automatic control systems (fcs)...automatic control systems (fcs) lecture-6 time domain analysis of...

Automatic Control Systems (FCS)

Lecture-6Time Domain Analysis of 2nd Order Systems

Introduction• We have already discussed the affect of location of pole and zero on the

transient response of 1st order systems.

• Compared to the simplicity of a first-order system, a second-order systemexhibits a wide range of responses that must be analyzed and described.

• Varying a first-order system's parameters (T, K) simply changes the speedand offset of the response

• Whereas, changes in the parameters of a second-order system canchange the form of the response.

• A second-order system can display characteristics much like a first-ordersystem or, depending on component values, display damped or pureoscillations for its transient response.

Introduction• A general second-order system (without zeros) is

characterized by the following transfer function.

22

2

2 nn

n

sssR

sC

)(

)(

)2()(

2

n

n

sssG

Open-Loop Transfer Function

Closed-Loop Transfer Function

Introduction

un-damped natural frequency of the second order system,which is the frequency of oscillation of the system withoutdamping.

22

2

2 nn

n

sssR

sC

)(

)(

n

damping ratio of the second order system, which is a measureof the degree of resistance to change in the system output.

Example#1

42

42

sssR

sC

)(

)(

• Determine the un-damped natural frequency and damping ratioof the following second order system.

42 n

22

2

2 nn

n

sssR

sC

)(

)(

• Compare the numerator and denominator of the given transferfunction with the general 2nd order transfer function.

sec/radn 2ssn 22

422 222 ssss nn 50.

1 n

Introduction

22

2

2 nn

n

sssR

sC

)(

)(

• The closed-loop poles of the system are

1

1

2

2

nn

nn

Introduction

• Depending upon the value of , a second-order system can be setinto one of the four categories:

1

1

2

2

nn

nn

1. Overdamped - when the system has two real distinct poles ( >1).

-a-b-cδ

jω

Introduction

• According the value of , a second-order system can be set intoone of the four categories:

1

1

2

2

nn

nn

2. Underdamped - when the system has two complex conjugate poles (0 < <1)

-a-b-cδ

jω

Introduction


1

1

2

2

nn

nn

3. Undamped - when the system has two imaginary poles ( = 0).

-a-b-cδ

jω

Introduction


1

1

2

2

nn

nn

4. Critically damped - when the system has two real but equal poles ( = 1).

-a-b-cδ

jω

Time-Domain Specification

11

For 0< <1 and ωn > 0, the 2nd order system’s response due to a unit step input looks like


12

• The delay (td) time is the time required for the response toreach half the final value the very first time.


13

• The rise time is the time required for the response to rise from 10%to 90%, 5% to 95%, or 0% to 100% of its final value.

• For underdamped second order systems, the 0% to 100% rise time isnormally used. For overdamped systems, the 10% to 90% rise time iscommonly used.

13


14

• The peak time is the time required for the response to reach the first peak of the overshoot.

1414


15

The maximum overshoot is the maximum peak value of theresponse curve measured from unity. If the final steady-statevalue of the response differs from unity, then it is common touse the maximum percent overshoot. It is defined by

The amount of the maximum (percent) overshoot directlyindicates the relative stability of the system.


16

• The settling time is the time required for the response curveto reach and stay within a range about the final value of sizespecified by absolute percentage of the final value (usually 2%or 5%).

S-Plane

δ

jω

• Natural Undamped Frequency.

n

• Distance from the origin of s-plane to pole is naturalundamped frequency inrad/sec.

S-Plane

δ

jω

• Let us draw a circle of radius 3 in s-plane.

3

-3

-3

3

• If a pole is located anywhere on the circumference of the circle thenatural undamped frequency would be 3 rad/sec.

S-Plane

δ

jω

• Therefore the s-plane is divided into Constant NaturalUndamped Frequency (ωn) Circles.

S-Plane

δ

jω

• Damping ratio.

• Cosine of the angle betweenvector connecting origin andpole and –ve real axis yieldsdamping ratio.

cos

S-Plane

δ

jω

• For Underdamped system therefore, 900 10

S-Plane

δ

jω

• For Undamped system therefore,90 0

S-Plane

δ

jω

• For overdamped and critically damped systems therefore,

0

1

S-Plane

δ

jω

• Draw a vector connecting origin of s-plane and some point P.

P

45

707045 .cos

S-Plane

δ

jω

• Therefore, s-plane is divided into sections of constant damping ratio lines.

Example-2• Determine the natural frequency and damping ratio of the poles from the

following pz-map.

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0

-1.5

-1

-0.5

0

0.5

1

1.50.220.420.60.740.840.91

0.96

0.99

0.220.420.60.740.840.91

0.96

0.99

0.511.522.533.54

Pole-Zero Map

Real Axis (seconds-1)

Imagin

ary

Axis

(seconds-1

)

Example-3

• Determine the naturalfrequency and damping ratio ofthe poles from the given pz-map.

• Also determine the transferfunction of the system and statewhether system isunderdamped, overdamped,undamped or critically damped.

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-3

-2

-1

0

1

2

3

0.420.560.7

0.82

0.91

0.975

0.5

1

1.5

2

2.5

3

0.5

1

1.5

2

2.5

3

0.140.280.420.560.7

0.82

0.91

0.975

0.140.28

Pole-Zero Map

Real Axis (seconds-1)

Imagin

ary

Axis

(seconds-1

)

Example-4

• The natural frequency of closedloop poles of 2nd order system is 2rad/sec and damping ratio is 0.5.

• Determine the location of closedloop poles so that the dampingratio remains same but the naturalundamped frequency is doubled.

42

4

2 222

2

sssssR

sC

nn

n

)(

)(-2 -1.5 -1 -0.5 0 0.5 1

-3

-2

-1

0

1

2

3

0.280.380.5

0.64

0.8

0.94

0.5

1

1.5

2

2.5

3

0.5

1

1.5

2

2.5

3

0.080.170.280.380.5

0.64

0.8

0.94

0.080.17

Pole-Zero Map

Real Axis

Imagin

ary

Axis

Example-4• Determine the location of closed loop poles so that the damping ratio remains same

but the natural undamped frequency is doubled.

-8 -6 -4 -2 0 2 4-5

-4

-3

-2

-1

0

1

2

3

4

5

0.5

0.5

24

Pole-Zero Map

Real Axis

Imagin

ary

Axis

S-Plane

1

1

2

2

nn

nn

Step Response of underdamped System

222222 2

21

nnnn

n

ss

s

ssC

)(

• The partial fraction expansion of above equation is given as

22 2

21

nn

n

ss

s

ssC

)(

22 ns

22 1 n

2221

21

nn

n

s

s

ssC )(

22

2

2 nn

n

sssR

sC

)(

)(

22

2

2)(

nn

n

ssssC

Step Response


• Above equation can be written as

2221

21

nn

n

s

s

ssC )(

22

21

dn

n

s

s

ssC

)(

21 nd• Where , is the frequency of transient oscillationsand is called damped natural frequency.

• The inverse Laplace transform of above equation can be obtainedeasily if C(s) is written in the following form:

2222

1

dn

n

dn

n

ss

s

ssC

)(


2222

1

dn

n

dn

n

ss

s

ssC

)(

22

2

2

22

111

dn

n

dn

n

ss

s

ssC

)(

222221

1

dn

d

dn

n

ss

s

ssC

)(

tetetc dt

dt nn

sincos)(

21

1


tetetc dt

dt nn

sincos)(

21

1

ttetc ddtn

sincos)(21

1

n

nd

21

• When 0

ttc ncos)( 1


ttetc ddtn

sincos)(21

1

sec/. radn and if 310

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8


ttetc ddtn

sincos)(21

1


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4


0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

b=0

b=0.2

b=0.4

b=0.6

b=0.9


0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

wn=0.5

wn=1

wn=1.5

wn=2

wn=2.5

Time Domain Specifications of Underdamped system

Time Domain Specifications (Rise Time)

ttetc ddtn

sincos)(21

1

equation above in Put rtt

rdrdt

r ttetc rn

sincos)(21

1

1)c(tr Where

rdrdt

tte rn

sincos21

0

0 rnte

rdrd tt

sincos

21

0


as writen-re be can equation above

0

1 2

rdrd tt

sincos

rdrd tt

cossin

21

21rd ttan

21 1

tanrd t


21 1

tanrd t

n

n

drt

21 11

tan

drt

b

a1 tan

Time Domain Specifications (Peak Time)

ttetc ddtn

sincos)(21

1

• In order to find peak time let us differentiate above equation w.r.t t.

ttettedt

tdcd

ddd

tdd

tn

nn

cossinsincos)(

22 11

tttte dd

dddn

dntn

cossinsincos22

2

11

0

tttte dn

dddn

dntn

cossinsincos2

2

2

2

1

1

1

0


tttte dn

dddn

dntn

cossinsincos2

2

2

2

1

1

1

0

0

1 2

2

tte ddd

ntn

sinsin

0 tne 0

1 2

2

tt dddn

sinsin

0

1 2

2

d

nd t

sin


0

1 2

2

d

nd t

sin

0

1 2

2

d

n

0tdsin

01 sintd

d

t

,,, 20

• Since for underdamped stable systems first peak is maximum peaktherefore,

dpt

Time Domain Specifications (Maximum Overshoot)

pdpd

t

p ttetc pn

sincos)(

21

1

1)(c

1001

1

12

pdpd

t

p tteM pn

sincos

equation abovein Put d

pt

100

1 2

dd

ddp

dn

eM

sincos

Time Domain Specifications (Maximum Overshoot)

100

1 2

dd

ddp

dn

eM

sincos

100

1 2

1 2

sincosn

n

eM p

1000121

eM p

10021

eM p

equation above in Put 21-ζωω nd

Time Domain Specifications (Settling Time)

ttetc ddtn

sincos)(21

1

12 nn

n

T

1

Real Part Imaginary Part

Time Domain Specifications (Settling Time)

n

T

1

• Settling time (2%) criterion• Time consumed in exponential decay up to 98% of the input.

ns Tt

44

• Settling time (5%) criterion• Time consumed in exponential decay up to 95% of the input.

ns Tt

33

Summary of Time Domain Specifications

ns Tt

44

ns Tt

33 100

21

eM p

21

nd

pt21

nd

rt

Rise Time Peak Time

Settling Time (2%)

Settling Time (4%)

Maximum Overshoot

Example#5• Consider the system shown in following figure, where

damping ratio is 0.6 and natural undamped frequency is 5rad/sec. Obtain the rise time tr, peak time tp, maximumovershoot Mp, and settling time 2% and 5% criterion ts whenthe system is subjected to a unit-step input.

Example#5

ns Tt

44

10021

eM p

dpt

drt

Rise Time Peak Time

Settling Time (2%) Maximum Overshoot

ns Tt

33

Settling Time (4%)

Example#5

drt

Rise Time

21

1413

n

rt.

rad 9301 2

1 .)(tan

n

n

str 550

6015

9301413

2.

.

..

Example#5

nst

4

dpt

Peak TimeSettling Time (2%)

nst

3

Settling Time (4%)

st p 78504

1413.

.

sts 331560

4.

.

sts 1560

3

.

Example#5

10021

eM p

Maximum Overshoot

1002601

601413

.

..

eM p

Example#5Step Response

Time (sec)

Am

plit

ude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

0.2

0.4

0.6

0.8

1

1.2

1.4

Mp

Rise Time

Example#6• For the system shown in Figure-(a), determine the values of gain K

and velocity-feedback constant Kh so that the maximum overshootin the unit-step response is 0.2 and the peak time is 1 sec. Withthese values of K and Kh, obtain the rise time and settling time.Assume that J=1 kg-m2 and B=1 N-m/rad/sec.

Example#6

Example#6

Nm/rad/sec and Since 11 2 BkgmJ

KsKKs

K

sR

sC

h

)()(

)(

12

22

2

2 nn

n

sssR

sC

)(

)(

• Comparing above T.F with general 2nd order T.F

Kn K

KK h

2

1 )(

Example#6

• Maximum overshoot is 0.2.

Kn K

KK h

2

1 )(

2021

.ln)ln(

e

• The peak time is 1 sec

dpt

245601

1413

.

.

n

21

14131

n

.

533.n

Example#6

Kn K

KK h

2

1 )(

963.n

K533.

512

533 2

.

.

K

K

).(.. hK512151224560

1780.hK

Example#6963.n

nst

4

nst

3

21

n

rt

str 650. sts 482.

sts 861.

Example#7When the system shown in Figure(a) is subjected to a unit-step input,the system output responds as shown in Figure(b). Determine thevalues of a and c from the response curve.

)( 1css

a

Example#8Figure (a) shows a mechanical vibratory system. When 2 lb of force(step input) is applied to the system, the mass oscillates, as shown inFigure (b). Determine m, b, and k of the system from this responsecurve.

Example#9Given the system shown in following figure, find J and D to yield 20%overshoot and a settling time of 2 seconds for a step input of torqueT(t).

Example#9

Step Response of critically damped System ( )

• The partial fraction expansion of above equation is given as

22

n

n

ssR

sC

)(

)(

22

n

n

sssC

)(

Step Response

22

2

nnn

n

s

C

s

B

s

A

ss

211

n

n

n ssssC

)(

teetct

nt nn 1)(

tetc ntn

11)(

1

Step Response of overdamped and undamped Systems

• Home Work

72

Example 10: Describe the nature of the second-order systemresponse via the value of the damping ratio for the systems withtransfer function

Second – Order System

128

12)(.1

2

sssG

168

16)(.2

2

sssG

208

20)(.3

2

sssG

Do them as your own revision

automatic control systems (fcs)...automatic control systems (fcs) lecture-6 time domain analysis of...

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