Rectangular Waveguides

Wave guide

The main advantage of wave-guides is

their relatively low level of radiation

losses (since the electric and agnetic

fields are completely enclosed by a

conducting wall) compared to

transmission lines.

Waveguide components

Figures from: www.microwaves101.com/encyclopedia/waveguide.cfm

Rectangular waveguide

Waveguide to coax adapter

E-teeWaveguide bends

More waveguides

http://www.tallguide.com/Waveguidelinearity.html

Uses

To reduce attenuation loss High frequencies High power

Can operate only above certain frequencies Acts as a High-pass filter

Normally circular or rectangular We will assume lossless rectangular

Rectangular WG

Need to find the fields components of the em wave inside the waveguide Ez Hz Ex Hx Ey Hy

We’ll find that waveguides don’t support TEM waves

http://www.ee.surrey.ac.uk/Personal/D.Jefferies/wguide.html

Rectangular Waveguides: Fields inside

Using phasors & assuming waveguide filled with

lossless dielectric material and walls of perfect conductor,

the wave inside should obey…

ck

HkH

EkE

22

22

22

where

0

0

Then applying on the z-component…

2

22

2

2

2

2

2

:obtain we wherefrom

)()()(),,(

:Variables of Separation of methodby Solving

0

kZ

Z

Y

Y

X

X

zZyYxXzyxE

Ekz

E

y

E

x

E

''''''

z

zzzz

022 zz EkE

Fields inside the waveguide

0

0

0

:sexpression in the resultswhich

2

2

2

2222

2

ZZ

YkY

XkX

kkk

kZ

Z

Y

Y

X

X

''

y''

x''

yx

''''''

zz

yy

xx

ececzZ

ykcykcY(y)

xkcxkcX(x)

65

43

21

)(

sincos

sincos

22222yx kkkh

Substituting

zz

yy

xx

ececzZ

ykcykcY(y)

xkcxkcX(x)

65

43

21

)(

sincos

sincos

)()()(),,( zZyYxXzyxEz

zyyxxz

zyyxxz

zzyyxxz

eykBykBxkBxkBH

eykAykAxkAxkAE

z

ececykcykcxkcxkcE

sincossincos

,field magnetic for theSimilarly

sincossincos

:direction-in traveling waveat the lookingonly If

sincossincos

4321

4321

654321

Other componentsFrom Faraday and Ampere Laws we can find the

remaining four components:

22222

22

22

22

22

yx

zzy

zzx

zzy

zzx

kkkh

where

y

H

hx

E

h

jH

x

H

hy

E

h

jH

x

H

h

j

y

E

hE

y

H

h

j

x

E

hE

*So once we know Ez and Hz, we can find all the other fields.

Modes of propagationFrom these equations we can conclude: TEM (Ez=Hz=0) can’t propagate.

TE (Ez=0) transverse electric In TE mode, the electric lines of flux are

perpendicular to the axis of the waveguide

TM (Hz=0) transverse magnetic, Ez exists In TM mode, the magnetic lines of flux are

perpendicular to the axis of the waveguide.

HE hybrid modes in which all components exists

TM Mode

Boundary conditions: ,axE

,byE

z

z

0at 0

0at 0

Figure from: www.ee.bilkent.edu.tr/~microwave/programs/magnetic/rect/info.htm

zyyxxz eykAykAxkAxkAE sincossincos 4321

zjyxz eykxkAAE sinsin42

From these, we conclude: X(x) is in the form of sin kxx, where kx=m/a, m=1,2,3,… Y(y) is in the form of sin kyy, where ky=n/b, n=1,2,3,… So the solution for Ez(x,y,z) is

TM Mode

Substituting

222

sinsin

b

n

a

mh

where

eyb

nx

a

mEE zj

oz

22 k

TMmn

Other components are

x

E

h

jH

y

E

h

jH

y

E

hE

x

E

hE

zy

zx

zy

zx

2

2

2

2

zoy

zox

zoy

zox

eb

yn

a

xmE

a

m

h

jH

eb

yn

a

xmE

b

n

h

jH

eb

yn

a

xmE

b

n

hE

eb

yn

a

xmE

a

m

hE

sincos

cossin

cossin

sincos

2

2

2

2

0

sinsin

z

zjoz

H

eyb

nx

a

mEE

TM modes

The m and n represent the mode of propagation and indicates the number of variations of the field in the x and y directions

Note that for the TM mode, if n or m is zero, all fields are zero.

See applet by Paul Falstad

http://www.falstad.com/embox/guide.html

TM Cutoff

The cutoff frequency occurs when

Evanescent:

Means no propagation, everything is attenuated

Propagation:

This is the case we are interested since is when the wave is allowed to travel through the guide.

222

222

b

n

a

m

kkk yx

22

222

1

2

1or

0then When

b

n

a

mf

jb

n

a

m

c

c

0 and When 22

2

b

n

a

m

0 and When 22

2

j

b

n

a

m

Cutoff

The cutoff frequency is the frequency below which attenuation occurs and above which propagation takes place. (High Pass)

The phase constant becomes

2222 1'

f

f

b

n

a

m c

22

2

'

b

n

a

muf mnc

fc,mn

attenuation

Propagation

of mode mn

Phase velocity and impedance

The phase velocity is defined as

And the intrinsic impedance of the mode is

f

uu p

p

2

'

2

1'

f

f

H

E

H

E c

x

y

y

xTM

Summary of TM modes

Wave in the dielectric medium

Inside the waveguide

/'

'/' u

2

1'

f

fcTM

2

1

'

ffc

/

1'2

ff

u

c

p

2

1'

f

fc

fu /''

/1'/' fu

Related example of how fields look:

Parallel plate waveguide - TM modes

a

xmsinAEz

ztje

0 a x m = 1

m = 2

m = 3

xz a

Ez

TE Mode

Boundary conditions: ,axE

,byE

y

x

0at 0

0at 0

Figure from: www.ee.bilkent.edu.tr/~microwave/programs/magnetic/rect/info.htm

zjyxz eykxkBBH coscos31

From these, we conclude: X(x) is in the form of cos kxx, where kx=m/a, m=0,1,2,3,… Y(y) is in the form of cos kyy, where ky=n/b, n=0,1,2,3,… So the solution for Ez(x,y,z) is

zyyxxz eykBykBxkBxkBH sincossincos 4321

TE Mode Substituting

Note that n and m cannot be both zero because the fields will all be zero.

222

again where

coscos

b

n

a

mh

eyb

n

a

xmHH zj

oz

TEmn

Other components are

zoy

zox

zoy

zox

eb

yn

a

xmH

b

n

h

jH

eb

yn

a

xmH

a

m

h

jH

eb

yn

a

xmH

a

m

h

jE

eb

yn

a

xmH

b

n

h

jE

sincos

cossin

cossin

sincos

2

2

2

2

0

coscos

z

zjoz

E

eyb

nx

a

mHH

y

H

hH

x

H

hH

x

H

h

jE

y

H

h

jE

zy

zx

zy

zx

2

2

2

2

Cutoff

The cutoff frequency is the same expression as for the TM mode

But the lowest attainable frequencies are lowest because here n or m can be zero.

22

2

'

b

n

a

muf mnc

fc,mn

attenuation

Propagation

of mode mn

Dominant Mode

The dominant mode is the mode with lowest cutoff frequency.

It’s always TE10

The order of the next modes change depending on the dimensions of the guide.

Summary of TE modes

Wave in the dielectric medium

Inside the waveguide

/'

'/' u

2

1

'

ffc

TE

2

1

'

ffc

/

1'2

ff

u

c

p

2

1'

f

fc

fu /''

/1'/' fu

Variation of wave impedance

Wave impedance varies with frequency and mode

TE

TM’

fc,mn

Example:

Consider a length of air-filled copper X-band waveguide, with dimensions a=2.286cm, b=1.016cm operating at 10GHz. Find the cutoff frequencies of all possible propagating modes.

Solution: From the formula for the cut-off frequency

22

2

'

b

n

a

muf mnc

Example

An air-filled 5-by 2-cm waveguide has

at 15GHz What mode is being propagated? Find Determine Ey/Ex

V/m 50sin40sin20 zjz eyxE

Group velocity, ug

Is the velocity at which the energy travels.

It is always less than u’

s

m

f

fuu c

g rad/m

rad/s 1'

/

12

2'uuu gp

zoy e

a

xmH

ah

jE

sin

2

Group Velocity

As frequency is increased, the group velocity increases.  

Power transmission The average Poynting vector for the waveguide

fields is

where = TE or TM depending on the mode

zEE

HEHEHE

yx

xyyxave

ˆ2

Re2

1Re

2

1

22

***

P

a

x

b

y

yx

aveave dxdyEE

dSP0 0

22

2P

[W/m2]

[W]

Attenuation in Lossy waveguide

When dielectric inside guide is lossy, and walls are not perfect conductors, power is lost as it travels along guide.

The loss power is

Where c+d are the attenuation due to ohmic (conduction) and dielectric losses

Usually c >> d

zoave ePP 2

aveave

L Pdz

dPP 2

Attenuation for TE10

Dielectric attenuation, Np/m

Conductor attenuation, Np/m

2

12

'

ffc

d

2

10,

2

10,

5.0

1'

2

f

f

a

b

f

fb

R c

c

sc

Dielectric conductivity!

Waveguide Cavities

Cavities, or resonators, are used for storing energy

Used in klystron tubes, band-pass filters and frequency meters

It’s equivalent to a RLC circuit at high frequency

Their shape is that of a cavity, either cylindrical or cubical.

Cavity TM Mode to z

:obtain we wherefrom

)()()(),,(

:Variables of Separationby Solving

zZyYxXzyxEz

zkczkczZ

ykcykcY(y)

xkcxkcX(x)

zz

yy

xx

sincos)(

sincos

sincos

65

43

21

2222zyx kkkkwhere

TMmnp Boundary Conditions

,czEE

,axE

,byE

xy

z

z

0at ,0

0at 0

0at 0

From these, we conclude:kx=m/aky=n/bkz=p/c

where c is the dimension in z-axis

2222

2

sinsinsin

c

p

b

n

a

mk

where

c

zp

b

yn

a

xmEE oz c

Resonant frequency

The resonant frequency is the same for TM or TE modes, except that the lowest-order TM is TM110 and the lowest-order in TE is TE101.

222

2

'

c

p

b

n

a

mufr

Cavity TE Mode to z

:obtain we wherefrom

)()()(),,(

:Variables of Separationby Solving

zZyYxXzyxH z

zkczkczZ

ykcykcY(y)

xkcxkcX(x)

zz

yy

xx

sincos)(

sincos

sincos

65

43

21

2222zyx kkkkwhere

TEmnp Boundary Conditions

,byE

,axE

,czH

x

y

z

0at ,0

0at 0

0at 0

From these, we conclude:kx=m/aky=n/bkz=p/c

where c is the dimension in z-axis

c

yp

b

yn

a

xmHH oz

sincoscos

c

Quality Factor, Q

The cavity has walls with finite conductivity and is therefore losing stored energy.

The quality factor, Q, characterized the loss and also the bandwidth of the cavity resonator.

Dielectric cavities are used for resonators, amplifiers and oscillators at microwave frequencies.

A dielectric resonator antenna with a cap for measuring the radiation efficiency

Univ. of Mississippi

Quality Factor, Q

Is defined as

2233

22

101

2

TE modedominant For the

101 caaccab

abccaQTE

cof

where

101

1

LP

W

latione of oscily per cyclloss energ

storedge energy Time averaπQ

2

2

Example

For a cavity of dimensions; 3cm x 2cm x 7cm filled with air and made of copper (c=5.8 x 107)

Find the resonant frequency and the quality factor for the dominant mode.

Answer:

GHzfr 44.57

1

2

0

3

1

2

10322210

6

9106.1

)1044.5(

1

co

378,568

73737322

723732233

22

101

TEQ

GHzfr 97

0

2

1

3

1

2

10322210

110

Queries ?

Thank you