EELE 3332 – Electromagnetic IIChapter 12 

Waveguides

Prof. Hala J. El‐KhozondarIslamic University of Gaza

Electrical Engineering Department

2016 1

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Waveguides

Waveguides are used at high frequencies since they have larger

bandwidth and lower signal attenuation than transmission lines.

Waveguides are used at high power applications.

Waveguides can operate above certain frequencies. (act as high

pass filter).

Normally circular or rectangular.

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Waveguides 

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Waveguides 

Dr. Talal Skaik 2012 IUG

Rectangular waveguide Waveguide to coax adapter

E-teeWaveguide bends

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Waveguide Filter

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Transmission Lines, Waveguides  ‐ Comparison

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Transmission Lines, Waveguides  ‐ Comparison

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12.2 Rectangular WaveguidesAssume a rectangular waveguide filled with lossless dielectricmaterial and walls of perfect conductor, Maxwell equations inphasor form become,

2 2

2 2

E E 0 where

H H 0s s

s s

kk

k

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Rectangular Waveguides

2 2

2 2 22

2 2 2

2

Applying on z-component:0

0

Solving by method of Separation of Variables:( , , ) ( ) ( ) ( )

from where we obtain:

zs zs

zs zs zszs

z

'' '' ''

E k E

E E E k Ex y z

E x y z X x Y y Z z

X Y Z kX Y Z

1 2

3 4

cos sin cos sin

( , , ) ( ) ( ) ( ) ( )

x x

y y

zs

X(x) c k x c k xY(y) c k y c k y

E x y z X x Y y Z z Z z c

5 6

1 2 3 4 5 6

1 2 3 4

1

cos sin cos sin

Assume wave propagates along waveguide in direction:

cos sin cos sin

Similarly for the magnetic field,

cos

z z

z zzs x x y y

zzs x x y y

zs

e c e

E c k x c k x c k y c k y c e c e

z

E A k x A k x A k y A k y e

H B

2 3 4sin cos sin zx x y yk x B k x B k y B k y e

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Rectangular Waveguides

From Maxwell’s equations, we can determine the other components Ex , Ey , Hx , Hy .

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Other Components

12

‐

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Other Components

FromMaxwell’s equations, we can determine the other components Ex , Ey , Hx , Hy .

2 2

2 2

2 2

2 2

2 2 2 2 2

zs zsxs

zs zsys

zs zsxs

zs zsys

x y

E HjEh x h y

E HjEh y h x

E HjHh y h x

E HjHh x h y

whereh k k k

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

From these equations we notice that there are different field patterns,

each of these field patterns is called a mode.

• Ezs=Hzs=0 (TEM mode): transverse electromagnetic mode. Both E

and H are transverse to the direction of propagation. From previous

equations we notice that all field components vanish for Ezs=Hzs=0.

→Rectangular waveguide can’t support TEM mode.

• Ezs=0, Hzs≠0 (TE modes) transverse electricThe electric field is transverse to the direction of propagation.

• Ezs ≠ 0, Hzs= 0 (TM modes) transverse magneticThe magnetic field is transverse to the direction of propagation.

• Ezs ≠ 0, Hzs ≠ 0 (HE modes) hybrid modes

All components exist. 14

Modes of Propagation

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Transverse Magnetic (TM) mode 1 2 3 40, cos sin cos sin

0 at 0 (bottom and top walls)0 at 0 (left and right walls)

Applying boundary conditio

Boundar

ns at ( 0 and

yConditio s

0

n

zz zs x x y y

zs

zs

H E A k x A k x A k y A k y e

E y ,bE x ,a

y x

zs 1 3

0 0 2 4

zs

) to E 0

sin sin ( )

Applying boundary conditions at ( and ) to Esin 0, sin 0 , This implies that :

, 1, 2,3,...,

zzs x y

x y

x

y

A A

E E k x k y e E A A

y b x ak a k b

k a m mk b n n

0

1, 2,3,...

,

sin sin

x y

zzs

m nor k ka b

m x n yE E ea b

Tangential components are continuous

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Transverse Magnetic (TM) mode

• Other components are

2

2

2

2

zsx

zsy

zsx

zsy

EEh x

EEh y

EjHh y

EjHh x

sin sin , 0zzs o zs

m nE E x y e Ha b

2

2

2

2

cos sin

sin cos

sin cos

cos sin

zxs o

zys o

zxs o

zys o

m m x n yE E eh a a b

n m x n yE E eh b a b

j n m x n yH E eh b a bj m m x n yH E eh a a b

2 22 2 2 x y

m nwhere h k ka b

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Transverse Magnetic (TM) mode

2 2

2 22

2 22

Propagation constant: ,

,

h k

m nh ka b

m na b

•Each set of integers m and n gives a different field pattern or mode.•Integer m equals the number of half cycle variations in the x-direction.•Integer n is the number of half cycle variations in the y-direction.•Note that for the TM mode, if n or m is zero, all fields are zero. Hence,TM11 is the lowest order mode of all the TMmn modes.

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Example: Field configuration for TM21 mode

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Transverse Magnetic (TM) mode2 2

2m na b

2 22

2 2

,

then 0

1 1or 2

No propagation takes pla

The cuttoff occurs when

ce at this frequenc

:

y

c

cm n

m n ja b

m nfa b

2 22When and 0

No wave propagation at all. (everything is attenuated)So when

Evanescent m

, all field components will decay exponantially

odes :

.c

m na b

f f

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Transverse Magnetic (TM) mode

2 22 and 0

This is the case we are interested since is when the wave is allowed to travel t

Propaga

hrough the guide. , at a given operating fre

tion occurs

que

whe

nc

n

m n ja b

So

y f, only those modes with will propagate.f fc

fc,mn

attenuation Propagation

of mode mn

The cutoff frequency is thefrequency below which attenuationoccurs and above which propagationtakes place. (High Pass)

2 2

,

2 2

2 22

:

1 1 2

' 1 , where '2

T can be written in term

The cutoff Frequency is

he phase co s of asnst t :an

cm n

cmn

c

m nfa b

u m nor f ua b

f

m na b

2 2

2

2

1

' 1 , where ' / 'c

m na b

f k uf

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Transverse Magnetic (TM) mode

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Transverse Magnetic (TM) mode

2 2

2 2

2 2 2 ' ', but ' , ''

' 1 1

:- (varies with freq

The gu

uency)

1 ' 1

ide wavelength is:

Intrinsic Impedance

, w

g g

c c

x c cTM TM

y

uff f

f f

E f fH f f

' /

', ', ', and ' are parameters for unguided wave propagatingin the same dielectric medium ( , ) unbounded by the waveguide. (i.e. waveguide removed and entire space is filled with diele

here

u

ctric.)

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Transverse Electric (TE) modes

1 2 3 4

2

0, cos sin cos sin

0 at 0 (bottom and top walls)0 at 0 (le

Boundaft and right walls)

0 at

ryConditions

0

zz zs x x y y

xs

ys

zs zsxs

E H B k x B k x B k y B k y e

E y ,bE x ,a

j H HE y ,bh y y

2

0 1 3

0 at 0

From this we conclude

cos cos ( = )

zs zsys

zzs o

j H HE x ,ah x x

m x nH H y e H B Ba b

Tangential components are continuous

Other components are

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Transverse Electric (TE) modes

2

2

2

2

cos sin

sin cos

sin cos

cos sin

zxs o

zys o

zxs o

zys o

j n m x n yE H eh b a bj m m x n yE H eh a a b

j m m x n yH H eh a a bj n m x n yH H eh b a b

0, cos cos zz zs o

m nE H H x y ea b

2

2

2

2

zxs

zys

zxs

zys

HjEh y

HjEh x

HHh x

HHh y

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Example: Field configuration for TE32 mode

• The cutoff frequency is the same expression as for the TMmode

• For TE modes, (m,n) may be (0,1) or (1,0) but not (0,0). Bothm and n cannot be zero at the same time because this will forcethe field components to vanish.

• Hence, the lowest mode can be TE10 or TE01 depending on thevalues of a and b.

• It is standard practice to have a>b, thus TE10 is the lowestmode.

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22

2'

bn

amuf mnc

TE  modes ‐ Cuttoff

10 '/ 2cf u a

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TE  modes• The dominant mode is the mode with lowest cutoff frequency.

The cutoff frequency of the TE10 mode is lower than that of TM11

mode. Hence, TE10 is the dominant mode.

If more than one mode is propagating, the waveguide is overmoded.

Single mode propagation is highly desirable to reduce dispersion.

This occurs between cutoff frequency for TE10 mode and cuttoff

frequency of next higher mode.

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2

2 2

he phase constant is the same as TM mode:

The intrinsic impedance of the TE mod

T

' 1 , where ' / '

1 ' , w ' /

1 1

e is:

c

xTE TE

yc c

f uf

E hereH f f

f f

22Note that ' ' 1 c

TE TM TMff

TE  modes

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For TE mode, cos cos

For TE mode, cos

In the time domain: =Re

cos cos

,

j zzs o

j zzs o

j tz zs

z o

y

m nH H x y ea b

xH H ea

H H e

orxH H t z

aSimilarly

E

0

0

sin sin

sin sin

0

x

z x y

a xH t za

a xH H t za

E E H

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TE10 mode

Variation of the field components with x for TE10mode.

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TE10 mode

Field lines for TE10 mode

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TE/TM  modesWave in the dielectric medium Inside the waveguide

/'

'/' u

2

1

'

ffc

TE

2

'

1 cff

/

1'2

ff

uc

p

2

1'

ffc

fu /''

/1'/' fu

2

, ' 1 cTM

ff

2 2

2 2 2 2

2 2

The cutoff frequency is given by:

' 1 , u'=2 2

Hence

4 4 2.5 10 1 10

cmnr r

cmn

u m n c cfa b

c m n c m nfa b

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Example 12.1

• Example: A rectangular waveguide with dimensions a=2.5 cm,b=1 cm is to operate below 15.1 GHz. How many TE and TMmodes can the waveguide transmit if the guide is filled with amedium characterized by σ=0, ε=4 ε0, µr=1? Calculate thecutoff frequencies of the modes.

2 2

2 2

01 01

02 02

03 03

10 10

20 20

4 2.5 10 1 10For TE mode ( =0, =1), 7.5 GHz For TE 15 GHzFor TE 22.5 GHz

For TE 3 GHzFor TE 6 GH

c mnc m nf

m n fcfcfc

fcfc

30 30

40 40

50 50

60 60

zFor TE 9 GHzFor TE 12 GHzFor TE 15 GHzFor TE 18 GHz

fcfcfcfc

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Example 12.1 ‐ solution

2 2

2 2

11 11 11

21 21 21

31 31 31

4 2.5 10 1 10

For TE , TM modes , 8.078 GHz For TE , TM modes , 9.6 GHz For TE , TM modes , 11.72 GHz

cmnc m nf

fcfcfc

41 41 41

12 12 12

Modes with cutoff frequencies les

For TE , TM modes , 14.14 GHz

s than or equal 15.1 GHzwill be tran

For TE

smitt

, TM modes , 15

ed. (11 TE modes

.3

an

GHz

d 4 TM

modes)

fcfc

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Example 12.1 ‐ solution

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Example 12.1 ‐ solution

Cutoff frequencies of rectangular waveguide with a 2.5b; for Example 12.1.

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Example 12.3

• Example: in a rectangular waveguide for which a=1.5 cm,b=0.8 cm, σ=0, µ=µ0, ε=4ε0.

• Determine:

• (a) the mode of operation.

• (b) the cutoff frequency

• (c) the phase constant β.

• (d) the propagation constant γ.

• (e) the intrinsic wave impedance η.

1132sin cos sin 10 A/mxx yH t z

a b

13 13 13

2 2

2 2

13 2 2

( ) the guide is operating at TM or TE . Suppose we choose TM .

' 1( ) , '2 2

1 3 28.57 GHz4 1.5 10 0.8 10

(c) ' 1

cmnr r

c

c

a

u m n c cb f ua b

cHence f

ff

13

2 2 2

11

211

8

2 2

TM

1 1

2 10 50 GHz

10 4 28.571 1718.81 rad/m3 10 50

( ) = 1718.81/ m

377 28.57( ) ' 1 1 154.7 50

rc c

c

r

f ff c f

f f

d j j

fef

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Example 12.3 ‐ Solution

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