Closed Loop Transfer Function - Electrical Control Engineering - Exam, Exams of Electrical Engineering

Main points of this exam paper are: Closed Loop Transfer Function, Input Voltage, Voltage Across, Transfer Function, Circuit Components, Pole-Zero Diagram, System Poles, Unit Step, Steady State Value, Damped

Typology: Exams

2012/2013

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CORK INSTITUTE OF TECHNOLOGY
INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ
Summer Examination 2001/11
Module Title: Electrical Control Engineering
Module Code: ELEC 7003
School: Electrical and Electronic Engineering
Programme Title: Bachelor of Engineering in Electrical Engineering Year 3
Programme Code: EELEC_7_Y3
External Examiner(s): Mr. G. Beecher, Dr. M. Duffy.
Internal Examiner(s): Mr. N. Canty.
Instructions: Answer any three questions
Duration: 2 Hours
Sitting: Summer 2011
Requirements for this examination:
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have received the
correct examination. If in doubt please contact an Invigilator.
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CORK INSTITUTE OF TECHNOLOGY

INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ

Summer Examination 2001/

Module Title: Electrical Control Engineering

Module Code: ELEC 7003

School: Electrical and Electronic Engineering

Programme Title: Bachelor of Engineering in Electrical Engineering – Year 3

Programme Code: EELEC_7_Y

External Examiner(s): Mr. G. Beecher, Dr. M. Duffy.

Internal Examiner(s): Mr. N. Canty.

Instructions: Answer any three questions

Duration: 2 Hours

Sitting: Summer 2011

Requirements for this examination:

Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have received the

correct examination.

If in doubt please contact an Invigilator.

Q

Consider the RLC circuit shown in Fig. 1 below, where the input voltage, vi   t , is the DC supply

voltage. The output voltage, vo   t , is the voltage across the capacitor C. The circuit can be

represented by the following differential equation;

   

  itdt dt c

dit vi t Rit L ()

   

(a) Show that the system transfer function can be given by the following equation;

 

 

LC

s L

R

s

LC

V s

V s

i

o 1

2  

10 marks

(b) Given that the value of the circuit components are as follows; R = 15Ω, L = 5H and C = 0.02F,

evaluate the system transfer function. 5 marks

(c) Draw a pole-zero diagram to show the location of the system poles and zeros 5 marks

Fig.

vi   t

vo   t

Q

(a)

Sketch the location of the poles and zeros on pole-zero diagrams for the following transfer functions;

(i) 5

s

s (ii)  2 5 

2 sss

6 marks

(b) Consider the closed loop control system shown below in Fig. 3.

Fig. 3

(i) Calculate the closed loop transfer function

 

R   s

Ys 10 marks

(ii)Assuming   s

R s

 , calculate the steady state value of the output, yss , using the Final Value

Theorem (FVT) 4 marks

R(s) (^) U(s) Y(s)

E(s)

G(s)

1

H(s)

s

12

C(s)

Q

Consider the closed loop control system shown below in Fig. 4 where C(s) is a PI Controller.

Fig. 4

(a) Show that the closed loop transfer function is given by

 

 

7

2 p i

p i

K K

s s

K s K

Rs

Ys

10 marks

(b) Determine values for Kp and Ki such that the closed loop system will have a critically

damped response with a 1% settling time, Ts  1 %= 18 seconds. Assume  

n

Ts

10 marks

R(s) (^) U(s) Y(s)

E(s)

C(s) G(s)

1

H(s)

s

K

K

i p  7 1

s

Laplace Transform Pairs

f(t) F(s)

1 Unit impulse^   t 1

2 Unit Step^1   t s

3 t 2

s

4  1 !

1

n

t

n

( n^ = 1,2,3,…) n s

5

n t ( n = 1,2,3,…) 1

ns

n

6

at e

sa

7

at te

  

2

sa

8  

n at t e n

 

1

1!

( n^ = 1,2,3,…)  

n sa

9

n at t e

 ( n = 1,2,3,…)  

1

 

n s a

n

10 sin  t 2 2 

s

11 cos  t 2 2 s  

s

12 sinh  t 2 2 

s

13 cosh  t 2 2 s  

s

14 ^ 

at e a

 1 

ssa

15 ^ 

at bt e e b a

   

 (^) sa  (^) sb

16 ^ 

bt at be ae b a

   

s a  s b

s

 

17 ^  

atbt be ae ab a b

ssa  sb

18 ^ 

at at e ate a

  1  

2

2

s sa

19 ^ 

at at e a

  1 

2

s  s  a 

2

(^20) e t

at

sin 

(^22)

sa

21 e t

at

cos 

2 2

s a

s a

22

e (^) n t

n (^) nt 2 2

sin 1 1

2 2

2

(^2) n n

n

s   s

23

   

e (^) n t nt^2 2

sin 1 1

 

2 1 1 tan

0  ^  )

2 2 s (^2) ns n

s

24

   

e (^) n t nt^2 2

sin 1 1

 

2 1 1 tan

0  ^  )

2 2

2

(^2) n n

n

ss   s

25 1 cos  t

2 2

2

ss

26  t^ sin^  t

2 2 2

3

s s

27 sin^  t^  t cos^  t

2 22

3 2

s

28 t^  t

sin 2

2 22 s  

s

29 t cos^  t

2 22

2 2

s

s

30

 1 t 2 t 

2 1

2 2

cos cos

   

2 2

2

2 2

2 2 1

2

s   s  

s

31 ^  t  t^  t 

sin cos 2

2 22

2

s  

s