Transverse resonance method
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Transcript Transverse resonance method
ECE 5317-6351
Microwave Engineering
Fall 2011
Fall 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
Z 0 , 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
Z 0 , 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
L 2
0
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
Z 0 X L 2 X L1
X L1 X L 2 Z 02
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 , kz
z
kz k
x
b
We choose a reference
plane at z = 0+.
mn
z
Z 0 ZTE
,TM
m,n
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 , kz
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:
m 0,1, 2,
or
n 0,1, 2,
f mnp
c
2
1
r r
m n p
a b h
2
The lowest mode is the TE101 mode.
2
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
Q
R
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
k x1
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:
Z 01TM
The reference plane is
chosen at the interface.
Z 00TM
x
Z 01TM
k x1
1
Z 00TM
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
jZ 01TM tan(k x1h) Z 00TM
j
k x1
1
tan(k x1h)
kx0
0
k x1
r j tan(k x1h)
kx0
22
TMx Surface-Wave Solution (cont.)
Letting
kx 0 j x 0 ,
We have
x 0 kz 2 k02
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