ECE-202 Exam II

March 10, 2013

Name: __________________________________ (Please print clearly)

Student ID: _________________

CIRCLE YOUR DIVISION

Section 2021, 7:30 MWF Section 2022, 3:30 MWF

Prof. DeCarlo Prof. DeCarlo

INSTRUCTIONS

There are 12 multiple choice worth 5 points each and

there is one workout problem worth 40 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 14

Multiple Choice. 5 points each totaling 60 points. 1. (Poles and zeros, transfer functions) The pole-zero plot of a transfer function of an active circuit is

A bounded input which will make the zero-state response unbounded (increasing with time) is: (1) δ (t) (2) u(t) (3) cos (ω1 −ω2 )t + 30

o( )u(t)

(4) sin(ω1t − 30o )u(t) (5) cos(ω2t + 30

o )u(t) (6) sin (ω1 +ω2 )t( )u(t) (7) two of above (8) none of above ANSWER: (5). 2. (Stability, admittance) In the circuit below, R = 2 Ω and C = 0.5 F, the COMPLETE range of K for which the input IMPEDANCE is BIBO stable is: (1) K < 0 (2) K > 0 (3) K <1 (4) K >1 (5) −1< K <1 (6) K > −1 (7) 1≤ K ≤ 2 (8) none of above

Exam 2, ECE-202, Sp 14

Solution 2. Iin = GVin +Cs Vin − KVin( ) implies

Iin = [Cs(1− K )+G]Vin ⇒ Zin =VinIin

= 1C(1− K )s +G

= 1

s + GC(1− K )

.

Stability implies s = − GC(1− K )

< 0⇒1− K > 0⇒ K <1 .

3. (Sinusoidal steady state analysis) A linear circuit with transfer function

H (s) =

Vout (s)Vin(s)

= 10 4− s

s2 + s+ 20 has voltage input vin(t) = 2cos(4t) V. Then

vout,ss(t) has magnitude

and phase (in rads) equal to:

(1) 10, −45o (2) 10, 90o (3) 10, −90o

(4) 20, −45o (5) 20, 45o (6) 20, 90o

(7) 20, −90o (8) 10, 45o (9) none of these

Solution 3. H ( j4) = 10 4− 4 j

−16+ 4 j + 20= 10 4− 4 j

4+ 4 j. H (4 j) = 10 and ∠H (4 j) = −45o − 45o = −90o .

vout,ss(t) = 20cos 4t − 90o( ) V. ANSWER: (7).

4. (Op amp circuit design from transfer function) The circuit below having transfer function

Hold (s) = − 2s+ 0.4

s+1 is to be frequency scaled by K f = 10 and magnitude scaled so that the largest

capacitor is 40 mF. If C1,old = 4 F, the value of

R2,new (in Ω) after magnitude and frequency scaling is:

(1) 2.5 (2) 2 (3) 0.4 (4) 4 (5) 5 (6) 0.2 (7) 10 (8) 0.05 (9) None of above

Exam 2, ECE-202, Sp 14

Solution 4:

H (s) = −Yin(s)Yf (s)

= −C1s+G1C2s+G2

= −

C1C2

s+G1C2

s+G2C2

= −

4C2

s+G1C2

s+G2C2

= − 2s+ 0.4s+1

. Thus

C2 = 2 F ⇒ G2 = 2, R2 = 0.5 Ω . C1,new =C1,oldKmK f

⇒ Km = 410 × 0.04

= 10 . R2,new = 5 Ω.

ANSWER: (5). 5. (Graphical Convolution) Consider the functions f (t) and h(t) (both shown below). Suppose

y(t) = f (t)∗h(t) . Then y(0)=: (1) 6 (2) 8 (3) 36 (4) 4 (5) 12 (6) 24 (7) 16 (8) none of these

01−

( )f t

3

4

0

( )h t

1−

3

t t3

Solution 5. (Graphical Convolution) Do a flip and shift on h(t) to produce h(0−τ ) on the τ − axis . This would overlap f (τ ) for 0 ≤ τ ≤ 2 . The product h(0−τ ) f (τ ) is nonzero only for 0 ≤ τ ≤ 2 and for this interval is a rectangle of height 12 and a base of 2. Hence, AREA = 24= y(0) . ANSWER: (6). 6. (Frequency Response, Filtering) The circuit below is of what general type: (1) Low Pass (2) High Pass (3) Band Pass (4) Band Reject (5) none of above

ANSWER: (3)

Exam 2, ECE-202, Sp 14

7. (Initial/Final value theorems) The s-domain response of a circuit is given by Vout (s) =(16s + 32)s(2s + 4)

.

The vout (∞) and vout (0+ ) are (respectively in V):

(1) 0, ∞ (2) 2, 2 (3) 16, 8 (4) 4, 4 (5) 4, 8 (6) 8, 4 (7) 16, 4 (8) 8, 8 (9) none of above Solution 7. vout (∞) = lim

s→0sVout (s) = 8 . vout (0

+ ) = lims→∞

sVout (s) = 8

8. (Switched capacitors) For the circuit shown below, C1 = 1 F, C2 = 4 F, vout (0

− ) = 10 , and vin(t) = 20u(t) V. The switch moves from position A to position B at t = 1 sec, where it remains until your graduation day. Then vout (2) = (in V): (1) 10 (2) 20 (3) 12 (4) 60 (5) 50 (6) 6 (7) 18 (8) none of above

Solution 8: vC1 (1

− ) = 20 V and vC2 (1− ) = 10 . Thus

C1vC1 (1− )+C2vC2 (1

− )⎡⎣ ⎤⎦ ×e−s

(C1 +C2 )s= 605se−s =Vout (s) V, vout (1

+ ) = 12 V.

9 . (Frequency response, transfer functions) The phase response of a You’re-nuts-Donuts coffee filter is shown below. (The major numbers on the x-axis are in order: 100, 101, 102, 103 .) The transfer function which has a phase response that best approximates the one shown is:

(1) 10 − ss + 500

(2) s +10s +100

(3) s + 400s +10

(4) s +10s + 400

(5) s +10s + 500

(6) 10 − ss + 400

(7) s + 500s +10

(8) none of above

Exam 2, ECE-202, Sp 14

ANSWER: (4) 10. (Switching circuits) In the circuit below Rs = 0.1 Ω, C = 1 F, and the switch moves from position A to position B at t = 1 second. If the input vin(t) = 10u(t)V, then vC (t) = (in V): (1) 5(1+ e−2t )u(t) (2) 5(1+ e−2(t−1))u(t −1) (3) 5(1− e−2t )u(t)

(4) 5(1− e−2(t−1))u(t −1) (5) 10(1+ e−2t )u(t) (6) 10(1+ e−2(t−1))u(t −1)

(7) 10(1− e−2t )u(t) (8) 10(1− e−2(t−1))u(t −1) (9) none of above

Solution 10. The time constant for t <1 sec is 0.1 sec, which means that the capacitor is fully charged at t = 1 s. Using the current source model for the capacitor, we see that the impedance seen by the

Exam 2, ECE-202, Sp 14

current source is going to be Z(s) = 1

s + ss +1

= s +1s(s + 2)

. Hence

VC (s) = CvC (1− )e−s s +1

s(s + 2)= 10e−s s +1

s(s + 2)= 5e

−s

s+ 5e

−s

s + 2. Hence vC (t) = 5(1+ e

−2(t−1))u(t −1) V.

11. (Convolution) The impulse response of a circuit is h(t) shown below. Suppose the input is

�

vin (t) = −π sin(πt)u(t) . The value of the convolution at t = 1 seconds is: (1) 1 (2) 2 (3) –1 (4) –4 (5) –3 (6) –2 (7) 0 (8) none of above

Solution 11. The derivative of h(t) is

�

h'(t) = δ(t) + δ(t −1). Further

�

vin (q)0

t∫ dq = cos(πq)u(q)]0

t = cos(πt) −1[ ]u(t) . Thus the convolution is

cos(πt)−1[ ]u(t)+ cos(π(t −1)−1[ ]u(t −1)( )t=1 = −2 . 12. (Convolution) The convolution of f (t) = 4u(t +1) with the impulse response h(t) = u(t +1)− 2r(t) is y(t) . The value of y(1) = : (1) 4 (2) 12 (3) –3 (4) –4 (5) –16 (6) –12 (7) 8 (8) none of above Solution 12. y(t) = 4u(t +1)*u(t +1)− 8u(t +1)*r(t) = 4r(t + 2)− 4(t +1)2u(t +1) . Thus

y(1) = 12 −16 = −4 .

Exam 2, ECE-202, Sp 14

Workout problem: (40 points): The circuit below is a boost converter charging circuit for a super capacitor represented by the capacitor on the circuit. The capacitor voltage at t = 0 is zero. The switch has been closed for a very long time and opens at t = 0. Subsequently, the switch closes at t = 1 s, opens at t = 2 and remains open. For the analysis to be done below assume R1 = 10 Ω,

L = 1 H, R2 = 5 Ω, C = 150

= 0.02 F (super cap), and vin(t) = 20u(t +10) V (to represent a 20 V

source). (a) (3 pts) Compute iL (0

− ) . (b) (17 pts) Compute VC (s) and vC (t) valid for the interval 0 ≤ t <1 . Specify vC (1

− ) . (c) (2 pts) For this problem is it necessary to know iL (1

− ) to compute iL (2− ) after the switch opens at

t = 2 sec? (d) (2 pts) What is the value of vC (2

− )? (e) (16 pts) Compute VC (s) and vC (t) valid for t ≥ 2 sec. Remark: one should note that the designer of this particular boost converter circuit should receive career counseling. ☺

Solution Workout. R1 = 10; R2 = 5; L = 1; C = 1/50; (a) % Time constant L/R1 for RL circuit when switch closed is: >> tau1 = L/R1 tau1 = 1.0000e-01 % Therefore, 100 time constants have passed since source turn on and inductor looks % like a short circuit. >> iL0 = 20/R1 iL0 = 2 (b) >> H = (1/(C*s))/(R1 + R2 + L*s + 1/(C*s)); >> H = collect(H) H =50/(s^2 + 15*s + 50) >> Vc = collect(H*(20/s + L*iL0)) Vc =100/(s^2 + 5*s)

Exam 2, ECE-202, Sp 14

>> vc = ilaplace(Vc) vc =20 - 20*exp(-5*t) vC (t) = 20(1− e

−5t )u(t), 0 ≤ t <1 After 5 time constants, vC (1

− ) = 20(1− e−5 )u(1) ≅ 20 V. (c) NO because there are 10 time constants over which the inductor is fully charged to 2 A. Hence iL (2

− ) = 2 A. (d) Because of the diode, no current flows after the switch closes. Hence vC (2

− ) ≅ 20 V. (e) >> Vc = collect(H*(20/s + L*iL0 - 20/s))+20/s Vc = 100/(s^2 + 15*s + 50) + 20/s Hence >> vc = ilaplace(Vc) vc = 20*exp(-5*t) - 20*exp(-10*t) + 20 Shifting this expression to the right by 2 units we have: vC (t) = 20u(t − 2)+ 20e

−5(t−2)u(t − 2)− 20e−10(t−2)u(t − 2) valid for t ≥ 2