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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013 - Electromagnetics and Applications
Fall 2005
Lecture 15 - Dielectric Waveguides
Prof. Markus Zahn
November 3, 2005
I. TM Solutions
We are considering solutions with
∂
=0
∂y
⎧ �
�
−α(x−d) ej(ωt−kz z)
⎪
x ≥ d
⎨
Re
�
A2 e
�
j(ωt−k
z)
z
Ez (x, z, t) =
Re
(A1 sin(kx x) + B1 cos(kx x)) e
|x| ≤ d
⎪
⎩
�
+α(x+d) j(ωt−kz z) �
x ≤ −d
Re A3 e
e
kx2 + kz2 = ω 2 �µ
(in dielectric)
−α2 + kz2 = ω 2 �0 µ0
(in free space)
For propagation in the dielectric and evanescence in free space
kz2 = ω 2 �µ − kx2 = ω 2 �0 µ0 + α2
kz2 < ω 2 �µ, kz2 > ω 2 �0 µ0 ⇒ ω 2 �0 µ0 < kz2 < ω 2 �µ
1
A. Odd Solutions [Ez (x, z, t) = −Ez (−x, z, t)]
⎧
−α(x−d)
⎪
⎨A2 e
Êz (x) = A1 sin(kx x)
⎪
⎩
−A2 eα(x+d)
x≥d
|x| ≤ d
x ≤ −d
¯ ⇒ ∂Êx − jkz Êz = 0
�·E
∂x
⎧ jk
−α(x−d) x > d
z
⎪
⎨− α A2 e
z
⇒
Êx =
−
jk
kx A1 cos(kx x) |x| < d
⎪
⎩
jkz
−
α A2 eα(x+d)
x < −d
¯
¯ = −µ ∂ H ⇒ Ĥx = 0, Ĥz = 0
�×E
∂t
�
�
1
∂Êz
Ĥy = −
−jkz Êx −
jωµ
∂x
⎧ jω�
−α(x−d) x ≥ d
0
⎪
⎨− α A2 e
=
− jω�
kx A1 cos(kx x) |x| ≤ d
⎪
⎩
jω�
−
α 0 A2 eα(x+d)
x ≤ −d
Boundary Conditions
Ez (x = d+ ) = Ez (x = d− ) ⇒ A1 sin(kx d) = A2
�0
�
jω�
jω�
Hy (x = d+ ) = Hy (x = d− ) ⇒
+
� A2 = +� A1 cos(kx d)
kx
α
A1
1
kx �0
�0 kx
⇒α=
=
=
tan(kx d)
A2
sin(kx d)
α� cos(kx d)
�
Critical condition for a guided wave occurs when α = 0. At this point
kx d = nπ, kz2 = ω 2 �0 µ0
� nπ �2
d
+ ω 2 �0 µ0 = ω 2 �µ ⇒ ω 2 =
(nπ/d)2
, n = 1, 2, 3, . . .
�µ − �0 µ0
For real frequencies (ω 2 > 0), �µ > �0 µ0
B. Even Solutions [Ez (x, z, t) = +Ez (−x, z, t)]
⎧
−α(x−d)
⎪
⎨B2 e
Êz (x) = B1 cos(kx x)
⎪
⎩
α(x+d)
B2 e
x≥d
|x| ≤ d
x ≤ −d
∂Êx
− jkz Êz = 0
∂x
⎧ jk
−α(x−d) x > d
z
⎪
⎨
− α B2 e
ˆx =
jkz B1 sin(kx x)
⇒
E
|x| < d
k
⎪
⎩
jkxz
α(x+d)
x < −d
α B2 e
2
⎧ jω�
−α(x−d)
0
⎪
⎨− α B2 e
jω�
Ĥy =
kx B1 sin(kx x)
⎪
⎩
jω�0
α(x+d)
α B2 e
x≥d
|
x| ≤ d
x ≤ −d
Boundary Conditions
Ez (x = d+ ) = Ez (x = d− ) ⇒ B2 = B1 cos(kx d)
jω�0
jω�
Hy (x = d+ ) = Hy (x = d− ) ⇒ −
B2 =
B1 sin(kx d)
α
kx
B2
�α
�0 kx
= cos(kx d) = −
sin(kx d) ⇒ α = −
cot(kx d)
B1
�0 kx
�
π
Critical Condition: α = 0 ⇒ kx d = (2n + 1) , kz2 = ω 2 �0 µ0
2
�
�2
2
ω =
(2n+1)π
2d
n = 0, 1, 2, . . .
�µ − �0 µ0
II. TE Solutions
A. Odd Solutions
⎧
−α(x−d)
⎪
⎨A2 e
Ĥz =
A1 sin(kx x)
⎪
⎩
−A2 eα(x−d)
x≥d
|x| ≤ d
x ≤ −d
¯ = ∂Hx + ∂Hz ⇒
∂Ĥx − jkz Ĥz = 0
� ·
H
∂x
∂z
∂x
⎧ jk
−α(x−d)
z
⎪
⎨− α A2 e
z
Ĥx =
−
jk
k A1 cos(kx x)
⎪
⎩
jkxz
−
α A2 eα(x+d)
¯
¯ = �
∂E ⇒
Êx = 0, Êz = 0
� ×
H
∂t
∂Ĥz
jω�Êy = −jkz Ĥx −
∂x
⎧ jωµ
−α(x−d)
0
⎪
⎨ α A2 e
Êy = jωµ
kx A1 cos(kx x)
⎪
⎩
jωµ
α(x+d)
0
α A2 e
x>d
|x| < d
x < −d
x≥d
|x| ≤ d
x ≤ −d
Boundary Conditions
� 0
�
jωµ
jωµ
Êy (x = d+ ) = Êy (x = d− ) ⇒ � A2 = � A1 cos(kx d)
kx
α
Ĥz (x = d+ ) = Ĥz (x = d− ) ⇒ A2 = A1 sin(kx d)
A2
µα
µ0 kx
= sin(kx d) =
cos(kx d) ⇒ α =
tan(kx d)
A1
µ0 kx
µ
3
B. Even Solutions
⎧
−α(x−d)
⎪
x≥d
⎨B2 e
Ĥz =
B1 cos(kx x) |x| ≤ d
⎪
⎩
B2 eα(x+d)
x ≤ −d
⎧ jk
−α(x−d) x > d
z
⎪
⎨− α B2 e
jkz
Ĥx =
kx B1 sin(kx x)
|x| < d
⎪
⎩ jkz
α(x+d)
x < −d
α B2 e
⎧ jωµ
−α(x−d)
0
⎪
x≥d
⎨ α B2 e
jωµ
Êy = −
kx B1 sin(kx x) |x| ≤ d
⎪
⎩
jωµ0
−
α B2 eα(x+d) x ≤ −d
� 0
�
jωµ
jωµ
Êy (x = d+ ) = Êy (x = d− ) ⇒ � B2 = −� B1 sin(kx d)
kx
α
Ĥz (x = d+ ) = Ĥz (x = d− ) ⇒ B2 = B1 cos(kx d)
B2
αµ
µ0 kx
= cos(kx d) = −
sin(kx d) ⇒ α = −
cot(kx d)
B1
µ0 kx
µ
4