −component. Then eqn (5) becomes
 t = 0 on 
̂ · 
(9)
However equation (4) has two components. The transverse component gives
 = 0 on 
(10)
 t = 0 on 
̂ × 
(11)
while the − component gives
Next we put these components into Maxwell’s equations. The “divergence”
equations are scalar equations, so let’s start with them:
µ
¶
 ·
 = 0 =   + ∇
t
t ·
∇

and evaluating the −derivative, we get
t ·
t = 0
 + ∇
(12)
t ·
t = 0
 + ∇
(13)
Similarly:
We separate the curl equations into transverse and −components. Take the
dot product of Faraday’s law with ̂:
³
´
´
³
 ×
 =  = ̂ · ∇
t ×
t
̂ · ∇
and also the cross product:
³
´
 ×
 = ̂ ×  

̂ × ∇
4
(14)
Let’s investigate the triple cross product on the left. Since ̂ is a constant, we
may move it through the ∇ operator in the BAC-CAB rule:
³
´
³
´ ³
´
 ×
 =∇
 ̂ · 
 − ̂ · ∇
 
 = ̂ ×  
t
̂ × ∇
The derivative   times ̂ appears in both terms in the middle, and so
cancels, leaving:
t
 t  −  
 t = ̂ × 
∇

and evaluating the −derivative, we get
 t  − 
 t = ̂ × 
t
∇
(15)
Similarly, from Ampere’s law, we have the transverse component:
 t  −  
 t = −̂ × 
t
∇
(16)
³
´
t ×
t
− = ̂ · ∇
(17)
and the −component
Equations (12), (13), (17) and (14) show that the longitudinal components
 t and 
t.
 and  act as sources of the transverse fields 
Now we can simplify by looking at the normal modes of the system.
Transverse Electric (TE) (or magnetic) modes.

In these modes there is no longitudinal component of :
 ≡ 0 everywhere
Thus boundary condition (10) is automatically satisfied. The remaining boundary conditions are (9) and (11), and we can find the version that we need by
taking the dot product of ̂ with equation (16):
³
´
 t  − ̂ · 
 t = −̂ · ̂ × 
t
̂ · ∇
The second term is zero on  (eqn 9), and we rearrange the triple scalar product
on the right, leaving:
´
³

 t = 0 on 
= ̂ · ̂ × 


where we used the other boundary condition (11) for 
Transverse Magnetic (TM) (or electric) modes.

In these modes there is no longitudinal component of :
 ≡ 0 everywhere
5
(18)
Thus the boundary condition (18) is trivially satisfied, and we must impose the
remaining condition (10)
 = 0 on S.
Since the Maxwell equations are linear, we can form superpositions of these
two sets of modes to obtain fields in the guide with non-zero longitudinal com and 
 These modes are the result of the constructive interponents of both 
ference mentioned above.
Transverse electromagnetic (TEM) modes
In these modes both  and  are zero everywhere. Then from (12) and
 t and ∇
t ·
 t are zero everywhere. This means we can express 
t
t ×
(14), ∇
as the gradient of a scalar function Φ that satisfies Laplace’s equation in two
dimensions. The boundary condition (11) becomes
µ
¶
Φ

̂ × ∇Φ = ̂ × ̂
= 0 on 

where  is a coordinate parallel to the surface  Since ̂ and ̂ are perpendicular,
̂ × ̂ is not zero, and so Φ = constant on  and therefore is constant everywhere
 t = 0 Thus these modes cannot exist inside a
inside the volume  making 
hollow guide. They may exist, and in fact become the dominant modes, inside a
guide with a separate inner boundary, like a coaxial cable. We will not consider
them further here.
Now let’s see how the equations simplify for the TE and TM modes..
TM modes
We start by finding an equation for   Since  ≡ 0 equation (16) simplifies
to:
t
−
t

t
= −̂ × 

t
=
̂ × 

and we substitute this result back into equation (15).
³
´
 t  = ̂ ×  ̂ × 
t + ∇
t
− 

2
 t −   
t
 t  = 
∇
 ¶
µ
2
t
=  1 − 2  

  back into equation (12):
and finally we substitute this result for 
 t 
∇
¢
2
 1 − 2 
¡
¢
∇2t  +  2  − 2 
t·
 + ∇
¡
6
= 0
= 0
(19)
(20)
or
with
2
∇2t  +  2  = 0
(21)
 2 ≡  2  − 2
(22)
Equation (21) is the defining differential equation for   Once we have solved
 t from equation (20) and then 
 t from equation (19).
for   we can find 
TE modes
The argument proceeds similarly. We start with equation (15) with  = 0,
to get:
 t = −  ̂ × 
t

(23)

and substitute into equation (16)
´
³
t + ∇
 t  = −̂ × −  ̂ × 
t
− 
µ
¶
2


 t  =   −
t
(24)
∇


Then finally from equation (13) we have:

 t · ³ ∇t  ´
 + ∇
2
  −  
= 0
∇2t  +  2 
= 0
which is the same differential equation that we found for  in the TM modes.
The solutions are different because the boundary conditions are different. Thus
the solution for the two modes proceeds as follows:
assumed
TM modes
 ≡ 0
TE modes
 ≡ 0
differential equation
∇2t  +  2  = 0
∇2t  +  2  = 0
boundary condition
 = 0 on S
next find
then find
 =

t =

 
 2 ∇t 

 ̂
t
×


= 0 on S
 =

This is an eigenvalue/
eigenfunction problem.
 
 2 ∇t 
t
 t = −  ̂ × 


The differential equation plus boundary condition is an eigenvalue problem
that produces a set of eigenfunctions  (or   ) and a set of eigenvalues
   The wave number  is then determined from equation (22):
2 =  2  −  2
2 This
equation is the Helmholtz equation.
7
(25)
√
Clearly if   is greater than   =  where  is the wave phase speed
in unbounded space,  becomes imaginary and the wave does not propagate.
There is a cut-off frequency for each mode, given by
 =   
If   is the lowest eigenvalue for any mode, the corresponding frequency   is the
cutoff frequency for the guide, and waves at lower frequencies cannot propagate
in the guide.
A few things to note: the wave number  is always less than the freespace value  and thus the wavelength is always greater than the free-space
wavelength. The phase speed in the waveguide is
 =

1
1
1
 √ =
=√ p

 1 −  2  2

and we can differentiate eqn (25) to get
2

 = 2

Then the group speed in the guide is

1
2
1
=√
=
=
 =

2


r
1−
 2
1
 √ =
2

Thus information travels more slowly than if the wave were to propagate in free
space.
TM modes in a rectangular wave guide
We use Cartesian coordinates with origin at one corner of the guide. Let
the guide have dimensions  in the -direction by  in the −direction. Let the
interior be full of air so 0 = 0 ' 1 Then the differential equation for 
is (21)
¶
µ 2
2

2
+
+

 = 0
2  2
As usual we look for a separated solution, choosing  =  ()  () to obtain:
 00
 00
+
+ 2 = 0


Each term must separately be constant, so we have:
 00
= −2

 00
= − 2

and
−2 −  2 +  2 = 0
8
The boundary condition is  = 0 on  so:
 = 0 at  = 0 and  = 
and
 = 0 at  = 0 and  = 
Thus the appropriate solutions are  = sin  and  = sin  with eigenvalues
chosen to fit the second boundary condition in each coordinate:
=


and  =


Thus, putting back the dependence on  and  we have
 =  sin
and
 2 =

 −
sin



³  ´2
+
(26)
³  ´2
(27)


Notice that the lowest possible values of  and  are 1 in each case, since taking
 or  = 0 would render  identically zero. Thus the lowest eigenvalue is
r
1
1
 11 = 
+ 2
2


and the cutoff frequency for the TM modes is:
r
r
1

2
1
 c,TM =  11 = 
+ 2 =
1+ 2
2




Jackson solves for the TE modes (pg 361). The eigenvalues are the same,
but in this case it is possible for one (but not both) of  and  to be zero,
leading to a lower cutoff frequency for the TE modes:
 TE =


assuming    This would be the cutoff frequency for the guide.
In the TM mode, the remaining fields are (eqns 20 and 26):
t

=
=
 

 −
∇t  sin
sin

2


³


 

 ´ −

(28)
̂
cos
sin
+
̂
sin
cos


2






and (eqn 19)
t

³ 



 

 ´ −
̂
×

̂
cos
sin
+
̂
sin
cos


2
2






³
´



 

 −
=  2 2 
(29)
̂ cos
sin
−
̂ sin
cos

 






=
9
where (eqns 25 and 27)
=
r
 2  −
³  ´2

−
³  ´2

As usual, the physical fields are given by the real part of each mathematical
 t ∝ sin ( − ). You should verify that
expression, so that, for example, 
these fields satisfy the boundary conditions at  = 0  and at  = 0 
Power
The power transmitted by the waves in the guide is:
 () = 1 
 ×


0
where here we must take the real, physical fields. Usually we are interested in
the time-averaged Poynting flux, which is given by
 ×
∗
  = Re 1 

20
(30)
where the fields on the right are the complex functions we have just found.
Proof of this result:
 =
 0 − = ̂0  − and similarly for 
 where 0 and 0 are
If 
real, then
³ ´
³ ´
 = 1 Re 
 × Re 


0
and the time average is
  = ̂ × ̂ 0 0 cos ( − ) cos ( − ) 
 
0
0 0
= ̂ × ̂
(cos  cos  + sin  sin ) (cos  cos  + sin  sin ) 
0
¿
0 0 £
cos  cos  cos2 +
=
̂ × ̂
0
¤®
(cos  sin  + cos  sin  ) cos  sin  + sin  sin  sin2 
0 0
0 0
= ̂ × ̂
(cos  cos  + sin  sin  ) = ̂ × ̂
cos ( −  )
20
20
1 
∗
= Re
×
20
so the two results are the same. (See also Notes 2 page 11.)
10
 are:
Using the solution (28, 29), the components of 
¢
1 ¡
 ∗ −  ∗
20
∙³
¸



 ´2 ³ 

 ´2
1 

(−)

+
cos
sin
sin
cos
Re


20  2
2  2






∙
¸
³
³
´
´
2
2
2






  
+
cos
sin
sin
cos
20  4 2






p
∙³
¸
³
´
2
2
2
2
2


   −  



 ´2
+
cos
sin
sin
cos
20
4
2






q
¡
¡
¢
¢
2 ∙
¸
2
 2
³ 
2
− 
0 

 ´2 ³ 

 ´2
 2 − 

+
cos
sin
sin
cos
h¡ ¢
¡  ¢2 i2
2






 2
+


   = Re
=
=
=
=
Check the dimensions!    is positive for all values of  and  showing that
power is propagating continuously along the guide in the positive -direction.
The transverse component  is:
¢
¢
1 ¡
1 ¡
 ∗ −  ∗ = Re
− ∗
   = Re
20
20
µ
¶





1

−0 sin
sin
(−) 2 2 0
cos
sin
= Re
20


 



= 0
Because there is no real part, the time averaged power flowing across the guide
is zero. Power sloshes back and forth, but there is no net energy transfer.
Fields in a parallel plate wave guide
By simplifying the shape of the guide even more, we can demonstrate how
the wave modes are formed by reflection of waves at the guide walls. Let this
guide exist in the region 0 ≤  ≤  with infinite extent in  For the TM modes,
the equation to be satisfied is:
with
¡ 2
¢
∇t +  2  = 0
 = 0 at  = 0 and  = 
Because the region is infinite in the −direction, the appropriate solution has
no −dependence:
 −
 =  sin
(31)


with

 =

Thus the cutoff frequency for these modes is

 TM =  1  =
(32)

11
Then the other components of the fields are:


=
=
 
 
 −
̂
∇  = 2
 cos

2
 


 −
̂
 cos



(33)
and



  = −    cos  − ̂
̂ × 
2

2 

 −
 
̂
= − 2
 cos

 

=
(34)
Let’s look at the electric field first. We write the sine and cosine as combinations of complex exponentials. From (31),
µ 
µ 
¶
¶

 − −
− + −
−
=  

−
 = 
2

2
and from (33)
 =



µ
 + −
2
¶
−
Thus we can write the electric field as a superposition
´
³
 total = 1 
 2
 1 + 

2
where the two superposed fields are
 1 =  (̂ −  ̂)  exp ( + ) −


and
 2 =  (̂ +  ̂)  exp ( − ) −


Similarly:
 t = −   ̂

2 
with
µ
 + −
2
¶
− =
´
1 ³
 2
1 + 
2
 12 = −   ̂ exp [ ( ± )] −

2
Now define the four vectors
1 = ̂ −  ̂;
2 = ̂ +  ̂
and
1 =  ̂ + ̂;
2 = − ̂ + ̂
12
Then for  = 1 2 the vectors are perpendicular:
 ·  = 0
and
¡
¢
2
1 × 1 = (̂ +  ̂) × (̂ −  ̂) = −̂  2 + 2 = −̂ 2 = 2 × 2

where we used equation (25). The vectors are shown in the diagram below.
The two electric field components are then:
 1 = 1  exp (1 · ) −


while
and
 2 = 2  exp (2 · ) −


 1
1 × 

 1
= − 2 ̂ exp [ ( + )] = 


consistent with (2) and (34).
 1  
 2 and 
 2 ) has the form of
 1 and 
Each of these sets of fields (
a free-space wave propagating in the direction given by the vectors 1 and 2
respectively and with wave number
p

|1 | = |2 | =  2 + 2 =

. These waves are moving across the guide at an angle given by
tan  =



= ± = ±q
2



− 2
2
that is, the waves are reflecting off the plates at  = 0  as shown in the
diagram. When the angle  becomes 2 the wave ceases to propagate along
the guide, but just bounces back and forth. This happens when tan  → ∞ or
=


This gives the cut-off frequency (32) we found before.
See also Lea and Burke pages 1058-1060.
13