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ece202 e2 sp2013.pdfECE202 Exam II March 5, 2013
ece202_e2_sp2013.pdfECE202 Exam II March 5, 2013
ece202 e2 sp2013.pdfECE202 Exam I...
ece202_e2_sp2013.pdfECE202 Exam II March 5, 2013
Page 4
Exam 2, ECE202, 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, ECE202, 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 MeyerDeCarlo sleepinducing 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, ECE202, 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, ECE202, 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, ECE202, Sp 13
Page 9
Exam 2, ECE202, Sp 13
Page 10
Exam 2, ECE202, 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 JanusJonelle 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 sworld 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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