ece202 e2 sp2008 sol.pdf-EE-202 Exam II ...
ece202_e2_sp2008_sol.pdf-EE-202 Exam II March 3, 2008
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ece202 e2 sp2008 sol.pdf-EE-202 Exa...
ece202_e2_sp2008_sol.pdf-EE-202 Exam II March 3, 2008
Page 4
Exam 2, ECE 202, Sp 08
5.
A circuit with transfer function
H
(
s
)
=
(100
+
s
)(100
!
s
)
s
2
+
s
+
(100)
2
is excited by the input
v
in
(
t
)
=
10 cos(100
t
!
45
o
)
V, then the magnitude of the output in SSS is:
(1)
100
(2) 200
(3) 300
(4) 400
(5) 50
(6)
1000
(7) 2000
(8)
none of above
Solution 5
: (7)
10
!
H
(
j
100)
=
10
!
(100
+
j
100)(100
"
j
100)
"
100
2
+
j
100
+
(100)
2
=
10
!
100
2
(
)
2
100
=
2000
6.
What is
v
out
(
t
)
in sinusoidal steady state for the circuit below if
v
in
(
t
)
=
10 cos(2
t
)
u
(
t
)
V?
(1)
5
2
cos(2t
!
90
o
)V
(2)
5
2
cos(2t)V
(3)
5
2
cos(2t
!
45
o
)V
(4)
5
2
cos(2t
+
45
o
)V
(5)
10 cos(2t
!
45
o
)V
(6)
10cos(2t)V
(7)
10 cos(2t
+
45
o
)V
(8)
None of these
Solution 6:
(3)
H
(
s
)
=
4
4
+
3
s
+
4
s
.
V
out
=
10
!
H
(
j
2)
=
40
4
+
j
6
"
j
2
=
10
1
+
j
=
5
2
#"
45
o
7
. The following is the magnitude frequency response of a transfer function H(s).

Page 5
Exam 2, ECE 202, Sp 08
0
1
2
3
4
5
6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Magnitude of H(jw)
TextEnd
Then the best candidate for H(s) is:
(1)
(
s
2
+
2
2
)(
s
2
+
2.5
2
)
(
s
+
0.4)
2
+
1.5
2
(
)
(
s
+
0.6)
2
+
2.5
2
(
)
(2)
(
s
2
+
1.87
2
)(
s
2
+
2.15
2
)
(
s
+
0.375)
2
+
1.5
2
(
)
(
s
+
0.62)
2
+
2.5
2
(
)
(3)
(
s
+
j
1.8)
2
(
s
!
j
2.2)
2
(
s
+
1)
2
(
s
+
4)
2
(4)
(
s
2
+
3.5)(
s
2
+
4.6)
(
s
+
j
2)
2
(
s
!
j
2)
2
(5)
(
s
2
+
1.8)(
s
2
+
2.2)
(
s
+
2)
4
(6)
(
s
2
+
4)(
s
2
+
9)
[(
s
+
1)
2
+
3][(
s
+
1)
2
+
8]
(7)
(
s
+
2)
4
(
s
+
1)
2
(
s
+
3)
2
(8)
none of above
Solution 7:
(2)
8.
The transfer function that best meets the phase response plot below is
H
(
s
)
=
:

Page 6
Exam 2, ECE 202, Sp 08
10
0
10
1
10
2
10
3
10
4
0
10
20
30
40
50
60
Phase in degrees
TextEnd
(1)
s
s
+
300
(2)
s
+
1000
s
+
10
(3)
s
!
100
s
+
1000
(4)
s
+
1000
s
+
100
(5)
s
+
100
s
+
1000
(6)
s
!
100
s
+
100
(7)
s
!
1000
s
+
100
(8)
s
!
100
s
!
1000
S
OLUTION
8.
(5)
»w = logspace(0,4,1000);
»n = [1
100];
»d = [1 1000];
»h = freqs(n,d,w);
»semilogx(w, angle(h)*180/pi)
»grid
»ylabel('Phase in degrees')

Page 7
Exam 2, ECE 202, Sp 08
9.
Consider the following magnitude frequency response of a transfer function H(s).
0
2
4
6
8
10
12
0
0.2
0.4
0.6
0.8
1
1.2
1.4
| H(jw) |
TextEnd
The type of frequency response is:
(1)
low pass
(2)
high pass
(3)
band pass
(4)
band reject
(5)
none of the above
S
OLUTION
9
.
(3)
10.
A circuit has
H
(
s
)
=
4
s
(
s
+
6)
2
+
12
.
If it is excited by the input
v
in
(
t
)
=
48
r
(
t
)
V, then the value of
the output for very very large t is:
(1) 1
(2) 2
(3) 0
(4) 4
(5) 1/12
(6)
16
(7)
(8)
none of above
S
OLUTION
10
.
(4)
lim
s
!
0
s
4
s
(
s
+
6)
2
+
12
"
48
s
2
#
\$
%
%
&
(
(
=
2
11.
The step response,
v
out
(
t
)
=
(in V):
(1)
1
+
2
e
0.5
t
(
)
u
(
t
)
(2)
1
+
2
e
!
0.5
t
(
)
u
(
t
)
(3)
1
!
2
e
0.5
t
(
)
u
(
t
)
(4)
1
+
e
0.5
t
(
)
u
(
t
)
(5)
1
!
e
0.5
t
(
)
u
(
t
)
(6)
1
!
e
!
0.5
t
(
)
u
(
t
)

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