EE-202
Exam III
April 16, 2009
(Please print clearly)
INSTRUCTIONS
There are 13 multiple choice worth 5 points each and there is 1 workout problem worth 35 points.
This is a closed book, closed notes exam. No scrap paper or calculators are permitted. A transform table will be handed out separately.
Carefully mark your multiple choice answers on the scantron form. Work on multiple choice problems and marked answers in the test booklet will not be graded.
Nothing is to be on the seat beside you.
When the exam ends, all writing is to stop. This is not negotiable.
No writing while turning in the exam/scantron or risk an F in the exam.
All students are expected to abide by the customary ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course.
Communicating with any of your classmates, in any language, by any means, for any reason, at any time between the official start of the exam and the official end of the exam is grounds for immediate ejection from the exam site and loss of all credit for this exercise.
Do not open, begin, or peek inside this exam until you are instructed to do so

EE-202, Ex 3 Sp 09 page 2
MULTIPLE CHOICE.
1. The resonant frequency of the circuit ( R = 2 Ω , C =
1
32
F, L
1
= 0.25
H, L
2
= 1 H, M =
3
H) below
8 is (in rad/s):
(1) 2 2 (2) 2
(5) 8 (6) 48
(3) 32
(7) 64
(4) 4
(8) None of above
Solution 1.
Z ( s ) = R + ( L
1
+ L
2
+ 2 M ) s +
1
Cs
= 2 + L eq s +
1
Cs
.
!
r
=
1
= 4 rad/s.
L eq
C
ANSWER: (4)
2. An active circuit has a bandpass type transfer function
H ( s )
= s
2
10 s
+
2.5
s
+
13
2
The magnitude of the transfer function at
!
m
is:
(1)
10
13
(5) 5 (6)
5
144
(2) 2 (3)
(7)
10
144
10
12
(4) 4
(8) None of above
Solution 2.
H m
ANSWER: (4)
=
4 .
3. Referring again to problem 2, the approximate value of the half power frequency
!
2
(1) 11.25 (2) 15.5 (3) 13 (4) 15.25
(5) 10.5 (6) 14.25 (7) None of above
(in rad/s) is:
Solution 3.
!
m
ANSWER: (6)
=
13
2 =
13 rad/s. B !
=
2.5
rad/s.
!
m
+
0.5
B !
=
13
+
1.25
=
14.25
rad/s.

EE-202, Ex 3 Sp 09 page 3
4.
In the bandpass circuit below L = 0.1
H, C =
1
40
F, R s
= 0.2
(approximate) bandwidth, B !
, and approximate peak frequency
!
Ω m
, and R p
= 20 Ω
are (in rad/s):
. The
(1) (2, 400) (2) (2, 20) (3) (4 , 10) (4) (4, 20)
(5) (2, 10) (6) (4, 400) (7) (0.25, 20) (8) None of above
Solution 4.
!
m
" !
r
=
LC
1
=
20 rad/s. Q coil
(
!
r
)
=
!
r
L
R s
=
10 . R eq
=
20 * 20
20
+
20
=
10 Ω .
B !
=
1
R eq
C
=
4 rad/s.
ANSWER: (4)
5. Refering again to the circuit of problem 4, the approximate values of the half power frequencies,
(
!
1
, !
!
2
) , are (in rad/s):
(1) (18, !
22) (2) (8, !
12) (3) (6, !
14)
(4) (16, !
24)
(7) (9, !
11)
(5) (10, !
14)
(8) None of above
(6) (19, !
21)
Solution 5.
!
1,2
ANSWER: (1)
"
20 !
1
=
18, !
22 rad/s.
6.
If G
=
1 S and C
=
0.1
F, then the value of L in H which makes the circuit resonant at
!
r is:
(1) 1
(7) 0.02
(2) 0.05 (3) 0.5
(8) none of above
(4) 0.01 (5) 0.1 (6) 0.2
=
10 rad/s

EE-202, Ex 3 Sp 09 page 4
Solution 6 . Z in
( j
!
) = j
!
L +
1
G + j
!
C
= j
!
#
$%
L "
G
2
C &
'(
+
G
2
G
+ !
2
C
2 + !
2
C
2
. Hence
L
=
G
2
C
+ !
2
C
ANSWER: (2)
2
=
1
0.1
+
1
=
0.05
H.
7.
Reconsider the circuit of problem 6. At resonance, the impedance reduces to a pure resistance, R , which is equal to (in Ω ):
(1) 1
(7) 0.2
(2) 2 (3) 1.25
(8) none of above
Solution 7 . R
=
G
2 +
G
!
2
C
2
=
1
2
=
0.5
Ω .
ANSWER: (5)
(4) 0.4 (5) 0.5 (6) 4

EE-202, Ex 3 Sp 09 page 5
CIRCUIT FOR PROBLEMS 8, 9, AND 10.
The transfer function of the Sallen and Key circuit below is H cir
( s ) = s
2 +
2
1
Q s
+
1
:
(6) 0.6
This circuit is to realize H
LP
( s )
=
80 s
2 +
5 s
+
100
.
8. The value of Q needed for the first stage of realization is:
(1) 0.5 (2) 2
(7) none of above
(3) 16 (4) 0.2 (5) 5
Solution 8.
!
0
= 10 rad/s. Q
=
!
0
5
=
2 .
ANSWER: (2)
9. The final value of C
2
to realize H
LP
( s ) is C
2 f
=
(in F):
(1) 1
(6) 0.05
(2) 2
(7) 16
(3) 0.25 (4) 4
(8) none of above
(5) 0.5
Solution 9. C
2
(5)
"#
!
5
3
, !
2.5
$
%&
=
1
Q
=
0.5
F. C
2 f
(6)
"#
!
8
5
, !
8
3
%&
$
=
C
2
!
0
=
0.5
10
=
0.05
F.
ANSWER: (6)
10. If input attenuation is used to adjust the DC gain using a combo of R
A then the values of R
A
and R
B
(in ohms) are respectively:
(1)
"#
!
2.5, !
5
3
%&
$
(2)
(
1.2, !
4
)
(3)
( )
(4)
(
4, !
and R
B
1.2
)
as shown below,
(7)
"#
!
8
3
, !
8
5
%&
$
(8) none of above

EE-202, Ex 3 Sp 09 page 6
Solution 10: 2
!
=
0.8
!
"
!
!
=
0.4
. 1
=
R
R
A
A
R
B
+
R
B
= !
R
A
!
"
!
R
A
=
2.5
Ω , R
B
=
!
1 " !
R
A
=
5
3
Ω .
ANSWER: (1)
11. Suppose the circuit below is a NLP prototype ( C
=
2 F and L
=
1
H) that is to be a HP filter
2
(1) 1 with 3 dB down point at
!
c
=
400 2 becomes a capacitor of value (in mF):
(6) 0.1
(2) 2
(7) 4
(3) 0.5
rad/s and a source resistance equal to 10
(4) 0.25
(8) none of above
(5) 5
Ω . The inductor
ANSWER: (4)
12. . The circuit shown below consists of a 100 Ω resistor in parallel with a REAL 10 !
µ
F capacitor having a Q cap
(
!
)
=
10 at
!
= 1000 rad/sec. (Replace the capacitor by its non-ideal equivalent.)
The input admittance of this combination (given in mhos or Siemens) is:
(1) 0.010
+
(10
µ
) s (2) 0.011
+
(10
µ
) s (3) 0.101
+
(10
µ
) s (4) 0.110
+
(10
µ
) s
(5) 0.01
+
1
(10
µ
) s
(6) 100 +
(10
1
µ
) s
(7) 100 + (10
µ
) s (8) None of these

EE-202, Ex 3 Sp 09 page 7
Solution 12: 10
= !
RC
=
10
3 "
10
# 5
R !
$
!
R
=
1000 Ω . G eq
=
1
100
+
1
1000
= 0.011
mhos.
ANSWER: (2)

EE-202, Ex 3 Sp 09 page 8
WORKOUT PROBLEM (40 PTS) (This problem is to be done using the observable canonical form using at most 5 op amps. If you use the controllable canonical form instead, your maximum points for a completely correct answer is 20 pts.)
Use the observable canonical form biquad realization technique to H sec1
( s )
=
V out
V in
= s
2
0.2
s
2
+
4 s
+
8
as follows:
(i) (6 pts) Construct the differential equation (time domain) in v out
( t ) and v in
( t ) associated with
H nuts
( s ) .
( s
2 +
4 s
+
8
)
V out
=
0.2
s
2
V in
!!
!
!!
!!
out
+
4 v !
out
+
8 v out
=
0.2
!!
in
(ii) (5 pts) Use the D k
and D
!
k
notation as per the class examples to put the differential equation of part (i) into the proper form for constructing the observable canonical biquad realization, i.e., obtain an expression for v out
in terms of v in
, v out
, and D
!
k
. v out
=
0.2
v in
!
D
!
1
(
4 v out
)
!
D
!
2
(
8 v out
)
(iii) (9 pts) Given your (correct) answer to (ii), define the variable x
1
( t ) as per the class room derivation, and construct and draw an op amp circuit for v out
( t ) in terms of v in
( t ) and x
1
( t ) . Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded. v out
=
0.2
v in
!
D
!
1
(
4 v out
)
!
D
!
2
(
8 v out
)
=
0.2
v in
+ x
1
( t )
Thus x
1
(t)
= !
D
!
1
(
4v out
)
!
D
!
2
(
8v out
)
and
!
v out
= !
0.2
v in
!
x
1
( t )

EE-202, Ex 3 Sp 09 page 9
(iv) (10 pts) Given your (correct) definition of x
1
, properly define the x
2
, and construct and draw an op amp circuit whose output is x
1
( t ) . Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded.
Dx
1
( t ) = !
4 v out
!
D
!
1
(
8 v out
)
= !
0.8
v in
!
4 x
1
( t ) + x
2
( t )
Thus
!
x !
1
= x
1
= " 0.8
!
v in
" 4
!
x
1
!
x
2
(v) (10 pts) Construct the op amp circuit whose output is
± x
2
( t ) depending on the sign needed for the input to the circuit for x
1
( t ) . Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded.
Dx
2
(t) = !
8v out
= !
( in
+ x
1
)
= !
1.6v in
!
8x
1
Thus
!
x !
2
= x
2
= "
1.6
!
v in
"
8
!
x
1

EE-202, Ex 3 Sp 09 page 10
Original Circuit Exact
Equivalent Circuit at
ω
0
Approximate
Equivalent circuit, for high Q,
(Q
L
> 6 and Q
C
> 6) and
ω within (1 ± 0.05)
!
0