ECE 201A
Homework 3 solution
1.
We can describe the problem with a transmission line equivalent circuit, which is
The propagation constants of the equivalent transmission lines,
0
β
and
1
β
, are the components of the
k
vector in
z
direction, i.e.,
0
0
0
2
cos
z
i
k
π
β
θ
λ
=
=
and
1
1
0
2
cos
z
t
k
n
π
β
θ
λ
=
=
The characteristics impedances of the equivalent transmission lines,
0
Z
and
1
Z
, are the wave impedances
in
z
direction, i.e.,
0
0
0
cos
z
i
Z
Z
η
θ
=
=
and
0
1
1
0
z
Z
Z
n
η
λ
=
=
since the wave is TE polarized.
Since regions I and III are infinite in extend so are the equivalent
transmission lines.
Since we see the characteristic impedance looking into an infinitely long transmission
line, the equivalent circuit becomes
i
θ
t
θ
i
H
G
i
E
G
0
ε
0
μ
RegionI
1
ε
0
μ
RegionII
0
ε
0
μ
RegionIII
d
z
x
1
0
r
ε
ε ε
=
r
n
ε
=
.
.
.
0
Z
0
Z
1
Z
0
β
0
β
1
β
d

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To find the power reflected back to region I, let’s find the reflection coefficient
Γ
at the first junction.
0
0
in
in
Z
Z
Z
Z
−
Γ =
+
, where
0
0
1
1
1
0
1
0
1
tan
tan
in
in
in
Z
Z
Z
jZ
d
Z
Z
Z
Z
Z
jZ
d
β
β
−
+
Γ =
=
+
+
.
Then
(
)
(
)
2
2
2
2
1
0
1
0
1
1
1
0
1
0
1
2
2
2
2
0
1
1
1
0
1
0
1
0
1
1
0
1
tan
tan
tan
tan
tan
2
tan
j Z
Z
d
Z Z
jZ
d
Z Z
jZ
d
Z Z
jZ
d
Z Z
jZ
d
Z Z
j Z
Z
d
β
β
β
β
β
β
−
+
−
−
Γ =
=
+
+
+
+
+
Power reflection coefficient is
2
Γ
, so
(
)
(
)
2
2
2
2
2
1
0
1
2
2
2
2
2
2
0
1
1
0
1
tan
4
tan
Z
Z
d
Z Z
Z
Z
d
β
β
−
Γ
=
+
+
For total transmission into region III
2
0
Γ
=
.
Because if there is no reflected power to region I, all the
power must be transmitted to region III, since all the media are lossless. This requires either
0
1
Z
Z
=
,
which is the trivial solution or
1
d
m
β
π
=
, which makes
1
tan
0
d
β
=
.
1
0
2
cos
t
n
d
d
m
π
β
θ
π
λ
=
=
, which
yields
0
2 cos
t
m
d
n
λ
θ
=
.
t
θ
is found using the Snell’s law, hence
1
sin
sin
sin
sin
i
t
t
i
n
n
θ
θ
θ
θ
=
⇒
=
2
2
2
2
2
sin
sin
cos
1
sin
1
i
i
t
t
n
n
n
θ
θ
θ
θ
−
=
−
=
−
=
Therefore
0
0
2
2
2
2
sin
2
sin
2
i
i
m
m
d
n
n
n
n
λ
λ
θ
θ
=
=
−
−
.
.
0
Z
0
Z
1
Z
0
β
1
β
d
in
Z