ECE 382 Fall 2016

Final ExamDecember 14, (Wednesday)

Name:

ID #:

• This is a 120 minute duration exam.

• Closed books, closed notes, no calculators, no crib-sheets.

• There are 15 multiple choice questions each worth 10 points (Part I), and five workout

problems each worth 10 points (Part II).

• Use a No. 2 pencil to fill in the computer score sheet.

• Keep your scan form hidden when you are not marking the answer.

• Turn in the entire test and the computer score sheet. Your score on Part I will be

based solely on what you marked on the computer score sheet.

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Part I (Multiple Choice Questions)

1. Consider a system consisting of a motor driving a load shown in Figure 1.

Assume that viscous friction of the combination of the motor, load, and gear referred

to the motor shaft is negligible. The moment of inertia of the load is Il = 50 kg m2.

The moment of inertia of the motor is Im = 9.5 kg m2. The gear ratio is N1/N2 = 1/10.

The torque Tm, in N·m, that the motor has to produce in order for the load to rotate

at the angular velocity ωl = θl = 3t rad/sec is:

(A) 150 (B) 200 (C) 250 (D) 300 (E) None of the above

Figure 1: System for Problem 1.

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2. Consider the negative unity feedback closed-loop system whose open-loop trans-

fer function is

G(s) =K(s+ 2)

s(s+ 1)2.

The closed-loop system is BIBO stable if and only if

(A) 0 < K < 1 (B) K > 0 (C) 0 < K < 2

(D) K > 1 (E) None of the above

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3. When solving this problem, you may wish to use the graph shown in Figure 2.

A negative unity feedback system has a forward transfer function

G(s) =25

s(s+ 10ζ).

Select ζ so that the step response of the closed-loop system has a maximum percent

overshoot of 10%. Then, the settling time, ts, using the 5% criterion is (in seconds):

(A) 43

(B) 1 (C) 2 (D) 5 (E) 34

0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

60

70

80

90

100

ζ

Mp(%

)

Figure 2: Mp versus ζ curve.

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4. We wish the dominant complex conjugate poles of a negative unity feedback

closed-loop system to be located at −2±j2. The deficiency angle of the uncompensated

closed-loop system at the upper dominant pole is 45◦. We use a PD compensator in

forward path. The transfer function of the compensator is

Gc(s) = Kc (1 + Tds) .

To achieve the design goal with this compensator, we choose Td to be:

(A) 1/4 (B) 1/2 (C) 2 (D) 4 (E) None of the above

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5. The step response c(t) of a first order system is such that

dc(t)

dt

t=0+

=1

15sec−1.

The time the step response reaches 95% of the final value is (in seconds):

(A) 15 (B) 30 (C) 145

(D) 160

(E) None of the above

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6. For the system shown in Figure 3, find the gain K so that Kv = 4√

2 sec−1.

Then, the phase margin, PM, (in degrees) of the closed-loop system is:

(A) 30◦ (B) −30◦ (C) 45◦ (D) −45◦ (E) None of the above

Figure 3: A block diagram of the system for Problem 6.

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7. Find Kc so that the dominant complex conjugate poles of the closed-loop

system shown in Figure 4 are located at −2± j√

3. The controller’s gain Kc is:

(A) 11 (B) 12 (C) 14 (D) 16 (E) None of the above

Figure 4: Closed-loop system with a compensator for Problem 7.

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8. Consider the following state-space model of a dynamical system:x1

x2

x3

=

0 1 0

0 0 1

0 0 0

x1

x2

x3

+

0

0

1

r

y =[

2 1 0

]x1

x2

x3

+ 5r.

The transfer function Y (s)/R(s) is equal to:

(A) s3

5s3+s+2 (B) 3s3+s+2s3 (C) 5s2+3

s3

(D) 5s3

3s2+2 (E) None of the above

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9. In Figure 5, Bode plots of an open-loop transfer function are shown. The gain

margin, in dB, of the unity feedback closed-loop system is about:

(A) -20 (B) 20 (C) 10 (D) -10 (E) None of the above

Frequency (rad/sec)

Ph

ase

(d

eg);

Magnitu

de (

dB

)

Bode Diagrams

−60

−40

−20

0

20

100

101

−260

−240

−220

−200

−180

Figure 5: Bode plots for Problem 9.

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10. For the closed-loop system, shown in Figure 6, the settling time, using the 5%

criterion, is (in secs):

(A) 0.8 (B) 5/4 (C) 15/16 (D) 0.5 (E) None of the above

Figure 6: Closed-loop system for Problem 10.

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11. The value of p for which the complex-conjugate closed-loop poles in the system

shown in Figure 7, are located at s = −2± j2 is about:

(A) −1 (B) −2 (C) −4 (D) −6 (E) None of the above

Figure 7: Closed-loop system for Problem 11.

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12. Which one of the following statements is incorrect?

(A) Phase-lag slows down transient response.

(B) Phase-lag compensator is used to decrease error constant while maintaining the

desired dominant poles in the s-plane.

(C) Phase-lead compensator requires additional amplifier.

(D) Phase-lead compensator increases system bandwidth.

(E) Phase-lead compensator is applied when fast transient response is desired.

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13. The input, e(t), to the type A proportional-plus-integral-plus-derivative (PID)

controller in standard form is the unit step function. The controller’s parameters are:

Kp = 10, Ti = 2 sec, and Td = 3 sec. The output u(t) of the controller at t = 2 sec is:

(A) 15 (B) 20 (C) 25 (D) 30 (E) 35

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14. Consider the negative unity feedback closed-loop system whose open-loop trans-

fer function is

G(s) =200

s(s+ 102).

The cutoff frequency, ωb, of the closed-loop system is about (in rad/sec)

(A) 200 (B) 102 (C) 2 (D) 1 (E) None of the above

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15. In Figure 8, Bode diagrams are shown of the uncompensated open-loop sys-

tem with the gain adjusted for the desired error coefficient. It is required that the

gain crossover frequency of the compensated system, using a lead compensator, be

30 rad/sec. The time constant T of the lead compensator was selected to be 1/6. The

transfer function of the resulting lead compensator is

(A)16s+11

100s+1(B)

1100s+116s+1

(C)16s+11

150s+1

(D)1

150s+116s+1

(E) 6s+1150s+1

100

101

102

−181

−180.8

−180.6

−180.4

−180.2

−180

−179.8

−179.6

−179.4

−179.2

−179

Ph

ase

(deg)

Bode Diagram

Frequency (rad/s)

−40

−30

−20

−10

0

10

20

30

40

System: sysFrequency (rad/s): 30Magnitude (dB): −19.1M

agnitude (

dB

)

Figure 8: Bode diagrams for Workout Problem 3.

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Part II (Workout Problems)

1. Consider the armature controlled DC motor characterized by the following parameters:

• viscous friction of the combination of the motor, load, and gear referred to the motor

shaft is negligible,

• moment of inertia of the combination of the motor, load, and gear referred to the motor

shaft Ieq = 0.1 kg·m2,

• armature inductance is negligible,

• the stall torque is τs = 0.1 Nm,

• when the armature voltage ea = 24 V and the motor is allowed to run unloaded, it

accelerates and eventually reaches a steady-state rotational speed of 240 rad/sec.

Find the transfer function ω(s)/Ea(s), where ω(s) is the Laplace transform of the motor

angular shaft velocity and Ea(s) is the Laplace transform of the armature voltage ea.

ω(s)Ea(s)

=

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Name:

Blank Page—To be used to show your work, if needed.

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2. Find the steady-state error for the system in Figure 9 for r(t) = (5t + sin t)1(t), where

1(t) is the unit step.

Hint: When using the final value theorem, you need to show clearly that its assumptions

are satisfied.

e(∞) =

Figure 9: Closed-loop system for Workout Problem 2.

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3. In Figure 10, Bode diagrams are shown of the uncompensated open-loop system with

the gain adjusted for the desired error coefficient. It is required that the gain crossover

frequency of the compensated system, using a lag compensator, be 1.57 rad/sec. Select

the corner frequency corresponding to the zero of the compensator one decade below the

new gain crossover frequency. Find the transfer function of the resulting lag compensator.

Gc(s) =

−10

0

10

20

30

40

50

System: sysFrequency (rad/s): 1.57Magnitude (dB): 20

Magnitude (

dB

)

10−1

100

101

−180

−135

−90

Pha

se (

deg)

Bode Diagram

Frequency (rad/s)

Figure 10: Bode diagrams for Workout Problem 3.

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4. For the negative unity feedback system with the plant transfer function

Gp(s) =2

s(s+ 2)2,

use one of attached tables to determine the parameters of the type A PID compensator.

Gpidstd(s) =

————————————————————————–

Table 1: Ziegler-Nichols tuning formulas based on the step response of the plant (open-loop

method).

Controller KP TI TD

P TL

∞ 0

PI 0.9TL

L0.3

0

PID 1.2TL

2L 0.5L

Table 2: Ziegler-Nichols tuning formulas based on critical gain Kcr and critical period Pcr

(closed-loop method).

Controller KP TI TD

P 0.5Kcr ∞ 0

PI 0.45KcrPcr

1.20

PID 0.6Kcr 0.5Pcr 0.125Pcr

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Blank Page—To be used to show your work, if needed.

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5. The input, e(t), to the PIDF controller in standard form is the unit step function. The

controller’s parameters are: Kp = 2, Ti = 2 sec, Td = 3 sec, and N = 100. Find the output

u(t) of the controller at t = 10 sec.

u(t)|t=10 =

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Blank Page—To be used to show your work, if needed.

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