ECE 5317-6351
Microwave Engineering
Fall 2012
Prof. David R. Jackson
Dept. of ECE
Notes 7
Waveguides Part 4:
Rectangular and Circular
Waveguide
1
Rectangular Waveguide
One of the earliest waveguides.
Still common for high power and high
microwave / millimeter-wave applications.
PEC
y
b
, ,
o
z
It is essentially an electromagnetic pipe
with a rectangular cross-section.
Single conductor No TEM mode
x
o
a
For convenience
a b.
the long dimension
lies along x.
2
TEz Modes
Recall
H z x, y, z hz x, y e
jk z z
b
where
2
k
c hz x, y 0
2
2
x y
2
PEC
y
, ,
2
o
z
x
o
a
kc k 2 k z2
1/2
Subject to B.C.’s:
and
H z
Ex 0
y
@ y 0, b
H z
Ey 0
x
@ x 0, a
3
TEz Modes (cont.)
2 2
2
h
x
,
y
k
hz x, y
z
c
2
2
x y
(eigenvalue problem)
Using separation of variables, let hz x, y X x Y y
d2X
d 2Y
2
Y
X
k
c XY
2
2
dx
dy
Must be a constant
1 d 2 X 1 d 2Y
2
k
c
X dx 2 Y dy 2
1 d2X
2
k
x
X dx 2
and
where k x2 k y2 kc2
1 d 2Y
2
k
y
Y dy 2
dispersion relationship
4
TEz Modes (cont.)
Hence,
X ( x)
Y ( y)
hz x, y ( A cos k x x B sin k x x)(C cos k y y D sin k y y )
Boundary Conditions:
A
D0
and
B
B0
and
hz
0
y
hz
0
x
n
ky
b
m
kx
a
m x n y
hz x, y Amn cos
cos
a
b
@ y 0, b
A
@ x 0, a
B
n 0,1,2,...
m 0,1,2,...
m n
kc2
a
b
2
and
2
5
TEz Modes (cont.)
Therefore,
m
H z Amn cos
a
n
x cos
b
y e
jk z z
k z k 2 kc2
m n
k
a b
2
2
2
From the previous field-representation equations,
we can show
n jk z z
x sin
y e
b
j m
m n jk z z
Ey
A
sin
x cos
y e
mn
2
kc a
a b
Ex
j n
m
A
cos
mn
kc2b
a
jk z m
m
A
sin
mn
kc2 a
a
n
x cos
y e
b
jk n
m n
H y z2 Amn cos
x sin
y e
kc b
a
b
Hx
jk z z
jk z z
Note:
m = 0,1,2,…
n = 0,1,2,…
But m = n = 0
is not allowed!
(non-physical solution)
H zˆ A00e
jkz
; H 0
6
TEz Modes (cont.)
Lossless Case c
k zmn k 2 k
mn 2
c
m n
k2
a b
2
2
TEmn mode is at cutoff when k kcmn
f
mn
c
2
m n
2 a b
1
2
Lowest cutoff frequency is for TE10 mode (a > b)
We will
revisit this
mode.
f
10
c
1
2a
Dominant TE mode
(lowest fc)
7
TEz Modes (cont.)
At the cutoff frequency of the TE10 mode (lossless waveguide):
f
10
c
1
2a
cd cd
10
f
fc
a
f fc
cd
1
2a
2a
d / 2
For a given operating wavelength (corresponding to f > fc) , the dimension
a must be at least this big in order for the TE10 mode to propagate.
Example: Air-filled waveguide, f = 10 GHz. We have that a > 3.0 cm/2 = 1.5 cm.
8
TMz Modes
Recall
PEC
y
Ez x, y, z ez x, y e
jk z z
b
, ,
o
where
z
2
2
2 2
y
x
2
e
x
,
y
k
c ez x, y
z
Subject to B.C.’s:
Ez 0
x
o
kc k k
2
2 1/2
z
a
(eigenvalue problem)
@ x 0, a
@ y 0, b
Thus, following same procedure as before, we have the following result:
9
TMz Modes (cont.)
X ( x)
Y ( y)
ez x, y ( A cos k x x B sin k x x)(C cos k y y D sin k y y )
Boundary Conditions:
A
C 0
and
B
A0
and
m
ez Bmn sin
a
ez 0
n
b
m
kx
a
ky
n
x sin
y
b
@ y 0, b
A
@ x 0, a
B
n 0,1,2,...
m 0,1,2,...
m n
kc2
a b
2
and
2
10
TMz Modes (cont.)
Therefore
m
Ez Bmn sin
a
n
x sin
b
y e
k z k 2 kc2
jk z z
m n
k
a
b
2
2
From the previous field-representation equations,
we can show
j c n
m
B
sin
mn
kc2b
a
n jk z z
x cos
y e
b
j c m
m n jk z z
Hy
B
cos
x sin
y e
mn
2
kc a
a b
Hx
jk z m
m
B
cos
mn
kc2 a
a
n
x sin
y e
b
jk n
m n
E y z2 Bmn sin
x cos
y e
kc b
a b
Ex
m=1,2,3,…
n =1,2,3,…
jk z z
Note: If either m or n is
zero, the field becomes
a trivial one in the TMz
case.
jk z z
11
TMz Modes (cont.)
Lossless Case c
mn k k
2
f cmn
mn 2
c
m n
k2
a b
2
2
m n
2 a b
1
2
(same as for
TE modes)
2
Lowest cutoff frequency is for the TM11 mode
2
1 1
f c11
2 a b
1
2
Dominant TM mode
(lowest fc)
12
Mode Chart
PEC
y
Two cases are considered:
b
, ,
o
b < a/2
z
x
o
a
a>b
Single mode operation
TE10
TE 20 TE 01
TE11
TM11
f
The maximum band for single
mode operation is 2 fc10.
ba/2
b > a/2
Single mode operation
TE10 TE 01 TE 20
f
TE11
TM11
f cmn
2
m n
2 a b
1
2
13
Dominant Mode: TE10 Mode
For this mode we have
m 1, n 0, kc
b
, ,
a
o
Hence we have
H z A10 cos
a
Hx j
Ey
PEC
y
xe
z
A10 sin
a
o
a
jk z z
kz a
j a
x
x e
A10 sin
a
E10
xe
2
kz k 2
a
jk z z
jk z z
E y E10 sin
a
A10
j a
xe
jk z z
E10
Ex Ez H y 0
14
Dispersion Diagram for TE10 Mode
Lossless Case c
f fc
vg slope
2
1
c10
kz k
a
2
g
2
(“Light line”)
v p slope
d
Group velocity: vg
d
Phase velocity: v p
Velocities are slopes on the
dispersion plot.
15
Field Plots for TE10 Mode
PEC
y
Top view
b
, ,
o
z
z
a
x
x
o
a
E
H
y
y
b
b
a
End view
x
z
Side view
16
Field Plots for TE10 Mode (cont.)
PEC
y
b
, ,
Top view
o
z
o
a
Js
x
a
z
x
H
y
y
b
b
a
End view
x
z
Side view
17
Power Flow for TE10 Mode
Time-average power flow in the z direction:
a b
1
*
P Re E H zˆ dydx
2 0 0
Note:
10
ab
2x
sin
dydx
2
a
0 0
a b
a b
1
Re E y H x* dydx
2 0 0
a 3 A10 b
Re k z
2
4
2
In terms of amplitude of the field amplitude, we have
ab
2
P
Re k z E10
4
10
A10
j a
E10
For a given maximum electric field level (e.g., the
breakdown field), the power is increased by
increasing the cross-sectional area (ab).
18
Attenuation for TE10 Mode
Pl (0)
Recall c
2 P0
Pl (0)
Rs
2
P0 P10
(calculated on previous slide)
PEC
y
2
Js d
b
C
J s nˆ H on conductor
, ,
o
z
Side walls
@ x 0 : J sside xˆ H
x 0
@ x a : J sside xˆ H
ˆ z yA
ˆ 10 e jkz z
yH
x a
x
o
C
a
H z A10 cos x e jk z z
a
ka
H x j z A10 sin x e jk z z
a
ˆ z yA
ˆ 10 e jkz z
yH
J syside A10 e jkz z
19
Attenuation for TE10 Mode (cont.)
Top and bottom walls
@ y 0 : J sbot yˆ H
y 0
@ y b : J stop yˆ H
b
, ,
y b
o
J stop J sbot
z
(since fields of this mode
are independent of y)
R
Pl (0) 2 s
2
b
b
J
side 2
s
0
R
dy s
2
a
a
0
2
0
2
k z2 a 3 a
2
Rs A10 b
2
2
2
2
x
C
o
a
H z A10 cos x e jk z z
a
ka
H x j z A10 sin x e jk z z
a
2
J stop dx
Rs J syside dy Rs J sxtop J sztop
0
PEC
y
dx
J sztop H x
A10 sin x e jkz z
a
A10 cos x e jk z z
a
J sztop j
J sxtop
J sxtop H z
kz a
20
Attenuation for TE10 Mode (cont.)
Assume f > fc
kz
(The wavenumber is taken as that
of a guide with perfect walls.)
PEC
y
b
Pl (0) Rs A10
2
, ,
2 a3 a
b
2
2
2
o
z
ab
2
P10
E
10
4
c
E10
x
o
C
a
j aA10
Simplify, using 2 k 2 kc2
Pl (0)
2 P10
kc10
a
Final result:
c
Rs
a 3b k
2
3 2
2
b
a
k [np/m]
21
Attenuation in dB/m
Let z = distance down the guide in meters.
c dB/m 20log10 e
c z
/ z
c z 20log10 (e) / z
Attenuation
[dB/m]
PEC
y
b
, ,
o
z
x
o
a
8.686 c
Hence
dB/m = 8.686 [np/m]
22
Attenuation for TE10 Mode (cont.)
Brass X-band air-filled waveguide
2.6 10
7
[S/m]
X band : 8 12 [GHz]
(See the table on the next slide.)
23
Attenuation for TE10 Mode (cont.)
Microwave Frequency Bands
Letter Designation
Frequency range
L band
1 to 2 GHz
S band
2 to 4 GHz
C band
4 to 8 GHz
X band
8 to 12 GHz
Ku band
12 to 18 GHz
K band
18 to 26.5 GHz
Ka band
26.5 to 40 GHz
Q band
33 to 50 GHz
U band
40 to 60 GHz
V band
50 to 75 GHz
E band
60 to 90 GHz
W band
75 to 110 GHz
F band
90 to 140 GHz
D band
110 to 170 GHz
(from Wikipedia)
24
Modes in an X-Band Waveguide
a 2.29cm (0.90")
b 1.02cm (0.40")
Mode
TE10
fc [GHz]
TE20
13.10
TE01
14.71
Standard X-band waveguide (WR90)
X band : 8 12 [GHz]
6.55
TE11
16.10
TM 11
16.10
TE30
19.65
TE21
19.69
TM 21
19.69
1"
b
a
0.5"
50 mil (0.05”) thickness
25
Example: X-Band Waveguide
Determine and g at 10 GHz and
6 GHz for the TE10 mode in an airfilled waveguide.
a = 2.29cm
0 , 0
b = 1.02cm
@ 10 GHz
2
2 10
2
8
3
10
0.0229
a
2
10
2
158.25 [rad/m]
2
2
g
0.0397
158.25
g 3.97 [cm]
26
Example: X-Band Waveguide (cont.)
@ 6 GHz
2
2 6 10
2
k z
8
3
10
0.0
22
9
a
j 55.04 [1/m]
2
9
2
55.04 [np/m]
g
2
478.08 [dB/m]
Evanescent mode: = 0; g is not defined!
27
Circular Waveguide
TMz mode:
a
2 Ez 0 , kc2 Ez 0 , 0
kz2 k 2 kc2
z
The solution in cylindrical coordinates is:
J n (kc ) sin(n )
Ez 0 ,
Yn (kc ) cos(n )
Note: The value n must be an integer to have unique fields.
28
Plot of Bessel Functions
1
1
0.8
J n (0) is finite
n=0
0.6
n=1
0.4
Jn (x)
n=2
J 0( x)
J 1( x)
0.2
J n( 2 x)
0
0.2
0.4
0.403 0.6
0
1
2
3
0
J n ( x) ~
2
n
cos x
,
x
2 4
4
5
x
x
x
6
7
8
9
10
10
1
J n ( x) ~ x n n n 0,1, 2,...., x 0
2 n!
29
Plot of Bessel Functions (cont.)
0.521
1
n=0
n=1
0
1
n=2
Yn (0) is infinite
2
Yn (x)
Y0( x)
Y1( x)
3
Yn( 2 x)
4
5
6
6.206
7
0
0
Yn ( x) ~
1
2
3
4
5
x
2
n
sin x
, x
x
2 4
6
7
8
9
10
x
10
Y0 ( x) ~
2 x
ln , 0.5772156, x 0
2
n
2
Yn ( x) ~ (n 1)! , n 1, 2,3,....., x 0
x
1
30
Circular Waveguide (cont.)
Choose (somewhat arbitrarily)
cos(n )
J n ( kc )
Ez , , z
cos(n ) e
Yn (kc )
jk z z
The field should be finite on the z axis
Yn (kc ) is not allowed
Ez ,, z cos(n ) J n (kc ) e
jkz z
31
Circular Waveguide (cont.)
B.C.’s:
Jn(x)
Ez a,, z 0
Hence
J n (kc a) 0
Sketch for a typical value of n (n 0).
Note: Pozar uses
the notation pmn.
xn3
xn1
kc a xnp
Note: The value xn0 = 0 is not included
since this would yield a trivial solution:
x
xn2
kc
xnp
a
J n xn 0 J n 0 0
a
This is true unless n = 0,
in which case we cannot
have p = 0.
32
Circular Waveguide (cont.)
TMnp mode:
Ez , , z cos(n ) J n xnp e
a
xnp
2
kz k
a
jk z z
n 0,1, 2
2
p 1, 2,3,.........
33
Cutoff Frequency: TMz
kz2 k 2 kc2
At f = fc :
k kc
kz 0
2 f c
fc
TM
xnp
a
xnp
a
cd
xnp
2 a
cd
c
r
34
Cutoff Frequency: TMz (cont.)
xnp values
p\n
0
1
2
3
4
5
1
2.405
3.832
5.136
6.380
7.588
8.771
2
5.520
7.016
8.417
9.761
11.065 12.339
3
8.654
10.173 11.620 13.015 14.372
4
11.792 13.324 14.796
TM01, TM11, TM21, TM02, ……..
35
TEz Modes
Proceeding as before, we now have that
H z ,, z cos(n ) J n (kc ) e
Set
jkz z
E a,, z 0
E
j H z
kc2
(From Ampere’s law)
H z
0
a
Hence
J n (kc a) 0
36
TEz Modes (cont.)
J n (kc a) 0
Jn' (x)
Sketch for a typical value of n (n 1).
x'n3
x'n1
x
x'n2
kc a xnp
kc
xnp
a
p 1, 2,3,.....
We don’t need to consider
p = 0; this is explained on
the next slide.
37
TEz Modes (cont.)
H z , , z cos(n ) J n xnp e
a
Note: If p = 0
We then have, for p = 0:
jk z z
0
xnp
n0
J n xnp
J n 0 0
a
n0
J 0 xnp
J 0 0 1
a
H zˆ e
jkz z
p 1, 2,
zˆ e
jkz
(trivial solution)
(nonphysical solution)
(The TE00 mode is not physical.)
38
Cutoff Frequency: TEz
kz2 k 2 kc2
kz 0
kc k
2 f c
Hence
fc
TE
xnp
a
xnp
a
cd
xnp
2 a
cd
c
r
39
Cutoff Frequency: TEz
x´np values
p\n
0
1
2
3
4
5
1
3.832
1.841
3.054
4.201
5.317
5.416
2
7.016
5.331
6.706
8.015
9.282
10.520
3
10.173 8.536
9.969
11.346 12.682 13.987
4
13.324 11.706 13.170
TE11, TE21, TE01, TE31, ……..
40
TE11 Mode
The dominant mode of circular waveguide is the TE11 mode.
Electric field
Magnetic field
(From Wikipedia)
TE10 mode of
rectangular waveguide
TE11 mode of
circular waveguide
The mode can be thought of as an evolution of the TE10 mode of
rectangular waveguide as the boundary changes shape.
41
TE01 Mode
The TE01 mode has the unusual property that the conductor attenuation
decreases with frequency. (With most waveguide modes, the conductor
attenuation increases with frequency.)
The TE01 mode was studied extensively as a candidate for longrange communications – but eventually fiber-optic cables became
available with even lower loss. It is still useful for some high-power
applications.
42
TE01 Mode (cont.)
c
TE11
TM01
TE21
Pl (0)
c
2 P0
TM11
P0 = 0 at cutoff
TE01
fc, TE11
fc, TM01
fc, TE21 fc, TE01
f
43
TE01 Mode (cont.)
Practical Note:
The TE01 mode has only an azimuthal ( -directed) surface current on the
wall of the waveguide. Therefore, it can be supported by a set of
conducting rings, while the lower modes (TE11 ,TM01, TE21, TM11) will not
propagate on such a structure.
(A helical spring will also work fine.)
44
TE01 Mode (cont.)
From the beginning, the most obvious application of waveguides had been as a
communications medium. It had been determined by both Schelkunoff and Mead,
independently, in July 1933, that an axially symmetric electric wave (TE01) in circular
waveguide would have an attenuation factor that decreased with increasing frequency
[44]. This unique characteristic was believed to offer a great potential for wide-band,
multichannel systems, and for many years to come the development of such a system
was a major focus of work within the waveguide group at BTL. It is important to note,
however, that the use of waveguide as a long transmission line never did prove to be
practical, and Southworth eventually began to realize that the role of waveguide would
be somewhat different than originally expected. In a memorandum dated October 23,
1939, he concluded that microwave radio with highly directive antennas was to be
preferred to long transmission lines. "Thus," he wrote, “we come to the conclusion that
the hollow, cylindrical conductor is to be valued primarily as a new circuit element, but
not yet as a new type of toll cable” [45]. It was as a circuit element in military radar that
waveguide technology was to find its first major application and to receive an enormous
stimulus to both practical and theoretical advance.
K. S. Packard, “The Origins of Waveguide: A Case of Multiple Rediscovery,”
IEEE Trans. MTT, pp. 961-969, Sept. 1984.
45