1) Rectangular waveguides can transmit electromagnetic waves above a certain cutoff frequency, acting as a high-pass filter. They support transverse electric (TE) and transverse magnetic (TM) modes of propagation. 2) For TM modes, the electric field is transverse to the direction of propagation, while the magnetic field has a longitudinal component. The modes are denoted TMmn, with m and n indicating the number of half-wavelength variations across the width and height. 3) For TE modes, the magnetic field is entirely transverse, while the electric field has a longitudinal component. These modes are denoted TEmn, with m and n having the same meaning as in the TM case.

1. Rectangular Waveguides
Dr. S. Cruz-Pol
INEL 6216
University of Puerto Rico
Mayagüez

2. Waveguide components
Figures from: www.microwaves101.com/encyclopedia/waveguide.cfm
Rectangular waveguide Waveguide to coax adapter
E-teeWaveguide bends

3. More waveguides
http://www.tallguide.com/Waveguidelinearity.html

4. 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

5. 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

6. 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
µεω22
22
22
where
0
0
=
=+∇
=+∇

7. Then applying on the z-component…
2
2
2
2
2
2
2
2
:obtainwewherefrom
)()()(),,(
:VariablesofSeparationofmethodbySolving
0
k
Z
Z
Y
Y
X
X
zZyYxXzyxE
Ek
z
E
y
E
x
E
''''''
z
z
zzz
−=++
=
=+
∂
∂
+
∂
∂
+
∂
∂
022
=+∇ zz EkE

8. Fields inside the waveguide
0
0
0
:sexpressionin theresultswhich
2
2
2
2222
2
=−
=+
=+
−=+−−
−=++
ZZ
YkY
XkX
kkk
k
Z
Z
Y
Y
X
X
''
y
''
x
''
yx
''''''
γ
γ
zz
yy
xx
ececzZ
ykcykcY(y)
xkcxkcX(x)
γγ −
+=
+=
+=
65
43
21
)(
sincos
sincos
22222
yx kkkh +=+= γ

9. Substituting
zz
yy
xx
ececzZ
ykcykcY(y)
xkcxkcX(x)
γγ −
+=
+=
+=
65
43
21
)(
sincos
sincos
)()()(),,( zZyYxXzyxEz =
( )( )( )
( )( )
( )( ) z
yyxxz
z
yyxxz
zz
yyxxz
eykBykBxkBxkBH
eykAykAxkAxkAE
z
ececykcykcxkcxkcE
γ
γ
γγ
−
−
−
++=
++=
+
+++=
sincossincos
,fieldmagneticfor theSimilarly
sincossincos
:direction-intravelingwaveat thelookingonlyIf
sincossincos
4321
4321
654321

10. Other components
From Faraday and Ampere Laws we can find the
remaining four components:
22222
22
22
22
22
yx
zz
y
zz
x
zz
y
zz
x
kkkh
where
y
H
hx
E
h
j
H
x
H
hy
E
h
j
H
x
H
h
j
y
E
h
E
y
H
h
j
x
E
h
E
+=+=
∂
∂
−
∂
∂
−=
∂
∂
−
∂
∂
=
∂
∂
−
∂
∂
−=
∂
∂
−
∂
∂
−=
γ
γωε
γωε
ωµγ
ωµγ
*So once we know
Ez and Hz, we can
find all the other
fields.

11. Modes of propagation
From 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

12. TM Mode
 Boundary
conditions: ,axE
,byE
z
z
0at0
0at0
==
==
Figure from: www.ee.bilkent.edu.tr/~microwave/programs/magnetic/rect/info.htm
( )( ) z
yyxxz eykAykAxkAxkAE γ−
++= sincossincos 4321
( )( ) zj
yxz eykxkAAE β−
= sinsin42
From these, we conclude:
X(x) is in the form of sin kxx,
where kx=mπ/a, m=1,2,3,…
Y(y) is in the form of sin kyy,
where ky=nπ/b, n=1,2,3,…
So the solution for Ez(x,y,z) is

13. TM Mode
 Substituting
22
2
sinsin






+





=












= −
b
n
a
m
h
where
ey
b
n
x
a
m
EE zj
oz
ππ
ππ β
22
k+= γ

14. TMmn
 Other components are
x
E
h
j
H
y
E
h
j
H
y
E
h
E
x
E
h
E
z
y
z
x
z
y
z
x
∂
∂
−=
∂
∂
=
∂
∂
−=
∂
∂
−=
2
2
2
2
ωε
ωε
γ
γ
z
oy
z
ox
z
oy
z
ox
e
b
yn
a
xm
E
a
m
h
j
H
e
b
yn
a
xm
E
b
n
h
j
H
e
b
yn
a
xm
E
b
n
h
E
e
b
yn
a
xm
E
a
m
h
E
γ
γ
γ
γ
πππωε
πππωε
πππγ
πππγ
−
−
−
−


















−=


















=


















−=


















−=
sincos
cossin
cossin
sincos
2
2
2
2
0
sinsin
=












= −
z
zj
oz
H
ey
b
n
x
a
m
EE βππ

15. 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

16. 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.
( )
µεω
ππ
γ
2
22
222
−





+





=
−+=
b
n
a
m
kkk yx
22
22
2
1
2
1
or
0thenWhen






+





=
=+=





+





=
b
n
a
m
f
j
b
n
a
m
c
c
ππ
µεπ
βαγ
ππ
µεω
0andWhen
22
2
==





+





< βαγ
ππ
µεω
b
n
a
m
0andWhen
22
2
==





+





> αβγ
ππ
µεω j
b
n
a
m

17. Cutoff
 The cutoff frequency is the frequency
below which attenuation occurs and above
which propagation takes place. (High Pass)
 The phase constant becomes
222
2
1' 





−=





−





−=
f
f
b
n
a
m c
β
ππ
µεωβ
22
2
'






+





=
b
n
a
mu
f mnc
fc,mn
attenuation Propagation
of mode mn

18. Phase velocity and impedance
 The phase velocity is defined as
 And the intrinsic impedance of the mode
is
f
u
u
p
p ===
β
π
λ
β
ω 2
'
2
1' 





−=−==
f
f
H
E
H
E c
x
y
y
x
TM ηη

19. Summary of TM modes
Wave in the dielectric
medium
Inside the waveguide
εµη /'=
µεωωβ == '/' u
2
1' 





−=
f
fc
TM ηη
2
1
'






−
=
f
fc
λ
λ
βω
β
ω
/
1'
2
=






−
=
f
f
u
c
p
2
1' 





−=
f
fc
ββ
fu /''=λ
µελβω /1'/' === fu

20. Related example of how fields look:
Parallel plate waveguide - TM modes





 π
=
a
xm
sinAEz
( )ztj
e β−ω
0 a x
m = 1
m = 2
m = 3
xz a
Ez

21. TE Mode
 Boundary
conditions: ,axE
,byE
y
x
0at0
0at0
==
==
Figure from: www.ee.bilkent.edu.tr/~microwave/programs/magnetic/rect/info.htm
( )( ) zj
yxz eykxkBBH β−
= coscos31
From these, we conclude:
X(x) is in the form of cos kxx,
where kx=mπ/a, m=0,1,2,3,…
Y(y) is in the form of cos kyy,
where ky=nπ/b, n=0,1,2,3,…
So the solution for Ez(x,y,z) is
( )( ) z
yyxxz eykBykBxkBxkBH γ−
++= sincossincos 4321

22. TE Mode
 Substituting
 Note that n and m cannot be both zero
because the fields will all be zero.
22
2
againwhere
coscos






+





=












= −
b
n
a
m
h
ey
b
n
a
xm
HH zj
oz
ππ
ππ β

23. TEmn
 Other components are
z
oy
z
ox
z
oy
z
ox
e
b
yn
a
xm
H
b
n
h
j
H
e
b
yn
a
xm
H
a
m
h
j
H
e
b
yn
a
xm
H
a
m
h
j
E
e
b
yn
a
xm
H
b
n
h
j
E
γ
γ
γ
γ
πππβ
πππβ
πππωµ
πππωµ
−
−
−
−


















=


















=


















−=


















=
sincos
cossin
cossin
sincos
2
2
2
2
0
coscos
=












= −
z
zj
oz
E
ey
b
n
x
a
m
HH βππ
y
H
h
H
x
H
h
H
x
H
h
j
E
y
H
h
j
E
z
y
z
x
z
y
z
x
∂
∂
−=
∂
∂
−=
∂
∂
−=
∂
∂
−=
2
2
2
2
γ
γ
ωµ
ωµ

24. 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
mu
f mnc
fc,mn
attenuation Propagation
of mode mn

25. 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.

26. Summary of TE modes
Wave in the dielectric
medium
Inside the waveguide
εµη /'=
µεωωβ == '/' u
2
1
'






−
=
f
fc
TE
η
η
2
1
'






−
=
f
fc
λ
λ
βω
β
ω
/
1'
2
=






−
=
f
f
u
c
p
2
1' 





−=
f
fc
ββ
fu /''=λ
µελβω /1'/' === fu

27. Variation of wave impedance
 Wave impedance varies with
frequency and mode
ηTE
ηTM
η’
η
0 fc,mn

28. 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
mu
f mnc

29. Example
An air-filled 5-by 2-cm waveguide has
at 15GHz
 What mode is being propagated?
 Find β
 Determine Ey/Ex
( ) ( ) V/m50sin40sin20 zj
z eyxE β
ππ −
=

30. Group velocity, ug
 Is the velocity at which
the energy travels.
 It is always less than u’






=











−=
∂∂
=
s
m
f
f
uu c
g
rad/m
rad/s
1'
/
1
2
ωβ
( )2
'uuu gp =
z
oy e
a
xm
H
ah
j
E γππωµ −












−= sin2
http://www.tpub.com/content/et/14092/css/14092_71.htm

31. Group Velocity
 As frequency is increased,
the group velocity increases.

32. Power transmission
 The average Poynting vector for the waveguide
fields is
 where η = ηTE or ηTM depending on the mode
[ ] [ ]
z
EE
HEHEHE
yx
xyyxave
ˆ
2
Re
2
1
Re
2
1
22
***
η
+
=
−=×=P
∫ ∫∫ = =
+
=⋅=
a
x
b
y
yx
aveave dxdy
EE
dSP
0 0
22
2η
P
[W/m2
]
[W]

33. 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 α=αc+αd are the attenuation due to ohmic
(conduction) and dielectric losses
 Usually αc >> αd
z
oave ePP α2−
=
ave
ave
L P
dz
dP
P α2=−=

34. Attenuation for TE10
 Dielectric attenuation, Np/m
 Conductor attenuation, Np/m
2
12
'






−
−=
f
fc
d
ση
α














+






−
−=
2
10,
2
10,
5.0
1'
2
f
f
a
b
f
f
b
R c
c
s
c
η
α
Dielectric
conductivity!

35. 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.

36. Cavity TM Mode to z
:obtainwewherefrom
)()()(),,(
:VariablesofSeparationbySolving
zZyYxXzyxEz =
zkczkczZ
ykcykcY(y)
xkcxkcX(x)
zz
yy
xx
sincos)(
sincos
sincos
65
43
21
+=
+=
+=
2222
zyx kkkkwhere ++=

37. TMmnp Boundary Conditions
,czEE
,axE
,byE
xy
z
z
0at,0
0at0
0at0
===
==
==
From these, we conclude:
kx=mπ/a
ky=nπ/b
kz=pπ/c
where c is the dimension in z-axis
µεω
πππ
πππ
2
222
2
sinsinsin
=





+





+





=


















=
c
p
b
n
a
m
k
where
c
zp
b
yn
a
xm
EE oz c

38. 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
'






+





+





=
c
p
b
n
a
mu
fr

39. Cavity TE Mode to z
:obtainwewherefrom
)()()(),,(
:VariablesofSeparationbySolving
zZyYxXzyxHz =
zkczkczZ
ykcykcY(y)
xkcxkcX(x)
zz
yy
xx
sincos)(
sincos
sincos
65
43
21
+=
+=
+=
2222
zyx kkkkwhere ++=

40. TEmnp Boundary Conditions
,byE
,axE
,czH
x
y
z
0at,0
0at0
0at0
==
==
==
From these, we conclude:
kx=mπ/a
ky=nπ/b
kz=pπ/c
where c is the dimension in z-axis


















=
c
yp
b
yn
a
xm
HH oz
πππ
sincoscos
c

41. 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.

42. A dielectric resonator antenna
with a cap for measuring the
radiation efficiency
Univ. of Mississippi

43. Quality Factor, Q
 Is defined as
( )
( ) ( )[ ]2233
22
101
2
TEmodedominantFor the
101
caaccab
abcca
QTE
+++
+
=
δ cof
where
σµπ
δ
101
1
=
LP
W
latione of oscily per cyclloss energ
storedge energyTime avera
πQ
π2
2
=
=

44. Example
For a cavity of dimensions; 3cm x 2cm x 7cm filled with
air and made of copper (σc=5.8 x 107
)
 Find the resonant frequency and the quality factor
for the dominant mode.
Answer:
GHzfr 44.5
7
1
2
0
3
1
2
103
22210
=





+





+




⋅
=
6
9
106.1
)1044.5(
1 −
⋅=
⋅
=
coσµ
δ
( )
( ) ( )[ ] 378,568
73737322
72373
2233
22
101
=
+⋅++⋅
⋅⋅+
=
δ
TEQ
GHzfr 9
7
0
2
1
3
1
2
103
22210
110 =





+





+




⋅
=