Badr University in Cairo

Faculty of Engineering and

Technology
Microwave Circuit
ECO 311
Lect. 3
Prof. Dalia Elsheakh
Dalia-Mohamed@buc.edu.eg
 Related example of how fields look:
Parallel plate waveguide - TM modes
 m x  e j(t − z )
Ez = A sin 
Ez  a 
m=1
0 a x

m=2

m=3
z a x
 Badr University in Cairo
Faculty of Engineering and
Technology
Microwave Circuit
Lect. 4

Prof. Dalia Elsheakh

Dalia-Mohamed@buc.edu.eg
 Chapter Contents
2.1 Introduction
2.2 Field Relationships in General Waveguides
2.3 General Solution for TE and TM waves
2.4 TE Modes in Rectangular Waveguide
2.5 Attenuation due to Dielectric Loss
2.6 TM Modes in Rectangular Waveguide
2.7 Power Transmitted in Rectangular Waveguide
BASICS ON WAVEGUIDES

DEFINING A WAVEGUIDE
RECTANGULAR WAVEGUIDES
EM FIELD CONFIGURATION WITHIN THE WAVEGUIDE
2.3 TE Modes Rectangular Waveguide (Continued)

Fig. 2 Photograph of Ka-band (WR-28) rectangular waveguide components. Clockwise

from top: a variable attenuator, and E-H (magic) tee junction, a directional coupler, an
adaptor to ridge waveguide, an E-plane swept bend, an adjustable short, a sliding
matched load and X-band slotted waveguide.
 ADVANTAGES
➢ Waveguides have several advantages over two-wire and coaxial
transmission lines.
➢ Skin effect tends to increase the effective resistance of the conductor.
Although energy transfer in coaxial cable is caused by electromagnetic field
motion, the magnitude of the field is limited by the size of the current-carrying
area of the inner conductor.
➢ The small size of the center conductor is even further reduced by skin effect
and energy transmission by coaxial cable becomes less efficient than by
waveguides.
➢ Dielectric losses are also lower in waveguides than in two-wire and coaxial
transmission lines.
➢ The dielectric in waveguides is air, which has a much lower dielectric loss
than conventional insulating materials. However, waveguides are also subject
to dielectric breakdown caused by standing waves.
➢ Power-handling capability is another advantage of waveguides.
Waveguides can handle more power than coaxial lines of the same size
because power-handling capability is directly related to the distance between
conductors.
 DISADVANTAGES
 Physical size is the primary lower-frequency limitation of
waveguides. The width of a waveguide must be approximately a half
wavelength at the frequency of the wave to be transported.

 Waveguides are difficult to install because of their rigid, hollow-

pipe shape. Special couplings at the joints are required to assure
proper operation.

 Costs are more and decrease the practicality of waveguide systems

at any other than microwave frequencies.
2.2 Field Relationships in General Waveguides (Continued)
• We assume time-harmonic fields with eit time dependence and the wave propagating
along the z-axis. The electric and magnetic fields can be written as:
 
E ( x, y, z ) = [e ( x, y ) + aˆ z ez ( x, y )]e − jz (2.1)
 
H ( x, y, z ) = [h ( x, y ) + aˆ z hz ( x, y )]e − jz (2.2)

 
• Where e ( x, y) and h ( x, y ) represent the transverse (âx, ây) electric and
magnetic field components.
• While ez(x, y) and hz(x, y) are the longitudinal electric and magnetic field
components.
• Later in this chapter, ez(x, y) and hz(x, y) are assumed to be separable in x and
y variables to simplify the solution of partial differential equations by applying
the method of separation of variables by letting hz(x, y) = X(x)Y(y) as will
explained in the next section.
• In (2.1) and (2.2) the wave is propagating in the +z direction and the –z
direction propagation can be obtained by replacing the propagation constant
 by -.
• If conductor or dielectric loss is present, the propagation constant will be
complex and j should then be replaced with  =  + j.
2.2 Field Relationships in General Waveguides (Continued)
• Assuming that the transmission line or wave guide region is source free (ρv = 0, J = 0 and 
= 0). Under these conditions,
 Maxwell’s equations may be written in terms of and
only as: E H 
 B H 
xE = − = − = − jH (2.3a)
t  t 
  D E 
xH = J + = = jE (2.3b)
  v t t
.E = =0 (2.3c)
 
.H = 0 (2.3d)
• With an e-jz dependence, the three components (Ex, Ey and Ez)of (4.3a) can
be derived as follows:

aˆ x aˆ y aˆ z
H x aˆ x
     (2.4a)
CurlE = xE = = − j H y aˆ y
x y z
Ex Ey Ez
H z aˆ z
2.1 Field Relationships in General Waveguides (Continued)

• So Ecan be derived as follows :
  Ez  Ey 
H x aˆ
aˆ x  y − z  =− j x
 
  Ez  Ex 
− a y  x − z  =− j
ˆ H y aˆ y
 
 y 
 Ex 

a z  x y  =− j
ˆ y
− H z aˆ z
 
• Since the wave is propagating along the z-axis with an e-jz dependence, so:
e( x, y )e − jz − j z
z = − j  e( x, y )e
  Ez  Ey  
 −  = − jH  E z + jE y = − jH x
 y z  x y

E z
+ jE y = − jH x (2.5a)
y
Or
• Similarly, the other to components can be written as follows:
(2.5b)
 E z 
− + jE x  = − jH y
 x 
E y E
(2.5c)
− x = − jH z
x y
2.1 Field Relationships in General Waveguides (Continued)

• A similar equation for H can be written then With an e-jz dependence, the three
components (Hx, Hy and Hz)of (4.3 b) can be reduced to the following:

 
  D E 
xH = J + = = jE 
t t aˆ x aˆ y aˆ z
Hz Hz E x aˆ x
aˆ x  y − z  = jEx aˆ x     
  CurlH = xH =
Hz Hx
= j E y aˆ y
−aˆ y  x − z  = jE y aˆ y
x y z
 
Hy Hx Hx Hy Hz
aˆ z  x − y  = jEz aˆ z
E z aˆ z
 
• Since the wave is propagating along the z-axis with an e-jz dependence, so:
H z h( x, y )e − jz
=− jh( x, y)e − jz 
+ jH y = jE x z (2.6a)
y
 H z  (2.6b)
− + jH x  = jE y
 x 
H y H x (2.6c)
− = jE z
x y
2.2 Field Relationships in General Waveguides (Continued)
• The six equation of (2.5) and (2.6) can be solved for the four transverse components
(Ex, Ey, Hx, and Hy ) in terms of the longitudinal components Ez, and Hz (for example, Hx
can be derived by eliminating Ey from (2.5a) and(2.6b) as follows:
j E z H z
H x = 2 ( − ) (2.7a)
kc y x
− j E z H z
H y = 2 ( + ) (2.7b)
kc x y
− j E z H z
Ex = 2 ( +  ) (2.8a)
kc x y
j E H z
E y = 2 (−  z +  ) (2.8b)
kc y x
• Where the cutoff wave number kc is given by: kc2 = k 2 −  2
and k is wave number of the material filling the transmission line or waveguide:
k =   If dielectric loss is present,  is complex ( = o r(1 – j tan),
where tan is the loss tangent of the material.
• Equations (2.7) and (2.8) are very useful general results that can be applied to
a variety of wave guiding systems as will be shown in the following sections.
 Then applying on the z-component…

 Ez + k Ez = 0
2 2

 Ez  Ez  Ez
2 2 2
+ + + k 2
Ez = 0
x 2
y 2
z 2

Solving by method of Separation of Variables :

E z ( x, y, z ) = X ( x)Y ( y ) Z ( z )
from where we obtain :
X '' Y '' Z ''
+ + = −k 2
X Y Z
 Fields inside the waveguide

X '' Y '' Z ''

+ + = −k 2
X Y Z
− k x2 − k y2 +  2 = −k 2
h 2 =  2 + k 2 = k x2 + k y2
which results in the expressions :
X '' + k x2 X = 0
X(x) = c1 cos k x x + c2 sin k x x
Y +k Y =0
'' 2
y
Y(y) = c3 cos k y y + c4 sin k y y
Z '' −  2 Z = 0 Z ( z ) = c5ez + c6 e −z
 Modes of propagation
From these equations we can conclude:
• TEM (Ez=Hz=0) can’t propagate.

• TE (Ez=0) transverse electric

• In TE mode, the electric lines of flux are perpendicular to the
axis of the waveguide

• TM (Hz=0) transverse magnetic, Ez exists

• In TM mode, the magnetic lines of flux are perpendicular to
the axis of the waveguide.

• HE hybrid modes in which all components exists

 2.2 General Solution for TE and TM waves
2.2.1 TE waves
Transverse electric (TE) or H-waves are characterized by Ez = 0 and Hz ≠ 0. Equation (2.7)
and (2.8) are reduced to:

− j H z
j E z H z Hx =
Hx = ( −  )  kc2 x (2.9a)
kc2
y x
− j E z H z − j H z
H y = 2 ( + )  Hy = (2.9b)
kc x y kc2 y
− j E z H z − j H z
Ex = 2 ( +  )  E = (2.10a)
kc x y x
kc2 y
j E z H z
E y = 2 (−  +  )  j H z
kc y x Ey = 2 (2.10b)
kc x
• The cutoff wave number kc ≠ 0 and the propagation constant  is given by:
 = k 2 − kc2 is generally a function of frequency and the geometry of the
line or waveguide. To apply (2.9) and (2.10) one must first find Hz from
Helmholtz wave equation 2 E + k 2 E = 0 which can be written as:
z z
2.2 General Solution for TE and TM waves (Continued)
2.2.1 TE waves (Continued)
 2 H z +  2 H z = 0 or 2 H z + k 2 H z = 0
• Where k is given by k =   . In Cartesian coordinate system the wave
or Helmholtz equation has the form:
 2 2 2 2
(2.11)  2 + 2 + 2 + k  H z = 0
 x y z 
From equation (4.2) Hz can be written as:
(2.12) H z = hz ( x, y)e− jz
From equation (2.12) substitute in (2.11) which can be reduced to a two-
dimensional wave equation for Hz and is written as:
 2 2 2 2 − j z  2 2 2 − jz  2 
 2 + 2 + 2 + k hz ( x, y )e =  2 + 2 + k hz ( x, y )e +  2 hz ( x, y )e − jz
 x y z   x y   z 
 2 2 2 − j z − j z h( x, y )e − jz
=  2 + 2 + k hz ( x, y )e + (−  )hz ( x, y )e
2
=0  = − j h ( x , y ) e − j z
 x y  z
or
 2 2    2
 2
2 (2.13)
 2 + 2 + [k −  ] hz ( x, y ) = 0   2 + 2 + kc hz ( x, y ) = 0
2 2

 x y   x y 
2.2 General Solution for TE and TM waves (Continued)
2.2.1 TE waves (Continued)
Since kc = k 2 −  2 (2.13) must be solved subject to the boundary
conditions of the specific wave guide.
• The TE wave impedance can be found from (2.9) and (2.10) as
E x − E y  k 
ZTE = = = = ; =
Hy Hx    (2.14)

Which is seen to be frequency dependent. TE wave can be supported inside

closed conductors, as well as between two or more conductors

The procedure for analyzing TE waveguides can be summarized as follows:

Step #1 Solve the reduced Helmholtz equation (2.13) for Hz. This solution will contain
several unknown constants and the unknown cutoff wave number kc.
Step #2 use (2.9) and (2.10) to find the transverse fields (Ex, Ey, Hx and Hy) from Hz
which is calculated in step #1.
Step #3 Apply the boundary conditions to the appropriate field components to find the
unknown constants and the unknown cutoff wave number kc.
Step #4 Calculate the TE wave impedance by (2.14)
 2.3 TE Modes in Rectangular Waveguide
• Rectangular waveguides were one of the earliest types of transmission
lines used to transport microwave signals and are still used today for many
applications.
• A large variety of components such as couplers, detectors, isolators,
attenuators, and slotted lines are commercially available for various
standard waveguide bands from 1 GHz to over 220 GHz.
• Fig. 2 shows some of the standard rectangular waveguide components
that are available.
• Because of the recent trend toward miniaturization and integration, a lot
of microwave circuitry is currently fabricated using planner transmission
lines, such as microstrip and strip lines, rather than microwave.
• There is, however, a need for waveguides in many applications such as
high power systems, millimeter wave systems, and in some precision test
applications.
• The hollow rectangular wave guide can propagate TE and TM modes, but
not TEM waves, since only one conductor is present.
2.3 TE Modes Rectangular Waveguide (Continued)
• The geometry of a rectangular wave guide is shown in Fig. 3.
• It is assumed that the guide is filled with a material of permittivity  and permeability
.
• It is standard convention to have the longest side of the waveguide along the x-axis, so
that a > b.
• The TE modes are characterized by the fields with Ez = 0, while Hz described by
equation (2.2) must satisfy the reduced wave equation (2.13):
 2 2 
 2 + 2 + kc2 hz ( x, y ) = 0 (2.21)
 x y 
where H z = hz ( x, y)e− jz
and k c = k 2 −  2 is the cutoff wave number
• The partial differential equation (2.21) can be solved
by the method of separation of variables by letting:

hz ( x, y) = X ( x)Y ( y) (2.22) Fig. 3 Geometry of a

rectangular waveguide.
• X(x) = is a function of the x coordinate only
• Y(y) = is a function of the y coordinate only
2.3 TE Modes Rectangular Waveguide (Continued)
• By substituting (2.22) into (2.21) to obtain:
d2X d 2Y
Y 2
+ X 2
+ k c XY = 0
2
(2.23)
dx dy
Divide both sides of (4.23) by XY, we get
Y d2X X d 2Y 2 XY
2
+ 2
+ k c =0 
XY dx XY dy XY
(2.24)
1 d2X 1 d 2Y
2
+ 2
+ k c = 0
2

X dx Y dy
• Then solve by the usual separation of variable arguments. Since the sum of the
two terms of the left hand side of (2.24) is a constant and each term is
independently variable, it follows that each term must be a constant.
• So we define separation constants kx and ky such that:

k =k +k
2
c
2
x
2
y
(2.25)
• Thus (2.24) can be separated into two equations as follows:
d X2 d 2Y
+ + yY = 0
2
xX =0
k 2
2
k (2.27)
2 (2.26) dy
dx
 2.3 TE Modes Rectangular Waveguide (Continued)
• The general solution of these equations (2.26) and (2.27) are in the form:
X = A cos (k x x) + B sin (k x x) (2.28)
Y = C cos( k y y ) + D sin ( k y y ) (2.29)
• The total solution of the Helmholtz equation in rectangular coordinates for hz
can be then written as:
hz ( x, y ) = [ A cos (k x x) + B sin (k x x)][C cos( k y y ) + D sin (k y y )]
(2.30)
H z ( x, y, z ) = hz ( x, y)e− jz
• The propagation of the wave in the guide is conventionally assumed in the
positive z direction.
• To evaluate the constants in (2.30) [A, B, C, D, kx and ky], we must apply the
boundary condition of the field components in the waveguide.
• In general at the surface of a perfect conductor, the following boundary
conditions are valid: (2.31)
Etan gential = 0
• Thus, the boundary condition of the electric field
components tangential to waveguide walls are :

E x ( x, y , z ) = 0 at y = 0, b (2.32)

E y ( x, y, z ) = 0 at x = 0, a (2.33)
 2.3 TE Modes Rectangular Waveguide (Continued)
• We thus can not use hz of (2.30) directly, but must first use (2.10) to find ex
and ey from hz:
− j H z
Ex = (4.10 a) and hz ( x, y ) = [ A cos (k x x) + B sin (k x x)][C cos( k y y ) + D sin (k y y )]
kc2 y H z ( x, y, z ) = hz ( x, y)e− jz
− j − jz
E x ( x, y , z ) =
2
e k y [ A cos (k x x) + B sin (k x x)][−C sin( k y y ) + D cos (k y y )]
kc
j H z (2.34)
Ey = 2 (2.10 b) 
kc x
− j − jz
E y ( x, y , z ) = 2
e k x [− A sin (k x x) + B cos (k x x)][C cos(k y y ) + D sin (k y y )]
kc
• From (4.34) and (4.33a), at y = 0, Ex(x, 0, z) = 0, we get: (2.35)

− j − jz
E x ( x,0, z ) = e k y [ A cos (k x x) + B sin (k x x)][−C sin( 0) + D cos (0)] = 0
kc2
− j − jz
2
e k y [ A cos (k x x) + B sin (k x x)][ D ] = 0  D = 0
kc
• From (2.34) and (2.33a), at y = b Ex(x, b, z) = 0, we get:
− j − jz n
e k y [ A cos ( k x x ) + B sin ( k x x )][ −C sin( k y b )] = 0  k y =
kc2 b
2.3 TE Modes Rectangular Waveguide (Continued)
• From (2.34) and (2.33b), at x = 0, Ey(0, y, z) = 0, we get:
− j − jz
E y (0, y, z ) = 2
e k x [− A sin (0) + B cos (0)][C cos(k y y ) + D sin (k y y )] = 0
kc
− j − jz
2
e k x [ B ][C cos(k y y ) + D sin (k y y )] = 0  B = 0
kc
• From (2.34) and (2.33b), at x = a, Ey(a, y) = 0, we get:

− j − jz m
E y (a, y, z ) = 2 e k x [− A sin (k x a) ][C cos(k y y ) + D sin (k y y )] = 0  k x =
kc a
• Substitute for (D = 0 and ky = n/b for n = 0,1,2, ….) and (B = 0 and kx = m/a for
m = 0,1,2, ….), the final solution for Hz (2.30) is then given by:

− j z  mx   ny  − jz

H z ( x, y, z ) = hz ( x, y )e = Amn cos  cos e (2.36)
 a   b 
• Where Amn is an arbitrary constant composed of the remaining constants A and C
of (4.30) and kc is given by:

 m   n 
2 2

kc = k x + k y = 
2 2
 + 
 a   b 
2.3 TE Modes Rectangular Waveguide (Continued)
• The transverse field components of the TEmn mode can be found using (2.9) and
(2.10) in terms of Hz (2.36) we obtain:
jn mx ny − jz
E x ( x, y , z ) = 2
Amn cos( ) sin( )e (2.37)
kc b a b
− jm mx ny − jz
E y ( x, y , z ) = 2
Amn sin( ) cos( )e (2.38)
kc a a b
jm mx ny − jz
H x ( x, y , z ) = 2
Amn sin( ) cos( )e (2.39)
k a
c a b
jn mx ny − jz
H y ( x, y , z ) = 2
Amn cos( ) sin( )e (2.40)
kc b a b
• The propagation constant  is given by:
  m 
2
 n 
2

 = k − kc =   − kc =   − 
2 2 2 2 2
 +   (2.41)
 a   b  
• Which is seen to be real when:
 m   n 
2 2

k  kc and kc = k x2 + k y2 =   + 
 a   b 
2.3 TE Modes Rectangular Waveguide (Continued)
• There are three cases for the propagation constant  in the waveguide:

  m  
2
 n 
2

Case 1:  = 0  = k − kc =   − kc =   − 
2 2 2 2 2
 +   =0
 a   b  
• At a particular frequency c the value of  = 0 and there will be no wave propagation (evanescence) in the guide, in this
case.

K = Kc or c2  = kc2
• This is the critical condition for cutoff. Each mode (combination of m and n)
thus has a cutoff frequency fcmn is expressed as:
 m   n 
2 2

c2  = kc2  (2f c ) 2  = kc2 =   + 

 a   b 
mn mn

 m   n 
2 2
kc 1
f cmn = =   + 
2   2    a   b 
(2.42)
VDielectric  m   n 
2 2

=   + 
2  a   b  VDielectric is the wave velocity in the
dielectric filling the waveguide.
If it is air filled, VDielectric = c
2.3 TE Modes Rectangular Waveguide (Continued)
  m  
2
 n 
2

Case 2:  > 0  = k − kc =   − kc =   − 
2 2 2 2 2
 +   0
 a   b  
• The wave will be propagating in the waveguide, in this case.

K  Kc or  2   kc2
• And  is expressed as:
  m   n  
2
 f cmn 
2 2

 =   −   +   =   1 −  
2

 a   b    f  (2.43)

• Where f is operating or wave frequency. (2.43) states that the operating

frequency, f must be above the cutoff frequency, fcmn in order to a wave to
propagate in the guide.
• This is the pass-band of the waveguide. Within the pass-band the wave
parameters are explained in the next section.

attenuation Propagation
of mode mn

fc,mn
2.3 TE Modes Rectangular Waveguide (Continued)
Wave Parameters
1. The phase velocity vp is the velocity of a point of constant phase or
t - z = constant and is given by:
dz  1 VDielectric
vp = = = =
dt   f cmn 
2
 f cmn 
2
(2.44 a)
  1 −   1 −  
 f   f 
where fcmn is cutoff frequency and VDieclectric is the wave velocity in the dielectric
with permeability  and permittivity .
• The group velocity vg is the velocity is at which the energy consisting of a
frequency region of the spectrum propagates:
2
d 1  f cmn 
vg = = = VDielctric 1 −   (2.44b)
d d / d  f 
2 Dielectric
2. The wavelength  is: = =
  f cmn 
2 (2.45a)
1 −  
 f 
Another expression for the critical wavelength is as follows: Since k c2 = k x2 + k y2
 Group velocity, ug

• Is the velocity at which the

energy travels.
2
1 f   rad/s   m 
ug = = u' 1 −  c   rad/m  =  s  j     mx  −z
 /  f  Ey = −  
h2  a 
H o sin  e
 a 

• It is always less than u’

u p u g = (u ')
2
 Group Velocity

• As frequency is
increased, the group
velocity increases.
 2.3 TE Modes Rectangular Waveguide (Continued)
Wave Parameters
2
 2   m  2  n  2 2
  = + c =
  a   b 
and
 c mn 2 2
 mn  m n (2.45b)
  + 
 a  b
From (2.42) and (2.45), the mode characteristics for rectangular waveguide
(a > 2b) cmn and fcmn are shown in Fig. 3.
2
c = V  m   n 
2 2
mn 2
m n
2 f cmn = Dielectric   + 
  +  2  a   b 
 a  b

Fig. 3 the mode characteristics for rectangular waveguide c and fc.