2260
F 09
HOMEWORK #3 prob 2 solution
E
X
:
+
–
v
o
v
i
+
–
R
1
=
20 k
Ω
R
2
=
30 k
Ω
R
3
=
3 k
Ω
L
=
500
μ
H
a)
Determine the transfer function V
o
/V
i
.
Hint:
use a Thevenin equivalent on the
left side.
b)
Plot V
o
/V
i
 versus
ω
.
c)
Find the cutoff frequency,
ω
c
.
S
OL
'
N
:
a) The technique employed in this solution focuses on writing the transfer
function,
H
(
j
ω
), as a product of a realvalued scaling factor and a simple
RLC
circuit whose transfer function,
H
1
(
j
ω
), involves only one
L
, one
C
,
and one
R
.
H
(
j
ω
)
=
k
⋅
H
1
(
j
ω
)
where
k
≡
real number
The virtue of this approach is that cutoff frequencies are unaffected when
a transfer function is scaled by a real number.
Since the magnitude of the
transfer function at the cutoff frequency is defined relative to the
maximum magnitude of the transfer function, and the magnitudes are
scaled by the same real number at every frequency, the cutoff frequency is
unaffected by a real scaling factor multiplying a transfer function.
Thus,
the cutoff frequency of
H
(
j
ω
) is the same as the cutoff frequency of
transfer function
H
1
(
j
ω
).
This idea will be exploited in part (c), but it also
helps to clarify the derivation of the transfer function.
Our first step is to take a Thevenin equivalent of the input voltage,
V
i
, and
R
1
and
R
2
.
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+
–
V
V
i
+
–
R
1
R
2
⇒
+
–
V
+
–
R
Th
=
R
1
R
2
V
Th
=
V
i
R
2
R
1
+
R
2
We find
v
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 Spring '12
 Frequency

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