EE 541
University of Southern California Viterbi School of Engineering
J. Choma.
Midterm Examination Solutions
1
Fall Semester, 2011
U
niversity of
S
outhern
C
alifornia
USC Viterbi School of Engineering
Ming Hsieh Department of Electrical Engineering
EE 541:
MIDTERM EXAMINATION
21 October 2010
(SOLUTIONS)
12:30 to 1:55
Problem #1:
Solution
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(25%)
The lowpass network shown in Figure (E1) is to be designed so that the frequency
response implied by its transfer function,
H(s) = V
o
/V
i
, closely approximates that of a second
order Butterworth filter.
Moreover, the network is to be designed to deliver a
3dB
bandwidth of
B
(in
radians/sec
) while sustaining a zero frequency transfer function value that closely ap
proaches one.
R
1
L
1
C
2
R
2
V
o
V
i
Figure (E1)
(a).
What general constraint must resistance
R
2
satisfy to ensure a zero frequency transfer func
tion value that is very close to one?
An inspection of the network at hand confirms a low frequency transfer function value that equals
the resistive divider,
R
2
/(R
2
+ R
1
)
.
If this divider is to approach one, it is clearly necessary to in
voke the constraint,
R
2
>> R
1
; that is,
resistance
R
1
must be very small in comparison to resis
tance
R
2
.
(b).
In terms of resistance
R
2
and bandwidth
B
, how must inductance
L
1
and capacitance
C
2
be
selected?
The transfer function of the network in Figure (E1) is
2
o
22
2
i
11
2
21
2
2
2
2
12
R
V
1s
R
C
H(s)
R
V
sL
R
R
C
R
RR
.
LR
R
C
R
s
L
C
(E11)
Since
R
1
must be very small to satisfy the zero frequency transmission requirement,
2
1
2
1
,
L
s
L
C
R
(E12)
whose generic second order form is
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View Full DocumentEE 541
University of Southern California Viterbi School of Engineering
J. Choma.
Midterm Examination Solutions
2
Fall Semester, 2011
2
2
1
12
2
2
n
n
11
H(s)
.
L
s
s
1s
s
L
C
1
R
Q
ω
ω
(E13)
If the network is to deliver a Butterworth frequency response, the quality factor,
Q
, must be the
in
verse of the square root of two
, wherein the selfresonant frequency,
ω
n
, becomes the network
3dB
bandwidth.
We note that
n
1
B
ω
,
LC
(E14)
and
n1
1
22
2
ω
L
L
2.
QR
R
R
(E15)
It follows that
2
2
L2
R
C
.
(E16)
Moreover,
2
2
1
B
,
2R C
2RC
(E17)
which stipulates,
2
2
1
C.
2R B
(E18)
Using this result, (E16) becomes
2
1
2R
L.
B
(E19)
Problem #2:
Solution
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—
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(30%)
Consider the normalized impedance function,
2
1a
p
Z(p)
.
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 Choma
 Electrical Engineering, Frequency, Butterworth Filter, Chebyshev filter, Southern California Viterbi, California Viterbi School, University of Southern California Viterbi School of Engineering, J. Choma

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