ECE-202 Exam II March 5, 2013

Name: __________________________________

(Please print clearly)

Student ID: _________________

CIRCLE YOUR DIVISION

Section 2022, 8:30 MWF Section 2021, 12:30 MWF Prof. DeCarlo Prof. DeCarlo

Section 2023 1:30 TuTh

Prof. Meyer

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 13

Multiple Choice. 5 points each totaling 50 points.

1. A transfer function, H (s) = K (s − ??)

(s − ??) has gain constant, K = 12 , and has the pole zero plot shown

below. The magnitude of the frequency response at ω = 4 rads/sec is: (1) 1 (2) 2 (3) 0.33333 (4) 4 (5) 5 (6) 48 (7) 12 (8) 120 (9) none of these

Solution 1. H (s) = 12 s(s − 3)(s + 3)(s2 + 4)

. H ( j4) = 12 j4( j4 − 3)( j4 + 3)(−16 + 4)

⇒ H ( j4) = 12 × 4 × 55 ×12

= 4

2. The step response of a circuit is vout,step(t) = 50 1− e

−2(t−1)( )u(t −1) V. Then the impulse response at

t = 2 second is (in V): (1) 15 (2) 15+δ (t − 2) (3) 40+δ (t − 2) (4) 40 (5) 100 (6) 30 (7) 30+δ (t − 2) (8) 20 (9) none of these

x -1 -2 -3 -4 ex 0.4 0.15 0.05 0.02

Solution 2. dvout,step

dt= 50 2e−2(t−1)( )u(t −1)+ 50 1− e−2(t−1)( )δ (t −1) = 100e−2(t−1)u(t −1) . At t = 2

sec we have that dvout,step

dt(1) = 15 .

Exam 2, ECE-202, Sp 13

3. The step response of a newly developed circuit is vout (t) = 10u(t)− 8e−2tu(t) V. The transfer

function H (s) = Vout (s)Vin (s)

= :

(1) 20s(s + 2)

(2) 20s + 2

(3) −8ss + 2

(4) 10s + 20s + 2

(5) 8s − 20s + 2

(6) 2s + 20s + 2

(7) 8ss + 2

(8) none of above

Solution 3. H (s) =

10s− 8s + 21s

= 10 − 8ss + 2

= 10s − 8s + 20s + 2

= 2s + 20s + 2

.

4. In the circuit below, L = 0.5 H, C = 0.5 F, R = 2 Ω, and iin(t) = 25 A (dc) for all time. Suppose that switch S1 has been in position B for a long time and switch S2 has been open for all past time. At time t = 0 , switch S1 moves to position A and switch S2 closes on position B instantaneously. If vout (0

− ) = 0 , then at t = 2 seconds, vout (t) is closest to (in volts): (1) 1 (2) 2 (3) –2 (4) 4 (5) –4 (6) –6 (7) –1 (8) –15 (9) none of these

x -1 -2 -3 -4 ex 0.4 0.15 0.05 0.02

Solution 4. iL (0

− ) = 25 A. Using the voltage source equivalent circuit for the inductor for t > 0 we

have by voltage division Vout (s) = −

1Cs

1Cs

+ Ls + RLiL (0

− ) =− 1C

s2 + RLs + 1

LC

iL (0− ) . Hence,

Exam 2, ECE-202, Sp 13

Vout (s) =−50

s2 + 4s + 4= −50(s + 2)2

. Thus vout (t) = −50te−2tu(t)⇒ vout (2) = −100e−4 V which means

vout (s) ≅ −2 V. 5. A linear circuit with transfer function

H (s) = Vout (s)Vin(s)

= 5 3− 2s2s2 + 2s +11

has voltage input vin(t) = 2cos(2t) V. Then vout,ss (t) has magnitude and phase (in rads) equal to:

(1) 1, 2 tan−1 −43

⎛⎝⎜

⎞⎠⎟ (2) 5, −2 tan−1 4

3⎛⎝⎜

⎞⎠⎟ (3) 10, 2 tan−1 4

3⎛⎝⎜

⎞⎠⎟

(4) 25, 2 tan−1 −43

⎛⎝⎜

⎞⎠⎟ (5) 5, 2 tan−1 4

3⎛⎝⎜

⎞⎠⎟ (6) 10, −2 tan−1 4

3⎛⎝⎜

⎞⎠⎟

(7) 25, 2 tan−1 43

⎛⎝⎜

⎞⎠⎟ (8) 1, 2 tan−1 4

3⎛⎝⎜

⎞⎠⎟ (9) none of these

Solution 5. H ( j2) = 5 3− 4 j−8 + 4 j +11

= 5 3− 4 j3+ 4 j

. H (2 j) = 5 and

∠H (2 j) = tan−1 −43

⎛⎝⎜

⎞⎠⎟ − tan

−1 43

⎛⎝⎜

⎞⎠⎟ = −2 tan−1 4

3⎛⎝⎜

⎞⎠⎟ . vout,ss (t) = 10cos 2t − 2 tan

−1 43

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

V.

6. The circuit below is to be frequency scaled so that the value of its frequency response at ω = 10 rad/s occurs at 200 rad/s. Further, the circuit is to be magnitude scaled so that the smallest capacitor is 0.01 F. The value of the largest capacitor after frequency and magnitude scaling is: (1) 0.1 (2) 0.2 (3) 0.01 (4) 0.4 (5) 0.02 (6) 0.04 (7) 0.05 (8) none of above

Exam 2, ECE-202, Sp 13

Solution 6. 0.01= 2KmK f

⇒ Km = 200K f

= 20020

= 10 . Clargest =4200

= 0.02 .

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

Solution 7. At s = 0 the top inductor is a short whereas the top cap is a short at s = ∞. Conversely, the vertical capacitor is an open at s = 0 while the vertical inductor is open at s = ∞. There is a point at which the parallel top branch is infinite impedance stopping current flow and there is a point where the vertical branch has zero impedance making the voltage output zero—both of these frequencies are eliminated making the circuit a band-reject filter. 8. The phase response of the famous Meyer-DeCarlo sleep-inducing lecture filter is shown below. The transfer function that has a phase response which best approximates the one shown is:

(1) − s + 20s + 400

(2) 20 − ss + 400

(3) s + 20s + 650

(4) 20 − ss + 650

(5) s +100s + 800

(6) 100 − ss + 800

(7) − s +10s + 500

(8) none of above

Exam 2, ECE-202, Sp 13

Solution 8. From the graph, at ω = 0 the phase is 0o and at ω = ∞ the phase is −180o . At approximately at ω = 20 the phase is −45o . Finally at ω = 400 the phase is −135o . ANSWER: 2. 9. In the circuit below, vin(t) = 6 V for all t. C1 = 5 F and C2 = C3 = 10 F. At time t = 0− , the switch,

S, is positioned as indicated (neither at A nor B) and vout (0− ) = 10 V. At t = 0+ , the switch moves to A

and at time t = 1 second it moves to B. vout (t) = at t = 2 seconds in V (at least approximately): (1) 10 (2) –14 (3) –4 (4) 4 (5) –5 (6) –6 (7) 14 (8) none of these

Exam 2, ECE-202, Sp 13

Solution 9. Ceq = 5 F and vCeq(1− ) = 10 V. H (s) = −C1s

Ceqs= −1 . By superposition,

Vout (s) =−6s+ 10s= 4s

in which case vout (2) = 4

10. Consider the functions ( )f t and ( )h t (both shown below). Suppose ( ) ( ) ( )y t f t h t= ∗ . Then (4) :y =

(1) 1 (2) 2 (3) 3 (4) 4 (5) 12 (6) 6 (7) 8 (8) none of these

01−

( )f t

3

4

0

( )h t

1−

3

t t3

Solution 10. Do a flip and shift on h(t) to produce h(4 −τ ) on the τ − axis . This would overlap f (τ ) for 1≤ τ ≤ 3 . The product h(4 −τ ) f (τ ) is nonzero only for 1≤ τ ≤ 3 and for this interval is a right triangle of height 12 and a base of 2. Hence, AREA = 12= y(4) .

Exam 2, ECE-202, Sp 13

Workout Problem 1: (20 points) Workout Problem 1: (a) (8 pts) Show that the following is true:

21( ) ( ) ( ) ( )2t u ty t u t r t= ∗ =

(b) (12 points) Consider h(t) and f (t) shown below: (i) (4 pts) Express the ( )h t and ( )f t as a sum of possibly shifted steps and ramps;

(ii) (8 pts) Using the result of part (a) of this problem, a similar class result for u(t)*u(t) , and the general time shift theorem for convolution, etc. compute ( ) ( ) ( )y t h t f t= ∗ . A different approach will result in a 30% point reduction.

Solution Approach 1. Direct integration

y(t) =−∞

+∞∫ r(τ )u(t −τ )dτ =

−∞

+∞∫ τu(τ )u(t −τ )dτ =

0

+∞∫ τu(t −τ )dτ

Case 1: t < 0 . Here u(t −τ ) = 0 implying that y(t) = 0 , t < 0 .

Case 2: t ≥ 0 . Here u(t −τ ) =1, τ ≤ t0, τ > t

⎧⎨⎪

⎩⎪ . Therefore,

y(t) =0

t∫ τdτ = 1

2τ 2

0

t= 12t2, t ≥ 0 . Combining cases 1 and 2, u(t)*r(t) = 1

2t2u(t) .

Approach 2. Convolution algebra: Let h(t) = u(t) and f (t) = r(t) = tu(t) , then use

y(t) = h(t)(1) ∗ f (t)(−1) .

h(t)(1) = du(t)dt

= δ (t)

Exam 2, ECE-202, Sp 13

f (t)(−1) =−∞

t∫ τu(τ )dτ =

0

t∫ τu(τ )dτ = 1

2τ 2u(τ )

0

t= 12t2u(t)

Thus,

y(t) =−∞

+∞∫ δ (t −τ ) 1

2τ 2u(τ )dτ =

t−t+

∫12τ 2u(τ )δ (t −τ )dτ = 1

2τ 2u(τ )

τ=t= 12t2u(t)

(b) f (t) = 4u(t +1)− 4u(t − 2) and h(t) = 2r(t + 2)− 4u(t + 2) . So

y(t) = 8u(t +1)*r(t + 2)−16u(t +1)*u(t + 2)− 8u(t − 2)*r(t + 2)+16u(t − 2)*u(t + 2) = 4(t + 3)2u(t + 3)−16r(t + 3)− 4t2u(t)+16r(t)

Exam 2, ECE-202, Sp 13

Workout problem 2: (30 points) The switch S1 has been in position A for a very long time during which vin(t) = 100 V (dc). Switch S2 is open. A Genie named Janus-Jonelle has secretly placed a 55 V

initial voltage on C2 using a high frequency impulse RAY—i.e., vC2(0− ) = 55 V. For convenience and

convention, all voltages are measured from top to bottom as plus to minus.

At t = 0 switch S1 moves to position B where it remains forever. At t = 1 switch S2 moves to position C where it remains forever. The values of the components are: C1 = 0.5 F, C2 = 0.4 F, R1 = 1 Ω, and R2 = 5 Ω.

(a) (2 pts) Find vC1(0− )

(b) (8 pts) Determine the voltages across the capacitors at 0 ≤ t <1 second. (c) (2 pts) Draw an equivalent circuit in the s-world using ONLY the VOLTAGE source models of a charged capacitor so that there is no confusion of what the current IC2(s) happens to be. (d) (8 pts) Write a single node equation in the unknown Vout (s) and solve for Vout (s) in terms of k. (e) (4 pts) For what range of values of k is the response bounded. (f) (6 pts) If k = 2 , find vout (t) and identify the transient and steady state parts of the response. Solution. (a) vC1(0

− ) = 100 V.

(b) C1vC1(0− )+C2vC2(0

− ) = (C1 +C2 )vC (0+ ) in which case

vC (0+ ) = C1vC1(0

− )+C2vC2(0− )

(C1 +C2 )= 50 + 22

0.9= 720.9

= 80 V.

(c) Draw

(d) Each capacitor is modeled as an ideal cap in series with a voltage source of value 80se−s . The

resulting node equation is:

C1s Vout (s)−80se−s⎛

⎝⎜⎞⎠⎟ +C2s Vout (s)−

80se−s⎛

⎝⎜⎞⎠⎟ − kC2s Vout (s)−

80se−s⎛

⎝⎜⎞⎠⎟ + 0.2Vout (s) = 0

C1 +C2(1− k)( )s + 0.2⎡⎣ ⎤⎦Vout (s) = 80 C1 +C2(1− k)( )e−s = 8e−s

Exam 2, ECE-202, Sp 13

or

Vout (s) =

80.5 + 0.4(1− k)( )

s + 0.20.5 + 0.4(1− k)( )

e−s

(e) 0.5 + 0.4(1− k) > 0⇒ 0.9 > 0.4k⇒ 94> k .

(f) Vout (s) =

80.5 + 0.4(1− k)( )

s + 0.20.5 + 0.4(1− k)( )

e−s = 80s + 2

By inspection, vout (t) = 80e

−2(t−1)u(t −1) V.