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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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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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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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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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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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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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(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

202ex2sp04sols

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