AN0015
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:
R2
37 Ω
R1
150 Ω
R3
150 Ω
ZL
50 Ω
ZIN
Z
ζ – 1
ζ + 1
∆
in
--------
-----------
S
=
Γ
Z
= Z
=
where ζ =
and is called the normalized input impedance.
11
in
L
o
Z
o
Solving for Zin:
||
||
Z
= R (R + (R R )) = 49.78Ω
in
1
2
3
L
Calculating Γin:
0.996 – 1
----------------------
= 0.00203
Γ
=
in
0.996 + 1
Now S11 may be expressed in dB:
S
= 20 × LOG(Γ ) = –53dB (matched in a 50Ω system)
11
in
(ii) S21:
V1
VL
ZS=Z0
R2
37 Ω
R1
150 Ω
R3
150 Ω
VS
ZIN
ZL=Z0
V
L
------
S
= 2 ×
21
V
S
where:
||
V
V
L
V
R
Z
Z
15
L
1
3
L
in
+ Z
37.5
49.78
1
4
------
------ ------ ------------------------------------ --------------------
---------------------- ------------------------- --
=
×
=
×
=
||
V
s
V
V
(R Z ) + R
Z
37.5 + 37 49.78 + 50
1
S
3
L
2
in
s
Therefore, S21 may be expressed in dBs:
V
L
1
2
--
------
S
= 20 × LOG 2 ×
= 20 × LOG
= –6dB
21
V
S
15-36
Copyright 1997-2002 RF Micro Devices, Inc.
RFMD [ RF MICRO DEVICES ]
