22 Solution
a. The 3 dB bandwidth is found using the roots of

F
(
ω
)

=

F
(
ω
o
)

√
2
=
1
√
2
.
Using the identity

1
1
±
j

=
1
√
2
it is easy to see that the 3 dB frequencies are the roots of
Q
(
ω
ω
o
−
ω
o
ω
)=
±
1
.
Or
ω
2
±
ω
o
Q
ω
−
ω
2
o
=0
.
The roots are:
∓
ω
o
2
Q
±
ω
o
2
Q
°
1+4
Q
2
Only two of these 4 roots are positive numbers. The 3 dB bandwidth is the magnitude of the diFerence
between the two positive (or negative) roots. The two positive roots are
−
ω
o
2
Q
+
ω
o
2
Q
°
Q
2
and
+
ω
o
2
Q
+
ω
o
2
Q
°
Q
2
The diFerence between these two roots is
ω
o
Q
.H
encethe3dBbandw
id
thi
s
ω
o
Q
.
b.
T
(
ω
G
1
−
AG
+
jQ
(
ω
ω
o
−
ω
o
ω
)
.
Using the same reasoning that was described in part a., we can immediately see that the 3 dB frequencies
are the roots of
Q
(
ω
ω
o
−
ω
o
ω
±
(1
−
AG
)
.
Or
ω
2
±
ω
o
Q
(1
−
AG
)
ω
−
ω
2
o
.
The two positive roots are
−
ω
o
2
Q
(1
−
AG
)+
ω
o
2
Q
°
(1
−
AG
)
2
+4
Q
2
and
+
ω
o
2
Q
(1
−
AG
ω
o
2
Q
°
(1
−
AG
)
2
Q
2
The diFerence between these two roots is
ω
o
Q
(1
−
AG
)
. Hence the 3 dB bandwidth is
ω
o
Q
(1
−
AG
)
. The
bandwidth goes to zero as
AG
→
1
.
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 Spring '08
 Staff
 Frequency, Hertz, Decibel, 3 dB bandwidth

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