E
Electrodynamics
H
Metal Waveguides
Dr. rer. nat. Muldarisnur
JURUSAN FISIKA UNIVERSITAS ANDALAS
Metal Waveguide Components
Rectangular waveguide
Waveguide to coax adapter
Waveguide bends
E-tee
Electrodynamics
Dr. rer. nat. Muldarisnur
2
More Waveguides
Electrodynamics
Dr. rer. nat. Muldarisnur
3
Rectangular Waveguides
 Need to find the fields
components of the EM wave
inside the waveguide
Ex , E y , Ez , H x , H y , H z
 We’ll find that waveguides don’t
support TEM waves
http://www.ee.surrey.ac.uk/Personal/D.Jefferies/wguide.html
Electrodynamics
Dr. rer. nat. Muldarisnur
4
Rectangular waveguides: Field Distribution
Then applying on the z-component…
Using phasors & assuming waveguide filled with
‒ lossless dielectric material and
2 E  k 2 E  0
‒ walls of perfect conductor
2 H  k 2 H  0
where
the wave inside should obey…
 2 Ez  k 2 Ez  0
k 2   2  c
 2 Ez  2 Ez  2 Ez
2



k
Ez  0
2
2
2
x
y
z
Solving by method of Separation of Variables:
Ez ( x, y, z )  X ( x)Y ( y ) Z ( z )
Electrodynamics
X '' Y '' Z ''
 
 k 2
X
Y
Z
Dr. rer. nat. Muldarisnur
5
Rectangular waveguides: Field Distribution
X '' Y '' Z ''
 
 k 2
X
Y
Z
2
2
2
2
k x  k y  k z  k
k k k k
2
z
2
2
x
2
y
Which results in the expressions:
X ''  k x2 X  0
X(x)  c1 cos k x x  c2 sin k x x
Y ''  k y2Y  0
Y(y)  c3 cos k y y  c4 sin k y y
Z k Z 0
Z ( z )  c5e  c6e
''
Electrodynamics
2
z
kz z
Dr. rer. nat. Muldarisnur
 kz z
6
Rectangular waveguides: Field Distribution
X(x)  c1 cos k x x  c2 sin k x x
E z ( x, y, z )  X ( x)Y ( y ) Z ( z )
Y(y)  c3 cos k y y  c4 sin k y y
Z ( z )  c5e k z  c6e  k z
z
z
Ez   c1 cos k x x  c2 sin k x x   c3 cos k y y  c4 sin k y y   c5e k z  c6e  k z 
z
z
If only looking at the wave traveling in  z -direction:
Ez   A1 cos k x x  A2 sin k x x   A3 cos k y y  A4 sin k y y  e  k z
z
Similarly for the magnetic field,
H z   B1 cos k x x  B2 sin k x x   B3 cos k y y  B4 sin k y y  e  k z
z
Electrodynamics
Dr. rer. nat. Muldarisnur
7
Rectangular waveguides: Field Distribution
Other components (x,y)
From Faraday and Ampere Laws we can find the remaining four
components:
i  H z k z Ez 
Ex  2

2 
k  k z  y  x 
i  H z k z Ez 
Ey   2

2 
k  k z  x  y 
i  k z H z Ez 
Hx  2

2 
k  k z   x
y 
i  k z H z Ez 
Hy  2

2 
k  k z   y
x 
So once we know Ez and Hz, we can find all the other fields.
Electrodynamics
Dr. rer. nat. Muldarisnur
8
Propagation Modes
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
Electrodynamics
Dr. rer. nat. Muldarisnur
9
Ez
TM Mode
  A cos k x  A sin k x   A cos k
1
x
2
x
3
Boundary conditions:
k z
y

A
sin
k
y
e
y
4
y 
z
E z  0 at y  0,b
E z  0 at x  0,a
From these, we conclude:
X ( x) is in the form of sin k x x
Where k  mp m  1, 2, 3,...
x
a
Y ( y ) is in the form of sin k y y
where ky=np/b, n=1,2,3,…
So the solution for Ez(x,y,z) is
Ez  A2 A4  sin k x x   sin k y y  e  k z
z
Electrodynamics
Dr. rer. nat. Muldarisnur
10
TM Mode : TMmn
 mp   np   k z
Ez  Eo sin 
x  sin 
y e
 a   b 
Hz  0
z
Other components are

ik z mp
 mp   np   k z

E
cos
x  sin 
y e

 k2  k2 a o
 a   b 

z
i  k z Ez 
ik z np
 mp 
 np   k z
Ey   2
 2
Eo sin 
x  cos 
y e

2 
2
k  k z   y 
k  kz b
 a 
 b 
Ex 
i  k z Ez
k 2  k z2   x
z
z
i  Ez
Hx  2
k  k z2  y

i
  k2  k2

z
i  Ez 
i
Hy  2
 2
2 
k  k z  x  k  k z2
Electrodynamics
np
 mp 
 np   k z
Eo sin 
x  cos 
y e
b
 a 
 b 
z
mp
 mp   np   k z
Eo cos 
x  sin 
y e
a
 a   b 
z
Dr. rer. nat. Muldarisnur
11
TM Mode : TMmn
 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
Electrodynamics
Dr. rer. nat. Muldarisnur
12
TM Mode: Cut-off
kz 
k
2
x
 k y2   k 2
 mp   np 
2
 
 
   
 a   b 
2
2
 The cutoff frequency occurs when
 mp   np 
When c 2  
  
 a   b 
2
2
1 1  mp   np 
or f c 

  
2p   a   b 
2
then k z    i  0
2
mp   np 
 Evanescent: When 2  
  
 a   b 
2
2
k z   and   0
Means no propagation, everything is attenuated
mp   np 
 Propagation: When 2  
    k z  i and   0
 a   b 
This is the case we are interested since is when the wave is allowed to
travel through the guide.
2
Electrodynamics
2
Dr. rer. nat. Muldarisnur
13
TM Mode: Cut-off
Propagation
attenuation
of mode mn
fc,mn
 The cut-off frequency is the frequency below which attenuation
occurs and above which propagation takes place. (High Pass)
2
f c , mn 
1 1 m n
   
2p   a   b 
2
 The phase constant becomes
 fc 
 mp   np 
2
     
 
   ' 1   
 a   b 
 f 
2
Electrodynamics
2
Dr. rer. nat. Muldarisnur
2
14
Field Distribution
 mpx  i t z 
Ez  Asin 
e
 a 
Ez
m=1
0
a x
m=2
m=3
z
Electrodynamics
a x
Dr. rer. nat. Muldarisnur
15
TE Mode
H z   B1 cos k x x  B2 sin k x x   B3 cos k y y  B4 sin k y y  e  kz z
E x  0 at y  0,b
Boundary conditions:
E y  0 at x  0,a
From these, we conclude:
X ( x) is in the form of cos k x x
mp
where k x 
m  0, 1, 2, 3,...
a
Y ( y ) is in the form of cos k y y
where k y  np m  0, 1, 2, 3,...
b
So the solution for Ez(x,y,z) is
H z  B1 B3  cos k x x   cos k y y  e  k z
z
Electrodynamics
Dr. rer. nat. Muldarisnur
16
TE Mode (Temn) : Field Distribution
 mp   np   k z
H z  H o cos 
x  cos 
y e
 a   b 
Ez  0
z
Other components are
i  H z 
i np
 mp   np   k z
Ex  2


H
cos
x  sin 
y e
o


2 
2
2
k  k z  y 
k  kz b
 a   b 
z
i  H z
Ey   2

k  k z2  x
i mp

 mp 
 np   k z
H o sin 
x  cos 
y e
 2
2
 k  kz a
 a 
 b 
i  k z H z 
ik z mp
 mp 
 np   k z
Hx  2


H
sin
x
cos
y e
o




2 
2
2
k  k z   x 
k  kz a
 a 
 b 
z
z
i  k z H z
Hy  2
k  k z2   y
Electrodynamics

ik z np
 mp   np   k z
  k 2  k 2 b H o cos  a x  sin  b y  e

 


z
z
Dr. rer. nat. Muldarisnur
17
TE Mode: Cut-off
Propagation
attenuation
of mode mn
fc,mn
 The cutoff frequency is the same expression as for the TM
mode
2
f c , mn
1 1 m n

   
2p   a   b 
2
 But the lowest attainable frequencies are lower because here
n or m can be zero.
Electrodynamics
Dr. rer. nat. Muldarisnur
18
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.
Electrodynamics
Dr. rer. nat. Muldarisnur
19
Waveguide Cavities
 Cavities, or resonators, are used for storing energy
 It’s equivalent to a RLC circuit at high frequency
 Their shape is that of a cavity, either cylindrical or cubical.
Electrodynamics
Dr. rer. nat. Muldarisnur
20
Cavity Mode to z
Solving by Separation of Variables:
Ez ( x, y, z )  X ( x)Y ( y ) Z ( z )
TM
TE
X(x)  c1 cos k x x  c2 sin k x x X(x)  c1 cos k x x  c2 sin k x x
Y(y)  c3 cos k y y  c4 sin k y y Y(y)  c3 cos k y y  c4 sin k y y
Z ( z )  c5 cos k z z  c6 sin k z z Z ( z )  c5 cos k z z  c6 sin k z z
k 2  k x2  k y2  k z 2
Electrodynamics
k  k  k  kz
2
2
x
Dr. rer. nat. Muldarisnur
2
y
2
21
TMmnp Boundary Conditions
From these, we conclude:
E z  0 at y  0 ,b
E z  0 at x  0,a
E y  E x  0, at z  0 ,c
kx=mp/a
ky=np/b
kz=pp/c
where c is the dimension in z-axis
 mpx   npy   ppz 
E z  Eo sin 
 sin 
 sin 

a
b
c

 
 

where
c
Electrodynamics
 mp   np   pp 
2
k2  
 
 
   
 a   b   c 
2
Dr. rer. nat. Muldarisnur
2
2
22
TEmnp Boundary Conditions
H z  0 at z  0 ,c
E y  0 at x  0 ,a
E x  0, at y  0,b
From these, we conclude:
mp
np
pp
ky 
kz 
a
b
c
m  0,1,2,3,... n  0,1,2,3,... p  0,1,2,3,...
kx 
where c is the dimension in z-axis
 mpx   npy   ppy 
H z  H o cos
 cos
 sin 

a
b
c

 
 

c
where
 mp   np   pp 
2
k2  
 
 
   
 a   b   c 
2
Electrodynamics
Dr. rer. nat. Muldarisnur
2
2
23
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.
2
2
1 1 m n  p
fr 
     
2p   a   b   c 
Electrodynamics
2
Dr. rer. nat. Muldarisnur
24
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.
Time average energy stored
Q  2π
loss energy per cycle of oscillation
W
 2p
PL
Electrodynamics
Dr. rer. nat. Muldarisnur
25