ece-202 exam 2 march 1, 2017 - weeklyjoys...ece-202, exam 2 spring 17 9 11. the circuit below is to...

ECE-202 Exam 2 March 1, 2017

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Section 2021 Section 2022

DeCarlo MWF 7:30-8:30 DeCarlo TTH 1:30-2:45

INSTRUCTIONS

There are 13 multiple choice worth 5 points each. There is one workout problems worth 35 points.

This is a closed book, closed notes exam. No scrap paper or calculators are

permitted. Nothing is to be on the seat beside you. Carefully mark your multiple choice answers on the scantron form and also on the text booklet as you will need to put the scantron inside the test booklet and turn both in at the end of the exam.

When the exam ends, all writing is to stop. No writing while turning in the exam/scantron or risk an F in the exam. Again, turn in the exam booklet with the scantron inside the exam. There are two exam forms and this is necessary for the proper grading of your scantron. 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. The professor reserves the right to move students around during the exam. Do not open, begin, or peek inside this exam until you are instructed to do so.

ECE-202, Exam 2 Spring 17 2

MULTIPLE CHOICE

1. A transfer function of the form

�

H(s) = K (s− z1)(s− z2)!(s− p1)(s− p2)!

has pole zero plot given below

with the pole at “–1” having multiplicity 2. H (2) = 10 . The value of K = : (1) 1 (2) 10 (3) 36 (4) 48 (5) 54 (6) 16 (7) 18 (8) 9 (9) none of above

Solution 1.

H (s) = Ks(s−1) s2 +1( )

(s+1)2 (s−1)2 +1( ) . H (2) = 10 = K 10

18⇒ K = 18

Answer: (7) 2. The transfer function that best meets the phase response plot below is H (s) = :

(1) s+ 30s+800

(2) s+ 40s+800

(3) −s+ 30−s+800

(4) s+ 20s+ 700

(5) s+ 30

s+ 900 (6)

−s+ 40−s+800

(7) −s+ 30−s+ 900

(8) −s+ 20−s+ 700


(9) none of the above

101 102 103−70

−60

−50

−40

−30

−20

−10

frequency in rad/s

phas

e in

deg

rees

Answer: (3) 3. The transfer function that best matches the magnitude frequency response below is:

(1)

10s(s2 +1002)(s+1)((s+1)2 +1002)

(2)

10s(s2 +1002)(s+1)((s+10)2 +1002)

(3)

10s(s2 +1002)(s+10)((s+10)2 +1002)

(4)

10s(s2 +100)(s+1)((s+1)2 +100)

(5)

10s(s2 +100)(s+1)((s+10)2 +100)

(6)

10s(s2 +100)(s+10)((s+10)2 +100)


(7)

10s(s2 +100)(s− 2)((s− 2)2 +100)

(8)

10s(s2 −100)(s+1)((s+1)2 +100)

(9)

10s(s2 −100)(s−1)((s+10)2 +100)

(10) none of above

0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

10

frequency in rad/s

Mag

nitu

de R

espo

nse

Answer: (4) 4. In the op am circuit below, R1 = R2 = 1 Ω, C1 = 0.1 F, C2 = C3 = 1 F, and R4 = 0.01 Ω. The magnitude frequency response is of which type: (1) Band pass (2) Band reject (3) Low pass (4) High pass (5) No pass (6) Everything pass (7) none of above


Answer: (2) 5. The Laplace transform of the complete response of an RLC circuit is given by

Vout (s) =12s− 12s +10

+ 2iL (0− )

s2 + 4− vC (0

− )s +10

. The transient part of the zero-input response and the

steady state part of the zero-state response are:

(1) −vC (0− )e−10tu(t), 12u(t) (2) −12e−10tu(t), iL (0

− )sin(2t)u(t) (3) 12e−10tu(t), iL (0

− )sin(2t)u(t) (4) −12e−10tu(t), 2iL (0− )sin(2t)u(t)

(5) −vC (0− )e−10tu(t), 12e−10tu(t) (6) vC (0

− )e−10tu(t), 12u(t) (7) none of above

Answer: (1)


6. The convolution, y(t) = h(t)* f (t) , where h(t) = 2u(t +1) and f (t) = r(t + 2)− r(t +1) , results in y(−1) = ( y(t) evaluated at t = −1 ): (1) 0 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 (8) 8 (9) none of above Solution 6. From class notes and HW,

y(t) = h(t)* f (t) = 2 (t + 3)2

2u(t + 3)− 2 (t + 2)2

2u(t + 2)

= (t + 3)2u(t + 3)− (t + 2)2u(t + 2) Hence, y(−1) = 4−1= 3. Answer: (3) 7. For t ≥ 0 the convolution of f (t) = 2u(−t) with h(t) = 4e−4tu(t) is (only for t ≥ 0 )

y(t) = f (t)* h(t) = :

(1) 0 (2) 8u(t) (3) 4e−4tu(t)

(4) −4e−4tu(t) (5) −2e−4tu(t) (6) 2e−4tu(t)

(7) 4u(t) (8) e−4tu(t) (9) none of above

Solution 7. y(t) = 8 u(τ − t)

−∞

∞

∫ e−4τu(τ )dτ = 8 e−4τ dτt

∞

∫ = −2e−4τ ⎤⎦t

∞= 2e−4tu(t)

Answer: (6)

8. The transfer function of a complex circuit is given by H (s) =

Vout (s)Vin(s)

= 32 2 (4+ s)(4− s)s2 +8s+ 48

.

The circuit is excited by an input vin(t) = 0.5cos(4t + 45o )u(t) V. The magnitude and phase (in

degrees), respectively, of vout (t) in SSS at is: (1) 32, − 45o (2) 16, − 45o (3) 16, 45o

(4) 32, 0o (5) 16, 90o (6) 32, 45o


(7) 16, 0o (8) 16 2, − 45o (9) none of above

Solution 8. H ( j4) = 32 2 ( j4+ 4)(4− 4 j)

−16+ 32 j + 48= 2 ( j4+ 4)(4− j4)

1+ j. Hence

H ( j4) = 2 4 2 × 4 2

2= 32 and ∠H ( j4) = 45− 45− 45= −45o . Hence

vout ,ss(t) = 16cos(4t)u(t) V. Answer: (7) 9. For the circuit below, C1 = 1 F, C2 = 4 F, C3 = 1 F, vin(t) = 20u(t +1.01) V,

vC1(−1.1− ) = vC2(−1.1− ) = 0 , and vout (0) = 9 . Switch operation: (i) For −1≤ t <1, the switch is in position A. (ii) At t = 1 s, the switch moves to position B where it remains forever and ever.

Then vout (3) = (in volts): (1) 0 (2) 20 (3) 10 (4) 4 (5) 5 (6) 16 (7) 20 (8) 8 (9) 9 (10) none of above

Solution 9. By voltage-division,

VC2(s) =

14s

14s

+ 1s

20s= 4

s which is valid for –1 ≤ t < 1.

On the other hand, when 1 ≤ t, by conservation of charge,

C2 × 4+C3 × 4 = C1 +C2( )vout (3) ,

which implies that vout (3) =

C2 × 4+C3 × 4C2 +C3

= 16+ 95

= 5 V.

Answer: (5)


10. The circuit below is to be analyzed using modified loop analysis. Suppose C = 2 F, gm = 2

mhos, R1 = R2 = R3 = R4 = 2 Ω, L = 2 H, vC (0− ) = 2 V and iL(0− ) = 0 . The constraint equation in the modified loop analysis equations is:

(1) 2VC = I2 − I1 (2) − I1 +

s−1s

⎛⎝⎜

⎞⎠⎟

I2 =4s

(3) I1 −

s+1s

⎛⎝⎜

⎞⎠⎟

I2 =2s

(4) I1 −

s+1s

⎛⎝⎜

⎞⎠⎟

I2 =−2s

(5) − I1 +

s−1s

⎛⎝⎜

⎞⎠⎟

I2 =2s

(6) − I1 +

s−1s

⎛⎝⎜

⎞⎠⎟

I2 =−4s

(7) − I1 +

s−1s

⎛⎝⎜

⎞⎠⎟

I2 =−2s

(8) I1 −

s−1s

⎛⎝⎜

⎞⎠⎟

I2 = 0 (9) I1 +

s−1s

⎛⎝⎜

⎞⎠⎟

I2 =−4s

(10) none of above

Solution 10. gmVC = I2 − I1 . Hence,

gm

I2Cs

+vC (0− )

s

⎛

⎝⎜

⎞

⎠⎟ = I2 − I1 ⇒− I1 + 1−

gmCs

⎛⎝⎜

⎞⎠⎟

I2 =gmvC (0− )

s.

Hence − I1 +

s−1s

⎛⎝⎜

⎞⎠⎟

I2 =4s

. Answer: (2)


11. The circuit below is to be analyzed using modified node analysis. All circuit element parameters have the value of 1 unit EXCEPT gm = 2 Ω. vC2(0− ) = 2 = vout (0

− ) V. Then the constraint equation for the modified node equations is: (1) VC1 −VC2 = Id + 2IC2 (2) VC1 − (1+ 2s)VC2 − Id = −2

(3) VC1 − (1− 2s)VC2 − Id = −4 (4) VC1 − (1+ 2s)VC2 + Id = −2

(5) VC1 − 3sVC2 − Id = −4 (6) VC1 − (1− 2s)VC2 + Id = −4

(7) VC1 − (1+ 2s)VC2 − Id = 4 (8) VC1 − (1+ 2s)VC2 − Id = −4 (9) None of above

Solution 11. VC1 −VC2 = R2Id + gmIC2 . IC2 = C2sVC2 −C2vC2(0− ) . Therefore,

VC1 −VC2 = R2Id + gm C2sVC2 −C2vC2(0− )( ) . Equivalently,

VC1 − 1+ gmC2s( )VC2 − R2Id = −gmC2vC2(0− ) or VC1 − (1+ 2s)VC2 − Id = −2vC2(0− ) = −4 Answer: (8)

12. The circuit below has transfer function H (s) =

VoutVin

. It is known that L = 1 H

and R1 = R2 = 1 Ω. The COMPLETE range of rm for which the circuit is BIBO stable is:

(1) rm > −2 (2) rm > 2 (3) rm > −1 (4) rm >1 (5) rm <1 (6) rm < 2 (7) rm < −1 (8) rm < −2 (9) none of above


Solution 12.

1s+1

(Vin −Vout ) = IR AND

1s+1

(Vout −Vin )+ Vout − rmIR( ) = 0 .

Hence

−1s+1

(Vin −Vout )−rm

s+1(Vin −Vout )+Vout = 0

(1+ rm)s+1

1+(1+ rm)

s+1⎛⎝⎜

⎞⎠⎟

=VoutVin

= H (s) =1+ rm

s+ (2+ rm)( ) Thus, rm > −2 . Answer (1)

13. The transfer function of the circuit below is: H (s) = 2s+ 0.1

s+ 0.1. In the realization achieved by

a student named Micro Henry Farad, C1 = C2 = 1 F. Remembering his 7:30 am 202 lecture, he

reasoned that he had to magnitude scale to obtain proper impedance levels. The parts store only

had, 1 mF capacitors, named after his daughter, Millie, and all manner of resistor values. The

magnitude scaled resistor value is in kΩ:

(1) 1 (2) 2 (3) 20 (4) 4 (5) 5 (6) 10

(7) 0.01 (8) 8 (9) none of above


Solution 13. C1sVin = (C2s+G)(Vout −Vin )⇒ (C1 +C2)s+G( )Vin = (C2s+G)Vout . Hence,

H (s) =

(C1 +C2)s+GC2s+G

= 2s+ 0.1s+ 0.1

. Thus R = 10 Ω. Km = 1

1×10−3 = 103 . Hence Rnew = 10 kΩ.


Workout. (35 pts) In the circuit below, R1 = 1 Ω, R2 = 1.5 Ω, L = 1 H, C = 0.2 F. Suppose

iL(0− ) = 0 and vC (0− ) = 0 . For 0 ≤ t <10 , L vin(t)⎡⎣ ⎤⎦ =

20s(s+ 2)

= 10s− 10

s+ 2 and for t ≥10 s,

vin(t) = 0 . The switch S is in position A for 0 ≤ t <10 and moves to position B at t = 10 s where

it remains.

(a) (14 pts) Valid for 0 ≤ t <10 s, compute Vout (s) and vout (t) .

(b) (5 pts) Compute iL(10− ) . (Correct = 5 pts; Incorrect = 0 pts)

(c) (4 pts) Draw the equivalent circuit valid for 10 ≤ t s.

(d) (12 pts) Valid for 10 ≤ t s. Compute the equivalent impedance seen by the source, say

Zeq(s) (6 pts), Vout (s) (3 pts) and vout (t) (3 pts).

Solution Workout:

(i) Vout (s) = 20

s(s+ 2)× s

s+1= 20

(s+1)(s+ 2)= 20

s+1− 20

s+ 2 and

vout (t) = 20 e−t − e−2t( )u(t) V.

(ii) Clearly, vin(t) = 10u(t)−10e−2tu(t) V which reaches a constant 10 volts very quickly.

Hence, at t = 10− , the inductor looks like a short, and iL(10− ) = 10 A.

(iii) For t ≥10 s, the equivalent circuit is:


(iv) The impedance seen by the source is:

(6 pts)

Zeq(s) = 1

0.4+ 0.2s+ 1s

= s0.2s2 + 0.4s+1

= 5ss2 + 2s+5

= 5s(s+1)2 + 22

(3 pts) Hence, Vout (s) = −Zeq(s)

iL(10− )e−10s

s

⎛

⎝⎜

⎞

⎠⎟ =

−50e−10s

(s+1)2 + 22 and (3 more pts)

vout (t) = −25e−(t−10) sin(2(t −10))u(t −10) V.

ece-202 exam 2 march 1, 2017 - weeklyjoys...ece-202, exam 2 spring 17 9 11. the circuit below is to...

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