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# Control systems

## 1.

Control SystemsDynamic Response: Dynamic Response

Analysis, Steady State Error

Md Hazrat Ali

Department of Mechanical Engineering,

School of Engineering,

Nazarbayev University

## 2.

By failing to prepare, you arepreparing to fail.

Benjamin Franklin

## 3.

Contents-Review of Previous Lectures

-System Response Analysis

## 4. Review

• Once transfer function is obtained, we can startto analyze the response of the system it

represents .

• A block diagram is a convenient tool to visualize

the systems as a collection of interrelated

subsystems that emphasize the relationships

among the system variables.

• Signal flow graph and Mason’s gain formula are

used to determine the transfer function of the

complex block diagram.

## 5. Review-Block Diagram

• Three Elementary Block Diagrams– Series connection

– Parallel connection

– Negative Feedback connection

## 6. Negative feedback :Single-loop gain

The gain of a single-loop negative feedback system isgiven by the forward gain divided by the sum of 1 plus the

loop gain.

Franklin et.al- pp.122

## 7. Review-Block Diagram

## 8. Table 2.6 (continued) Block Diagram Transformations

Review-Block DiagramTable 2.6 (continued) Block Diagram Transformations

## 9. Review-Block Diagram

Practice: Find the transfer function of thefollowing block diagram

Y(s)

2s 4

2

R (s) s 2s 4

## 10. Review-Block Diagram

Practice:## 11. Review-Block Diagram

## 12.

Time domain and frequency domain## 13.

## 14.

## 15.

## 16.

## 17. Poles and Zeros

H (s) Km

i 1

n

i 1

( s zi )

( s pi )

K is the transfer gain

The roots of numerator is called zeros of the system. Zeros

correspond to signal transmission-blocking properties.

The roots of denominator are called poles of the system. Poles

determine the stability properties and natural or unforced

behavior of the system.

Poles and zeros can be complex quantities.

zi=pi, cancellation of pole-zero may lead to undesirable system

properties.

## 18. System Response: Complex system

## 19.

## 20.

Key points: Effect of Poles and Zeros## 21. System Response

Example:Consider the following transfer function

Y (s)

2s 1

2

R ( s ) s 3s 2

Determine:

• Poles and Zeros?

21

## 22. System Response: Effect of pole location

## 23. System Response

Example:Consider the following transfer function

Y ( s)

s 3

2

R ( s ) s 5s 6

Determine:

• Poles and Zeros?

23

## 24. Time-Domain Specification

Test Input Signals• To measure the performance of a system we use

standard test input signals. This allows us to compare

the performance of our system for different designs.

• The standard test inputs used are the step input, the

ramp input, and the parabolic input.

• A unit impulse function is also useful for test signal

purpose.

## 25. Time-Domain Specification

Test Input Signalsstep input

ramp input

parabolic input

A, t 0

r (t )

0, t 0

A

R( s)

s

At , t 0

r (t )

0, t 0

A

R( s) 2

s

At 2 , t 0

r (t )

0, t 0

2A

R( s) 3

s

The step input is the easiest to generate and evaluate and is

usually chosen for performance tests.

## 26. Time-Domain Specification

ExampleThe transfer function:

9

G (s)

s 10

The system response to a unit step input (A=1):

y (t ) 0.9 1 e 10t

y ( ) 0.9

## 27. System Response

Example 2:Consider the following transfer function

Y ( s)

7s 1/ 2

2

R ( s ) s 5s 6

Determine:

• impulse response

## 28.

First Order System Response## 29.

First Order System ResponseExample 1:

Consider the following transfer function

Y (s)

1

H (s)

R(s) s

Determine:

• impulse response(response when r(t) is

impulse function)

•Classify stability

## 30.

First Order System Response- Impulse response1/

## 31.

Standard Second Order System• Let us consider the following closed-loop system:

• The TF of the closed-loop system:

G(s)

K

T ( s)

2

1 G ( s ) s ps K

• Utilizing the general notation of 2nd Order System:

T (s) 2

s 2 n s n2

2

n

• Where n is natural frequency and is damping ratio

## 32.

Standard Second Order Systemd n 1

2

cos 1

0 1

## 33. Figure 3.24 Graphs of regions in the s-plane delineated by certain transient requirements: (a) rise time; (b) overshoot; (c) settling time; (d) composite of all three requirements

Transformation of the specification to the splaneFigure 3.24 Graphs of regions in the s-plane delineated by certain transient requirements: (a) rise time; (b) overshoot;

(c) settling time; (d) composite of all three requirements

## 34. Transformation of the specification to the s-plane

Example 3.25Find allowable regions in the s-plane for the poles

transfer function of system if the system response

requirements are tr ≤ 0.6, Mp <= 10% and ts <= 3 sec.

## 35.

Standard Second Order SystemMp?

## 36.

Standard Second Order SystemStandard Second Order System

Poles (roots) location of the second order complex

system.

s1,2 n j 1 2 n n j d

where

d is the "damped natural frequency".

## 37.

Standard Second Order System• Classification of Type Response of 2nd Order Systems

Undamped: =0

Critical damped: =1

Under-damped: 0< <1

Over-damped: >1

## 38.

Standard Second Order SystemAs decreases, the response becomes increasingly

oscillatory.

## 39. Time-Domain Specification

• Standard performance measures are usually defined in termof the step response of a 2nd order systems:

## 40. Time-Domain Specification

• Standard performance measures are usually definedin term of the step response of a 2nd order systems:

– Rise time, Tr : time needed from 0 to 100% of fv for

underdamped systems and Tr1 from 1090% of fv for

overdamped systems.

– The settling time ts is the time it takes the system transient

to decay.

– The overshoot Mp is the maximum amount of the system

overshoots its final value divided by its final value.

– The peak time tp, is the time it takes the system to reach

the maximum overshoot.

## 41. Time-Domain Specification

-Rise Time, Tr-A precise analytical relationship between rise time

and damping ratio cannot be found. However, it

can be found using numerically using computer.

A rough estimation of the rise time is as follows

1.8

tr

n

## 42. Time-Domain Specification

-Maximum Overshoot, MpMaximum overshoot (in percentage) is defined as

Mp

' Peak Value ' ' Final Value '

100%

' Final Value '

M p 100e

1 2

## 43. Time-Domain Specification

-Peak Time TpTp is found by differentiating y(t) and finding thefirst zero crossing after t=0.

Tp

d n 1 2

## 44. Time-Domain Specification

-Settling Time Ts-For a second order system, we seek to determine the

time Ts for which the response remains within

certain percentage (1%, 2% ) of the final value.

For 1% settling time

4.6.6

4

Ts

T

s

n n

For 2% settling time

4

Ts

n

## 45. Time-Domain Specification

Exercise # 1Find Tr, Tp, Mp and Ts for the following transfer

function:

25

T (s) 2

s 30 s 225

45

## 46. Time-Domain Specification

Exercise # 2Find Tr, Tp, Mp and Ts for the following transfer

function:

19

T (s) 2

s 24 s 19 / 3

46

## 47. Exercise # 3

If the system response requirements are tr = 0.6, Mp =10% and ts = 3 sec.

Find:

n ,

47

## 48. Exercise # 4

Problem# If the system response requirements are tr =0.6, Mp = 10% and ts = 3 sec.

Find:

n ,

For 1% settling time

1.8

tr

n

4.6

Ts

n

48

## 49. Time-Domain Specification

Exercise # 5Find Tr, Tp, Mp and Ts for the following transfer

function:

1

T (s) 2

s 15s 100

## 50. Time-Domain Specification

Exercise # 6Find Tr, Tp, Mp and Ts for the following transfer

function:

5

T (s) 2

s 30 s 225

50

## 51. Time-Domain Specification

Exercise # 7Find Tr, Tp, Mp and Ts for the following transfer

function:

3

T (s) 2

s 24 s 9

51

## 52. Figure - Multiple-loop feedback control system.

Example - Block diagramFind TF from the given block diagram

Figure - Multiple-loop feedback control system.

## 53. Figure 2.27 Block diagram reduction of the system of Figure 2.26.

Quiz # 4- Answer to Q1Figure 2.27 Block diagram reduction of the system of Figure 2.26.

## 54. System Response

Find TF from the given block diagramConsider the following transfer function

Y (s)

2s 1

H (s)

2

R ( s ) s 3s 2

Determine:

i) Impulse response graphically

ii) Classify stability

## 55. Impulse response

Answer## 56. Midterm Exam

March 4, 2016, Friday,Time:8.00-9.00

Venue-6.141 & 5.103

Topics- Cover Until February

## 57.

## 58. Tell me, I will forget! Show me, I may remember! Involve me, I will understand!

Benjamin Franklin## 59. Further Reading

Franklin, et. al., Chapter 3Section 3.1-3.6

Richard C. Dorf et.al, Chapter 3

Additional notes are uploaded on moodle