This document discusses different types of waveguides, including rectangular waveguides, circular waveguides, coaxial lines, optical waveguides, and parallel-plate waveguides. It describes the different modes of wave propagation including TEM, TE, TM, and HE modes. Cutoff frequencies and wavelengths are defined for rectangular and parallel-plate waveguides. Dominant TE10 mode is described for rectangular waveguides.

1. Waveguides
Mr. HIMANSHU DIWAKAR
Assistant Professor
GETGI
JIT 1

2. JIT 2

3. Basic waveguides
1. Waveguide
Rectangular waveguide Circular waveguide Coaxial line
Optical waveguide Parallel-plate waveguide
JIT 3

4.  Transverse Electro Magnetic (TEM) wave:
Here both electric and magnetic fields are directed
components. (i.e.) E z = 0 and Hz = 0
 Transverse Electric (TE) wave: Here only the electric field is purely transverse to the direction
of propagation and the magnetic field is not purely transverse. (i.e.) E z = 0, Hz ≠ 0
 Transverse Magnetic (TM) wave: Here only magnetic field is transverse to the direction of
propagation and the electric field is not purely transverse. (i.e.) E z ≠ 0, Hz = 0.
 Hybrid (HE) wave: Here neither electric nor magnetic fields are purely transverse to the
direction of propagation. (i.e.) E z ≠ 0, Hz ≠ 0.
4
Types of Modes
JIT

5. Transmission line
 Voltage applied between conductors(E: vertically between
the conductors)
 Interior fields: TEM (Transverse ElectroMagnetic) wave
(wave vector indicates the direction of wave propagation
as well as the direction of power flow)
1. Waveguide
JIT 5

6. Waveguide
 A waveguide is a structure that guides waves, such
as electromagnetic waves or sound waves They enable a signal to
propagate with minimal loss of energy by restricting expansion to
one dimension or two
 Zigzag reflection, waveguide mode, cutoff frequency
 k|||| du kk
JIT 6

7. JIT 7

8. The electric and magnetic wave equations in frequency
domain is given by
𝛻2
𝐸 = 𝛾2
𝐸
𝛻2
𝐻 = 𝛾2
𝐻
𝛾2
= 𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀)
For a loss less dielectric or perfect conductor
𝛾2
= −𝜔2
𝜇𝜀
The above equations are like Helmholtz equations
𝛻2
𝐸𝑧𝑠 = 𝛾2
𝐸𝑧𝑠
Let 𝐸𝑧𝑠=X(x).Y(y).Z(z)
Be the solution of above equations
JIT 8

9. Cont’d
So separation equation
-𝑘 𝑥
2
−𝑘 𝑦
2
−𝑘 𝑧
2
= 𝛾2
On Solving the above equations
𝐸𝑧𝑠 = 𝐴. sin 𝑘 𝑥. 𝑥 + 𝐵. cos(𝑘 𝑥. 𝑥) × 𝐶. sin 𝑘 𝑦. 𝑦 + 𝐷. cos(𝑘 𝑦. 𝑦)
× 𝐸. sin 𝑘 𝑧. 𝑧 + 𝐹. cos(𝑘 𝑧. 𝑧)
The propagation of wave in guide is conventionally assumed in +ve Z
direction.
JIT 9

10. Here the propagation constant 𝛾𝑔 in guide differs from intrinsic
propagation constant 𝛾
Let 𝛾𝑔
2 = 𝛾 2 + 𝑘 𝑥
2
+ 𝑘 𝑦
2
And 𝛾𝑔
2
= 𝛾 2
+ 𝑘 𝑐
2
Where 𝑘 𝑐= 𝑘 𝑥
2
+ 𝑘 𝑦
2
is cutoff wave number.
For a lossless dielectric 𝛾2 = −𝜔2 𝜇𝜀
So 𝛾𝑔= ±𝑗 𝜔2 𝜇𝜀 − 𝑘 𝑐
2
So there are three cases….
JIT 10
Cont’d

11. Case 1
If 𝜔2
𝜇𝜀 = 𝑘 𝑐
2
𝛾𝑔= 0 no propagation
This is critical condition for cutoff propagation
𝜔𝑐=
1
𝜇𝜖
𝑘 𝑥
2
+ 𝑘 𝑦
2
𝑓𝑐=
1
2𝜋 𝜇𝜖
𝑘 𝑥
2
+ 𝑘 𝑦
2
JIT 11

12. JIT 12
Case 2
If 𝜔2
𝜇𝜀 > 𝑘 𝑐
2
𝛾𝑔= ±𝑗𝛽𝑔 = ±𝑗𝜔 𝜇𝜖 1 −
𝑓𝑐
𝑓
2
This shows that operating frequency should be greater than critical
frequency to propagate the wave in wave in wave guide

13. JIT 13
Case 3
If 𝜔2
𝜇𝜀 < 𝑘 𝑐
2
𝛾𝑔= ±𝛼 𝑔 = ±𝜔 𝜇𝜖
𝑓𝑐
𝑓
2
− 1
This shows that if operating frequency is below the cutoff frequency the
wave will decay exponentially wrt a factor -𝛼 𝑔 𝑧 and there will no wave
propagation
There for the solution of Helmholtz equation in rectangular co-ordinates
is given by
𝐸𝑧𝑠 = 𝐴. sin 𝑘 𝑥. 𝑥 + 𝐵. cos(𝑘 𝑥. 𝑥) × 𝐶. sin 𝑘 𝑦. 𝑦 + 𝐷. cos(𝑘 𝑦. 𝑦)
× 𝑒−𝑗𝛽

14. Rectangular waveguide
  WR (Waveguide Rectangular) series
- EIA (Electronic Industry Association) designation
WR-62
- Size: 1.58 cmx0.79 cm
- Recommended range: 12.4-18.0 GHz
- Cutoff: 9.486 GHz
2/
cm/10054262inch/10062
ab
.a



JIT 14

15. Waveguide modes
 TE (Transverse Electric) mode
- E parallel to the transverse plane of the waveguide
- In waveguide Wave propagates in +ve Z direction
- TEmn in characterized by Ez=0
- In other words the z component of magnetic field must exist
in order to have energy transmission in the guide.
 TM (Transverse Magnetic) mode
- H is within the transverse plane of the waveguide
JIT 15

16. TE modes in rectangular waveguides
𝛻2 × 𝐸𝑠 = −𝑗𝜔𝜇𝐻𝑠
𝛻2 × 𝐻𝑠 = −𝑗𝜔𝜇𝐸𝑠
Therefore the magnetic field in +ve Z direction ie. The solution of above
partial differential equations
H0z is the amplitude constant, so field equations in rectangular
waveguides
JIT 16
mode:)sin()sin(),(
mode:)cos()cos(),(
mn
zj
nmz
mn
zj
nmz
TMeybxayxE
TEeybxayxH
mn
mn






222
and,,where nmmnnm bak
b
n
b
a
m
a  

zj
nmzz
mn
eybxaHH 
 )cos()cos(0

17. Utilization
Transmission of power
3. Waveguide
JIT 17

18. TE and TM modes
  Hz and Ez fields: TE and TM modes
 Non-TEM modes: Hz = Ez = 0
 Concept of a dominant mode: TE10 mode
mode:)sin()sin(),(
mode:)cos()cos(),(
mn
zj
nmz
mn
zj
nmz
TMeybxayxE
TEeybxayxH
mn
mn






Boundary condition enforcements
: PEC (Perfect Electric Conductor)
222
and,,where nmmnnm bak
b
n
b
a
m
a  

3. Waveguide
JIT 18

19. Cutoff wave no kc = 𝑘 𝑥
2
+ 𝑘 𝑦
2
Where 𝑘 𝑥=𝑎 𝑚 and 𝑘 𝑦=𝑏 𝑛
kc =
𝑚𝜋
𝑎
2
+
𝑛𝜋
𝑏
2
= 𝜔𝑐 𝜇𝜀
fc =
1
𝜇𝜀
𝑚𝜋
𝑎
2
+
𝑛𝜋
𝑏
2
Propagation constant as discussed earlier
𝛾𝑔= ±𝑗 𝜔2 𝜇𝜀 − 𝑘 𝑐
2
So from case-1 and case-2 propagation constant or phase constant
𝛽𝑔 = 𝜔 𝜇𝜖 1 −
𝑓𝑐
𝑓
2
JIT 19

20. And attenuation constant
𝛼 𝑔 = 𝜔 𝜇𝜖
𝑓𝑐
𝑓
2
− 1
We know that 𝑓𝑐 =
𝑐
𝜆 𝑐
𝑎𝑛𝑑 𝜆 𝑐 =
𝑐
𝑓𝑐
So cutoff frequency
𝜆 𝑐 =
𝑐
𝑐
2𝜋
𝑚
𝑎
2
+
𝑛
𝑏
2
𝜆 𝑐 =
2
𝑚
𝑎
2
+
𝑛
𝑏
2
JIT 20

21. Dominant mode: TE10 mode







a
k

 2
10
.0and1where
)/cos(
)cos()cos(),(
10
10
01





nm
eax
eybxayxH
zj
zj
z




3. Waveguide
JIT 21

22. Parallel-plate waveguide
2. Parallel-plate
Phase front: out of phase
Phase front: in phase (guided mode)
JIT 22

23. Wavenumbers
22
mm kk 
mediumcnonmagnetiandLossless
c
n
c
k r
rooo

 
indexrefrectiveaiswheren
2. Parallel-plate
JIT 23

24.  mdkdk mm 2
d
m
km


Reflections
2. Parallel-plate
JIT 24

25. 0 
TE and TM modes
2. Parallel-plate
TM modeTE mode
JIT 25

26. Cutoff frequency
22
22
1111
11
2
coscoscoscoscos





































 
nd
cm
k
kd
m
kkk
nd
m
nd
cm
kd
m
k
k
kk
mm
m
mmm







nd
cm
m cm

 :mode,forfrequencycutoff
2
1 








 cm
m
c
n
propagate.notdoesmodetheandimaginaryis,If
propagate.willmodetheandconstantphasevalued-real,If
mcm
mcm




2. Parallel-plate
JIT 26

27. m
ndc
cm
cm
22
:hwavelengtcutoff 



2
1
2







cm
m
n





Cutoff wavelength
2. Parallel-plate
JIT 27

28. TE mode representation
   
 
waves.planedownwardandupwardtheof
ionsuperpositthefromresultingpatternceinterferentheisfieldmodeTEThe
cutoff)abovemode(TE
)cos(sin)Re(),(
sinsin2)(
,and
'
0
'
000
r
0
r
0
ztxkEeEtzE
exkEexkjEeeeEE
zxkk
eEeEE
mm
tj
ysy
zj
m
zj
m
zjxjkxjk
ys
zxzmxmzmxm
jj
ys
mmmmm










aaraakaak du
kk du
2. Parallel-plate
JIT 28

29.    
)cutoffbelowmodeTE(
)cos(sin),(andsin
1
2
1,If
'
0
'
0
22
texkEtzEexkEE
n
c
n
j
z
my
z
mys
cm
cmcm
cm
mmcm
mm








 














TE mode representation
infinity.approachesas90gapproachin
increases,anglewavethe,decreased)isor(cutoffbeyondincreasedisAs
guide.down theprogressforwardnomakingarethey
forth;andbackreflectingjustarewavesplanetheand0),(cutoffAt
o
cm


  m
cm
cm
m




 cos
2. Parallel-plate
JIT 29

30. Phase and group velocity
mm
c
n
k 

 sinsinm 
m
pm
n
c
v


sin
velocityPhase
m

m
cm
m
gm
n
c
n
c
d
d
v 




sin1:velocityGroup
2







.relativityspecialaofprincipletheenot violat:
mediumin thelightofspeedtheexceedmayThis
2. Parallel-plate
JIT 30

31. Field analysis
cnkk /where22
 ss EE
):variation-,0(0
)ofcomponenta(onlymodesTE
2
2
2
2
2
2
2
2
2
zj
ysysysysys
m
ezE
y
EkE
z
E
y
E
x
y













E
zj
mys
m
exfEE 
 )(0
0)()(
)( 2
2
2
2
 xfk
dx
xfd
mm
m
 0)(
)(
, 2
2
2
22
2  xfk
dx
xfd
kk mm
m
mm








d
xm
xfdxxE
xkxkxf
my
mmm

sin)(.and0atzerobemust:BC
),sin()cos()(


2. Parallel-plate
JIT 31

32. Characteristics of TE mode
cavityresonantldimensionaOne
2
sinsincutoff,At
2
2
cutoff,At/2and0
.2isshiftphasetrip-roundNet
walls.conductingebetween thdownandupbouncessimplywaveThe
zero.isguidein theincidenceofanglewaveplanethecutoff,At
00 













cm
ysys
cm
cm
cmmm
xn
EE
d
xm
EE
n
m
d
n
d
m
nkk
m







zj
ys
m
e
d
xm
EE  






 sin0
2. Parallel-plate
JIT 32

33. Field representations
x
zj
mmz
zj
mmx
ys
z
ys mm
exkEjexkEk
z
E
x
E
y
j
zx
aaaaE
E
HE
H
s
s
ss
s












)sin()cos(
,ofcomponentaOnly
modeTEaforofcomponentsand
00
zj
m
m
zs
zj
m
m
xs
mm
exkE
k
jHexkEH 

 
 )cos(,)sin( 00
2. Parallel-plate
JIT 33

34.    
 1cossinand
)(cos)(sin||
||
22222
00222/1220
**



AAkk
EkE
xkxkk
E
HHHH
mm
mmmm
zszsxsxsxs
*





s
sss
H
HHH
Intrinsic impedance

2. Parallel-plate
JIT 34

35. Circular waveguide
  WC (Waveguide Circular) series
 Hz and Ez fields: TE and TM modes
3. Waveguide
JIT 35

36. JIT 36
 A microstrip is constructed with a flat conductor suspended over a ground
plane. The conductor and ground plane are separated by a dielectric.
 The surface microstrip transmission line also has free space (air) as the
dielectric above the conductor.
 This structure can be built in materials other than printed circuit boards,
but will always consist of a conductor separated from a ground plane by
some dielectric material.
Microstrip transmission line
4. Tx line

37. Circular waveguide
Circular waveguides offer implementation advantages
over rectangular waveguide in that installation is much simpler.
When forming runs for turns and offsets - particularly when large radii
are involved - and the wind loading is less on a round cross-section,
meaning towers do not need to be as robust.
JIT 37

38. For a circular waveguide of radius a,
we can perform the same sequence
of steps in cylindrical coordinates as
we did in rectangular coordinates to
find the transverse field components
in terms of the longitudinal (i.e. Ez,
Hz) components.
JIT 38

39. The scalar Helmholtz equation in cylindrical co-ordinate is given by
Using the method of separation of variables, the solution of above equation
is assumed
𝛹 = 𝑅 𝑟 . ∅ ∅ . 𝑍 𝑧 −−−− −(1)
Substituting (1) into (a) and solving this equation for
𝑅 𝑟 , ∅ ∅ 𝑎𝑛𝑑 𝑍 𝑧
So here also
𝑘 𝑐
2
+ 𝛾2
= 𝛾𝑔
2
This is also called as characteristic equation of Bessel’s equations.
For a lossless guide
𝛽𝑔 = ± 𝜔2 𝜇𝜖 − 𝑘 𝑐
2
JIT 39
(a)

40. The total solution of Helmholtz equation in cylindrical co-ordinate
JIT 40

41. TE Modes in Circular Waveguides
It is commonly assumed that the waves in a circular waveguide are
propagating in the positive z direction. Here in this mode 𝐸𝑧 = 0, so
𝐸 𝑥, 𝐸 𝑦, 𝐻 𝑥, 𝐻 𝑦 𝑎𝑛𝑑 𝐻𝑧.
After substituting boundary conditions the final solution is
𝛹 = 𝛹0. 𝐽 𝑛 𝑘 𝑐. 𝑟 cos 𝑛∅ . 𝑒−𝑗𝛽 𝑔 𝑧
For a lossless dielectric, Maxwell’s equations
𝛻 × 𝐸 = −𝑗𝜔𝜇𝐻𝑠
𝛻 × 𝐻 = 𝑗𝜔𝜇𝐸𝑠
JIT 41

42. In cylindrical co-ordinates, the components of E and
H fields can be expressed as
JIT 42

43. When the differentiation 𝜕/𝜕𝑧 is replaced by (−𝑗𝛽𝑔)and the z
component 𝐸𝑧 by zero, the TE mode equations in terms of 𝐻𝑧in
circular waveguide are expressed as 𝐸 𝑍 = 0
JIT 43
Where 𝑘 𝑐
2
= 𝜔2 𝜇𝜖 − 𝛽𝑔

44. The permissible value of kc can be written as
𝑘 𝑐 =
𝑋 𝑛𝑝
′
𝑎
Where 𝑋 𝑛𝑝
′
is a constant
And from above table 𝑋 𝑛𝑝
′
=1.841 for TM11 MODE
JIT 44

45. The final equations for the E and H fields can be
written as
JIT 45

46. Where Zg = Er/ 𝐻∅, = - 𝐸∅/ Hr has been replaced for the wave
impedance in the guide and where n = 0,1,2,3,... And p = 1, 2, 3, 4,....
The first subscript n represents the number of full cycles of field
variation in one revolution through 2𝜋 rad of ∅.
The second subscript p indicates the number of zeroes of 𝐸∅.
The mode propagation constant is determined by
𝛽𝑔 = ± 𝜔2 𝜇𝜖 −
𝑋 𝑛𝑝
′
𝑎
JIT 46

47. The cutoff wave number of a mode is that for which the mode
propagation constant vanishes. Hence
𝑘 𝑐 =
𝑋 𝑛𝑝
′
𝑎
= 𝜔𝑐 𝜇𝜖
So
𝑓𝑐 =
𝑋 𝑛𝑝
′
2𝜋𝑎 𝜇𝜖
And the phase velocity for TE modes is
𝑣𝑔 =
𝜔
𝛽𝑔
=
𝑣 𝑝
1 −
𝑓𝑐
𝑓
2
𝑤ℎ𝑒𝑟𝑒 𝑣 𝑝 =
1
𝜇𝜖
=
𝑐
𝜇 𝑟 𝜖 𝑟
JIT 47

48. The wavelength and wave impedance for TE modes in a circular guide
are given, respectively, by
JIT 48
𝒁 𝒈 =
𝑬 𝒙
𝑯 𝒚
=

49. TM Modes in Circular Waveguides
The TMnp modes in a circular guide are characterized by Hz = 0.
However, the z component of the electric field E, must exist in order to
have energy transmission in the guide.
Consequently, the Helmholtz equation for Ez in a circular waveguide is
given by
Its solution is given in Eq.
Which is subject to the given boundary conditions.
JIT 49
(A)

50. JIT 50

51. Similarly
On differentiating equation (A) wrt z and substituting the result in
above equations yield the field equations of TMnp modes in a
circular waveguide:
JIT 51

52. JIT 52

53. Some of the TM-mode characteristic equations in the circular guide
are identical to those of the TE mode, but some are different. For
convenience, all are shown here:
JIT 53

54. It should be noted that the dominant mode, or the mode
of lowest cutoff frequency in a circular waveguide, is
the mode of TEnp that has the smallest value of the
product, kc .a = 1. 841, as shown in above Table.
JIT 54

55. JIT 55
Thank you