11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a...

109
1 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium

Transcript of 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a...

Page 1: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

11

Engineering Electromagnetics

EssentialsChapter 9

Waveguides: solution of the wave equation For a wave in a bounded medium

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2

Objective

Topics dealt with

To solve wave equation and study propagation of electromagnetic wave through bounded media, namely hollow-pipe waveguides, and predict their characteristics

Solution of the wave equation of a rectangular waveguide in transverse electric (TE) mode for all the field components

Characteristic equation or dispersion relation of a rectangular waveguide excited in transverse electric (TE) mode in terms of wave frequency, wave phase propagation constant and waveguide cutoff frequency

Dependence of waveguide cutoff frequency on waveguide dimensions with reference to a waveguide excited in TE mode

Characteristic parameters namely phase velocity, group velocity, guide wavelength and wave impedance and their dependence on wave frequency relative to waveguide cutoff frequency

Evanescent mode in a waveguide

Dimension-wise and mode-wise operating frequency criteria

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33

Excitation of a rectangular waveguide in transverse magnetic (TM) mode

Significance of mode numbers vis-à-vis field pattern

Cylindrical waveguide

Inability of a hollow-pipe waveguide to support a TEM mode

Power flow and power loss in a waveguide

Power loss per unit area, power loss per unit length and attenuation constant there from

Background

Maxwell’s equations (Chapter 5), electromagnetic boundary conditions at conductor-dielectric interface (Chapter 7), basic concepts of electromagnetic power flow (Chapter 8) and those of circuit theory

3

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44

A ‘waveguide’ is a hollow metal pipe used to ‘guide’ the transmission of an electromagnetic ‘wave’ from one point to another.

It is a microwave-frequency counterpart of the lower-frequency connecting wire.

It is thus essentially a type of transmission line, some of the other types being the two-wire line, the coaxial cable, the parallel-plate line, the stripline, the microstrip line, etc.

We have seen that a transverse electromagnetic wave (TEM) mode of propagation is supported by a free-space medium with a phase velocity equal to the velocity of light c.

Can a TEM mode be supported by a hollow-pipe waveguide? Would the phase velocity of a wave propagating through such a waveguide be the same as or different from c?

We can answer to such question and many others concerning the characteristics of a waveguide with the help of electromagnetic analysis of the waveguide.

Such an analysis is based on setting up the wave equation in electric and magnetic fields in the waveguide and solving them with the help of relevant electromagnetic boundary conditions.

We take up here for analysis two types of waveguides, namely rectangular and cylindrical or circular waveguides which have rectangular and circular cross sections respectively perpendicular to the direction of wave propagation.

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Rectangular waveguide

Rectangular waveguide can be treated in rectangular system of coordinates in view of the rectangular symmetry of its geometry. The waveguide will be considered for both TE and TM modes of excitation. (We will see later the inability of the hollow-pipe waveguides to support TEM mode of propagation).

The TE mode is also known as the H mode since it is associated with a non-zero value of the axial magnetic field:

The TM mode is also known as the E mode since it is associated with a non-zero value of the axial electric field:

.0,0 zz EH

.0,0 zz HE

a

b

Z

Y

X

Rectangular waveguide with its right side wall located at x = 0; left side at x = a; bottom wall at y = 0 and top wall at y = b.

mode) (TM )0,0(

mode) (TE )0,0(

2

2

002

2

2

2

2

2

2

2

002

2

2

2

2

2

zzzzzz

zzzzzz

HEt

E

z

E

y

E

x

E

EHt

H

z

H

y

H

x

H

Wave equations have already been introduced in Chapter 6.

Wave equations

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66

(rewritten)

a

b

Z

Y

X

Rectangular waveguide with its right side wall located at x = 0; left side at x = a; bottom wall at y = 0 and top wall at y = b.

TE mode

mode) (TE )0,0(2

2

002

2

2

2

2

2

zzzzzz EH

t

H

z

H

y

H

x

H

)(exp ztj

jt /jz /

zzzz

zzzz

HHjHjjt

H

HHjHjjz

H

2222

2

2222

2

))((

))((

Field quantities varying as

zzzz HH

y

H

x

H 200

22

2

2

2

0)( 222

2

2

2

zzz Hk

y

H

x

H

constant)n propagatio

space-(free

)( 2/100k

(wave equation rewritten)

(wave equation)

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x

XY

x

H z

XYH z

Field solutions

Let us use the method of separation of variables for solving wave equation

0)( 222

2

2

2

zzz Hk

y

H

x

H (wave equation rewritten)

X is a function of only x

2

2

2

2

x

XY

x

H z

0)( 222

2

2

2

XYky

YX

x

XY

2

2

2

2

y

YX

y

H z

Y is a function of only y

y

YX

y

H z

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a

b

Z

Y

X

0)( 222

2

2

2

XYky

YX

x

XY (rewritten)

)(11 22

2

2

2

2

ky

Y

Yx

X

X

Dividing by XY and rearranging terms

Equates a function of x alone in its left-hand side with a function of y alone on its right-hand side. This can happen only when the individual functions of x and y are constants.

22

21A

x

X

X

222

2

2

)(1

Aky

Y

Y

22222

2

)(1

BkAy

Y

Y

A2 is a constant

B2 is a constant

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)(exp)sincos)(sincos( 4321 ztjByCByCAxCAxCH z

ByCByCY sincos 43 AxCAxCX sincos 21

(rewritten) (rewritten)22

21A

x

X

X

2

2

21B

y

Y

Y

Solution

C1 and C2 are constants

C3 and C4 are constants

XYH z

Invoking the factor exp j(t - z) which was otherwise understood

(transverse components of the electric field and magnetic field components)

Maxwell’s equations

yxyx HHEE , and ,

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1010

t

EH

t

HE

0

0

j

t

)(exp

as vary quantities Field

ztj

)(

)(

0

0

zzyyxx

zxy

yzx

xyz

zzyyxx

zxy

yzx

xyz

aEaEaEj

ay

H

x

Ha

x

H

z

Ha

z

H

y

H

aHaHaHj

ay

E

x

Ea

x

E

z

Ea

z

E

y

E

(Maxwell’s equation)

Expanding the curl in rectangular system

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1111

)(

)(

0

0

yyxx

zxy

yzx

xyz

zzyyxx

zxy

yx

xy

aEaEj

ay

H

x

Ha

x

H

z

Ha

z

H

y

H

aHaHaHj

ay

E

x

Ea

z

Ea

z

E

)(

)(

0

0

zzyyxx

zxy

yzx

xyz

zzyyxx

zxy

yzx

xyz

aEaEaEj

ay

H

x

Ha

x

H

z

Ha

z

H

y

H

aHaHaHj

ay

E

x

Ea

x

E

z

Ea

z

E

y

E

0zE

(TE mode)

(rewritten)

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)(

)(

0

0

yyxx

zxy

yzx

xyz

zzyyxx

zxy

yx

xy

aEaEj

ay

H

x

Ha

x

H

z

Ha

z

H

y

H

aHaHaHj

ay

E

x

Ea

z

Ea

z

E

(rewritten)

x component

y component

yzx

xy

Ejx

H

z

H

Hjz

E

0

0

j

z

)(exp

as vary quantities Field

ztj

yz

x

xy

Ejx

HHj

HjEj

0

0

)/( 0 yx EH

yzy Ejx

HEj 0

0

2

Rearranging terms

00

0022

x

HEj zy

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)(

)(

0

0

yyxx

zxy

yzx

xyz

zzyyxx

zxy

yx

xy

aEaEj

ay

H

x

Ha

x

H

z

Ha

z

H

y

H

aHaHaHj

ay

E

x

Ea

z

Ea

z

E

(rewritten)

y component

x component

jz

)(exp

as vary quantities Field

ztj

)/( 0 xy EH

xzx Ejy

HEj 0

0

2

Rearranging terms

00

0022

y

HEj zx

Similarly

xyz

yx

Ejz

H

y

H

Hjz

E

0

0

xyz

yx

EjHjy

H

HjEj

0

0

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00

0022

y

HEj zx

0

0

0022

x

HEj zy

(recalled) (recalled)

x

H

k

jE zy

22

0

2/100 )( k

(recalled)

y

H

k

jE zx

220

y

HE

x

HE

zx

zy

We will recall the above relations of proportionality in the electromagnetic boundary conditions at the walls of the waveguide in the analysis to follow.

Electromagnetic boundary condition at the waveguide wall:

0)(2 zzyyxxnn aEaEaEaEa

0)( 12 EEan

(subscript 1 referring to the conducting waveguide wall region 1 and subscript 2 to the free-space region 2 inside the waveguide with unit vector directed from region 1 to 2)

mode) (TE 0zE

0)(2 yyxxnn aEaEaEa

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xn aa

ax

: wallsideLeft

0)( yyxxx aEaEa

0)( yyxxx aEaEa

(electromagnetic boundary condition at the left and right side walls of the rectangular waveguide)

0)(2 yyxxnn aEaEaEa

0zyaE

xn aa

x

0

: wallsideRight

0 zyaE

axbyEy at )0(0 0at )0(0 xbyEyx

HE zy

axbyx

H z

at )0(0 0at )0(0

xbyx

H z

(recalled)

(left side wall) (right side wall)

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)ofvaluesallfor(

0)sincos)()(( 432

y

ByCByCACx

H z

)cossin)(sincos(

)sincos)(cossin(

4321

4321

ByBCByBCAxCAxCy

H

ByCByCAxACAxACx

H

z

z

axbyx

H z

at )0(0 0at )0(0

xbyx

H z

)(exp)sincos)(sincos( 4321 ztjByCByCAxCAxCH z (recalled)

(left side wall) (right side wall)

ISituation

0,02 AC

IISituation

0,0 2 CA

This can happen only if

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0)sincos)(cossin)(( 4321

ByCByCAxACAxAC

x

H z

ISituation

0,02 AC axbyx

H z

at )0(0 (left side wall)

)ofvaluesallfor(

0)sincos(sin 431

y

ByCByCAxAC

)(exp)sincos)(sincos( 4321 ztjByCByCAxCAxCH z

0

I) (Situation 0,0

2

1

C

AC (left side wall at x = a)

0zH The finding contradicts with the TE-mode condition (Situation I):

(Situation I)

0zH

0,0 21 CC

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18

0

I) (Situation 0,0

2

1

C

AC leads to the contradiction with the TE-mode condition

(left side wall at x = a)

0 besides 0,0 21 CCA

Therefore, in order to avoid this contradiction in Situation I let us put

(Situation I)

(left side wall at x = a)

)ofvaluesallfor(

0)sincos)(cossin)(( 4321

y

ByCByCAxACAxACx

H z

0sin Aa

(rewritten)(left side wall at x = a)

...),3,2,1( mmAa

(m = 0 that makes A = 0 is excluded to avoid contradiction with A ≠ 0 taken here)I)Situation(...),3,2,1( m

a

mA

I)Situation(...),3,2,1(with

)(exp)sincos)(cos( 431

ma

mA

ztjByCByCAxCH z

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19

Similarly we can proceed recalling Situation II: 0,0 2 CA

)(exp)sincos)(( 431 ztjByCByCCH z II)Situation()ofvaluesallfor( y

)(exp)sincos)(sincos( 4321 ztjByCByCAxCAxCH z

(recalled)

Tacitly expressed as

)(exp)sincos)(cos( 431 ztjByCByCAxCH z

II)Situation()0( nginterpreti ma

mA

II)Situation()0(with

)(exp)sincos)(cos( 431

ma

mA

ztjByCByCAxCH z

19

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20

I)Situation(...),3,2,1(with

)(exp)sincos)(cos( 431

ma

mA

ztjByCByCAxCH z

II)Situation()0(with

)(exp)sincos)(cos( 431

ma

mA

ztjByCByCAxCH z

(rewritten)

(recalled)Combining

...),3,2,1,0(with

)(exp)sincos)(cos( 431

ma

mA

ztjByCByCAxCH z

)ofvaluesallfor( y

)ofvaluesallfor( y

20

We have so far considered electromagnetic boundary conditions at the left and right sidewalls of the rectangular waveguide. Let us next turn towards electromagnetic boundary conditions at the top and bottom walls of the rectangular waveguide.

Page 21: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

212121

yn aa

y

0

: wallBottom

0)( yyxxy aEaEa

0)( yyxxy aEaEa

0)(2 yyxxnn aEaEaEa

0zxaE

yn aa

by

: wallTop

0 zxaE

0at )0(0 yaxEx byaxEx at )0(0y

HE zx

0at )0(0

yax

y

H z byaxy

H z

at )0(0

(recalled)

(bottom wall) (top wall)

(electromagnetic boundary condition at the bottom and top walls of the rectangular waveguide)

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22

0at )0(0

yax

y

H z byaxy

H z

at )0(0

(bottom wall) (top wall)

)cossin)(sincos( 4321 ByBCByBCAxCAxCy

H z

...),3,2,1,0(with

)(exp)sincos)(cos( 213

nb

nB

ztjAxCAxCByCH z

Hence using the same procedure as followed earlier, which starts from the boundary conditions at the left and right side walls, we can obtain, now starting from the boundary conditions at the bottom and top walls, the following expression:

)ofvaluesallfor( x

Page 23: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

23

...),3,2,1,0(with

)(exp)sincos)(cos( 213

nb

nB

ztjAxCAxCByCH z

)ofvaluesallfor( x

We have obtained the following:

...),3,2,1,0(with

)(exp)sincos)(cos( 431

ma

mA

ztjByCByCAxCH z

)ofvaluesallfor( y

When combined, these two expressions for Hz, while remembering that the former is valid for all values of y and that the latter for all values of x, prompt us to take the factor cosAx from the former and the factor cosBy from the latter in the field expression to write it as

...),3,2,1,0,(,with )(expcoscos0 nmb

nB

a

mAztjByAxHH zz

)(exp)cos)(cos( 31 ztjByCAxCHH zz

310 Putting CCH z

...),3,2,1,0,(, nmb

nB

a

mA

Page 24: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

2424

...),3,2,1,0(...);,3,2,1,0(with

)(expcoscos0

nb

nBm

a

mA

ztjByAxHH zz

...),3,2,1,0,()(exp)cos()cos(0 nmztjyb

nx

a

mHH zz

...),3,2,1,0,()(exp)cos()sin(0

nmztjy

b

nx

a

mH

a

m

x

Hz

z

)(exp)sin()cos(0 ztjyb

nx

a

mH

b

n

y

Hz

z

...),3,2,1,0,(, nmb

nB

a

mA

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2525

x

H

k

jE zy

22

0

y

H

k

jE zx

220

...),3,2,1,0,()(exp)cos()sin(0

nmztjy

b

nx

a

mH

a

m

x

Hz

z

)(exp)sin()cos(0 ztjyb

nx

a

mH

b

n

y

Hz

z

...),3,2,1,0,()(exp)cos()sin(0220

nmztjy

b

nx

a

mH

a

m

k

jE zy

...),3,2,1,0,()(exp)sin()cos(0220

nmztjy

b

nx

a

mH

b

n

k

jE zx

(rewritten)

(rewritten)

(rewritten)

(rewritten)

Page 26: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

2626

...),3,2,1,0,()(exp)cos()sin(0220

nmztjy

b

nx

a

mH

a

m

k

jE zy

2222 )( BkA

(rewritten)

(rewritten)

(introduced earlier while solving the wave equation by the method of separation of variables)

2222 BAk

...),3,2,1,0,()(exp)cos()sin(0220

nmztjy

b

nx

a

mH

a

m

BA

jE zy

...),3,2,1,0,()(exp)sin()cos(0220

nmztjy

b

nx

a

mH

b

n

k

jE zx

...),3,2,1,0,()(exp)sin()cos(0220

nmztjy

b

nx

a

mH

b

n

BA

jE zx

Page 27: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

2727

...),3,2,1,0,(, nmb

nB

a

mA

...),3,2,1,0,()(exp)cos()sin(0220

nmztjy

b

nx

a

mH

a

m

BA

jE zy

...),3,2,1,0,(

)(exp)cos()sin(0220

nm

ztjyb

nx

a

mH

a

m

bn

am

jE zy

...),3,2,1,0,(

)(exp)sin()cos(0220

nm

ztjyb

nx

a

mH

b

n

bn

am

jE zx

...),3,2,1,0,()(exp)sin()cos(0220

nmztjy

b

nx

a

mH

b

n

BA

jE zx

Page 28: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

2828

yx EH0

xy EH0

...),3,2,1,0,()(exp)cos()sin(0220

nmztjyb

nx

a

mH

a

m

b

n

a

m

jE zy

(rewritten)

...),3,2,1,0,(

)(exp)cos()sin(022

nm

ztjyb

nx

a

mH

a

m

bn

am

jH zx

...),3,2,1,0,()(exp)sin()cos(0220

nmztjyb

nx

a

mH

b

n

bn

am

jE zx

...),3,2,1,0,(

)(exp)sin()cos(022

nm

ztjyb

nx

a

mH

b

n

bn

am

jH zy

(rewritten)

Page 29: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

2929

Field expressions of a rectangular waveguide excited in the TEmn mode put together

...),3,2,1,0,()(exp)cos()cos(0 nmztjyb

nx

a

mHH zz

...),3,2,1,0,()(exp)sin()cos(0220

nmztjyb

nx

a

mH

b

n

bn

am

jE zx

...),3,2,1,0,()(exp)cos()sin(0220

nmztjyb

nx

a

mH

a

m

bn

am

jE zy

...),3,2,1,0,()(exp)cos()sin(022

nmztjyb

nx

a

mH

a

m

bn

am

jH zx

...),3,2,1,0,()(exp)sin()cos(022

nmztjyb

nx

a

mH

b

n

bn

am

jH zy

0zE

(m and n are the mode numbers of the rectangular waveguide excited in the TEmn mode)

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3030

Characteristic equation or dispersion relation of a rectangular waveguide excited in the TE mode

...),3,2,1,0,(22

22

nm

b

n

a

mk

222ckk

...),3,2,1,0,( Putting

2/122

nm

b

n

a

mkc

(recalled)

222

2

ckc 022222 ckc c

frequency cutoff thebeing number, wavecutoff ,/ ccc ck

light of velocity thebeing )/(1

constant,n propagatio space-free ,/)(2/1

00

2/100

c

ck

02222 cc (dispersion relation)

(- relationship with reference to a rectangular waveguide excited in the TE mode)

Page 31: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

31

Dispersion relation

2 2 2 2 0cc

cc -

line

cc

g

k

k

2

2

2

222ckk

222222

cg

222

111

gc

versus dispersion characteristics (dispersion curve) of the waveguide which meets the c-line at , the latter shown as a dotted line having positive and negative slopes of magnitude equal to the velocity of light c for the positive and the negative values of respectively. The intercept of the dispersion curve with the axis gives the cutoff frequency c.

(recalled)

(characteristic equation of the waveguide)

Dispersion curve: a hyperbola

Page 32: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

32

cc f

c

...),3,2,1,0,(

2

2

2/122

2/122

2/122

nm

bn

amf

c

b

n

a

mcf

b

n

a

mc

cc

c

c

2/122

b

n

a

mkc

2c

cf

ck cc

(cutoff wave number)

(cutoff frequency)

(cutoff wavelength)

2

222

cc

02222 cc

cc

2/122 )(

2 2 2 2 0cc

cc -

line

In general, there will be two values of for a given value of one positive and the other negative.

Page 33: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

33

Waveguide a (inch) b (inch) c (mm) fc(=c/2a) (GHz)

WR-2300 23 11.5 1168.4 0.257

WR-340 3.4 1.7 172.72 1.737

WR-284 2.84 1.34 144.27 2.08

WR-137 1.372 0.62 69.70 4.30

WR-112 1.122 0.497 57.0 5.26

WR-90 0.9 0.4 45.72 6.56

WR-3 0.034 0.017 1.73 173.7

Typical waveguide dimensions and corresponding cutoff frequencies (fc) and cutoff wavelengths (c) for the TE10 mode for typical commercial rectangular waveguide types

WaveguideWaveguide modes in order of increasing cutoff frequency (fc) shown in parenthesis against the

modes

WR-2300

TE10 (0.257 GHz)

TE20/ TE01 (0.51 GHz)

TE11 (0.61 GHz)

WR-284

TE10 (2.08 GHz)

TE20 (4.16 GHz)

TE01 (4.41 GHz)

TE11 (4.87 GHz)

WR-90

TE10 (6.56 GHz)

TE11 (9.88 GHz)

TE20 (13.1 GHz)

TE01 (14.76 GHz)

Modes in order of increasing cutoff frequency for typical commercial rectangular waveguide types

Page 34: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

34

2/1222/122ph)()( cc

c

c

v

g

k

2

2 2

g

Characteristic parameters

cc

2/122 )(

2/1

2

22/1

2

22/122

ph

1

1

1

1

)(

f

fc

v

ccc

2/1

2

2

ph

1

1

ffc

vk

c

g

Wave phase velocity:

Guide wavelength:

ck

phv

2/1

2

2

ph

1

1

f

fc

v

c

g

(recalled)

(rewritten)

Page 35: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

35

Wave group velocity

We have already defined the phase velocity of a wave vph that involves the phase propagation constant , which, in turn, occurs in the representation of a wave associated with a physical quantity, for instance, the electric field:

)(exp0 ztjEE

velocityphase /

/constantn propagatio phase

amplitude field

ph

ph

0

v

v

E

However, the phase velocity has a meaning only with reference to an infinite monochromatic wave train.

In order to convey information, we have to resort to a group of wave trains at different frequencies the form a finite wave train or a wave packet, say, in the form of a modulated wave such that the modulation envelope contains the information to be conveyed.

The velocity of the wave packet or group of waves is called the group velocity of the wave, which also represents the velocity with which the energy is transported.

We may then find the group velocity as the velocity of the amplitude of the group of waves in the wave packet. Through any medium, which, in general, is dispersive, the wave components of the group travel with different phase velocities.

In what will follow, we are going to show, considering two components of the wave packet, that the wave group velocity vg can be found from the relation:

gv

Page 36: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

36

])()[(exp])()[()[exp(exp0 ztjztjztjEE

])()[([exp])()[([exp 00 ztjEztjEE

constant zt

Let us consider two components of the wave packet one at - and the other at + , which are separated in frequency by 2 with phase propagation constants - and + respectively.

Considering the magnitudes of the two electric field components to be the same for the sake of simplicity

sincosexprelation theof In view jj

)(exp)cos(2 0 ztjztEE

0dt

dz

The amplitude of the electric field of the combination of the wave components of the group has the wave-like variation with distance and time appearing in the cosine function.

Putting the argument of the cosine function as constant in order to find the wave group velocity vgDifferentiating

dt

dz

gvFor small values of in the limit (wave group velocity)

Page 37: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

3737

2ph cvv g

ph

222

v

cccvg

02222 cc

Group velocity of the waveguide:

gv

2 2 2 2 0cc

cc -

line

The group velocity vg of the waveguide can be found from the relation

phv

022 2 c

The slope of the - dispersion curve of the waveguide at any point represents the group velocity of the waveguide at the point while the slope of the line joining the point and the origin of the curve represents the phase velocity of the waveguide.

Taking the partial derivative

2/1

2

2

ph

1

1

ffc

v

c

2/1

2

2

phph

11

f

f

c

vv

c

c

vcg

Page 38: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

38

You will surely enjoy an illustrative example in which we will find the length of the waveguide of cutoff wavelength 69.70 mm that will ensure that a signal at 8.6 GHz emerging out of the guide is delayed by 1 μs with respect to the signal that propagates outside the waveguide.

s 10s 1 6delay

GHz 3.4Hz103.4107.69

103 93

8

cc

cf

m/s 103 8c

s 10)0516.0()103333.0103849.0(11 888

delay

llcv

lg

8

2/1

2

2

1031

f

fv cg

(time delay)

(given) m 107.69 3c

2/1

2

2

1

f

f

c

vcg

m/s 10598.210325.01 882/1 gv

m/s 103 (given); GHz 3.4 and GHz 6.8 8 cff c

(time delay) (given)

km 938.1m 1038.1910)0516.0(

10 28

6

ll = waveguide length

(required waveguide length)

Page 39: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

3939

2/100

ph

)(

1

c

v

2/1002/1

0

0

0

0

0

0

)(

WZ

2/122

2/122

transverse

transverse

)(

)(

yx

yxW

HH

EE

H

EZ

Wave impedance:

xy

yx

EH

EH

0

0

(recalled)

0

2/122

20

2

2

2/122

transverse

transverse

)])([(

)(

xy

yxW

EE

EE

H

EZ

2/1

2

2

ph

01

1

ffc

vZ

c

W

2/1

2

2

ph

1

1

f

fc

v

c

Page 40: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

4040

2/1

2

2

ph

01

1

ffc

vZ

c

W

2/1

2

2

1

f

f

c

vcg

2/1

2

2

ph

1

1

f

fc

v

c

2/1

2

2

1

1

f

fc

g

Waveguide characteristic parameters put together:

)(

0

ph

f

Z

cv

cv

W

g

g

)(

0

ph

c

W

g

gff

Z

v

v

gv c

pv c

g

0WZ

cf f

p gv c

0g Wv c Z

1 2

1

2

0WZ

Page 41: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

41

Evanescent modecc

2/122 )( (recalled)

For frequencies above the cutoff frequency: >c, the phase propagation constant becomes positive real, which occurring in the phase factor exp j(t-z) and, in turn, in the expressions for the RF quantities, refers to a ‘propagating’ wave through the waveguide along positive z.

Taking the upper sign

cc

2/122 )( Positive real for c

Positive or negative real for c cc

jj

cj

cjj

cc

cc

2/1222/122

2/1222/122

)()()(

])(1[()(

)()((

Let us now take frequencies less than the cutoff frequency: <c and take the phase factor as exp j(t-z) :

)exp(exp

))(

(exp)exp()exp()exp()(exp2/122

ztj

zc

tjztjztj c

Phase factor occurring in field expressions taking the positive lower sign in :

cc

2/122 )(

Non-propagating evanescent mode )( c

Page 42: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

42

)exp(exp

))(

(exp)exp()exp()exp()(exp2/122

ztj

zc

tjztjztj c

(Non-propagating evanescent mode)

Phase factor occurring in field expressions

Recalled

)( c

...),3,2,1,0,()(exp)sin()cos(0220

nmztjyb

nx

a

mH

b

n

bn

am

jE zx

(recalled)

cc

2/122 )(

...),3,2,1,0,(exp)exp()sin()cos(0220

nmtjzyb

nx

a

mH

b

n

bn

am

jE zx

)( c

)exp(expby )(exp Replacing ztjztj

)( c

The field amplitude decays exponentially with the axial distance z as exp(-z) for <c. is known as the attenuation constant in this non-propagating mode called the evanescent mode (corresponding to non-propagating mode vanishing).

Power consideration in this mode is taken up later.

Page 43: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

4343

)exp()exp()sin()cos(0220 tjzy

b

nx

a

mH

b

n

bn

am

jE zx

)exp()exp()sin()cos(0220* tjzy

b

nx

a

mH

b

n

bn

am

jE zx

(rewritten)

Taking complex conjugate

Following the same procedure we can write

)exp()exp(cossin0220 tjzy

b

nx

a

mH

a

m

bn

am

jE zy

)exp()exp(cossin0220* tjzy

b

nx

a

mH

a

m

bn

am

jE zy

)( c

)( c

Page 44: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

44

xy

yx

Ej

H

Ej

H

0

0

xy

yx

EH

EH

0

0

(recalled)

j

)Re(2

1

)Re(2

1)PRe()P(

**

*componentcomplexcomponentaverage

xyyx

zzz

HEHE

HE

2jj

*

0

*

*

0

*

xy

yx

Ej

H

Ej

H

(z-component of average Poynting vector)

)( c

)( c

Page 45: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

45

)(Re2

1

)Re(2

1)P(

**

0

**componentaverage

yEEEEj

HEHE

yxx

xyyxz

)Re(2

1

)Re(2

1)PRe()P(

**

*componentcomplexcomponentaverage

xyyx

zzz

HEHE

HE

(z-component of average Poynting vector)

)( c (rewritten)

*

0

*

*

0

*

xy

yx

Ej

H

Ej

H

)( c

Page 46: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

46

)2exp(cossin

)2exp(sincos

2220

2

220*

2220

2

220*

zyb

nx

a

mH

a

m

bn

am

EE

zyb

nx

a

mH

b

n

bn

am

EE

zyy

zxx

Real quantities

0

)(Re2

1)P( **

0componentaverage

yEEEEj yxxz

imaginarypurely is )( **

yEEEEj yxx

)( c

)( c

The average Poynting vector being thus zero, we infer that there would be no power flow in the waveguide in the evanescent mode below cutoff (<c). The waveguide in this mode acts as a reactive load such that the power oscillates back and forth between the source and the waveguide. .

Page 47: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

4747

Let us take up a simple numerical example to show that a rectangular waveguide of cross-sectional dimensions 2.5 cm × 1 cm can support typically the propagating mode TE10 and the evanescent mode TE01 at 10 GHz operating frequency.

2/122

2

b

n

a

mcfc

GHz 6 Hz 106105.22

103 92

8

cf

mode TE

m/s103

m 100.1

m 105.2

0,1

10

8

2

2

c

b

a

nm

(given) GHz 10 fcff

mode) (TE

m/s103

m 100.1

m 105.2

1,0

01

8

2

2

c

b

a

nm

GHz 15 Hz 1015100.12

103 92

8

cf

cff

(Evanescent mode) (Propagating mode)

2/12

2

2

2

8

100.1105.22

103

nmfc

mode) (TE01mode) (TE10

Thus, the waveguide can support non-propagating evanescent mode TE01 and the propagating mode TE10

Page 48: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

48

What will be the nature of the attenuation constant versus frequency plot of a waveguide in the evanescent mode?

2222cc

)()( 2/122

cc

c

Attenuation constant in evanescent mode

(recalled)

The plot of versus becomes an ellipse. The intercept of the ellipse on the axis (ordinate), corresponding to = 0, is found to be c//c. Similarly, the intercept of the ellipse on axis (abscissa), corresponding to = 0, is found to be c.

1)/( 2

2

2

2

ccc

1

2

2

2

2

b

y

a

x

/,

,

bca

yx

cc

(equation of an ellipse)

Page 49: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

4949

TM mode

2

2

002

2

2

2

2

2

t

E

z

E

y

E

x

E zzzz

0)( 222

2

2

2

zzz Hk

y

H

x

H

)(exp)sincos(

)sincos(

43

21

ztjyBCyBC

xACxACEz

The method of analysis of the rectangular waveguide excited in the TM mode closely follows that of the waveguide excited in the TE mode the latter already presented. We can then obtain the TM-mode expressions using the same steps as that followed while obtaining the TE-mode expressions as follows.

)0,0( zz HE)0,0( zz EH

TE-mode expressions already developed in steps

Corresponding TM-mode expressions that follow

2

2

002

2

2

2

2

2

t

H

z

H

y

H

x

H zzzz

0)( 222

2

2

2

zzz Ek

y

E

x

E

)(exp)sincos(

)sincos(

43

21

ztjByCByC

AxCAxCH z

Page 50: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

50

TE-mode expressions already developed

Corresponding TM-mode expressions that follow

xbyExyEyxEyaxE

z

z

z

z

allfor)walltop(0allfor)0wallbottom(0

allfor)0sidewallright(0allfor)sidewallleft(0

)(expsinsin0 ztjyBxAEE zz )(expcoscos0 ztjByAxHH zz

xbyE

xyE

yxE

yaxE

x

x

y

y

allfor)walltop(0

allfor)0wallbottom(0

allfor)0sidewallright(0

allfor)sidewallleft(0

...),3,2,1,0,(

nm

b

nB

a

mA

...),3,2,1,0,(

nm

b

nB

a

mA

)(exp

)cos()cos(0

ztj

yb

nx

a

mHH zz

)(exp

)cos()cos(0

ztj

yb

nx

a

mEE zz

Page 51: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

51

TE-mode expressions already developed

Corresponding TM-mode expressions that follow

2222 )( BkA

2222

b

n

a

mk

222ckk

...),3,2,1,0,(

2/122

nm

b

n

a

mkc

02222 cc

2222 )( BkA

2222

b

n

a

mk

222ckk

...),3,2,1,0,(

2/122

nm

b

n

a

mkc

022222 ckc c 022222 ckc c

02222 cc

Page 52: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

5252

TE-mode expressions already developed

Corresponding TM-mode expressions that follow

...),3,2,1,0,(

2

2

2/122

2/122

2/122

nm

bn

amf

c

b

n

a

mcf

b

n

a

mc

cc

c

c

...),3,2,1,0,(

2

2

2/122

2/122

2/122

nm

bn

amf

c

b

n

a

mcf

b

n

a

mc

cc

c

c

Page 53: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

53

Dominant mode of a rectangular waveguide

• The cutoff frequency of the TE10 mode is the lowest of all other modes.

• If we choose the operating frequency of the waveguide above the cutoff frequency of the TE10 mode, then the TE10 mode will propagate through the waveguide. Also, any next higher order mode will co-propagate if its cutoff frequency is lower than the operating frequency.

• For instance, let the operating frequency be 4.5 GHz for waveguide WR-284. This is higher than the cutoff frequencies of TE10, TE20, and TE01 modes , which are respectively 2.08 GHz, 4.16 GHz, and 4.41 GHz, but lower than the cutoff frequency 4.87 GHz of TE11. ( See slide number 33 for dimensions and cutoff frequencies of waveguide WR-284.)

• Therefore, at the operating frequency of 4.5 GHz, waveguide WR-284 will allow propagation of TE10, TE20, and TE01 but makes mode TE11 evanescent, attenuating it. • Thus, TE10 mode (which has the lowest cutoff frequency of all the modes) will be excited in a rectangular waveguide if the operating frequency is higher than its cutoff frequency irrespective of whether any other modes are exited or not. Hence the TE10 mode is called the dominant mode of a rectangular waveguide.

• If the operating frequency is chosen at 3 GHz for waveguide WR-284, which is above the cutoff frequency 2.08 GHz of the dominant mode TE10 but below the cutoff frequency 4.16 GHz of TE20, then only the mode TE10, which is the dominant mode, will be excited in the rectangular waveguide.

• Moreover, the cutoff frequency of the TE10 mode is the lowest of the cutoff frequencies of all the TE and TM modes taken together and continues to be called the dominant mode with due consideration to the excitation of the waveguide in both the TE and TM mode types.

Page 54: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

54

Starting from the expression for Ez and taking help from Maxwell’s equations, we can find the transverse electric and magnetic field components in the TM mode as we have obtained earlier in the TE mode. Thus, we obtain the field expressions for the TMmn mode as follows:

0zH

...),3,2,1,0,()(expsincos022

nmztjy

b

nx

a

mE

a

m

bn

am

jE zx

...),3,2,1,0,()(expsincos0220

nmztjyb

nx

a

mE

a

m

bn

am

jH zy

...),3,2,1,0,()(expcossin0220

nmztjyb

nx

a

mE

b

n

bn

am

jH zx

...),3,2,1,0,()(exp)cos()cos(0 nmztjyb

nx

a

mEE zz

...),3,2,1,0,()(expcossin022

nmztjy

b

nx

a

mE

b

n

bn

am

jE zy

(TMmn-mode field expressions of a rectangular waveguide)

Page 55: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

55

Field pattern and significance of mode numbers of a rectangular waveguide

mode) TE(

)(expsin

)(expsin

)(expcos

10

0

00

0

ztjxa

Haj

H

ztjxa

Haj

E

ztjxa

HH

zx

zy

zz

mode) TE(

)(expsin

)(expsin

)(expcos

01

0

00

0

ztjyb

Hbj

H

ztjyb

Hbj

E

ztjyb

HH

zy

zx

zz

Expressions for non-zero field components of a rectangular waveguide for typical lower-order modes TE10, TE10, TE20 and TM11:

Page 56: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

56

mode) TE(

)(exp2

sin2

)(exp2

sin2

)(exp2

cos

20

0

00

0

ztjxa

Haj

H

ztjxa

Haj

E

ztjxa

HH

zx

zy

zz

mode) TM(

)(expsincos

)(expcossin

)(expsinsin

)(expcossin

)(expsincos

11

0220

0220

0

022

022

ztjyb

xa

Ea

ba

jH

ztjyb

xa

Eb

ba

jH

ztjyb

xa

EE

ztjyb

xa

Eb

ba

jE

ztjyb

xa

Ea

ba

jE

zy

zx

zz

zy

zx

Page 57: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

57

a

b

TE10 mode

a

b

TE20 mode

a

b

x

yz

TM11 mode

Field pattern of typical lower-order rectangular waveguide modes

The field pattern of the rectangular waveguide has been more elaborately explained in the book. However, in what follows next, let us discuss the significance of the mode number vis-à-vis the field pattern.

Page 58: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

58

Significance of the TEmn-mode or TMmn-mode field patterns of a rectangular waveguide vis-à-vis the mode numbers m and n

It is easy to infer by examining field patterns that

• mode number m, the first suffix of TEmn or TMmn, indicates the number of maxima of any field component, electric or magnetic, along the broad dimension of the waveguide

 • mode number n, the second suffix of TEmn or TMmn, indicates the number of maxima of

any field component, electric or magnetic, along the narrow dimension of the waveguide

Alternatively, m and n can be interpreted as the numbers of half-wave field patterns across the broad and narrow dimensions of the waveguide respectively.

Page 59: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

59

Examples:

For TE10 mode (m = 1, n = 0), the first suffix being unity (m = 1), there is a single maximum of the field component along the waveguide broad dimension. The second suffix being zero (n = 0), there is no maximum of the field component along the waveguide narrow dimension. For TE20 mode (m =2, n = 0), the first suffix being 2 (m = 2), there exist two maxima of the field component along the waveguide broad dimension. The second suffix being zero (n = 0) there is no maximum of the field component along the waveguide narrow dimension.  Similar observations can be made by examining the field pattern of the TMmn mode, for instance, the TM11 mode as well.  

a

b

a

b

x

yz

TE10 mode

TE20 mode

TM11 mode

a

b

Later on we are going to make similar inferences on the significance of mode numbers vis-à-vis field pattern with respect to a cylindrical waveguide.

Page 60: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

60

Cylindrical waveguide

Cylindrical or circular waveguide can be treated in cylindrical system of coordinates (r,,z) in view of the circular symmetry of this type of waveguide.

X

Y

Z

O

Z

For a cylindrical waveguide it turns out that the dominant mode is the TE11 mode characterized by having the lowest cutoff frequency. Let us emphasize here mainly the TE11 mode for analysis, although the analytical approach presented here is rather general and is easily applicable to the analysis of other higher order TEmn and TMmn modes as well.

02

2

002

t

HH zz

011

2

2

002

2

2

2

2

t

H

z

HH

rr

Hr

rrzzzz

)0,0( zz EHWave equation for the TE mode

Expanding Laplacian in cylindrical coordinates

Page 61: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

61

011

2

2

002

2

2

2

2

t

H

z

HH

rr

Hr

rrzzzz

011

2

2

22

2

002

2

2

2

zzzzz H

rt

H

z

H

r

H

rr

H

01

)(1

2

2

22

002

2

2

z

zzz H

rH

r

H

rr

H

2

2

22

002

2

2 1)(

1

z

zzz H

rH

r

H

rr

H

(rewritten)

Carrying out the differentiation of the first term

In view of the field dependence exp j(t-z) which is understood, enabling us to put /t = j and /z = - j

Rearranging terms

Page 62: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

62

2

2

22

002

2

2 1)(

1

z

zzz H

rH

r

H

rr

H

RH z

2

2

2

2

2

2

2

2

)(

R

H

r

R

r

H

r

R

r

R

r

H

z

z

z

2

2

22

002

2

2 1)(

1

Rr

Rr

R

rr

R

(wave equation) (rewritten)

Let us use the method of separation of variables for solving the wave equation and put

alone offunction a is

and alone offunction a is

rR

Page 63: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

63

Rby Dividing

11

)(111

2

2

22

002

2

2

rr

R

Rrr

R

R

2/1222/12222/1200

2 )()/()( kc

1111

2

2

22

2

2

rr

R

Rrr

R

R

2

2

22

002

2

2 1)(

1

Rr

Rr

R

rr

R(rewritten)

1

2

222

2

22

r

r

R

R

r

r

R

R

r

2by gMultiplyin r

Page 64: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

64

1

2

222

2

22

r

r

R

R

r

r

R

R

r(rewritten)

2m

22

2

2222

22

1m

mrr

R

R

r

r

R

R

r

Equates a function of r alone on its left hand side with a function of alone on its right hand side. This can hold good only when the individual functions each become equal to a constant.

Putting this constant as

Page 65: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

65

2

22

2

2 11

r

m

r

R

Rrr

R

R

01

2

222

2

22

Rr

m

x

R

xx

R

2222

22

mrr

R

R

r

r

R

R

r

(rewritten)

0)(1

2

22

2

2

Rr

m

r

R

rr

R

2by Dividng r

x

r

rx

termsgRearrangin

In terms of a dimensionless quantity x

Page 66: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

66

RH z

Rfor Solve

22

2 1m

)()( xQYxPJR mm

01

1

2

222

2

22

22

2

Rr

m

x

R

xx

R

m

(rewritten)

for Solve

We can obtain the expression for the axial component of magnetic field Hz from the solution of its components R and :

01

2

222

2

22

Rr

m

x

R

xx

R

mDmC sincos

022

2

m

Well known solution

Bessel equation

Well known solution

C and D are constants

P and Q are constants.

Jm(x) is the mth order ordinary Bessel function of the first kind.

Ym(x) is the mth order Bessel function of the second kind.

Page 67: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

67

RH z

)sinsincos(cos mmM 0 as )( xxYm

mDmC sincos )()( xQYxPJR mm (rewritten)

(rewritten)

sin

cos

MD

MC

)cos( mM

]2)cos[(

])2(cos[)cos(

mm

mm

)( rPJR m

C

D

DCM

1

2/122

tan

)(

mM cos

The value of at angle would be the same as that at an angle + 2 (the radial coordinate r remaining unchanged)

a relation that shall hold good only for integral values of m, that is, for m = 0, 1, 2, 3, ….

)cos( mM (rewritten)

Choosing ψ = 0 that amounts to taking the reference of such that it corresponds to the maximum value = M at = 0

which in turn would make

0 as xR

which is an impossibility since the field cannot shoot up to infinity at x (= r) 0, that is at r 0 (axis of the waveguide).

In order to prevent this impossibility we must put the constant Q = 0 in the expression for R.

)()( xQYxPJR mm 0Q

)(xPJR m

rx

Page 68: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

68

mM

rPJR m

cos

)(

0zHPM

RH z

)(expcos)(0 ztjmrJHH mzz

(recalled)

mrPMJH mz cos)(

mrJHH mzz cos)(0

Invoking the understood factor )(exp ztj

Next we can find the rest of the field expressions with the help of the above axial component of magnetic field and the two of the following Maxwell’s equations:

t

EH

t

HE

0

0

(Maxwell’s equations)

Page 69: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

69

zz

rr

zr

zrr

z

at

Ha

t

Ha

t

H

aE

r

rE

r

ar

E

z

Ea

z

EE

r

0

)(1

1

t

H

z

EE

rrz

0

1

t

E

r

H

z

H zr

0

zz

rr

zr

zrr

z

at

Ea

t

Ea

t

E

aH

r

rH

r

ar

H

z

Ha

z

HH

r

0

)(1

1

t

H

r

E

z

E zr

0

t

E

z

HH

rrz

0

1

t

EH

0

t

HE

0 Maxwell’s equations

(-component)

(r-component)(r-component)

(-component)rHE 0 r

z EjHjH

r 0

1

Ejr

HHj zr 0

HEr 0

0zE

0zE (TE mode)

0zE

Page 70: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

70

Ejr

HHj

HE

zr

r

0

0

HE

EjHjH

r

r

rz

0

0

1

(recalled)

r

HjH

r

HjE

zr

z

2

20

zr

z

H

r

jE

H

r

jH

20

2

(recalled)

Eliminating Hr and E respectively

Eliminating Er and H respectively

mrJHH mzz cos)(0

)(expcos)(

)(expsin)(

0

0

ztjmrJHr

H

ztjmrmJHH

mzz

mzz

Taking partial derivatives

(recalled)

)()( rJdr

drJ mm

2/122

2/1200

2

)(

)(

k

Page 71: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

71

r

HjH

r

HjE

zr

z

2

20

zr

z

H

r

jE

H

r

jH

20

2

(rewritten) (rewritten)

)(expsin)(

)(expsin)(

020

02

ztjmrJHr

mjE

ztjmrmJHr

jH

mzr

mz

)(expcos)(

)(expcos)(

0

00

ztjmrJHj

H

ztjmrJHj

E

mzr

mz

)(expcos)(

)(expsin)(

0

0

ztjmrJHr

H

ztjmrmJHH

mzz

mzz

(rewritten)

Page 72: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

72

Field expressions of a cylindrical waveguide excited in the TEmn mode put together

)(expsin)(020 ztjmrJHr

mjE mzr

)(expcos)(00 ztjmrJH

jE mz

0zE

)(expcos)(0 ztjmrJHj

H mzr

)(expsin)(02ztjmrmJH

r

jH mz

)(expcos)(0 ztjmrJHH mzz

(m and n are the mode numbers of the cylindrical waveguide excited in the TEmn mode)

Page 73: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

73

arn Ea

02

arE

0

rn aa

zr

rr

aaa

aa

0

zzrr aEaEaEE

2

Electromagnetic boundary conditions at the waveguide wall of a cylindrical waveguide

Boundary condition at the interface between a conductor and a dielectric/free-space introduced in Chapter 7, here the interface being the conducting waveguide wall of radius r = a, the subscript 2 referring to the free-space region inside the waveguide

Unit vector directed from region 1 (here, conducting waveguide wall) to region 2 (here, free-space region inside the waveguide), thus being radially inward at the waveguide wall

arrrr aEaEa

0)(

0)( aJm

arzaE 0

mnXa

(electromagnetic boundary condition at the conducting wall of the waveguide)

)(expcos)(00 ztjmrJH

jE mz

has a number of zeros or roots corresponding to its multiple solutions

(corresponding to the nth zero or root, where Xmn is called the eigenvalue of the cylindrical waveguide excited in the mode TEmn)

Page 74: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

74

mnXa (corresponding to the nth zero or root, where Xmn is called the eigenvalue of the cylindrical waveguide excited in the mode TEmn)

(rewritten)

970.9,706.6,054.3

536.8,31.5,841.1

174.10,016.7,832.3

232221

131211

030201

XXX

XXX

XXX

mnXak 2/122 )(

2/1222/12222/1200

2 )()/()( kc

0222 ckk

mnXa

(a few lower order eigenvalues of the roots taken from easily available text on Bessel functions, the eigenvalue X11 = 1.841 corresponding to m =1, n =1 for the TE11 mode being the lowest of them)

Dispersion relation and cutoff frequency of a cylindrical waveguide

a

Xk mnc

ck

02222 cc

a

cXmnc

222 )(a

Xk mn

ck cc

(dispersion relation)

ca

X cmn

(cutoff frequency)

Page 75: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

75

02222 cc Dispersion relation of a cylindrical waveguide is the same as that of a rectangular waveguide with appropriate interpretation of the waveguide cutoff frequency.

(rewritten)

Obviously, the nature of the - dispersion curve of a a cylindrical waveguide is identical with that of a rectangular waveguide.

2 2 2 2 0cc

cc -

line

a

cXmnc

Dominant mode of a cylindrical waveguide

For a cylindrical waveguide, here considered as excited in the TE mode, we have seen that the TE11 mode has the lowest eigenvalue X11 = 1.841 and correspondingly the lowest cutoff frequency c = Xmnc/a = 1.841c/a.

In fact, the analysis in the TM mode (which, though presented earlier for a rectangular waveguide, has not been done here for a cylindrical waveguide) would reveal that, of all the modes of the TE and TM modes of a cylindrical waveguide taken together, the TE11 mode has the lowest cutoff frequency.

Therefore,

TE11 mode is the dominant mode in a cylindrical waveguide

TE10 is the dominant mode is in a rectangular waveguide.

Page 76: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

76

a

cfc 2

2

22

rect822

)area(cf

caaaba

cf

ca

2

841.1

2

222

circular4

)841.1()area(

cf

ca

a

cXfc

112

cf

ca

2

a

c

a

cXfc

2

841.1

211

In an illustrative example you will find it of interest to compare the cross-sectional area of a circular or cylindrical waveguide with that of a rectangular waveguide that has its wider dimension twice its narrower dimension and that has its cutoff frequency the same as that of the rectangular waveguide, taking both the waveguides excited in the dominant mode.

2/122

2

b

n

a

mcfc

(a = wider dimension; b = narrower dimension)

Rectangular waveguide Cylindrical waveguide

Dominant mode: m =1, n = 0

(cross-sectional area of the given rectangular waveguide in terms of the cutoff frequency, the latter being the same as that of the given circular or cylindrical waveguide in the example)

a

cXmnc

(recalled)

Dominant mode: m =1, n = 1

(replacing the symbol a with a/ to avoid the confusion with the symbol a for the narrow dimension of a rectangular waveguide)

)841.1( 11 X

16.2143.3

)841.1(2

)8/(

)4/()841.1(

)area(

)area( 2

22

222

rect

circular

c

c

fc

fc

Page 77: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

77

Field pattern and significance of mode numbers of a cylindrical waveguide

)()( 10 xJxJ

)(exp832.3

)(exp832.3

0

)(exp832.3

0

00

10

100

ztjra

JHH

ztjra

JHj

H

E

ztjra

JHj

E

E

zz

zr

z

z

r

Field expressions of a cylindrical waveguide for typical lower-order modes TE01, TE02 and TE11 (interpreting the TEmn-mode field expressions already deduced):

(recurrence relation taking help of)

(TE01 mode)

Page 78: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

78

)(exp016.7

)(exp016.7

0

)(exp016.7

0

00

10

100

ztjra

JHH

ztjra

JHj

H

E

ztjra

JHj

E

E

zz

zr

z

z

r

(TE02 mode)

)(expcos841.1

)(expsin841.1

)(expcos841.1

0

)(expcos841.1

)(expsin841.1

10

102

10

100

1020

ztjra

JHH

ztjra

JHr

jH

ztjra

JHj

H

E

ztjra

JHj

E

ztjra

JHr

jE

zz

z

zr

z

z

zr

(TE11 mode) )/841.1(

)/841.1()841.1/(

)/841.1(

2

1

1

arJ

arJra

arJ

(relation to be made use of)

Page 79: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

79

(TE01 mode) (TE02 mode)

(TE0n mode; n = 1, 2, 3)

D

C

B

A

(TE11 mode)

Field pattern of typical lower-order cylindrical waveguide modes

The field pattern of the cylindrical waveguide has been more elaborately explained in the book. However, in what follows next let us discuss the significance of the mode number vis-à-vis the field pattern of a cylindrical waveguide.

Page 80: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

Significance of mode numbers of a cylindrical waveguide  TE01 mode:mode number m = 0 no half-wave field patterns around the half circumferencemode number n =1 one field maximum across the waveguide radius approximately mid-way between the axis and the wall of the waveguide  TE02 mode: mode number m = 0 no half-wave field patterns around the half circumferencemode number n = 2 two maxima across the waveguide radius, between the axis and the wall of the waveguide, such that if one of the maxima is positive the other is negative.  TE11 mode: mode number m = 1 a single half-wave field pattern around the half circumference (or a single full-wave field pattern around the full circumference) of the waveguidemode number n =1 across the waveguide radius, a single maximum at the axis of the waveguide  Based on the observation for the TE11 mode, TE01 and the TE02 modes above, the meaning of the mode numbers of a cylindrical waveguide excited in the TEmn mode is

• m (azimuthal mode number) is the number of half-wave field patterns around the half waveguide circumference;  •n (radial mode number) is the number of maxima, positive and negative inclusive, across the waveguide radius.

 Such interpretation is valid for the TMmn mode as well.

D

C

B

A

(TE01 mode)

(TE02 mode)

(TE11 mode)

Page 81: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

81

The TEM mode is characterized by transverse electric and magnetic fields and no axial electric and magnetic field components (Ez = Hz = 0).

However, for continuous magnetic field lines to exist in a hollow-pipe waveguide, there should be an axial current present in the waveguide in the form of either conduction or displacement current in the axial direction.

The absence of a conductor does not allow such conduction current in the axial direction. Further, for the displacement current in the axial direction to exist in the waveguide, there should be an axial electric field, the time variation of which being responsible for such current.

However, the TEM mode does not permit the axial electric field. Therefore, a hollow-pipe waveguide cannot support the TEM mode.

On the other hand, a two-conductor structure comprising a hollow cylindrical waveguide with a conducting coaxial solid circular rod insert, known as a coaxial waveguide, can support the TEM mode.

Inability of a hollow-pipe waveguide to support a TEM mode:

Page 82: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

82

Power Flow and power loss in a waveguide

We can find power transmitted in the axial direction z of a rectangular waveguide by integrating the axial z-component of the average complex Poynting vector (see Chapter 8):

*average Re2/1P HE

ax

x

by

y

zdxdyHEP0 0

*)Re(2

1

***)( xyyxz HEHEHE

over the waveguide cross-sectional area (= ab) transverse to the axis z of the waveguide as follows:

ax

x

by

y

xyyx dxdyHEHEP0 0

** )Re(2

1

Page 83: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

83

)(expsin/

)(expsin/

)(expsin/

0

0*

0

00

ztjxa

Ha

jH

ztjxa

Ha

jH

ztjxa

Ha

jE

HE

zx

zx

zy

yx

ax

x

by

y

z dydxa

xH

aaP

0 0

220

0 sin2

1

ax

x

by

y

xyyx dxdyHEHEP0 0

** )Re(2

1

(power transmitted in the axial direction z of a rectangular waveguide)

(recalled)

Restricting the analysis to the dominant mode TE10

(TE10 mode)

)(22

1 20

0 ba

Haa

P z

bdya

dxa

xax

x

by

y

0 0

2 ;2

sin

20

3024

1zbHaP

(TE10 mode)

(power transmitted in the axial direction z of a rectangular waveguide)

Page 84: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

84

20

3024

1zbHaP

(power handling capability of the rectangular waveguide in the dominant TE10 mode)

Power handling capability of a waveguide

(power transmitted in the axial direction z of a rectangular waveguide)

a

EH

y

z0

maximun0

)(expsin00 ztjx

aH

ajE zy

Let us take a rectangular waveguide excited in the dominant TE10 mode.

(dominant-mode TE10 field expressions) (recalled)

Let us find an expression for the maximum permissible Pmaximum, that is, the power handling capability of the waveguide for a known magnitude of the maximum electric field amplitude

which the atmosphere of the inside of the waveguide can withstand before it breaks down.

2

maximun0

maximum )(4

1yEabP

maximunyE

(axial magnetic field amplitude in terms of the breakdown electric field)

Page 85: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

85

m.1054.27.17.1 and m 1054.24.34.3 22 ba

2

maximun0

2/122

maximum )(4

)/1(y

c Ec

ffabP

In an illustrative example let us calculate the maximum power handling capability of WR-340 rectangular waveguide, excited in the dominant mode, operating at 3 GHz frequency, taking the breakdown limit of air as 29 kV/cm and taking the waveguide dimensions as:

2

maximun0

maximum )(4

1yEabP

(rewritten)

2/122

ph

ph

)/1/(1/

/2/

ffcv

fv

c

(recalled)

GHz 737.1Hz10737.1 6 cf

a

cfc 2

Recalling the expression for the cutoff frequency of a rectangular waveguide in the dominant mode, we can calculate:

m 1054.24.34.3 2a (dominant mode: m =1, n = 0)

Page 86: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

86

m/s 103

H/m 1048

70

c

2252

maximun

2

24

4

)V/m()1029()(

335.0

m1029.37

1054.27.154.24.3

y

c

E

f

f

ab

MW 952.16 W10952.16

W10103143.344

29298155.029.37

6

94maximum

P

GHz 737.1Hz10737.1 6 cf(rewritten)

2

maximun0

2/122

maximum )(4

)/1(y

c Ec

ffabP

V/m1029kV/cm29 5

maximunyE

(given)

Hz103GHz3 6f

m1054.27.17.1

m 1054.24.34.32

2

b

a

(given)

(given)

Page 87: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

87

*LA 2

1ssS JJRP

sn JHa 2

Power loss per unit area and power loss per unit length of a rectangular waveguide

Power loss per unit area at a conducting surface here the interface between the conducting wall and the inside of the rectangular waveguide is given by (see Chapter 8)

density current surface

resistance surface

areaunit per losspower LA

s

S

J

R

P

Surface current density at any of the four waveguide walls can be found from the following electromagnetic boundary condition at the concerned waveguide wall:

Recalling (from Chapter 7) the electromagnetic boundary condition at the interface between region 2, here the free-space region inside the waveguide wall, and region 1, here the conducting wall region

Let us next find the surface current densities developed at the right side wall (x = 0), left side wall (x = a), bottom wall (y = 0) and top wall (y = b) respectively.

a

b

Z

Y

X

inside) space-(free 2region to wall)g(conductin

1region waveguide thefrom directed

runit vecto theis na

Page 88: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

88

zzxz aztjxa

Haztjxa

Haj

H

)(expcos)(expsin 002

zzyyxx aHaHaHH

2

sn JHa 2

mode) TE(

)(expcos

0

)(expsin

10

0

0

ztjxa

HH

H

ztjxa

Haj

H

zz

y

zx

xzxs aztjx

aH

ajaJ

)(expsin][ 0 sidewallright

(recalled)

zz aztjx

aH

)(expcos0

wall)side(right xn aa

(boundary condition) (recalled)

yzs aztjHJ

)(exp0 sidewallright

10cos

wall)side(right 0

0

x

aaa

aa

yzx

xx

Page 89: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

89

zzxz aztjxa

Haztjxa

Haj

H

)(expcos)(expsin 002

sn JHa 2

xzxs aztjx

aH

ajaJ

)(expsin][ 0 sidewallleft

wall)side(left xn aa

(boundary condition) (recalled)

yzs aztjHJ

)(exp0 sidewallleft

zz aztjx

aH

)(expcos0

1cos

wall)side(left

0

ax

aaa

aa

yzx

xx

Similarly, we can derive next an expression for the surface current density developed at the left side wall.

Page 90: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

90

zzxz aztjxa

Haztjxa

Haj

H

)(expcos)(expsin 002

sn JHa 2

xzys aztjx

aH

ajaJ

)(expsin][ 0 wallbottom

wall)(bottom yn aa

(boundary condition) (recalled)

xzzzs aztjxa

Haztjxa

Haj

J

)(expcos)(expsin 00 wallbottom

zz aztjx

aH

)(expcos0

wall)(bottom 0y

aaa

aaa

xzy

zxy

Next, let us derive next an expression for the surface current density developed at the bottom wall.

(TE10 mode)

Page 91: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

91

zzxz aztjxa

Haztjxa

Haj

H

)(expcos)(expsin 002

sn JHa 2

xzys aztjx

aH

ajaJ

)(expsin][ 0 walltop

wall)(top yn aa

(boundary condition) (recalled)

xzzzs aztjxa

Haztjxa

Haj

J

)(expcos)(expsin 00 walltop

zz aztjx

aH

)(expcos0

wall)(top by

aaa

aaa

xzy

zxy

Next, let us derive next an expression for the surface current density developed at the top wall.

Page 92: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

92

Expressions for surface current densities at the waveguide walls are required to find power loss per unit area PLA in terms of their surface resistance RS for each of these walls with the help of the expression:

xzzzs

xzzzs

yzs

yzs

aztjxa

Haztjxa

Haj

J

aztjxa

Haztjxa

Haj

J

aztjHJ

aztjHJ

)(expcos)(expsin

)(expcos)(expsin

)(exp

)(exp

00 walltop

00 wallbottom

0 sidewallleft

0 sidewallright

xzzzs

xzzzs

yzs

yzs

aztjxa

Haztjxa

Haj

J

aztjxa

Haztjxa

Haj

J

aztjHJ

aztjHJ

)(expcos)(expsin

)(expcos)(expsin

)(exp

)(exp

00 walltop*

00 wallbottom*

0 sidewallleft *

0 sidewallright *

Taking complex conjugate

(rewritten)

*LA 2

1ssS JJRP

(deduced in Chapter 8)

(TE10 mode)

Page 93: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

93

])(exp[])(exp[ 00*

yzyzss aztjHaztjHJJ

])(exp[])(exp[ 00*

yzyzss aztjHaztjHJJ

(right side wall)

(left side wall)

])(expcos)(expsin[

])(expcos)(expsin[

00

00*

xzzz

xzzzss

aztjxa

Haztjxa

Haj

aztjxa

Haztjxa

Haj

JJ

(bottom wall)

])(expcos)(expsin[

])(exp)(expcos)(expsin[

00

00*

xzzz

xxzzzss

aztjxa

Haztjxa

Haj

aztjaztjxa

Haztjxa

Haj

JJ

(top wall)

Using the expressions for the surface current densities and those of their complex conjugates already presented, which are involved in the expression for power loss per unit area PLA, we can write the following expressions with respect to all the four sides of the waveguide wall:

We can next simplify the above expressions before they can be used in the expression for PLA.

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94

xa

xa

aHR

aztjxa

Haztjxa

Haj

aztjxa

Haztjxa

Haj

RJJRP

zS

xzzz

xzzzSssS

222

222

0

00

00*

LA

cossin2

1

])(expcos)(expsin[

])(expcos)(expsin[2

1

2

1

20

00*

LA

2

1

])(exp[])(exp[2

1

2

1

zS

yzyzssS

HR

aztjHaztjHJJRP

(right side wall)

20

00*

LA

2

1

])(exp[])(exp[2

1

2

1

zS

yzyzssS

HR

aztjHaztjHJJRP

(left side wall)

(bottom wall)

xa

xa

aHR

aztjxa

Haztjxa

Haj

aztjxa

Haztjxa

Haj

RJJRP

zS

xzzz

xzzzSssS

222

222

0

00

00*

LA

cossin2

1

])(expcos)(expsin[

])(expcos)(expsin[2

1

2

1

(top wall)

Page 95: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

95

xa

xa

aHRPP

HRPP

zS

zS

222

222

0 walltopLA, wallbottomLA,

20sidewallleft LA,sidewallrightLA,

cossin2

1

2

1

We can then put together the expressions for power loss per unit area on all the four walls of the waveguide:

dydzHRdP zS2

0LL 2

1

From the expression for power loss per unit area, we can also find power loss per unit length of the waveguide as follows.

Element of power loss dPL over an element of strip of elemental thickness dz and elemental area dydz across the narrow dimension of the waveguide on its right sidewall (x = 0)

(right sidewall)

Power loss per unit length PLL across the narrow dimension of the waveguide on its right sidewall bHR

dzdyHRdPP

zS

z

z

by

y

zS

20

1

00

20LLsidewallrightLL,

2

1

2

1

(right sidewall)

dydzHRdP zS2

0LL 2

1

Element of power loss dPL over an element of strip of elemental thickness dz and elemental area dydz across the narrow dimension of the waveguide on its left sidewall (x = a)

(left sidewall)

bHR

dzdyHRdPP

zS

z

z

by

yzS

20

1

00

20LLsidewallleftLL,

2

12

1

(left sidewall) Power loss per unit length PLL across the narrow dimension of the waveguide on its left sidewall

Page 96: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

96

dxdzxa

xa

aHRdP zS

222

222

0LL cossin2

1Element of power loss dPL over an element of strip of elemental thickness dz and elemental area dxdz across the broad dimension of the waveguide on its bottom wall (y = 0)

(bottom wall)

Power loss per unit length PLL across the broad dimension of the waveguide on its bottom wall

1

00

222

222

0LLwallbottomLL, )cossin(2

1 z

z

ay

y

zS dzdxxa

xa

aHRdPP

x

ax

a

aHRP zS

222

222

0 wallbottomLA, cossin2

1

(bottom wall)

2

2cos1

cos

2

2cos1

sin

2

2

xax

a

xax

a

1

22

12

222

0wallbottomLL, aa

HRP zS

(using the relation

and evaluating the integral)

Page 97: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

97

dxdzxa

xa

aHRdP zS

222

222

0LL cossin2

1Element of power loss dPL over an element of strip of elemental thickness dz and elemental area dxdz across the broad dimension of the waveguide on its top wall (y = b)

(top wall)

Power loss per unit length PLL across the broad dimension of the waveguide on its top wall

1

00

222

222

0LLwalltopLL, )cossin(2

1 z

z

ay

yzS dzdxx

ax

a

aHRdPP

x

ax

a

aHRP zS

222

222

0 walltopLA, cossin2

1

(top wall)

2

2cos1

cos

2

2cos1

sin

2

2

xax

a

xax

a

1

22

12

222

0walltopLL, aa

HRP zS

(using the relation

and evaluating the integral)

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98

Power loss per unit length of the rectangular waveguide adding contributions of all the four walls

2

222

0

2

222

02

222

02

02

0LL

12

122

11

22

1

2

1

2

1

aabHR

aaHR

aaHRbHRbHRP

zS

zSzSzSzS

)(exp)exp()exp()exp(exp)exp( ztjzzjztjztj

)2exp()0()( zPzP

The above expression is useful in finding the expression for the attenuation constant of the waveguide.

j

Power loss caused by the finite resistivity of the waveguide wall may be accounted for by generalizing the dependence of field components as

propagation constant

*average Re2/1P HE

phase propagation constant attenuation constant

The amplitude of the field component decreases exponentially with z, with the factor exp (-z)

Power P (z) transmitted through the waveguide decreases exponentially with z, with the factor exp (-2z)

See for instance average complex Poynting or power density vector involving the electric and magnetic field components each decreasing with the factor exp (-z) making their product depending on the factor exp (-2z)

(power transmitted through the waveguide)

(TE10 mode)

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99

)2exp()0()( zPzP (rewritten)

)(2)2exp()0(2)(

zPzPdz

zdP

)(2

)(

zPdzzdP

LL

)(P

dz

zdP

Differentiating

)(2LL

zP

P

P

P

2LL

dropping the parenthesis from P(z)

Page 100: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

100

)(

1

2

12/1

2

2

2

2

0 a

cab

b c

c

cc

ba

aabRS

30

2

222 1

22

P

P

2LL (rewritten)

2

222

0LL 12

aabHRP zS

ba

aab

30

2

222 1

22

(recalled)

20

3024

1zbHaP

(recalled)

(TE10 mode)

(TE10 mode)

(TE10 mode)

1

sR

phv

(in terms of the conductivity and skin depth of the material of the waveguide wall; see Chapter 6)

(recalled)2/1

2

2

ph

1

1

c

c

v

(attenuation constant of the rectangular waveguide in the dominant TE10 mode)

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101

In a numerical example, let us calculate the attenuation constant of a rectangular waveguide (a = 0.9// and b = 0.4//; cutoff frequency = 6.56 GHz) made of aluminum ( = 35.4x106 mho/m) at the operating frequency 9 GHz in the dominant mode.

mho/m104.35

Hz1056.6

Hz109

m1054.24.04.0

m1054.29.09.0

6

9

9

2

2

cf

f

b

a

0

1

f

2/1

2

2

2

2

0

01

21

1

ff

ff

ab

f

bc

c

)(

1

2

12/1

2

2

2

2

0 a

cab

b c

c

cc

After a simple algebra and recalling

(skin depth)

(TE10 dominant mode) (recalled)

(given)

154.00177.068.80177.0 Np/m dB/m0177.0

ohm 3770 H/m 104 7

0

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102

2/1

2

2

2

2

01

2

1

c

cc

f

f

f

fab

f

f

b

01 f

2

f

fc

2/1

4/34/1

)1(

2

ab

G

cf

bG 0

0

1

In another interesting problem, let us find the value of the wave frequency relative to the cutoff frequency of a rectangular waveguide excited in the dominant mode that results in the minimum attenuation due to the finite conductivity of the material of the waveguide, in terms of the waveguide dimensions. Numerically appreciate the problem taking a = 1.8b and fc = 6.56 GHz.

)// nginterpreti (recalled ffcc

Page 103: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

103

2/1

4/34/1

)1(

2

ab

G

0df

d

d

d

df

d

0)1(2

2)1(

32

1

)1(

24/94/54/54/1

2/1

a

bab

f

G

df

d

c

(rewritten)

2

f

fc (rewritten)

At the frequency where the minimum attenuation of the waveguide takes place

0)1(2

2)1(

32

1 4/94/54/54/1

ab

ab

Quantity outside the parenthesis being non-zero for practical values of

)/( 22 c

Page 104: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

104

4/1by Dividing

0)63(2 2 abab 2

12)1(

3 2

a

b

a

b

abbaba

b

f

f

c 8)63()63(

42

b

abbaba

f

fc4

8)63()63( 22

4/14/94/54/5

2

12)1(

3

a

b

a

b

0)1(2

2)1(

32

1 4/94/54/54/1

ab

ab

(rewritten)

Can be rearranged as

Can be solved as

Page 105: 11 Engineering Electromagnetics Essentials Chapter 9 Waveguides: solution of the wave equation For a wave in a bounded medium.

105

abbaba

b

f

f

c 8)63()63(

42

(rewritten)

ba 8.1

56.6cf

bbbba 4.1168.1363

46.24.14)4.11(4.11

4222

bbb

b

f

f

c

(given)

24.148.188 bbbab

13.1646.256.6 f GHz

GHz

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106106

Waveguidea hollow pipe made of a conducting materialis extensively used for the transmission of power in the microwave frequency range.

Waveguide can support transverse electric (TE) mode, which is characterized by non-zero axial magnetic field and zero axial electric field

Waveguide can also support transverse magnetic TM mode, which is characterized by non-zero axial electric field and zero axial magnetic field.

Waveguide behaves as a high-pass filter supporting propagating waves above a cutoff frequency that is related to waveguide dimensions.

TE-mode and TM-mode field solutions for both the rectangular and cylindrical waveguides have been obtained.

Characteristic equation or dispersion relation of a waveguide can be found with the help of the field solutions and electromagnetic boundary condition that the tangential component of the electric field is nil at the conducting surface of the waveguide wall.

One and the same dispersion relation is obtained between the wave angular frequency and phase propagation constant of a waveguide for the TE and the TM modes involving their respective cutoff frequencies c.

- dispersion plots of Identical nature are generated for rectangular and cylindrical waveguides excited in TE or TM mode.

Summarising Notes

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107107

Cutoff frequency of the waveguide is the frequency corresponding to zero value of phase propagation constant (which can be identified as the point of intersection between the - dispersion plot and the abscissa, that is, -axis of the plot).

Cutoff frequency of the waveguide depends on the waveguide dimensions and waveguide mode chosen.

Characteristic parameters of a waveguide are guide wavelength, phase propagation constant, phase velocity, group velocity and wave impedance, each of them depending on the operating frequency relative to the cutoff frequency of the waveguide.

Evanescent mode is supported by a waveguide below its cutoff frequency associated with no component of the average Poynting vector in the direction of wave propagation, corresponding to no power flow in the waveguide.

Mode numbers m and n are subscripted in the nomenclatures TEmn and the TMmn representing respectively the transverse electric and transverse magnetic modes of the waveguide.

Dominant mode of a waveguide is characterized by the lowest value of the cutoff frequencies of all the TEmn and the TMmn modes of the waveguide.

Dominant mode of a rectangular waveguide is the mode TE10.

Dominant mode of a cylindrical waveguide is the mode TE11.

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108108

Mode numbers m and n of the TEmn and TMmn modes of a rectangular or cylindrical waveguide can be correlated with their respective field patterns across the waveguide cross section.

For a rectangular waveguide, the mode number m indicates the number of maxima (of any field component) along the broad dimension of the waveguide, while the mode number n indicates the number of maxima (of any field component) along the narrow dimension of the waveguide. Alternatively, you may interpret m and n as the numbers of half-wave field patterns across the broad and the narrow dimensions of the waveguide respectively.

For a cylindrical waveguide, the mode number m indicates the number of half-wave field pattern around the half circumference and n indicates the number of positive or negative maxima across the waveguide radius

Why a hollow-pipe waveguide cannot support transverse electromagnetic (TEM) mode, for which the axial electric field and the axial magnetic field are each nil, has been explained.

Expression for the power propagating through a rectangular waveguide above its cutoff frequency has been developed and hence the power handling capability of the waveguide has been found in terms of the breakdown voltage of the medium filling the hollow region of the waveguide for dominant-mode excitation of the waveguide.

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109109

Expression for the power loss per unit length of the walls of a rectangular waveguide due to the finite resistivity of the material making the walls has been developed for dominant-mode excitation of the waveguide.

Expression for the attenuation constant of a dominant-mode-excited rectangular waveguide has been developed using the expressions for propagating power and power loss per unit length of the waveguide.

Attenuation constant of a waveguide depends on the operating frequency and the waveguide dimensions which should be taken into consideration while choosing the waveguide mode and frequency for lower waveguide attenuation.

Readers are encouraged to go through Chapter 9 of the book for more topics and more worked-out examples and review questions.