JIT 1

WaveguidesMr. HIMANSHU DIWAKAR

Assistant ProfessorGETGI

JIT 2

JIT 3

Basic waveguides 1. Waveguide

Rectangular waveguide Circular waveguide Coaxial line

Optical waveguide Parallel-plate waveguide

JIT

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 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 7

JIT 8

The electric and magnetic wave equations in frequency domain is given by

For a loss less dielectric or perfect conductor

The above equations are like Helmholtz equations

Let =X(x).Y(y).Z(z)Be the solution of above equations

JIT 9

Cont’dSo separation equation-= On Solving the above equations

The propagation of wave in guide is conventionally assumed in +ve Z direction.

JIT 10

Here the propagation constant in guide differs from intrinsic propagation constant

Let And

Where is cutoff wave number.For a lossless dielectric

So So there are three cases….

Cont’d

JIT 11

Case 1

If no propagation

This is critical condition for cutoff propagation

JIT 12

Case 2

If

This shows that operating frequency should be greater than critical frequency to propagate the wave in wave in wave guide

JIT 13

Case 3If

This shows that if operating frequency is below the cutoff frequency the wave will decay exponentially wrt a factor - and there will no wave

propagationThere for the solution of Helmholtz equation in rectangular co-ordinates

is given by

JIT 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/100 54262inch/100 62

ab.a

JIT 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 16

TE modes in rectangular waveguides

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

mode:)sin()sin(),(

mode:)cos()cos(),(

mnzj

nmz

mnzj

nmz

TMeybxayxE

TEeybxayxHmn

mn

222 and,, where nmmnnm bakb

nba

ma

zjnmzz

mneybxaHH )cos()cos(0

JIT 17

Utilization

Transmission of power

3. Waveguide

JIT 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(),(

mnzj

nmz

mnzj

nmz

TMeybxayxE

TEeybxayxHmn

mn

Boundary condition enforcements: PEC (Perfect Electric Conductor)

222 and,, where nmmnnm bakb

nba

ma

3. Waveguide

JIT 19

Cutoff wave no kc =Where = and =

kc =fc =

Propagation constant as discussed earlier

So from case-1 and case-2 propagation constant or phase constant

JIT 20

And attenuation constant

We know that So cutoff frequency

JIT 21

Dominant mode: TE10 mode

ak 2

10

.0 and 1 where)/cos(

)cos()cos(),(10

1001

nmeax

eybxayxHzj

zjz

3. Waveguide

JIT 22

Parallel-plate waveguide 2. Parallel-plate

Phase front: out of phase

Phase front: in phase (guided mode)

JIT 23

Wavenumbers

22mm kk

medium cnonmagneti and Lossless

cn

ck r

rooo

index refrective a is where n

2. Parallel-plate

JIT 24

mdkdk mm 2d

mkm

Reflections 2. Parallel-plate

JIT 25

0

TE and TM modes 2. Parallel-plate

TM modeTE mode

JIT 26

Cutoff frequency

2222

1111

11

2coscoscoscoscos

ndcmk

kdmkkk

ndm

ndcm

kdm

kk

kk

mm

mmmm

ndcmm cm : mode,for frequency cutoff

2

1

cm

m cn

propagate.not does mode theandimaginary is , Ifpropagate. willmode theand constant phase valued-real , If

mcm

mcm

2. Parallel-plate

JIT 27

mndc

cmcm

22 :h wavelengtcutoff

2

12

cmm

n

Cutoff wavelength 2. Parallel-plate

JIT 28

TE mode representation

waves.plane downward and upward theofionsuperposit thefrom resultingpattern ceinterferen theis field mode TE The

cutoff) above mode (TE

)cos(sin)Re(),(

sinsin2)(

, and

'0

'000

r0

r0

ztxkEeEtzE

exkEexkjEeeeEE

zxkk

eEeEE

mmtj

ysy

zjm

zjm

zjxjkxjkys

zxzmxmzmxm

jjys

mmmmm

aaraakaak du

kk du

2. Parallel-plate

JIT 29

)cutoff below mode TE(

)cos(sin),( and sin

121 , If

'0

'0

22

texkEtzEexkEE

nc

nj

zmy

zmys

cm

cmcm

cmmmcm

mm

TE mode representation

infinity. approaches as 90 gapproachin

increases, angle wave the,decreased) is or ( cutoff beyond increased is Asguide. down the progress forward no making arethey

forth; andback reflectingjust are wavesplane theand 0 ),( cutoffAt

o

cm

m

cm

cmm

cos

2. Parallel-plate

JIT 30

Phase and group velocity

mm cnk sinsinm

mpm n

cv

sin

velocity Phasem

mcm

mgm n

cnc

ddv

sin1 : velocityGroup

2

.relativity special a of principle theenot violat :medium in thelight of speed theexceedmay This

2. Parallel-plate

JIT 31

Field analysis

cnkk / where22 ss EE

):variation- ,0( 0

) ofcomponent a(only modes TE

2

22

2

2

2

2

2

2zj

ysysysysysmezE

yEkE

zE

yE

x

y

E

zjmys

mexfEE )(0

0)()()( 2

22

2

xfkdx

xfdmm

m 0)()(

, 2

2

222

2 xfkdx

xfdkk mm

mmm

dxmxfdxxE

xkxkxf

my

mmm

sin)( . and 0at zero bemust :BC

),sin()cos()(

2. Parallel-plate

JIT 32

Characteristics of TE mode

cavityresonant ldimensiona One

2sinsin cutoff,At

22 cutoff,At /2 and 0

.2 isshift phase trip-roundNet walls.conducting ebetween thdown and up bouncessimply waveThe

zero. is guide in the incidence of angle waveplane thecutoff,At

00

cmysys

cm

cmcmmm

xnEEd

xmEE

nm

dnd

mnkk

m

zjys

med

xmEE

sin0

2. Parallel-plate

JIT 33

Field representations

xzj

mmzzj

mmxys

zys mm exkEjexkEk

zE

xE

yj

zx

aaaaE

EHE

H

s

s

ss

s

)sin()cos(

, ofcomponent aOnly

mode TE afor of components and

00

zjm

mzs

zjm

mxs

mm exkEk

jHexkEH

)cos( ,)sin( 00

2. Parallel-plate

JIT 34

1cossin and

)(cos)(sin||

||

22222

00222/1220

**

AAkk

EkExkxkk

E

HHHH

mm

mmmm

zszsxsxsxs*

s

sss

H

HHH

Intrinsic impedance

2. Parallel-plate

JIT 35

Circular waveguide

WC (Waveguide Circular) series Hz and Ez fields: TE and TM modes

3. Waveguide

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

JIT 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 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 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

Substituting (1) into (a) and solving this equation for So here also

This is also called as characteristic equation of Bessel’s equations.For a lossless guide

(a)

JIT 40

The total solution of Helmholtz equation in cylindrical co-ordinate

JIT 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 , so

After substituting boundary conditions the final solution is

For a lossless dielectric, Maxwell’s equations

JIT 42

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

JIT 43

When the differentiation is replaced by ()and the z component by zero, the TE mode equations in terms of in circular waveguide are expressed as

Where

JIT 44

The permissible value of kc can be written as

Where is a constantAnd from above table =1.841 for TM11 MODE

JIT 45

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

JIT 46

Where Zg = Er/ , = - / 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 rad of .

The second subscript p indicates the number of zeroes of .

The mode propagation constant is determined by

JIT 47

The cutoff wave number of a mode is that for which the mode propagation constant vanishes. Hence

So

And the phase velocity for TE modes is

JIT 48

The wavelength and wave impedance for TE modes in a circular guide are given, respectively, by

𝒁 𝒈=𝑬 𝒙

𝑯 𝒚=¿

JIT 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 byIts solution is given in Eq.

Which is subject to the given boundary conditions.

(A)

JIT 50

JIT 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 52

JIT 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 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 theproduct, kc .a = 1. 841, as shown in above Table.

JIT 55

Thank you