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Rectangular Waveguides. Dr. S. Cruz-Pol INEL 6216 University of Puerto Rico Mayag ü ez. Waveguide components. Waveguide to coax adapter. Rectangular waveguide. Waveguide bends. E-tee. Figures from: www.microwaves101.com/encyclopedia/waveguide.cfm. More waveguides. - PowerPoint PPT Presentation
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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
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
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
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 TM111 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
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