ECE 5317-6351
Microwave Engineering
Fall
Fall2011
2011
Prof. David R. Jackson
Dept. of ECE
Notes 13
Transverse Resonance
Method
1
Transverse Resonance Method
This is a general method that can be used to help us calculate
various important quantities:
 Wavenumbers for complicated waveguiding structures (dielectricloaded waveguides, surface waves, etc.)
 Resonance frequencies of resonant cavities
We do this by deriving a “Transverse Resonance Equation (TRE).”
2
Transverse Resonance Equation (TRE)
To illustrate the method, consider a lossless resonator formed by a
transmission line with reactive loads at the ends.
R
Z L1  jX L1
Z0
x
Z0
x = x0
Z L 2  jX L 2
x=L
R = reference plane at arbitrary x = x0
We wish to find the resonance frequency of this transmission-line resonator.
3
TRE (cont.)
R
Z L1  jX L1
Z0
Z0
x
x = x0
Z L 2  jX L 2
x=L
Examine the voltages and currents at the reference plane:
Il
R
Ir
+
Vr
-
+
Vl
x = x0
4
TRE (cont.)
Il
R
Ir
+
Vr
-
+
Vl
-
Z in
Define impedances:

r
V
Z in  r
I

Vl
Z in  l
I
x

x = x0
Z in
Boundary conditions:
V V
r
l
Hence:


Z in   Z in
Ir  Il
5
TRE (cont.)
R

Z in
Z in
TRE




Z in   Z in
or
Y in   Y in
Note about the reference
plane: Although the
location of the reference
plane is arbitrary, a
“good” choice will keep
the algebra to a
minimum.
6
Example
Derive a transcendental equation for the resonance frequency of this
transmission-line resonator.
L
Z0 , k  k0  r
Z L1  jX L1
Z L 2  jX L 2
x
We choose a reference plane at x = 0+.
7
Example (cont.)
L
Z L1  jX L1
R
Z0 , k  k0  r
Z L 2  jX L 2
x
Apply TRE:


Z in   Z in
  Z L 2  jZ 0 tan   L   
 Z L1    Z 0 
 
  Z 0  jZ L 2 tan   L   
8
Example (cont.)
  Z L 2  jZ 0 tan   L   
Z L1    Z 0 
 
  Z 0  jZ L 2 tan   L   

  jX L 2  jZ 0 tan   L   
jX L1    Z 0 
 
Z

j
jX
tan

L




  0
L2
 



  jX  jZ tan k L 
L2
0
0
r
jX L1    Z 0 
  Z  j  jX  tan k L 
L2
0
r
  0









jX L1 Z0  j  jX L 2  tan k0 L  r
  Z  jX
0
L2

 jZ0 tan k0 L  r

9
Example (cont.)
After simplifying, we have

tan k0 L  r

Z0  X L 2  X L1 

X L1 X L 2  Z02
Special cases:
X L1  X L 2  0
X L1  0, X L 2  
 k0 L  r  n
 k0 L  r   2n  1  / 2
10
Rectangular Resonator
Derive a transcendental equation for the resonance frequency of a
rectangular resonator.
z
PEC boundary
Orient so that
b < a <h
h
 r , r
y
a
x
b
The structure is thought of as supporting RWG modes bouncing back
and forth in the z direction.
We have TMmnp and TEmnp modes.
The index p describes the
variation in the z direction.
11
Rectangular Resonator (cont.)
z
We use a Transverse Equivalent Network (TEN):
PEC boundary
h
h
 r , r
y
a
Z0 , k z
z
kz  k
x
b
We choose a reference
plane at z = 0+.
mn
z
 m, n 
Z0  ZTE
,TM


Z in   Z in

Z in  0 (PEC bottom)

Hence
Z in  0
12
Rectangular Resonator (cont.)
Hence
z

Z in  jZ0 tan   h   0

PEC boundary
h

y

 m,n 
 m, n 
jZTE
tan
k
h 0
,TM
z


 r , r
a
x
b

tan k z m,n  h  0
h

kz
m,n
h  p , p  1,2
Z0 , k z

 m   n 
2
h k 
 
  p
 a   b 
2
2
z
13
Rectangular Resonator (cont.)
z
Solving for the wavenumber we have
 m   n   p 
k 
 
 

 a   b   h 
2
2
PEC boundary
2
 r , r
h
y
a
x
b
Hence
 m   n   p 
r  r  
 
 

 a   b   h 
2
2 f mnp 0 0
2
Note: The TMz and TEz
modes have the same
resonance frequency.
2
TEmnp mode:
or
f mnp
c

2
1
r  r
 m   n   p 

 
 

 a   b   h 
2
The lowest mode is the TE101 mode.
2
2
m  0,1, 2,
n  0,1, 2,
p  1, 2,
 m, n    0, 0 
c  2.99792458 108 [m/s]
14
Rectangular Resonator (cont.)
z
TE101 mode:
c 1
f101 
2 r  r
2
1 1
   
a h
PEC boundary
2
h
 r , r
y
a
x
b
x  z 
H z  x, y, z   H 0 cos 
 sin  
 a   h 
Note: The sin is used to ensure the boundary condition on the PEC top and bottom plates:
Hn  H z  0
The other field components, Ey and Hx, can be found from Hz.
15
Rectangular Resonator (cont.)
z
Practical excitation
by a coaxial probe
PEC boundary
h
 r , r
y
a
x
b
Lp (Probe inductance)
R
L C
Circuit model
Tank (RLC) circuit
16
Rectangular Resonator (cont.)
Z RLC
R



1  j 2 Q   1
 0 
Q  0
Q = quality factor of resonator
0 
U
Pdave
U  U E  U H  energy stored
1
LC
Pdave  average power dissipated
Lp (Probe inductance)
R
L C
R
Q
0 L
Circuit model
Tank (RLC) circuit
17
Rectangular Resonator (cont.)
Z RLC 
R


1  j 2 Q   1
 0 
Z RLC
RRLC
f
f0
X RLC
18
Grounded Dielectric Slab
Derive a transcendental equation for wavenumber of the TMx surface
waves by using the TRE.
x
h
 r , r
z
Assumption: There is no variation of the fields in the y direction,
and propagation is along the z direction.
19
Grounded Dielectric Slab
x
TMx
H
 r , r
E
z
Z
Z TM 
TM
01

kx1
1
Z
TM
00

kx0
 0
Ez
(defined for a wave going in the  x direction)
H y
20
TMx Surface-Wave Solution
R
h
TEN:
Z01TM
Z
The reference plane is
chosen at the interface.
TM
00
x
Z01TM 
kx1
Z00TM 
1
kx0
 0
k x1  k12  k z 2
1
2 2
k x 0  (k0  k z )   j k z 2  k0 2   j x 0
2
Z in  jZ01TM tan(kx1h)
Z in  Z00TM
21
TMx Surface-Wave Solution (cont.)
TRE:
Z in  Z in

jZ01TM tan(kx1h)   Z00TM

j
k x1
1
tan(k x1h)  
kx0
 0

 k x1 
 r   j   tan(k x1h)
 kx0 
22
TMx Surface-Wave Solution (cont.)
Letting
k x 0   j x 0 ,
We have
 x 0  k z 2  k0 2
 k x1 
r  
 tan(k x1h)
  x0 
or


 r k z 2  k0 2  k12  k z 2 tan  h k12  k z 2 


Note: This method was a lot simpler than doing the EM analysis and
applying the boundary conditions!
23