TEM RESONATORS
0
Voltages and Currents:
z
i
v+
-
Zo, c
D
v(t,z) = V+ cos(ωt - kz) + V- cos(ωt + kz)
⇒ V- = -V+
Boundary conditions: v(t,z) = 0 at z = 0
v(t,z) =
V [cos(ωt - kz) - cos(ωt + kz)] = 2V sin ωt sin kz *
+
⎛ V+ ⎞ +
⎛ V+ ⎞
i(t,z) = ⎜ Z ⎟ [cos(ωt - kz) + cos(ωt + kz)] = 2 ⎜ ⎟ cos ωt cos kz *
⎝ o⎠
⎝ Zo ⎠
Resonant Frequencies ωn:
⇒
2V+ sin(ωt) sin(kD) = 0
knD = nπ,
i(t,z)
n = 0, 1, 2,…
knD = (ωn/c)D = nπ
⇒
ωn = nπc/D = 2πc/λn
⇒ nλn/2 = D
v(z,t)
ω1
0
* cos α - cos β = -2 sin[(α + β)/2] sin [(α - β)/2]
cos α + cos β = 2 cos[(α + β)/2] cos [(α - β)/2]
D
z
L15-1
MORE TEM RESONANCES
Resonant Frequencies ωn:
ωn = nπc
d
0
nλ n
=d
2
DC Resonance (ω = 0):
Current flows in loop,
Voltages are zero
we = 0, wm ≠ 0 [J]
z
i
v+
-
Zo, c
i(t,z)
v(z,t)
d
ω2
0
d
λn = d
i
Open-Circuit Resonator:
nλn/2 = d,
λn = 2d/n
λ0 = ∞
λ1 = 2d
λ2 = d
λ3 = 2d/3
n = 0,1,2,… ⇒
ωn = nπc/d = 2πc/λn
→ ωo = 0
→ ω1 = πc/d
→ ω2 = 2πc/d
→ ω3 = 3πc/d
z
v+
-
Zo, c
d
v(z,t)
i(t,z)
0
ω2
z
λn = d
L15-2
ANOTHER TEM RESONANCE
Quarter-Wave Resonator:
i
D = λ/4, 3λ/4, 5λ/4,… ⇒
D
D = (λn/4)(2n - 1) [n = 1,2,3,…]
λn = 4D/(2n - 1)
i(t,z)
λ1 = 4D
λ2 = 4D/3
v+
-
Zo, c
ω2
0
v(z,t)
ω1
z
λ3 = 4D/5
λ4 = 4D/7
L15-3
ENERGY IN TEM RESONATORS
Electric Energy Density We = Cv2/2 [J/m]:
v(t,z) = Vo sin ωnt sin knz
We(t,z) = 1 CV02 (sin2 ωnt )( sin2 knz) [J/m]
2
= 1 CVo2 (1 − cos 2ωnt)(1 − cos 2knz) ≥ 0
2
i(t,z)
v(z,t)
ω1
0
z
D
Magnetic Energy Density Wm = Li2/2 :
Vo
=
i(t,z) Z cos ωnt cosknz
o
2
⎛ Vo ⎞
1
2
2
Wm(t,z) = 2 L ⎜ Z ⎟ cos ωnt cos knz [J/m]
⎝ o⎠
2
V
⎛
⎞
= 1 L ⎜ o ⎟ (1 + cos 2ωnt)(1 + cos 2knz)
2 ⎝ Zo ⎠
We(z,t)
2ω1
0
Wm(t,z)
z
D
Total Electric and Magnetic Energies
we and wm:
2
⎛V ⎞
We = 1 CVo2 (1 − cos 2knz)
Wm = 1 L ⎜ o ⎟ (1 + cos 2knz)
4
4 ⎝ Zo ⎠
2
V
⎛
⎞
Over n λ we have w e = w m because CVo2 = L ⎜ o ⎟ (recall Z o2 = L )
4
C
⎝ Zo ⎠
* < • > signifies time average of •.
L15-4
TEM RESONATOR LOSS AND Q
Distributed Losses:
RΔz LΔz
Pdiss (t,z) = i R + v G [watts / m]
2
v = V0cos ωt
2
i
GΔz
R (ohms m-1)
G (Siemens m-1)
CΔz
Pd [W] = ∫0 Pdiss dz
Δz
2
D
= ∫0 ( Vo Zo ) cos2ωn t ( cos2k n z ) Rdz
D
+ ∫0 Vo 2 sin2ωn t ( sin2k n z ) Gdz
D
GDVo
2
Pd = RD ( Vo Zo ) +
4
4
(for
)
d
2
[watts]
Example of Q Calculation (n =1):
ωo w T πcCZo2
C(L/C) πZo
QI = QL =
=
=π 1
=
Pd
RD
RD
RD
LC
ωo (n = 1)
wT = 2 we
Pd
= πc
D
= 1 CVo2D
4
2
V
⎛
⎞
= RD ⎜ o ⎟
4 ⎝ Zo ⎠
v(z)
R≠0
v +-
Zo,c
D
L15-5
LUMPED LOSSES IN TEM RESONATORS
Calculation of Q:
ωo w T
Q=
Pd
Iz=0
Perturbation Technique:
Find Pd using unperturbed v or i
Need (Δv/vo) << 1, (Δi/io) << 1, Γ ≅ Γo
Loss Computation Examples:
a Pd ≅ 1 | I z=0 |2 R a
2
b Pd ≅ 1 |Vz=d/2 |2
2Rb
V(z)
δ
i
d
a c
b
v+0
D
z=D
I = (Vo/Zo)cos(πz/D)
δ Z ,c
o
Use unperturbed V, I
1 |V sinkδ|2
2R c o
d Pd ≅ 1 |(Vo /Z o )sinkδ|2 Rd
2
c Pd ≅
When is a perturbation too large?
a Want Γz=0 ≅ -1 ≅ Zn − 1
Zn + 1
b Want Γ(z=D/2) ≅ 1
c Want Rc >> |-jZosin kδ|
⇒ Ra << Zo (Zan<< 1)
⇒ Rb >> Zo (Zbn >> 1)
⇒ Rc > ~Zo (if 2πδ/λ << 1)
L15-6
COUPLED TEM RESONATORS
Example: computation of Q:
Qint,ext,or loaded =
ωo = πc
D
Vo
δ
ωo w T
i
Zo,c v +D
R ohms m-1
Pd:int,ext,or loaded
C|Vo |2
w T = 2<w e > = 2D
8
|V(z)|
Zo,c
2
Pd;ext
D )2
Vo sin( πδ )
(
D
=
⇒ Qext ≅ δ
2Zo
2π
V
Pd;int = RD o
4 Zo
2
⇒ Qint =
πZo
RD
Example, perfect coupling:
Qint = Q ext ⇒ δ =
3
RD
2π 2 Z o
Re q
ωo 2ωo
R
Δω =
=
=2 =
[r/s]
QL
QI
L Leq
+ R
Th
- VTh
Ceq +
Req
Pd(ω)
Δω
Leq
1
1/2
ω
ωo
Resonance
L15-7
COUPLED TEM RESONATORS
Resonator equivalent circuit
ω
2ωo Re q
[r/s]
Δω = o =
=
QL
QI
Leq
1
ωo = πc =
D
LeqCeq
Zn (ω = ωo ) =
⇒ Γ at ωo
Re q Pd;ext QI
=
=
R Th Pd;int QE
3
Eqn
3
Unk
δ
Zo,c
- VTh Req
Imperfect coupling if QE ≠ QI
Lossless bandpass filter
If both lines are Zo and δ1 = δ2,
Then QI = QE ⇒ perfect coupling
and perfect match at ωo
i
Zo,c v +D
R ohms m-1
+ R
Th
Req -R Th
(Γ =
)
Req +R Th
Zo,c
|V(z)|
Ceq +
Leq
-
|V(z)|
δ
i δ
Zo,c v +D
Lossless
bandpass filter
Zo,c
L15-8
MIT OpenCourseWare
http://ocw.mit.edu
6.013 Electromagnetics and Applications
Spring 2009
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