ec 2305 unit-v

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EC2305TRANSMISSION LINES AND WAVEGUIDES


UNIT-IV

UNIT V WAVEGUIDES

Application of Maxwells equations to the rectangular waveguide, TM waves in Rectangular guide.

TE waves in Rectangular waveguide Cylindrical waveguides. The TEM wave in coaxial lines,

Excitation of wave guides. Guide termination and resonant cavities.

5.1 APPLICATION OF MAXWELLS EQUATIONS TO THE RECTANGULAR

WAVEGUIDE

A Waveguides is a uniform guiding structure which is used to transmit an electromagnetic wave

through it at microwave frequencies. A waveguide consists of a hollow metallic tube of any arbitrary but

uniform cross-section. But the most commonly used waveguides are of either rectangular or circular

cross- section. These simple structure are of either are less expensive to manufacture then other structureand have same electrical properties compared with others. The typical waveguide structures used

commonly are as shown in the fig. This waveguides is considered of two types one is rectangular wave

guides and second one is circular waveguides.

Analysis of the waveguides means to find the configuration of the electromagnetic fields within

the guide. Such configuration of the electromagnetic fields can be determined by solving the Maxwells

equations with appropriate boundary conditions and then obtaining the expressions for these fields for

different modes of the wave propagation within the guide.

A rectangular waveguide is a hollow metal pipe with rectangular cross-section of width a and

height b as shown in the fig. according to the standard convention, the longest side of the waveguide is

considered along x-axis, with condition a b>

Assume that four conducting boundaries of the waveguide encloses a dielectric with extends in

z-direction axially. Let be the permittivity and be the permeability of the dielectric enclosed by the

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waveguide boundaries. Also assume that both, the conductor and the dielectric are loss-free in the ideal

guide.

Maxwells equations will be solved to determine the electromagnetic field configurations in the

rectangular region.

Maxwells equations for a non-conducting rectangular region and given as

H j E

E j H

=

=

j a a a

j a a a

x y zx y z

x y zx y z

H E E E

E H H H

= + +

= + +

x y z

x y z

a a a

Hx y z

H H H

=

y yx xz zx y x

H HH HH Ha a a

x z z x x y

= + +

Equating x, y and z components on both sides,

(1)

yzx

x zy

y xz

HHj E

y z

H Hj E

z x

H Hj E

x y

=

=

=

x y z

x y z

Similarly

a a a

Ex y z

E E E

=

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y yx xz zx y x

E EE EE Ea a a

y z z x x y

= + +

Equating x, y and z components on both sides

(2)

yzx

x zy

y xz

EEj H

y z

E Ej H

z x

E Ej H

x y

=

=

=

The wave equation is given by

2 2

2 2

2

=( +j ) (j )

E E

H H

Where

=

=

For a non-conducting medium, it becomes

2 2

2 2

E E

H H

=

=

2 2 22

2 2 2

2 2 22

2 2 2

.....(3)

E E EE

x y z

H H HH

x y z

+ + =

+ + =

It is assumed that the propagation is in the z direction an the variation of field components in this z

direction may be expressed in the formZ

e

Where is propagation constant

=+j

If =0, wave propagates without attenuation.

If is real i.e. =0, there is no wave motion but only an exponential decrease in amplitude.

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0

y

y 0

H

H

z

yz

y

yz

y y

Let H e

H e H

=

= =

x

0

y

y

x

H, z

E

E

z

E

z

x

yz

y

y

x

Similarly H

Let E e

E

Similarly E

=

=

=

=

y

( )

( ) .....(4)

H( )

x

field

( )

( )

( )

zy x

zx y

xz

zy x

zx y

y xz

HH j E a

y

HH j E b

x

Hj E c

y

Magatic

EE j H a

y

EE j H b

x

E Ej H c

x y

+ =

=

=

+ =

=

=

.....(5)

2 22 2

2 2

2 22 2

2 2

2 22 2

2 2

( )

.....(6)

( )

Where and

E EE E a

x y

H HH H b

x y

E HE H

x x

+ + =

+ + =

= =

Solving the equation (4) and (5), the fields , ,x y xH H E and yE can be found out.

To solvexH

Consider equation (4b) Consider equation (5a)

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zx y

HH j E

x

=

1 ( )

zx y

zx y

HH j E

x

HjH E i

x

=

=

Consider equation (5a)

1( )

zy x

zy x

EE j H

y

EjE H ii

y

+ =

=

Substitute equation ii to equation i

( )

2

2 2

2

2 2

2 2

2 2

2 2

2 2 2

2

1 1

1

1

1

where h

h

z zx x

z zx x

z zx x

z zx

z zx

zx

E Hj jH H

y x

E HjH H

y x

E HjH H

y x

E HjH

y x

E HH j

y x

H jH

x

=

= +

+ =

+=

+ =

= +

= +

2h

zE

y

To solvey

E

Substitute equation i to equation ii

1 1z zy y

H Ej jE E

x y

=

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2

2 2

1z zy y

H EjE E

x y

= +

( )

2

2 2

2 2

2 2

2 2

2 2 2

2 2

1

1

where h

h h

z zy y

z zy

z zy

z zy

H EjE E

x y

H EjE

x y

H EE j

x y

E HjE

y x

+ =

+=

+ =

= +

= +

To solve xE


1( )

zy x

zx y

HH j E

y

HE H iii

j j y

+ =

= +

Consider equation (5b)

1( )

zx y

zx y

zy x

EE j H

x

EE j H

x

EH E iv

j j x

=

=

= +

Substitute equation iv to equation iii

2

2 2

1 1

1

z zx x

z zx x

E HE Ej j j x j y

E HE E

x j y

= + +

= +

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2

2 2

1z zx x

E HE E

x j y

+ = +

( )

2 2

2 2

2 2

2 2 2

2 2

1

where h

h h

z zx

z zx

z zx

E HE

x j y

E HE j

x y

E HjE

x y

+= +

+ =

= +

=

To solvey

H

Substitute equation iii to equation iv

( )

2

2 2

2

2 2

2

2 2

2 2

2 2

2 2

1 1

1

1

1

1

z zy y

z zy y

z zy y

z zy y

z zy

zy

H EH H

j j j y j x

H EH H

y j x

H EH H

y j x

H EH H

y j xH E

Hy j x

HH j

y

= + +

= +

+ = +

+ = +

+

= +

+ =

2 2 2

2 2

where h

h h

z

z zy

E

x

H EjH

y x

= +

=

2 2 2 2

2 2 2 2

h h h h

andh h h h

z z z zx y

z z z zx y

H E H Ej jH H

x y y x

E H E Hj jE E

x y y x

= + =

= = +

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The components of electric and magnetic field strength ( , ,x y x

H H E andy

E ) are expressed in

terms of Ez and Hz. It is observed that there must be z components of either E or H; otherwise all the

components would be zero. Although in general case Ez and Hz may be present at the same time, it is

convenient to divide the solutions into two sets. In the first case, there is a component of E in the direction

of propagation (Ez), but no component of H in this direction. Such waves are called E or Transverse

magnetic (TM) waves. In the second case, there is a component of H in the direction of propagation

(Hz), but no component of E in this direction. Such waves are called Hwaves or transverse Electric

(TE) waves.

5.2 TM WAVES IN RECTANGULAR GUIDE

There is a component of E in the direction of propagation (Ez), but no component of H in this direction.

Such waves are called E or Transverse magnetic(TM) waves.

For transverse magnetic (TM) wave the magnetic field component exists only in the direction transverse

to the direction of propagation wave i.e. Z-direction. There is no component of the magnetic field that

exists along the Z-direction. 0Z

H = But 0Z

E

The differential equation for the TM wave

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( )

2 22 2

2 2

2 22 2

2 2

2 2 2 2

2 2

2 2 2

2 22

2 2

0

0

where h

h 0 (1)

Z ZZ Z

Z ZZ Z

Z ZZ

Z ZZ

E EE E

x y

E EE E

x y

E E Ex y

E EE

x y

+ + =

+ + + =

+ + + =

= +

+ + =

The wave equations are partial differential equations that can be solved by the usual technique of

assuming a product, solution. This procedure leads to two ordinary differentials the solutions of which are

known. Nothing that

0 - zZ z

0

z

E (x, y, z) = E (x, y) e

E =XY

Where X is a function of x alone and Y is a function of y alone. In equation 1 we can write

2 22

2 2h 0 (2)

X YY X XY

x y

+ + =

Divide by XY in equation (2) we get

2 2

2

2 21 1 h 0 (3)X Y

X x Y y + + =

As X and Y are independent variable, it is possible to equate first and second terms in equation (3) to

other constants such that their addition is equal to2h

2 2 2i.e. h A B= +

22

2

2

22

1( )

Let ----------(4)

1 --------------------(b)

XA a

X x

Y BY y

=

=

Hence equation (4a) and (4b) can be written as

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22

2

22

2

0 ( )

----------(5)

Y=0--------------------(b)

XA X a

x

YB

y

+ =

+

Equation (5a) and (5b) are ordinary second order differential equations. Their solutions can be written

directly as

1 2

3 4

cos sin

cos sin

X C Bx C Bx

Y C Ay C Ay

= +

= +

Where1 2 3 4, , and CC C C are arbitrary constants. To evaluate these arbitrary constants the boundary

condition are used

( ) ( )

0

z

z 1 2 3 4

E =

E = cos sin cos sin (6)

XY

C Bx C Bx C Ay C Ay+ +

Boundary conditions

For the rectangular waveguide, the geometry using rectangular co-ordinate system is as shown in the

above fig. We have studied that along the walls of the waveguide, none of the component of the electric

field can exit. Hence we can write boundary conditions as

0 at =0, for all values of y from 0 to bzE x=

0 at =a, for all values of y from 0 to bz

E x=

0 at =0, for all values of from 0 to a and

0 at =b, for all values of from 0 to a

z

z

E y x

E y x

=

=

As there are four boundary conditions, the four arbitrary constants can be easily evaluated as follows

With third boundary condition substituted in equation (6) we can write

( )1 2 30= cos sinC Bx C Bx C +

Note that the term 1 2cos sinC Bx C Bx+ cannot be zero. Hence to have component

30, C must be zero

zE =

3 0C =

Substituting the value of arbitrary constant 3 0C = in equation (6) we get

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( ) ( )z 1 2 4E = cos sin sin (7)C Bx C Bx C Ay+

Now using first boundary condition in equation (7) we get

( )( )1 40 = 0 sinC C Ay+

Hence only the condition that will make 0 all y frome 0 to b is given byz

E for=

10C =

Substituting the value of arbitrary constant 1 0C = in equation (7) we get

z 2 4E = sin sinC C Bx Ay

2 4Substitute C=C C

zE = sin sin (8)C Bx Ay

Now using second boundary condition in equation (8) we get

0 = sin sinC Ba Ay

Now note that sinAy can not be zero, if C gives zero that no solutions exists, hence C is not zero, thus

to 0z

E = , the condition can be written as

sin 0Ba=

This indicates Ba must be multiple of

where m=0,1,2,...........Ba m

mB

a

=

=

Substituting the value of arbitrary constant B in equation (8) we get

zE = sin sin (9)m

C x Aya

Now using fourth boundary condition in equation (9) we get

0= sin sinm

C x Aba

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Since C can not zero (for the existing solution) and sin 0m

xa

for allx , the condition is given by

sin 0Ab=

This indicates Ab must be multiple of

where n=0,1,2,...........Ab n

nA

b

=

=

Substituting the value of arbitrary constant A in equation (9) we get

0

ZE = sin sin (10)

m nC x y

a b

The field equation of rectangular waveguides

2 2

0

2

=0h h

( )h

z zx z

zx

E HjE H

x y

EE i

x

=

=

Now different equation (10) with respect to x we get

cos sin ( )zE m m n

C x y iix a a b

=

Substituting equation (ii) to equation (i) we get

0

2cos sin (11)

hx

m m nE C x y

a a b

=


2 2

0

2

=0h h

( )h

z zy z

zy

E HjE H

y x

EE iv

y

= +

=

Now different equation (11) with respect to y we get

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0

2

sin cos ( )

Substituting equation (iv) to equation (iii) we get

sin cos (12)

h

z

y

E n m nC x y v

y b a b

n m nE C x y

b a b

=

=


2 2

0

2

0

2

=0h h

( )h

Substituting equation (v) to equation (vi) we get

sin cos (13)h

z zx z

zx

x

H EjH H

x y

EjH vi

y

j n m n

H C x yb a b

= +

=

=


2 2

0

2

=0h h

( )h

z zy z

zy

H EjH H

y x

EjH vii

x

=

=

Substituting equation (ii) to equation (vii) we get

0

2cos sin (14)

hy

j m m nH C x y

a a b

=

The TM waves in Rectangular waveguide the field equation is

0

Z

0 0

2 2

0 0

2 2

E = sin sin

cos sin sin cosh h

sin cos andh h

x y

x y

m nC x y

a b

m m n n m nE C x y E C x y

a a b b a b

j n m n j mH C x y Hb a b a

= =

= =

cos sinm nC x ya b

The wave equation is

0 z

y yE E e

= 0 zx x

E E e = 0 z

y yH H e

= 0 zx x

H H e = And 0 z

z zE E e

=

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z

2 2

2

E = sin sin

cos sin sin cosh h

sin cos andh

z

z z

x y

z

x y

m nC x y e

a b

m m n n m nE C x y e E C x y e

a a b b a b

j n m nH C x y e Hb a b

= =

= =

2cos sin

h

zj m m nC x y ea a b

Here no wave motion means 0

j

j

= + =

=

These expressions show how each of the components of electric and magnetic field strengths varies with

x and y. The variation with time and along the axis of the guide, that is electric field the z direction, is

shown by putting back intro each of these expressions the factorj z

e

z

2 2

2

E = sin sin

cos sin sin cosh h

sin cos andh

j z

j z j z

x y

j z

x

m n

C x y ea b

j m m n j n m nE C x y e E C x y e

a a b b a b

j n m nH C x y e

b a b

= =

=

2

cos sinh

j z

y

j m m nH C x y e

a a b

=

Characteristic ofmn

TM wave

Propagation constant

= +

=

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=

= +

= +

= +

= +

At lower freq. is < +

becomes real with value of, and =0. There is only

attenuation, without any propagation.

At higher frequencies value of becomes greater than

+

making .imaginary.

2 2

2

2 2

2

2 22

mFor ffc =0

a b

mFor f=fc 0

a b

n

nj

n

+ > =

+ < =

+ = =

Cut off frequency:-

f=fc

= +

= +

= + =

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= +

= +

= +

Phase constant:-

= + +

( )

=

=

=

= =

=

=

Cut-off wave length

=

=

1v

=

2 21 m

a b2c

nf

= +

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( ) ( )

=

+

= +

Guide wavelength:-

The guide wave length is defined as the distance traveled for the phase shift through 2radius.

2=

2=

g

g

2

2 2

cf f

1

=v

2 2=

g

c

v

f f

Take f out side we get

2=

1

g

c

v

ff

f

=v

f

2=

1

g

cf

f

= , =cc

v vf f

2=

1

g

c

Velocity of propagation:-

The phase velocity is defined asthe rate at which wave changes its phase as the wave propagates

inside the region between parallel planes.It is denoted by ! and its given by

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=

=

!

!


=

!

2 f =

=!

=

!

1=v

=

=

!

!

= , =cc

v vf f

Group velocity

The group velocity is defined as the velocity with which the energy propagates along a guide is called

group velocity

g

dv

d

=

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= + +

( )

=

=

=

= +

=

=

"

Differentiating the above expression with respect we get

=

=

Take out side we get

=

1=v

=

=

=

= , =cc

v vf f

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Multiply phase velocity and group velocity we get

2

1

p g

c

vv v

f

f

=

2

1 cf

vf

2

p gv v v =

Wave impedance or Characteristic impedance equation of TM wave

In transmission-line theory power is propagated along one axis only, and only one impedance constant is

involved. However, in the three dimensional wave propagation power may be transmitted along three

axes of the coordinate system and consequently three impedance constants must be defined.

Characteristic impedances are defined by the following ratios of electric to magnetic field

strengths for the positive directions of the coordinates.

( )x

o TM

y

EZ

H=

2 2

2

=0h h

h

z zx z

zx

E HjE H

x y

EE

x

=

=

2 2

2

=0h h

h

z zy z

zy

H EjH H

y x

EjH

x

=

=

( )o TMZ

=2h

z

E

x

2

h

j zE

x

( )o TMZj

=

In Rectangular waveguides =j =0

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0TM

jZ =

j

0TMZ

=

=

2 2

0

2c

TM

f fZ

=

Take out side we getf

2

0

0

2 1 c

TM

TM

ff

fZ

Z

=

=

2

1 cf

f

0TMZ

=

2

1 cf

f

2

01 c

TM

fZ

f

=

=

2

0 1 c

TM

fZ

f

=

= , =c

c

v vf f

Or

2

0 1TMc

Z

=

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5.3 TE WAVES IN RECTANGULAR GUIDE

There is a component of H in the direction of propagation (Hz), but no component of E in this

direction. Such waves are called Hwaves or transverse Electric (TE) waves.

For transverse electric (TE) wave the electric field component exists only in the direction

transverse to the direction of propagation wave i.e. Z-direction. There is no component of the electric

field that exists along the Z-direction. 0ZH But 0ZE =

The differential equation for the TE wave

( )

2 22 2

2 2

2 22 2

2 2

2 22 2

2 2

2 2 2

2 22

2 2

0

0

where h

h 0 (1)

Z ZZ Z

Z ZZ Z

Z ZZ

Z ZZ

H H H Hx y

H HH H

x y

H HH

x y

H HH

x y

+ + =

+ + + =

+ + + =

= +

+ + =

The wave equations are partial differential equations that can be solved by the usual technique of

assuming a product, solution. This procedure leads to two ordinary differentials the solutions of which are

known. Nothing that

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0 - z

Z z

0

z

(x, y, z) = H (x, y) e

=

H

H XY

Where X is a function of x alone and Y is a function of y alone. In equation 1 we can write

2 22

2 2h 0 (2)

X YY X XY

x y

+ + =

Divide by XY in equation (2) we get

2 22

2 2

1 1h 0 (3)

X Y

X x Y y

+ + =

As X and Y are independent variable, it is possible to equate first and second terms in equation (3) to

other constants such that their addition is equal to2

h

2 2 2i.e. h A B= +

22

2

22

2

1( )

Let ----------(4)1

--------------------(b)

XA a

X x

YB

Y y

=

=

Hence equation (4a) and (4b) can be written as

2

22

22

2

0 ( )----------(5)

Y=0--------------------(b)

XA X a

x

YB

y

+ =

+

Equation (5a) and (5b) are ordinary second order differential equations. Their solutions can be written

directly as

1 2

3 4

cos sin

cos sin

X C Bx C Bx

Y C Ay C Ay

= +

= +

Where1 2 3 4, , and CC C C are arbitrary constants. To evaluate these arbitrary constants the boundary

condition are used

( ) ( )

0

z

0

z 1 2 3 4

=

= cos sin cos sin (6)

H XY

H C Bx C Bx C Ay C Ay+ +

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Boundary conditions

For the rectangular waveguide, the geometry using rectangular co-ordinate system is as shown in the

above fig. We have studied that along the walls of the waveguide, none of the component of the magnetic

field can exit. In case of the TE wave the boundary conditions are slightly different than those for TM

wave. Because in TE wave, the component in the direction of propagation i.e. z-direction can not exit

(i.e. 0Z

E = ). So in case of the TE wave the boundary conditions are specified for the electric field

componentsxE and yE inx and y direction respectively. Thus the boundary conditions for the TE

wave are given as

0 at =0, for all values of y from 0 to b

0 at =a, for all values of y from 0 to b

0 at =0, for all values of from 0 to a and

0 at =b, for all values of from 0 to a

y

y

x

x

E x

E x

E y x

E y x

=

=

=

=

As there are four boundary conditions, the four arbitrary constants can be easily evaluated as follows. As

the field componentZ

E can not exist, we must apply the boundary conditions with the field components

xE and yE only. Consider the field equation of rectangular waveguides is

2 2 2 2and

h h h h

z z z zx y

E H E Hj jE E

x y y x

= = +

2 2

2

=0h h

( )h

z zx z

zx

E HjE Ex y

HjE i

y

=

=

Differentiating equation (6) with respect to y we get

( )( )1 2 3 4cos sin sin coszH C Bx C Bx AC Ay AC Ay

y

= + +

Substituting this value in equation (i) we get

( )( )1 2 3 42 cos sin sin cos (7)hxj

E C Bx C Bx AC Ay AC Ay

= + +

With third boundary condition substituted in equation (7) we can write

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( ) ( )1 2 420 cos sinh

jC Bx C Bx AC

= +

Note that the term 1 2cos sinC Bx C Bx+ cannot be zero. Hence to have component

30, C must be zero

xE =

40C =

Substituting the value of arbitrary constant4

0C = in equation (6) we get

( )( )z 1 2 3= cos sin cos (8)H C Bx C Bx C Ay+

Consider the field equation of rectangular waveguides is

2 2

2

=0h h

-------(ii)h

z z

y z

zy

E HjE E

y x

HjE

x

= +

=

Differentiating equation (8) with respect to x we get

( ) ( )1 2 3sin cos coszH BC Bx BC Bx C Ay

x

= +

Substituting this value in equation (ii) we get

( ) ( )1 2 32 sin cos cos (9)hyj

E BC Bx BC Bx C Ay

= +

With first boundary condition substituted in equation (9) we can write

( ) ( )2 320 cosh

jBC C Ay

=

Hence only the condition that will make 0 all y frome 0 to b is given byy

E for=

2 0C =

Substituting the value of arbitrary constant2

0C = in equation (8) we get

z 1 3H = cos cosC C Bx Ay

1 3Substitute C=C C

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zH =Ccos cos (10)Bx Ay


sin coszH

BC Bx Ay

x

=

Substituting this value in equation (ii) we get

2sin cos (11)

hy

jE BC Bx Ay

=

With second boundary condition substituted in equation (11) we can write

20 sin cos

h

jBC Ba Ay

=

Now note that cosAy can not be zero, if C gives zero that no solutions exists, hence C is not zero, thus

to 0y

E = , the condition can be written as

sin 0Ba=

This indicates Ba must be multiple of

where m=0,1,2,...........Ba m

mB

a

=

=

Substituting the value of arbitrary constant B in equation (10) we get

zH = cos cos (11)

mC x Ay

a


cos sinzH m

AC x Ayy a

=

Substituting this value in equation (i) we get

2cos sin (12)

hx

j mE AC x Ay

a

=

Now using fourth boundary condition in equation (12) we get

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20 cos sin

h

j mAC x Ab

a

=

Since C can not zero (for the existing solution) and cos 0m

xa

for allx , the condition is given by

sin 0Ab=

This indicates Ab must be multiple of

where n=0,1,2,...........Ab n

nA

b

=

=

Substituting the value of arbitrary constant A in equation (12) we get

0

ZH = cos cos

m nC x y

a b


2 2

0

2

=0h h

( )h

z zx z

zx

E HjE E

x y

HjE iv

y

=

=


cos sin (15)zH n m n

C x yy b a b

=

Substituting the equation (15) in equation (iv) we get

0

2cos sin (16)

hx

j n m nE C x y

b a b

=

2 2

0

2

=0h h

( )h

z zy z

zy

H EjH E

y x

HH v

y

=

=

Substituting the equation (15) in equation (v) we get

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0

2cos sin (17)

hy

n m nH C x y

b a b

=

2 2

0

2

=0h h

( )h

z zy z

zy

E HjE E

y x

HjE vi

x

= +

=


sin cos (18)zH m m n

C x yx a a b

=

Substituting the equation (18) in equation (vi) we get

02 sin cosh

y j m m nE C x ya a b

=

2 2

0

2

=0h h

( )h

z zx z

zx

H EjH E

x y

HH vii

x

= +

=

Substituting the equation (18) in equation (vii) we get

0

2 sin cos (19)hx

m m n

H C x ya a b

=

The TE waves in Rectangular waveguide the field equation is

0

Z

0 0

2 2

0

2

H = cos cos

cos sin sin cosh h

sin cos and

h

x y

x y

m nC x y

a b

j n m n j m m nE C x y E C x y

b a b a a b

m m nH C x y H

a a b

= =

=

0

2cos sin

h

n m nC x y

b a b

=


0 z

y yE E e = 0 zx xE E e

= 0 zy yH H e

= 0 zx xH H e = And 0 zz zH H e

=

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z

2 2

2

H = cos cos

cos sin sin cosh h

sin cosh

z

z z

x y

z

x

m nC x y e

a b

j n m n j m m nE C x y e E C x y e

b a b a a b

m m nH C x y ea a b

= =

=

2and cos sin

h

z

yn m nH C x y eb a b

=


j

j

= + =

=




e

z

2 2

2

H = cos cos

cos sin sin cosh h

sin cosh

j z

j z j z

x y

x

m n

C x y ea b

j n m n j m m nE C x y e E C x y e

b a b a a b

j m m nH C x y e

a a b

= =

=

2

and cos sinh

j z j z

y

j n m nH C x y e

b a b

=

Characteristic ofmn

TE wave


= +

=

=

= +

= +

= +

= +

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At lower freq. is < +

becomes real with value of, and =0. There is only

attenuation, without any propagation.

At higher frequencies value of

becomes greater than

+

making .imaginary.

2 2

2

2 2

2

2 2

2

mFor ffc =0

a b

mFor f=fc 0

a b

n

nj

n

+ > =

+ < =

+ = =

Cut off frequency:-

f=fc

= +

= +

= + =

= +

= +

= +

Phase constant:-

= + +

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( )

=

=

=

= =

=

=

Cut-off wave length

=

=

1v

=

2 21 m

a b2c

nf

= +

( ) ( )

=

+

=+

Guide wavelength:-


2=

2=

g

g

2 2 2cf f

1

=v

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2 2=

g

c

v

f f


2=

1

g

c

v

ff

f

= vf

2=

1

g

cf

f

= , =c

c

v vf f

2

=

1

g

c




=

=

!

!


=

!

2 f =

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=!

=

!

1

=v

=

=

!

!

= , =c

c

v vf f

Group velocity


group velocity

g

dv d

=

= + +

( )

=

=

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=

= +

=

=

"


=

=


=

1=v

=

=

=

= , =cc

v vf f


2

1

p g

c

vv v

f

f

=

2

1 cf

vf

2

p gv v v =

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Wave impedance or Characteristic impedance equation of TE wave






( )x

o TE

y

EZ

H=

2 2

2

E =0h h

h

z zx z

zx

E HjE

x y

HjE

y

=

=

2 2

2

E =0h h

h

z zy z

zy

H EjH

y x

HH

y

=

=

( )o TEZ

=2h

j zH

y

2h

zH

y

( )o TE

jZ

=

In Rectangular waveguides =j =0

0TE

jZ =

j

0TE

Z

=

=

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02 2

02

0

2


2 1

TE

c

TE

c

TE

Zf f

f

Z

ff

f

Z

=

=

=

2

0

1 c

TE

f

f

Z

=

2

02

1

1

c

TE

c

f

f

Z

f

f

=

=

02

1

TE

c

Z

f

f

=

Or

02

1

TE

c

Z

=

= , =cc

v vf f

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5.4 CYLINDRICAL WAVEGUIDES

The method of solution of the electromagnetic equations for guided of-circular cross section is similar to

that followed for rectangular guides however, in order to simplify the applications of the boundary

conditions it is expedient to express the field equations & the wave equations in the cylindrical co-

ordinate system.

Consider a hollow cylindrical of perfectly conducting material with finite radius a shown in fig.

The Maxwells equation of non-conducting medium is

H j E

E j H

=

=

After solving Maxwells equation the radial and tangential components of electric and magnetic fields can

be obtain. The radial component are H andE , the tangential components are H and E

In cylindrical co-ordinates in a non-conducting region Maxwells equations are.

j a a a (1)

j a a a (2)

z z

z z

H E E E

E H H H

= + +

= + +

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1 1z

z

a a a

Hz

H H H

=

1 1(2)z z z

H H H HH HH a a a

z z

= + +

Equating , and z components on both sides,

Equating Equation(2) and Equation(1)

1

(3)

1

z

z

z

HH j Ez

H Hj E

z

H Hj E

=

=

=

1 1z

z

a a a

Ez

E E E

=

1 1(4)z z z

E E E EE EE a a a

z z

= + +

Equating , and z components on both sides,Equating Equation(4) and Equation(2)

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1

(5)

1

z

z

z

EEj H

z

E Ej H

z

E E j H

=

=

=

The wave equation is given by

2 2

2 2

2

=( +j ) (j )

E E

H H

Where

=

=

For a non-conducting medium, it becomes

2 2

2 2

E E

H H

=

=

2 2 22

2 2 2 2

2 2 22

2 2 2 2

1 1

.....(6)1 1

E E E EE

z

H H H HH

z

+ + + + =

+ + + =


0 0 0 0

0 0 0 0

E and E

and

z z z z

z z z z

H H e H H e E e E e

H H E EH e H e E e E e

z z z z

H HH H

z z

= = = =

= = = =

= =

and

E EE E

z z

= =

Substituting the above expression in equation (3), equation (5) and equation (6)

Electric Field component

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1( )

( ) (7)

1 ( )

z

z

z

HH j E a

HH j E b

H H j E c

+ =

=

=

Magnetic Field component

1( )

( ) (8)

1 ( )

z

z

z

EE j H a

EE j H b

E E j H c

+ =

=

=

2 22 2

2 2 2

2 22 2

2 2 2

2 22 2

2 2

1 1

-------(8)1 1

Where and

E E EE E

H H HH H

r

E HE H

+ + + =

+ + + =

= =

Solving the equation (4) and (5), the fields , ,H H E and E can be found out.

To solve E


1( )

z

z

HH j E

HE H ij j

=

=


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1

1( )

z

z

EE j H

EH E ii

j j

+ =

=

Substitute equation ii to equation i

( )

2

2 2

2

2 2

2 2

2 2

2 2

1 1

1

1

1

z z

z z

z z

z z

z z

E HE E

j j j j

E HE E

j

E HE E

j

E HE j

E HE j

=

=

+ =

+=

+ = +

2 2 2

2 2

h

h h

z z

where

E HjE

= +

= +

To solve H

Substitute equation i to equation ii

2

2 2

1 1

1

z z

z z

E HH H

j j j j

E HH H

j

=

=

2

2 2

2 2

2 2

1

1

z z

z z

E HH H

jE H

Hj

+ =

+

=

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( )2 2

2 2 2

2 2

h

h h

z z

z z

E HjH

where

E HjH

+ =

= +

=

To solve H


1( )

z

z

EE j H

EH E iii

j j

=

= +


1

1( )

z

z

HH j E

HE H iv

j j

+ =

= +

Substitute equation iv to equation iii

( )

2

2 2

2

2 2

2 2

2 2

2 2

2 2 2

1 1

1

1

1

h

z z

z z

z z

z z

z z

H EH Hj j j j

H EH H

j

H EH H

j

H EH

j

H E

H j

where

H

= + +

= +

+ = +

+= +

+ =

= +

=2 2h h

z zH Ej

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5.5 TM WAVES IN CYLINDRICAL WAVEGUIDE

Transverse magnetic (TM) waves are in which the magnetic field strength H is entirely traverse.

It has an electric field strength Ezin the direction of propagation and no component of magnetic field Hz

in the same direction (Hz=0). But the electric field exists in all direction the wave equation is

2 2 0ZE E + =

2 22 2

2 2 2

1 1z z zz z

E E EE E

+ + + =

( )2 2

2 2

2 2 2

1 10 (1)z z z z

E E EE

+ + + + =

2 22

2 2 2

1 10 (2)z z z

z

E E Eh E

+ + + =

Where X () is a function of alone and Y () is a function of alone. Sub the expression for

ZE XY= in the wave equation gives.

SubstitutingZ

E XY=

2 22

2 2 20 (3)

X Y X X YY h XY

+ + + =

2Multiplying

XY

2 2 22 2

2 2

10 (4)

X X Yh

X X Y

+ + + =

In equation (4) the first, second and forth terms are functions of only while the third term is only the

function of only. It is obvious to get such terms because X and Y are independent Variables, we can

equate above independent terms to another constants. Hence we can write,

22 2

2

22

2

1 Where is constant

0

Y n nY

Yn Y

=

+ =

For above ordinary second order differential equation, the solution is given by

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cos sin Where A and B is constantY A n B n = +

2 22 2 2

20 (5)

X Xh n

X X

+ + =

Multiplying the above equation by X on both the Sides, we get

( )2

2 2 2 2

20 (6)

X Xh n X

+ + =

Since h is constant, we can modify equation (6) to get the standard form of the differential equation, we

get h =

( )( )

( )( )

( )2

2 2 2 2

20 (7)

X Xh h h n X

hh

+ + =

Above equation is standard Bessels differential equation. The solution of this equation is given by

n n

n

J ( h) Where C constant

and J ( h) is a solution or first type of order n

nX C

=

Hence the complete solution for zE can be obtained by substituting values of X and Y from equation (7)

we get

( )( )

n

n n n

J ( h) cos sinJ ( h) cos sin -------(8) Where A B B

Z

Z n

Z n n n n

E XY

E C A n B nE A n B n C A C

=

= += + = =

Form Trigonometric Properties

2 2 1

' '

' 2 2

' 1

cos sin cos tan

cos sin cos

Where =

= tan

nn n n n

n

n n n

n n n

n

n

BA n B n A B n

A

A n B n A n

A A B

Bn n

A

+ = +

+ =

+

Hence equation (8) can be modified as

0 ' '

nJ ( h)cos (9)z nE A n =

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We can write all the field components of the electric and magnetic field as follows

2 2

0

2

0h h

( )h

z zz

z

H EjE H

EE i

= =

=

Differentiating equation (9) with respect to we get

' ' 'n nn

' ' '

n

J ( h) J ( h)h cos J ( h)

hJ ( h) cos -------(10)

zn

zn

EA n

EA n

= =

=

Substituting the equation (10) in equation (i) we get

0

2hE

= ' hnA

' '

n

0 ' ' '

n

J ( h)cos

J ( h)cos (11)h

n

n

E A n

=

2 2

0

2

0h h

( )h

z zz

z

H EjH H

EjH ii

= =

=

Substituting the equation (10) in equation (ii) we get

0

2h

jH

= ' h

nA ' 'n

0 ' ' '

n

J ( h)cos

J ( h) cos (12)h

n

n

jH A n

=

2 2

02

0h h

( )h

z zz

z

E HjH H

EjH iii

= =

=


( )' 'nJ ( h)sin (13)z nE

nA n

=

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Substituting the equation (13) in equation (iii) we get

( )0 ' 'n2 J ( h)sin (14)h nj

H nA n

=

2 2

0

2

0h h

( )h

z zz

z

E HjE H

EE iv

= + =

=

Substituting the equation (13) in equation (iv) we get

( )0 ' 'n2 J ( h)sin (15)h nE nA n

=

The TM waves in cylindrical waveguide the field equation is

( )

( )

0 ' '

n

0 ' ' ' 0 ' '

n n2

0 ' ' 0 ' ' '

n n2

J ( h)cos

J ( h)cos J ( h)sinh h

J ( h)sin and J ( h)cosh h

z n

n n

n n

E A n

jE A n H nA n

jE nA n H A n

=

= =

= =


0 0 0 0 0 E and E and Ez z z z zz z

H H e H H e E e E e E e = = = = =

The TM waves in cylindrical waveguide the field equation is

( )

( )

' '

n

' ' ' ' '

n n2

' ' ' ' '

n n2

J ( h)cos

J ( h)cos J ( h)sinh h


z

z n

z z

n n

z z

n n

E A n e

jE A n e H nA n e

jE nA n e H A n e

=

= =

= =


j

j

= + =

=

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e

( )

( )

' '

n

' ' ' ' '

n n2

' ' ' ' '

n n2

J ( h)cos

J ( h) cos J ( h)sinh h


j z

z n

j z j z

n n

j z j z

n n

E A n e

j jE A n e H nA n e

j jE nA n e H A n e

=

= =

= =

Boundary Conditions:

0 all along the circumference of the cylindrical

0 at =a, for all values of varying from 0 to 2

z

z

E

E

=

=

' '

nJ ( h) cos z

z nE A n e =

If applying first boundary condition means all value should be zero

So apply second boundary condition

' '

n

n

0 J ( h)cos

J ( h)=0

z

nA a n e

a

=

For different orders of the Bessel function (i.e. for different n) this equation can be satisfied for different

values of a such values are called Eigen value

nm nm

nmnm

ah P

Ph

a

=

=

Order of the Bessel

function n1n

P 2n

P 3n

P

0 2.405 5.52 8.65

1 3.83 7.06 10.17

2 5.13 8.41 11.62

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Characteristic ofnm

TM wave


= +

=

=

nmnm

Ph

a=

=

#

At lower freq. is <

#

becomes real with value of, and =0. There is only attenuation,

without any propagation.


#

making .imaginary.

2

2

2

2

2

2

For ffc =0

For f=fc 0

nm

nm

nm

P

a

Pj

a

P

a

> =

< =

= =

Cut off frequency:-

f=fc

=

=

#

#

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= =

#

=

=

=

#

#

#

Cut-off Wave Length

=

=

1v

=

1

2

nmc

Pf

a

=

=

#

=

=

#

#

Phase constant:-

=

# #

( )

=

=

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=

= =

=

=

Guide wavelength:-


2=

2=

g

g

2 2 2cf f

1

=v

2 2=

g

c

v

f f


2=

1

g

c

v

ff

f

= v

f

2=

1

g

cf

f

= , =cc

v vf f

2=1

g

c

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=

=

!

!


=

!

2 f =

=!

=

!

1

=v

=

=

!

!

= , =c

c

v vf f

Group velocity


group velocity

g

dv

d

=

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=

# #

( )

=

=

=

= +

=

=

"


=

=


=

1=v

=

=

=

= , =cc

v vf f

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2

1

p g

c

vv v

f

f

=

2

1 cf

vf

2

p gv v v =







( )o TM

EZ

H

=

2 2

2

=0h h

h

z zz

z

H EjE H

EE

=

=

2 2

2

=0h h

h

z zz

z

H EjH H

EjH

=

=

( )o TMZ

=2 h

zE

2h

j zE

( )o TMZj

=

In Cylindrical waveguides =j =0

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0TM

jZ =

j

0TMZ

=

=

2 2

0

2c

TM

f fZ

=

Take out side we getf

2

0

0

2 1 c

TM

TM

ff

fZ

Z

=

=

2

1 cf

f

0TMZ

=

2

1 cf

f

2

0 1 c

TM

f

Z f

=

=

2

0 1 c

TM

fZ

f

=

= , =

c

c

v vf f

Or

2

0 1TMc

Z

=

5.6 TE WAVES IN CYLINDRICAL WAVEGUIDE

Transverse electric (TM) there is a component of H in the direction of propagation (Hz), but no

component of E in this direction. Such waves are called Hwaves or transverse Electric (TE) waves.

For transverse electric waves 0z

E = is identically zero & 0z

H is given by equation

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2 2 0zH H + =

2 22 2

2 2 2

1 1z z z

z z

H H HH H

+ + + =

( )2 2

2 2

2 2 2

1 10 (1)z z z z

H H HH

+ + + + =

2 22

2 2 2

1 10 (2)z z z

z

H H Hh H

+ + + =

Where X () is a function of alone and Y () is a function of alone. Sub the expression for

zH XY= in the wave equation gives.

Substituting zH XY=

2 22

2 2 20 (3)

X Y X X YY h XY

+ + + =

2

MultiplyingXY

2 2 22 2

2 2

10 (4)

X X Yh

X X Y

+ + + =

In equation (4) the first, second and forth terms are functions of only while the third term is only the

function of only. It is obvious to get such terms because X and Y are independent Variables, we can

equate above independent terms to another constants. Hence we can write,

22 2

2

22

2

1 Where is constant

0

Yn n

Y

Yn Y

=

+ =

For above ordinary second order differential equation, the solution is given by

cos sin Where A and B is constantY A n B n = +

2 22 2 2

20 (5)

X Xh n

X X

+ + =

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Multiplying the above equation by X on both the Sides, we get

( )2

2 2 2 2

20 (6)

X Xh n X

+ + =

Since h is constant, we can modify equation (6) to get the standard form of the differential equation, weget h =

( )( )

( )( )

( )2

2 2 2 2

20 (7)

X Xh h h n X

hh

+ + =

Above equation is standard Bessels differential equation. The solution of this equation is given by

n n

n

J ( h) Where C constant

and J ( h) is a solution or first type of order n

nX C

=

Hence the complete solution for zH can be obtained by substituting values of X and Y from equation (7)

we get

( )

( )

n

n n n

J ( h) cos sin

J ( h) cos sin -------(8) Where A B B

z

z n

z n n n n

H XY

H C A n B n

H A n B n C A C

=

= +

= + = =

Form Trigonometric Properties

2 2 1

' '

' 2 2

' 1

cos sin cos tan

cos sin cos

Where =

= tan

nn n n n

n

n n n

n n n

n

n

BA n B n A B nA

A n B n A n

A A B

Bn n

A

+ = +

+ =

+

Hence equation (8) can be modified as

0 ' 'nJ ( h)cos (9)z nH A n =

2 2

0

2

0h h

( )h

z zz

z

H EjE E

HjE i

= =

=

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' '

nJ ( h)sin (10)z

n

HnA n

=

Substituting the equation (10) in equation (i) we get

0 ' '

n2J ( h)sin (11)

h n

jE nA n

=

2 2

0

2

0h h

( )h

z zz

z

H EjH E

HH ii

= =

=

Substituting the equation (10) in equation (ii) we get

0 ' '

n2J ( h)sin (12)

h nH nA n

=

2 2

0

2

0h h

( )h

z zz

z

E HjE E

HjE iii

= + =

=


' ' 'n nn

' ' '

n

J ( h) J ( h)h cos J ( h)

hJ ( h)cos (13)

zn

zn

HA n

HA n

= =

=


0

2h

jE

= ' hnA

' '

n

0 ' ' '

n

J ( h)cos

J ( h)cos (14)h

n

n

jE A n

=

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2 2

0

2

0h h

( )h

z zz

z

E HjH E

HH iv

= =

=


2hH

= ' h

nA ' 'n

0 ' ' '

n

J ( h)cos

J ( h)cos (15)h

n

n

H A n

=

The TE waves in cylindrical waveguide the field equation is

0 ' '

n

0 ' ' 0 ' ' '

n n2

0 ' ' ' 0 ' '

n n2

J ( h)cos

J ( h)sin J ( h)cosh h

J ( h)cos and J ( h)sinh h

z n

n n

n n

H A n

jE nA n H A n

jE A n H nA n

=

= =

= =


0 0 0 0 0 E and E and Hz z z z z

z zH H e H H e E e E e E e

= = = = =

The TE waves in cylindrical waveguide the field equation is

' '

n

' ' ' ' '

n n2

' ' ' ' '

n n2

J ( h)cos

J ( h)sin J ( h)cosh h


z

z n

z z

n n

z z

n n

H A n e

jE nA n e H A n e

jE A n e H nA n e

=

= =

= =


j

j

= + =

=




e

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' '

n

' ' ' ' '

n n2

' ' ' ' '

n n2

J ( h)cos

J ( h)sin J ( h) cosh h


j z

z n

j z j z

n n

j z j z

n n

H A n e

j jE nA n e H A n e

j jE A n e H nA n e

=

= =

= =

Boundary Conditions:

We have discussed boundary conditions in case of the TM waves in which the electric field exits

along the direction of the propagation. But in case of the TE waves, the electric field components along

the z-axis are zero are specified for the tangential components of the electric field i.e. E . The boundary

condition can be specified as.

0 all along the circumference of the cylindrical

0 at =a, for all values of varying from 0 to 2

E

E

=

=

Using above boundary condition in equation E

' ' '

n0 J (ah)cosh

z

n

jA n e

=

For all values of, 'cos 0n

'

nJ (ah)=0

For different orders of the Bessel function (i.e. for different n) this equation can be satisfied for different

values of a such values are called Eigen value

'

'

nm nm

nmnm

ah P

Ph

a

=

=

Order of the Bessel

function n

'

1nP '

2nP '

3nP

0 3.83 7.01 10.17

1 1.841 5.33 8.53

2 3.05 6.73 9.97

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Characteristic ofnm

TM wave


= +

=

=

'

nmnm

Ph

a=

=

$#

At lower freq. is <

#

becomes real with value of, and =0. There is only attenuation,

without any propagation.


$#

making .imaginary.

2'

2

2'

2

2'

2

For ffc =0

For f=fc 0

nm

nm

nm

P

a

Pj

a

P

a

> =

< =

= =

Cut off frequency:-

f=fc

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=

=

$

$

#

#

= =

$

#

=

=

=

$

$

$

#

#

#

Cut-off Wave Length

=

=

1v

=

'1

2

nmc

Pf

a

=

=

$

#

=

=

$

$

#

#

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Phase constant:-

=

$ $

# #

( )

=

=

=

= =

=

=

Guide wavelength:-


2=

2=

g

g

2 2 2cf f

1

=v

2 2=g

c

v

f f


2=

1

g

c

v

ff

f

= vf

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2=

1

g

cf

f

= , =c

c

v vf f

2=

1

g

c




=

=

!

!


=

!

2 f =

=!

=

!

1

=v

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=

=

!

!

= , =cc

v vf f

Group velocity


group velocity

g

dv

d

=

=

$ $

# #

( )

=

=

=

= +

=

=

"


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=

=


=

1=v

=

=

=

= , =cc

v vf f


2

1

p g

c

vv v

f

f

=

2

1 cf

vf

2

p gv v v =







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( )o TE

EZ

H

=

2 2

2

0h h

h

z zz

z

H EjE E

HjE

= =

=

2 2

2

0h h

h

z zz

z

H EjH E

HH

= =

=

( )o TMZ

=2 h

j

zH

2h

zH

( )o TE

jZ

=

In Cylindrical waveguides =j =0

0TE

j

Z = j

0TEZ

=

=

02 2

02

2


2 1

TE

c

TE

c

Zf f

f

Z

fff

=

=

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0TEZ

=

2

0

1 c

TE

f

f

Z

=

2

02

1

1

c

TE

c

f

f

Z

f

f

=

=

02

1

TE

c

Zf

f

=

Or

02

1

TE

c

Z

=

= , =cc

v vf f

5.7 THE TEM WAVE IN COAXIAL LINES

The TEM mode (some times called the principal made) is approximate form of propagation encountered

on parallel-wire and co-axial lines at low frequencies. Actually, it exists only in theory on the dissipation

less form of those lines; but since the losses are small, the true condition closely approximates the TEM

field.

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The analyze the TEM wave in co-axial line it is desirable to start with all the fields varying in time and

propagating in z-direction. Co-axial cable structure is circular formant so consider circular wave equation

Electric Field component

1 ( )

( ) (1)

1( )

z

z

z

H H j E a

HH j E b

H Hj E c

+ =

=

=

Magnetic Field component

1 ( )

( ) (2)

1( )

z

z

z

E E j H a

EE j H b

E Ej H c

+ =

=

=

The TEM mode is special case of the TM mode with TEM mode 0z zH E= = it may also be established

that 0H E = = under these conditions equation (1) and (2) reduce to

(3)

10 (4)

H j E

H

=

=

(5)

10 (6)

E j H

E

=

=

Consider Equation (3) and (5)

( )

( )

jH E i

H E iij

=

=

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Multiply equation (i) and (ii) we get

2 2

2

jH E

j

j

H

=

=

j

2

2 2

=

(7)

E

E H

E H

E H

=

=

=

If the circumference of the inner conductor of radius a is chosen as bath, then if0

I is value of

instantaneous current in the inner conductor

0

0

0

2

and thus the magnetic field intensity is expressible in terms of the inner conductor current is

2

2

aH I

IH a

a

IH

=

= =

=

(8)

Substituting the equation (8) in equation (7) we get

0 (9)2

IE

=

Showing the electric field also to inversely proportional to the as expected from the cylindrical

symmetry of the situation

The voltage drop from the center conductor to the outer conductor of the co-axial line may be written in

terms of the maximum value as

0

0

1

2

ln (10)2

b

a

b

a

V E d

IV d

I bV

a

=

=

=

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The characteristics impedance of such an infinite length line is the ratio 00

I=IV V

I I

0

0

I

Z

=0

ln2

b

a

I

0

1ln (11)

2

bZ

a

=

Inserting the z-propagation and time functions given the instantaneous voltage as

0 ln2

j zI bV e

a

=

Taking the z-derivative of the voltage

0

0

0

0

ln z=02

ln wave =2

ln2

ln (12)2

j zIV bj ez a

IV bj TEM

z a

IV bj

z a

V j bI

z a

=

= =

=

=

In terms of the effective current and voltage

0

j z

oI I e

=

Taking the z-derivative of the voltage

0

0

z=0

(13)

j zo

o

Ij I e

z

Ij I

z

=

=

Consider Equation (10) we get

0 ln2

I bV

a

=

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0

2(14)

ln

VI

b

a

=

Multiply equation (14) and (13)

2 wave =

ln

2

ln

2(15)

ln

o

o

o

I Vj TEM

bz

a

I Vj

bz

a

I jV

bz

a

= =

=

=

The line at radio frequency inductor and capacitor value

2ln C=

2ln

bL

ba

a

=

0I=I

Vj LI

z

Ij CV

z

=

=

A common method of excitation of TM modes in a circular waveguide by co-axial line is shown. At the

end of the co-axial line a large magnetic field exists in the direction of propagation the magnetic field

from the co-axial line will excide the TM modes in the guide however, when the guide is connected to the

source by a co-axial a discontinuity problem. At the function will increase the eventually decrease the

power transmission; it is after necessary to place a turning device around the function in order suppress

the reflection.

5.8 EXCITATION METHODS

In order to launch a particular mode, a type of probe is chosen which will produce lines of E and

H that are roughly parallel to the lines of E and H for that mode possible methods for feeding rectangular

waveguides are shown.

Generally a guide is closed at one end by conducting wall. An exciting antenna rod is

inserted through the end or side of the guide. The end of the guide which is closed by a conducting wall

acts as reflector. By properly adjusting the distance between the rod and the end, we can achieve the

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transmitted wave and reflected wave in phase such that both the waves propagate as a single wave along

the guide.

Depending on the type of rod or probe used the lines of the electric and magnetic

fields are produced which are roughly parallel to the lines of the electric and magnetic fields for the

particular mode in the guide.

In figure the probe is parallel to that y axis and so produces lines of E in the y direction and

lines of H which the in the X z-plane. This is the perfect field configuration for the TE10mode.

In Figure the parallel produces fed with opposite phase tend to set up the TE20mode in figure the

probes which are parallel to the z-axis produces electric field liens in the XY plane for TE11mode.

In figure the probe parallel to the z-axis produce magnetic field lines in the x y plane. This is the

perfect field configuration for the TM11mode.

It is possible for several modes to exist simultaneously in waveguides, if the frequency is above

cut-off for these particular modes. However the waveguide dimensions are often chosen so that only the

dominant mode can exist.

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These exiting sources are used not only excite waves desired but for exciting higher ordered modes also.

By suitably selecting dimensions of the guide, we can get only one desired wave cut-off frequency. At the

same time other waves attenuate and cant travel through the guide.

Let us assume that a TEM wave exist inside a rectangular waveguide which is a single conductor system.

Existence of TEM wave means the magnetic field must the entirely in the transverse plane. For a

magnetic field, div 0H= that is the magnetic field lines must form closed loops in the x-y transverse

plane inside the waveguide. If use apply amperes circuit law tot his magnetic field, the lines of this

magnetic field aloud these closed path us must be equal to the current enclosed in the axial direction

current or a displacement orient in the z-direction existence of such displacement current will require an

axial component of electric field E2B~A if E2is present then this wave cannot a TEM wave. Further, if

instead of displacement current conduction current exists, then there should be a centre conductor to

provide return path, which is not the case in a rectangular waveguide. This argument holds good for any

ingle conductor waveguide. There no TEM wave can exist inside a single conductor waveguide.

5.9 GUIDE TERMINATION

We have already studied that to avoid reflection at the receiving end, the transmission lines are properly

terminated at the receiving end. Now wave guide is a form of transmission line. Hence to avoid reflection

losses, the wave guide must be properly terminated. Now this proper termination can be achieved by

using a termination which provides the wave impedance equal to the impedance of the transmitted mode

in the guide. Many of the times it is very difficult to get suitable value of the load, we have to use

different forms of coupling to the guide. The main application of these different forms of coupling is to

transform impedance to the desired value.

The most common from of the load that is coupled to the guide is free space. Generally it is that the

energy conveyed by the guide must be radiated or transmitted as a space wave. The horn antenna can be

used couple guide impedance to wave impedance. The impedance of the horn antenna at mouth is equal tothe intrinsic impedance of free space i.e. 377. Now this horn antenna is coupled to guide in such a way

that the connecting end of the antenna matches the impedance of the guide to the wave impedance of

377. To achieve this generally the area of the guide is expanded in appropriate direction. To achieve

good impedance transformation without reflection, the horn antenna which is several wavelengths long is

selected.

Some times in case co-axial lines strip lines, dissipative loads are used. For matching the impedance a

tapered lossy dielectric is placed at the end of shorted line. Such loads are again designed on the principal

of the tapered or exponential lines. By using dielectrics with considerable conductive, incident power is

absorbed without considerable reflection and radiation.

The simplest form of the dissipative load i.e. non-reflecting terminations is a wave guide entering a tank

of water at small angles. The guide then gets partially filled with wedge water. This wedge of water is of

the form approximately same as exponential taper. The water furnishes dielectric losses.

Similar to this, materials such as bearing porcelains or graphite bearing plastics, metallic bearing plastics

can be used in the wedge shapes to replace water.

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5.10 RESONATORS

Basically resonators are used for storing energy. At very high frequencies, say 100MHz and above, the

RLC circuit elements are inefficient when used as resonators because the circuit dimensions are

comparable with the operating wavelength. Due to this, unwanted radiation takes place. Hence at high

frequencies, the RLC resonant circuits are replaced by electromagnetic resonant cavities.

Resonators are divided in to there types

1. Rectangular cavity Resonator

2. Circular Cavity Resonator

3. Semicircular cavity Resonator

5.10.1 RECTANGULAR CAVITY RESONATOR

In general the rectangular waveguides are constructed from closed section of the waveguide, as

waveguide is the type of the transmission line. Usually the rectangular waveguide resonators are short

circuited at both the ends to avoid the radiation losses from open end of the waveguide. Due to short

circuited ends of the waveguide a cavity or closed box is formed. Within this cavity, both the energies,

electric and magnetic are stored. The power dissipation is observed at the metallic conducting walls of the

waveguides as well as in the electric inside the cavity. Through a small aperture or a small probe or a loop

such resonators are coupled.

The geometry of the rectangular cavity resonator is shown fig

The wave equations in the rectangular resonator should satisfy the boundary conditions of the zero

tangential of the electric field strength (E) at the four walls

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Transverse Electric ( )mnpTE mode

The magnetic field expression in the z-direction is given by

0 cos cos sinz

m n p

H H x y za b d

= Where

0,1,2,3...........m=represents the number of the

half wave periodicity in the x direction

Where m=0,1,2,3,..... represents the number of the half wave periodicity in the x direction

n=0,1,2,3,..... represents the number of the half wave periodicity in the y direction

p=1,2,3,........represents the number of the half wave periodicity in the z direction

The electric field in the z-direction is

0z

E =

The magnetic field in the x direction is

2

2

2 2 2

2

2

0

2

02

0

2

1

Where h

cos cos sin

cos cos cos

zx

x

x

x

HH

h x z

m n p

a b d

H m n pH x y z

h x z a b d

H p m n pH x y zh d x a b d

H pH

h d

=

= + +

=

=

=

sin cos cos

m m n px y z

a a b d

The magnetic field in the y direction is

2

2

2 2 2

2

1

Where h

zy

HH

h y z

m n p

a b d

=

= + +

2

0

2cos cos sin

y

H m n pH x y z

h y z a b d

=

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0

2

0

2

cos cos cos

cos sin cos

y

y

H p m n pH x y z

h d y a b d

H p n m n pH x y z

h d b a b d

=

=

The electric field in the x direction is

2

02

02

cos cos sin

cos sin sin

zx

x

x

HjE

h y

j m n pE H x y z

h y a b d

j n m n pE H x y z

h b a b d

=

=

=

The electric field in the y direction is

2

02

02

cos cos sin

sin cos sin

zy

y

y

HjE

h x

j m n pE H x y z

h y a b d

j m m n pE H x y z

h a a b d

=

=

=

Transverse magnetic ( )mnpTM mode

The magnetic field expression in the z-direction is given by

0sin sin cos

z

m n pE E x y z

a b d

=

Where 0,1,2,3...........m= represents the number of the

half wave periodicity in the x direction

Where m=0,1,2,3,..... represents the number of the half wave periodicity in the x direction

n=0,1,2,3,..... represents the number of the half wave periodicity in the y direction

p=1,2,3,........represents the number of the half wave periodicity in the z direction

The electric field in the z-direction is

0z

H =

The electric field in the x direction is

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2

2

2 2 2

2

1

Where h

zx

EE

h x z

m n p

a b d

=

= + +

2

0

2

2

0

2

0

2

sin sin cos

sin sin sin

cos sin sin

x

x

x

E m n pE x y z

h x z a b d

E p m n pE x y z

h d x a b d

E p m m n pE x y z

h d a a b d

=

=

=

The elec

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