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EE-202
Exam II
March 8, 2004
Name: __________________________________
Student ID: _________________
CIRCLE YOUR DIVISION
Division: 201-1 (morning) 201-2 (afternoon)
INSTRUCTIONS
There are 12 multiple choice worth 5 points each;
there is 1 workout problem worth 40 points.
This is a closed book, closed notes exam. No scrap
paper or calculators are permitted. A transform table and
properties table are attached at the back of the exam.
All students are expected to abide by the usual ethical
standards of the university, i.e., your answers must reflectonly 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.
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EE-202, Ex 2, page 2
MULTIPLE CHOICE.
1. A transfer function has H(s) = K(s !z1)(s !z2 )
(s ! p1)(s ! p2 )and has a gain of 2 at s = 0. Then
K = :
(1) 8 (2) 8 (3) 4 (4) 4
(5) 16 (6) 16 (7) 2 (8) None of above
2. The switch S1 has been closed for a long time and opens at t= 1! . Switch S2 closes
at t= 1+ . Thenvout
(2) = (in V):
(1) 10 (2) 20 (3) 10 (4) 40
(5) 5 (6) 20 (7) 40 (8) None of above
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EE-202, Ex 2, page 3
3. The circuit below has a normalized critical frequency !c,norm
= 1 rad/s. The circuit
must be frequency and magnitude scaled so that the new critical frequency is !c= 10
rad/s the largest capacitor is 0.2 F. The new value of the inductor is Lnew
= (in H):
(1) 1 (2) 2.5 (3) 0.01 (4) 0.2(5) 5 (6) 0.05 (7) 0.1 (8) none of above
4. If a circuit with H(s) = 2 (s+
2)(s!
2)(s + 4 + 2j)(s + 4 ! 2j)
= 2 s
2!
4(s + 4)
2+ 4
is excited by the
input vin(t) = 2 cos(2t+ 45
o) V, then the phase of the output sinusoid in SSS is:
(1) 45o (2) !45o (3) 90o (4) !90o
(5) 135o (6) !135o (7) 180o (8) none of above
5. For the circuit with transfer function of problem 4, the magnitude of the output
sinusoid in SSS is:
(1) 1 (2) 2 (3) 1 2 (4) 4
(5) 0.5 (6) 2 32 (7) 0.25 (8) none of above
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EE-202, Ex 2, page 4
6 A circuit has
H(s) = 2(s + 2)(s ! 2)
(s + 4 + 2j)(s + 4 ! 2j)= 2
s2! 4
(s + 4)2 + 4. If it is excited by the
input vin(t) = 20u(t) V, then the value of the output for very large t is:
(1) 2 (2) 2 (3) 4 (4) 4(5) 0.4 (6) 0.4 (7) 8 (8) 8
7. For the circuit below to be stable in the sense of BIBO, the complete range ofa must
be:(1) a > 2 (2) a < 2 (3) a > 0.5 (4) 1 > a
(5) a > !1 (6) a > 2 (7) a > 0 (8) None of above
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EE-202, Ex 2, page 5
8. The following is the magnitude frequency response of a transfer function H(s).
The transfer function leading to this magnitude response is:
(1)(s
2+ 4)(s
2+16)
(s +1)2+ 2!
"#$%&
(s +1)2+ 4!
"#$%&
(2)(s
2+ 2)(s
2+ 4)
(s +1)2+ 2!
"#$%&
(s +1)2+ 4!
"#$%&
(3)(s2 + 2)(s2 + 4)
(s +1)2 + 4!"#
$%&
(s +1)2 +16!"#
$%&
(4)(s2 + 4)(s2 +16)
(s +1)2 + 4!"#
$%&
(s +1)2 +16!"#
$%&
(5) (s2 ! 4)(s2 !16)
(s +1)2+ 2"
#$%&'
(s +1)2+ 4"
#$%&'
(6) (s2 ! 4)(s2 !16)
(s +1)2+ 4"
#$%&'
(s +1)2+16"
#$%&'
(7) None of above
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EE-202, Ex 2, page 6
9. The transfer function that best meets the phase response plot below is H(s) =:
10-2
100
102
104
-100
-80
-60
-40
-20
0
Phase
Response
TextEnd
(1)s ! 400
s + 2(2)
s !1000
s !1(3)
s +1000
s +1(4)
s ! 400
s ! 2
(5)s + 400
s + 2(6)
s +1
s +100(7)
s +100
s +1
(8) none of above
10. The step response of a linear circuit is y(t) = 5r(t!1) !10r(t! 3). The zero-state
response of this same circuit at time t = 3 seconds to an input x(t) = 2!(t"1) is:
(1) 1 (2) 2 (3) 0 (4) 5
(6) 10 (7) -5 (8) none of above
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EE-202, Ex 2, page 7
11. The output of a linear and relaxed circuit with input x(t) and impulse response h(t)
(shown below) at time t = 6 seconds is:
(1) 1 (2) 2 (3) 3 (4) 4 (5) 0
(6) 6 (7) 8 (8) None of the above
12. For h(t) and f(t) sketched below, the convolution y(t) = h(t)*f(t) has y(1) = :
(1) 1 (2) 2 (3) 3 (4) 4
(5) 0.5 (6) 0 (7) 0.25 (8) None of above
13. For h(t) and f(t) sketched below, the convolution y(t) = h(t)*f(t) is given by which
picture:
(1) (2)
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EE-202, Ex 2, page 8
(3) (4)
(5) (6)
(7) None of above
SOLUTIONS MC
Solution 1: H(s) = K(s ! 2)(s + 2)
(s + 2)2 + 4" H(0) = !2 = K
!4
8" K= 4 .
Solution 2: 0.25vC1
(1!) = 0.5v
out
(1+ ) " 5 = 0.5vout
(1+ ) " vout
(1+ ) = 10 V
Answer: (1)
Solution 3. 0.2 =4
10Km
! Km= 2 . Thus Lnew =
Km
KfLold= 0.1 H. Answer: (7)
Solution 4: !Output= !H(j2) + 45o = 45o +135o " 45o + 45o = 180o Answer: (7)
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EE-202, Ex 2, page 10
Hence, the answer is:
Answer: (5)
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EE-202, Ex 2, page 12
vC
(t) = 30(1! e!2t)u(t) V, 0 " t< 10
(ii) (5 pts) Vout
(s) =s
(s + 0.5)(s + 2)
Vin(s) =
30
(s + 0.5)(s + 2)
=20
s + 0.5
!
20
s + 2
. Hence
vout
(t) = 20(e!0.5t
! e!2t)u(t) V, 0 " t< 10
(c) By inspection, vC(10! ) " 30 V (2 pts) and i
L(10! ) = 60 A (3 pts). Exact numbers
are wrong here and only get 1 pt total because the student did not follow directions
and/or failed to exercise engineering judgment in approximating.
(d) (5 pts) Equivalent Circuit is:
(e) (10 pts) Writing a node equation for Vout
(s) yields:
30 !60
s
"
#$
%
&'e!10s
= s ! 2 +2
s
+ 4"
#$
%
&'Vout
(s) =s2+ 2s + 2
s
"
#$$
%
&''Vout
(s) (3 pts)
Hence
Vout
(s) =s
(s +1)2 +1230 !
60
s
"#$
%&'e!10s
= 30(s +1)e!10s
(s +1)2 +12! 90
e!10s
(s +1)2 +12 (4 pts)
Thus for 10 t ( 3 pts)
vout
(t) = 30e!(t!10) cos(t!10)u(t!10) ! 90e!
(t!10) sin(t!10)u(t!10) V
or equivalently
vout
(t) = 30e!(t!10) cos(t!10) ! 3sin(t!10)( )u(t!10)V