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final 3.pdfECE 232  Circuits and Systems
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final 3.pdfECE 232  Circuits and ...
final_3.pdfECE 232  Circuits and Systems
Page 1
ECE 232  Circuits and Systems II
Final (Spring 2011)
Please provide clear and complete answers. Don’t forget to specify the units of measure!
1. (2 points)
Consider the transfer function
(
) = 100
(
+ 1)
(
+ 10)(
+ 100)(
+ 1000)
a.
Draw the amplitude Bode plot for the transfer function.
Specify clearly the relevant
values on the two axes. In particular, what is (approximately)
max
?
b. What type of
ﬁ
lter is it? Using Bode’s approximation,
ﬁ
nd the
3
cuto
ﬀ
frequency (or
frequencies) (Hint: a linear increase of
20
for every decade is the same as an increase of
3
for every multiplicative increase by
√
2
).
t
[s]
t
[s]
1
2
2
x(t)
h(t)
1
2
1
2. (2 points)
Perform the convolution
(
)=
(
)
∗
(
)
of the two
signals
(
)
and
(
)
shown in the
ﬁ
gure on the left.
)
(
t
v
i
R
+

+

)
(
t
v
o
L
C
L
R
3. (3 points)
Consider the circuit
in the
ﬁ
gure on the left with
=1
Ω
= 100
= 100
and
= 10
Ω
a. Find the transfer function between
(
)
and
(
)
.
b. Find and sketch the amplitude response. What type of
ﬁ
lter is it? What is
max
and
what are the
3
cuto
ﬀ
frequencies?
c. Discuss the e
ﬀ
ect of
d.
Calculate the quality factor and relate this result to the position of the poles in the
complex plane.
1
Page 2
)
(
t
v
i
R
+

+

)
(
t
v
o
L
L
R
4. (3 points)
Consider the circuit in
the
ﬁ
gure on the left with
=1
Ω
,
=1
,
=1
Ω
a. Find the transfer function between
(
)
and
(
)
. What are the poles and zeros?
b. Sketch the amplitude and phase response. What type of
ﬁ
lter is it? What is
max
and
what is the
3
cuto
ﬀ
frequency?
c. Assume that the inductor has a current at time zero equal to
1
(with direction from
top to bottom) and that
(
)=
−
1000
(
)
. Find the Laplace transform of the output
(
)
Sol
.:
1.
We
ﬁ
st write
(
)
=
100
(1 +
)
10
·
100
·
1000(1 +
10
)(1 +
100
)(1 +
1000
)
=
1
10
4
(1 +
)
(1 +
10
)(1 +
100
)(1 +
1000
)
a. Using the approach seen in class, the Bode plots starts with a
20
slope until fre
quency
1
, then the slope is
40
until
10
, where it becomes
20
,
to become
0
at frequency
100
and
ﬁ
nally
−
20
at
1000
. Note
that
is approximately obtained between
100
and
1000
, so we get
max
'
(
100)
'
1
10
4
100
·
100
10
=0
1
b. The
ﬁ
lter is bandpass with central frequency between
100
and
1000
. The
3
cuto
ﬀ
frequencies are approximately
1
=
100
√
2
= 70
71
2
=
1000
√
2 = 1414
2
2.
We perform the convolution between
(
)
and
(
+ 1)
(that is,
(
)
shifted left by
1
second) as
˜
(
)=
(
)
∗
(
+1)
. The
ﬁ
nal result
(
)
will be obtained by moving the obtained
solution
˜
(
)
to the right by
1
second.
2
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