ECE-202 Exam II

March 5, 2012

Name: __________________________________ (Please print clearly)

Student ID: _________________

CIRCLE YOUR DIVISION

Morning Afternoon

INSTRUCTIONS

There are 10 multiple choice worth 5 points each and there is two workout problems worth 50 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.

Exam 2, ECE-202, Sp 12

Multiple Choice. 5 points each totaling 50 points. 1. The pole-zero plot of the transfer function of an active ECE 255 grade amplifier circuit is

If H (s) = 1at s = 0 , then the transfer function is:

(1)

�

(s2 +1)(s+ 2)(s+1)2 +1[ ](s−1) (2)

�

(s+1)2 +1[ ](s−1)(s2 +1)(s+ 2)

(3) (s2 +1)(s + 2)(s +1)2 +1⎡⎣ ⎤⎦(s +1)

(4) − (s2 +1)(s + 2)(s −1)2 +1⎡⎣ ⎤⎦(s +1)

(5)

�

− (s2 +1)(s+ 2)(s+1)2 +1[ ](s−1) (6)

�

−(s+1)2 +1[ ](s−1)(s2 +1)(s+ 2)

(7) − (s2 +1)(s + 2)(s −1)2 +1⎡⎣ ⎤⎦(s −1)

(8) none of above

Solution 1. H (s) = K (s2 +1)(s + 2)(s −1) (s +1)2 +1⎡⎣ ⎤⎦

and H (0) = K 2−1× 2

= −K = 1⇒ K = −1. Hence

H (s) = −(s2 +1)(s + 2)(s −1) (s +1)2 +1⎡⎣ ⎤⎦

2. A transfer function of the form

�

H(s) = K (s− z1)(s− z2)(s− p1)(s− p2)

has pole zero plot given below and

K = 3× 35 . The magnitude of the frequency response at ω = 4 rad/sec is: (1) –35 (2) 35 (3) 3 (4) –3 (5) 5 (6) –5 (7) –7 (8) none of above

Exam 2, ECE-202, Sp 12

Solution 2. H (s) = K s(s − 3)(s + 3) s2 + 4⎡⎣ ⎤⎦

. H ( j4) = K j4( j4 − 3)( j4 + 3) −16 + 4[ ] = K

4 j4 − 3−12 j4 + 3

= K3= 35

3. For t < 0 , the switch is not connected and all capacitor voltages are zero. At t = 0 , the switch moves to position A; and then moves to position B at t = 1 second. Then vout (1

+ ) = (in volts): (1) 12 (2) 2 (3) 3 (4) 4 (5) 8 (6) 6 (7) 10 (8) none of above

Soluiton 3. The capacitor that is connected to position A has an initial voltage of 12 V. Replacing the capacitor by its series equivalent circuit and applying voltage division implies that

Vout (s) =

12Cs

12Cs

+ 12Cs

+ 22Cs

× 12s= 14× 12s= 3s

4. The transfer function of a circuit is H (s) = Vout (s)Vin (s)

= 50 10 − ss + 500

. The circuit is excited by a

sinusoidal input vin (t) = 2cos(10t)u(t) V. The magnitude and phase of the STEADY STATE output are APPROXIMATELY (in V and degrees):

(1) 2 2, 46o (2) 2 2, –46o (3) 2 2, 44o

Exam 2, ECE-202, Sp 12

(4) 2 2, −44o (5) 2, 46o (6) 2, −46o (7) 2, 44o (8) 2, –44o (9) none of above

Solution 4. Vout ( j10) = 5010 − j10500 + j10

× 2∠0 . Vout ( j10) = 10010 − j10500 + j10

≅100 250

= 2 2 .

∠Vout ( j10) = ∠ 10 − j10( )  −  ∠ 500 + j10( ) ≅ −45o −1o = −46o . ANSWER (2) 5. The circuit below is to be frequency scaled so that the value of its frequency response at ω = 2 rad/s occurs at 20 rad/s. Further, the circuit is to be magnitude scaled so that the largest capacitor is 0.1 F. The value of the smallest resistor after frequency and magnitude scaling is: (1) 1 (2) 2 (3) 20 (4) 4 (5) 0.4 (6) 8 (7) 0.2 (8) none of above

Solution 5. 0.1= 4KmK f

⇒ Km = 40K f

= 4010

= 4 . Rsmallest = Km ×1= Km = 4 .

6. The circuit below is of what type: (1) Low Pass (2) High Pass (3) Band Pass (4) Band Reject (5) none of above

7. The response, vC (t) , to a step input for the LC circuit below, in which vC (0

− ) = 10 V and

iL (0− ) = 2 amps, contains:

Exam 2, ECE-202, Sp 12

(1) only a zero-state part (2) only a steady state part (3) only a zero-input part (4) only a transient part (5) a steady state and a transient part (6) only a forced response (7) only a natural response (8) None of the above

8. In the circuit below, R = 4 Ω. The zero-input response to the initial condition iL (0− ) = 0.5 A is

�

vout,zi (t) = (in V):

(1) 0.5e−2tu(t) (2) 2e−2tu(t) (3) −2e−2tu(t)

(4) 4e−2tu(t) (5) −0.5e−2tu(t) (6) −4e−2tu(t) (7) −2e−4tu(t) (8) none of above

Solution 8. Using the current source model of the inductor, arrow pointing down, we have that

Vout ,zi (s) = − R × 2sR + 2s

× iL (0− )

s= −4iL (0

− )s + 2

= −2s + 2

. vout (t) = −2e−2tu(t) V.

9. The impulse response, h(t) , of a particular circuit is approximately measured on a scope as illustrated below. The input f (t) (also below) is then applied to the circuit in which T = 1 and K = 2 . The value of the response, y(t) , at t = 4 is: (1) 1 (2) 2 (3) 0 (4) 4 (5) 5 (6) 6 (7) 7 (8) none of these

Exam 2, ECE-202, Sp 12

Solution 9. The center of f (4 −τ ) is at 4 and thus the area of the product is 4. 10. The phase response of a filter is shown below. The transfer function which has a phase response which best approximates the one shown is:

(1) s +1s +100

(2) 1− ss +100

(3) s −1s +100

(4) s +10s +100

(5) s −10s +100

(6) 10 − ss +100

(7) 10 − ss + 200

(8) none of above

ANSWER: (2)

Exam 2, ECE-202, Sp 12

Workout Problem 1: (20 points) Use the convolution algebra techniques and only the convolution algebra techniques to compute the convolution y(t) = h(t) * f1(t) where f1(t) = 16sin(2t)u(t) and h(t) is given in the figure below in which T = 1 and K = 2 . First express h(t) as a sum of scaled/shifted ramps.

Solution Workout 1: (a) (7 pts) h(t) = Kr(t +1)− Kr(t) . h(2)(t) = Kδ (t +1)− Kδ (t) .

(b) (8 pts) 16 sin(2τ )dτ0

t

∫ = −8cos(2τ )]0t u(t) = 8 1− cos(2t)[ ]u(t)

F(t) 8 1− cos(2τ )[ ]

0

t

∫ dτ = 8t − 4sin(2t)[ ]u(t)

(c) (5 pts) Thus

y(t) = KF(t +1)− KF(t)    = 8K(t +1)− 4K sin(2(t +1))[ ]u(t +1)− 8Kt − 4K sin(2t)[ ]u(t)

Exam 2, ECE-202, Sp 12

Workout Problem 2. Boost Converter Circuit: In the circuit below, R1 = 2 Ω, L = 0.5 H, R2 = 2 Ω,

and C = 18

F. Suppose vin(t) = 5u(t) V, iL (0− ) = 0 , and vC (0

− ) = 0 V. Suppose further that the switch

closes at t = 0 , and opens at t = 2 sec. At t = 4 sec, the switch closes again and opens at t = 6 sec and closes again and remains closed forever at t = 8 sec. (a) For 0 ≤ t <1 , compute the inductor current (from left to right) with the switch closed by first computing IL (s) .

(b) Compute approximately iL (2− ) = iL (2

+ ) . (c) Draw the equivalent circuit valid for 2 ≤ t ≤ 4 sec . (d) Compute VC (s) . (e) Compute vC (t) for 2 ≤ t ≤ 4 sec . (f) Determine the capacitor voltage for 4 ≤ t < 6 sec. (g) Determine the capacitor voltage for 6 ≤ t < 8 sec. (g) Plot vC (t) for 0 ≤ t ≤ 8 sec in MATLAB or MATHEMATICA. Hand plots not accepted.

Solution. (a) IL (s) =5

s(0.5s + 2)= 10s(s + 4)

= 2.5s

− 2.5s + 4

in which case iL (t) = 2.5(1− e−4t )u(t) A.

(b) iL (2− ) = iL (2

+ ) = 2.5 A. (c) (d) (i)

VC1(s) =

8s

4 + 0.5s + 8s

× 5se−2s = 16

s2 + 8s +16× 5se−2s

= 80s(s2 + 8s +16)

e−2s = 80s(s + 4)2

e−2s = 5s− 5s + 4

− 20(s + 4)2

⎡

⎣⎢

⎤

⎦⎥e

−2s

(ii) VC2(s) = 16 × 0.5 × 2.5

(s + 4)2e−2s = 20

(s + 4)2e−2s

(iii) VC (s) =VC1(s)+VC

2(s) = 5s− 5s + 4

⎡⎣⎢

⎤⎦⎥e−2s

(iv) vC (t) = 5 − 5e−4(t−2)⎡⎣ ⎤⎦u(t − 2)

Exam 2, ECE-202, Sp 12

(e) vC (5) = 5 V.