|
Γ
|
=
S
−
1
S
+
1
=
3
−
1
3
+
1
=
2
4
=
0
.
5
Z
L
=
R
L
−
jX
C
,
where
X
C
=
1
ω
C
.
Γ
=
Z
L
−
Z
0
Z
L
+
Z
0
|
Γ
|
2
=
·μ
Z
L
−
Z
0
Z
L
+
Z
0
¶μ
Z
∗
L
−
Z
0
Z
∗
L
+
Z
0
¶̧
|
Γ
|
2
=
Z
L
Z
∗
L
+
Z
2
0
−
Z
0
(
Z
L
+
Z
∗
L
)
Z
L
Z
∗
L
+
Z
2
0
+
Z
0
(
Z
L
+
Z
∗
L
)
Noting that:
Z
L
Z
∗
L
=(
R
L
−
jX
C
)(
R
L
+
jX
C
)=
R
2
L
+
X
2
C
,
Z
0
(
Z
L
+
Z
∗
L
)=
Z
0
(
R
L
−
jX
C
+
R
L
+
jX
C
)=
2
Z
0
R
L
,
|
Γ
|
2
=
R
2
L
+
X
2
C
+
Z
2
0
−
2
Z
0
R
L
R
2
L
+
X
2
C
+
Z
2
0
+
2
Z
0
R
L
.
Upon substituting
|
Γ
L
|
=
0
.
5,
R
L
=
75
Ω
, and
Z
0
=
50
Ω
, and then solving for
X
C
,
we have
X
C
=
66
.
1
Ω
.
Hence
C
=
1
ω
X
C
=
1
2
π
×
10
7
×
66
.
1
=
2
.
41
×
10
−
10
=
241 pF
.