Rectangular Waveguides

Dr. S. Cruz-PolINEL 6216University of Puerto RicoMayagüez

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

kZZ

YY

XX

zZyYxXzyxE

EkzE

yE

xE

''''''

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

kZZ

YY

XX

''

y''

x''

yx

''''''

zz

yy

xx

ececzZ

ykcykcY(y)xkcxkcX(x)

65

43

21

)(

sincossincos

22222yx kkkh

Substituting

zz

yy

xx

ececzZ

ykcykcY(y)xkcxkcX(x)

65

43

21

)(

sincossincos

)()()(),,( 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

wherey

Hhx

EhjH

xH

hyE

hjH

xH

hj

yE

hE

yH

hj

xE

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 00at 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

bn

amh

where

eyb

nxa

mEE zjoz

22 k

TMmn

Other components are

xE

hjH

yE

hjH

yE

hE

xE

hE

zy

zx

zy

zx

2

2

2

2

zoy

zox

zoy

zox

eb

yna

xmEa

mhjH

eb

yna

xmEb

nhjH

eb

yna

xmEb

nh

E

eb

yna

xmEa

mh

E

sincos

cossin

cossin

sincos

2

2

2

2

0

sinsin

z

zjoz

H

eyb

nxa

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 Falstadhttp://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

bn

am

kkk yx

22

222

121or

0then When

bn

amf

jb

na

m

c

c

0 and When 22

2

bn

am

0 and When 22

2

j

bn

am

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'

ff

bn

am c

22

2'

bn

amuf mnc

fc,mn

attenuation

Propagationof mode mn

Phase velocity and impedance

The phase velocity is defined as

And the intrinsic impedance of the mode is

fu

u pp

2

'

2

1'

ff

HE

HE c

x

y

y

xTM

Summary of TM modesWave in the dielectric medium

Inside the waveguide

/'

'/' u

2

1'

ffc

TM

2

1

'

ffc

/

1'2

ff

uc

p

2

1'

ffc

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 = 3xz a

Ez

TE Mode

Boundary conditions: ,axE

,byE

y

x

0at 00at 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

bn

amh

eyb

na

xmHH zjoz

TEmn

Other components are

zoy

zox

zoy

zox

eb

yna

xmHb

nhjH

eb

yna

xmHa

mhjH

eb

yna

xmHa

mhjE

eb

yna

xmHb

nhjE

sincos

cossin

cossin

sincos

2

2

2

2

0

coscos

z

zjoz

E

eyb

nxa

mHH

yH

hH

xH

hH

xH

hjE

yH

hjE

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'

bn

amuf mnc

fc,mn

attenuation

Propagationof 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 modesWave in the dielectric medium

Inside the waveguide

/'

'/' u

2

1

'

ffc

TE

2

1

'

ffc

/

1'2

ff

uc

p

2

1'

ffc

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'

bn

amuf 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’

sm

ffuu c

g rad/mrad/s 1'

/1

2

2'uuu gp

zoy e

axmH

ahjE

sin2

http://www.tpub.com/content/et/14092/css/14092_71.htm

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

Re21Re

21

22

***

P

a

x

b

y

yxaveave dxdy

EEdSP

0 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,

210,

5.0

1'

2f

fab

ff

b

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 SolvingzZyYxXzyxEz

zkczkczZ

ykcykcY(y)xkcxkcX(x)

zz

yy

xx

sincos)(

sincossincos

65

43

21

2222zyx kkkkwhere

TMmnp Boundary Conditions

,czEE,axE,byE

xy

z

z

0at ,00at 00at 0

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

where c is the dimension in z-axis

2222

2

sinsinsin

cp

bn

amk

wherec

zpb

yna

xmEE oz c

Resonant frequency

The resonant frequency is the same for TM or TE modes, except that the lowest-order TM is TM111 and the lowest-order in TE is TE101.

222

2'

cp

bn

amufr

Cavity TE Mode to z

:obtain we wherefrom)()()(),,(

:Variables of Separationby SolvingzZyYxXzyxH z

zkczkczZ

ykcykcY(y)xkcxkcX(x)

zz

yy

xx

sincos)(

sincossincos

65

43

21

2222zyx kkkkwhere

TEmnp Boundary Conditions

,byE

,axE,czH

x

y

z

0at ,0

0at 00at 0

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

where c is the dimension in z-axis

cyp

byn

axmHH oz

sincoscosc

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 caaccababccaQTE

cof

where

101

1

LPW

latione of oscily per cyclloss energstoredge energy Time averaπQ

2

2

ExampleFor 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 44.571

20

31

2103 22210

6

9106.1

)1044.5(1

co

378,568

7373732272373

2233

22

101

TEQ

GHzfr 970

21

31

2103 22210

110