2 / 6 page
15-36
AN0015
15
Copyright 1997-2002 RF Micro Devices, Inc.
Solution:
Notice the "PI" attenuator shown above is a symmetrical circuit, and therefore the knowledge learned from one port may
be applied to the other port.
(i) S11:
where
and is called the normalized input impedance.
Solving for Zin:
Calculating
Γ
in:
Now S11 may be expressed in dB:
(matched in a 50
Ω system)
(ii) S21:
where:
Therefore, S21 may be expressed in dBs:
R
2
37
Ω
R
1
150
Ω
R
3
150
Ω
Z
IN
Z
L
50
Ω
S
11
Γ
∆
in
Z
L
Z
o
ζ 1
–
ζ 1
+
------------
==
=
ζ
Z
in
Z
o
--------
=
Z
in
R
1
R
2
R
3
R
L
||
()
+
() 49.78Ω
=
||
=
Γ
in
0.996
1
–
0.996
1
+
----------------------
0.00203
==
S
11
20
LOG
Γ
in
()
53dB
–
=
×
=
R
2
37
Ω
R
1
150
Ω
R
3
150
Ω
Z
IN
Z
L=Z0
Z
S=Z0
V
L
V
1
V
S
S
21
2
V
L
V
S
-------
×
=
V
L
V
s
-------
V
L
V
1
-------
V
1
V
S
------
×
R
3
Z
L
||
R
3
Z
L
||
() R
2
+
-------------------------------------
Z
in
Z
in
Z
s
+
--------------------
37.5
37.5
37
+
----------------------
49.78
49.78
50
+
-------------------------
×
1
4
---
=
=
=
S
21
20
LOG
×
2
V
L
V
S
-------
×
20
LOG
×
1
2
---
6dB
–
==
=
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