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ece202 e2 sp2013.pdf
ece202_e2_sp2013.pdf
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ece202 e2 sp2013.pdf-ECE-202 Exam II March 5, 2013
ece202_e2_sp2013.pdf-ECE-202 Exam II March 5, 2013
ece202 e2 sp2013.pdf-ECE-202 Exam I...
ece202_e2_sp2013.pdf-ECE-202 Exam II March 5, 2013
Page 4
Exam 2, ECE-202, Sp 13
5.
A linear circuit with transfer function
H
(
s
)
=
V
out
(
s
)
V
in
(
s
)
=
5
3
−
2
s
2
s
2
+
2
s
+
11
has voltage input
v
in
(
t
)
=
5 cos(2
t
)
V.
Then
v
out
,
ss
(
t
)
has magnitude and phase (in rads) equal to:
(1) 1,
2 tan
−
1
−
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(2) 5,
2 tan
−
1
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(3)
10,
−
2 tan
−
1
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(4) 25,
2 tan
−
1
−
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(5) 5,
−
2 tan
−
1
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(6)
10,
2 tan
−
1
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(7)
25,
2 tan
−
1
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(8)
1,
2 tan
−
1
4
3
⎛
⎝
⎜
⎞
⎠
⎟
(9)
none of these
6.
The circuit below is to be frequency scaled so that the value of its frequency response at
ω
= 10 rad/s
occurs at 200 rad/s.
Further, the circuit is to be magnitude scaled so that the smallest capacitor is 0.02 F.
The value of the largest capacitor after frequency and magnitude scaling is:
(1) 0.1
(2) 0.2
(3) 0.01
(4) 0.4
(5) 0.02
(6) 0.04
(7) 0.05
(8)
none of above
Page 5
Exam 2, ECE-202, Sp 13
7.
The circuit below is of what type:
(1)
Low Pass
(2)
High Pass
(3)
Band Reject
(4)
Band Pass
(5)
none of above
8.
The phase response of the famous Meyer-DeCarlo sleep-inducing lecture filter is shown below.
The
transfer function that has a phase response which best approximates the one shown is:
(1)
s
+
100
s
+
800
(2)
100
−
s
s
+
800
(3)
−
s
+
10
s
+
500
(4)
20
−
s
s
+
650
(5)
−
s
+
20
s
+
400
(6)
20
−
s
s
+
400
(7)
s
+
20
s
+
650
(8)
none of above
Page 6
Exam 2, ECE-202, Sp 13
9.
In the circuit below,
v
in
(
t
)
=
6
V for all t.
C
1
=
5
F and
C
2
=
C
3
=
10
F.
At time
t
=
0
−
, the switch,
S, is positioned as indicated (neither at A nor B) and
v
out
(0
−
)
=
2
V.
At
t
=
0
+
, the switch moves to A
and at time
t
=
1
second it moves to B.
v
out
(
t
)
=
at
t
=
2
seconds in V (at least approximately):
(1) 10
(2) –14
(3)
–4
(4) 4
(5) –5
(6)
–6
(7)
14
(8)
none of these
10.
Consider the functions
()
ft
and
()
ht
(both shown below).
Suppose
()
()
()
yt
ft
ht
=
∗
.
Then
(4) :
y
=
(1) 1
(2) 2
(3)
3
(4) 4
(5) 8
(6)
6
(7)
12
(8)
none of these
0
1
−
()
ft
3
4
0
()
ht
1
−
3
t
t
3
Page 7
Exam 2, ECE-202, Sp 13
Workout Problem 1
:
(20 points)
(a)
(8 pts)
Show that the following is true:
2
1
()
()
()
()
2
tut
yt
ut
rt
=
∗
=
(b)
(12 points) Consider
h
(
t
)
and
f
(
t
)
shown below:
(i) (4 pts) Express the
()
ht
and
()
ft
as a sum of possibly shifted steps and ramps;
(ii) (8 pts)
Using the result of part (a) of this problem, a similar class result for
u
(
t
)*
u
(
t
)
, and the
general time shift theorem for convolution, etc. compute
()
()
()
yt
ht
ft
=
∗
.
A different approach will
result in a 30% point reduction.
Page 8
Exam 2, ECE-202, Sp 13
Page 9
Exam 2, ECE-202, Sp 13
Page 10
Exam 2, ECE-202, Sp 13
Workout problem 2: (30 points)
:
The switch S1 has been in position A for a very long time during
which
v
in
(
t
)
=
100
V (dc).
Switch S2 is open.
A Genie named Janus-Jonelle has secretly placed a 55 V
initial voltage on
C
2
using a high frequency impulse RAY—i.e.,
v
C
2
(0
−
)
=
55
V.
For convenience and
convention, all voltages are measured from top to bottom as plus to minus.
At
t
=
0
switch S1 moves to position B where it remains forever.
At
t
=
1
switch S2 moves to position
C where it remains forever.
The values of the components are:
C
1
=
0.5
F,
C
2
=
0.4
F,
R
1
=
1
Ω
,
and
R
2
=
5
Ω
.
(a)
(2 pts)
Find
v
C
1
(0
−
)
(b)
(8 pts)
Determine the voltages across the capacitors at
0
≤
t
<
1
second.
(c)
(2 pts)
Draw an equivalent circuit in the s-world using ONLY the VOLTAGE source models of a
charged capacitor so that there is no confusion of what the current
I
C
2
(
s
)
happens to be.
(d)
(8 pts)
Write a single node equation in the unknown
V
out
(
s
)
and solve for
V
out
(
s
)
in terms of k.
(e)
(4 pts)
For what range of values of k is the response bounded.
(f)
(6 pts)
If
k
=
2
, find
v
out
(
t
)
and identify the transient and steady state parts of the response.
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