1 Time Response. CHAPTER 3. 2 3.2 Poles and Zeros and System Response. Figure 3.1: (a) System...

1

Time Time ResponsRespons

e.e.

CHAPTER CHAPTER 33

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3.2 Poles and Zeros and System 3.2 Poles and Zeros and System Response.Response.

ssR

sssG

sRsGsCs

KsG

1)(

52)(

where)()()(

)(

55/35/2

5)5()2()(

)()(

ss

sB

sA

ssssC

ssKsC

t

t

eKKtc

etc

21

5

)(53

52)( Figure 3.1: (a) System showing input

and output; (b) Pole-zero plot of the system;

(c) Evolution of a system response.

3

The The transfer functiontransfer function,,

For a unit step of; 1/s,For a unit step of; 1/s,

Its response,Its response,

A, B and B’ are constant. For K=1 and A, B and B’ are constant. For K=1 and =1/a then,=1/a then,

Time constantTime constant (1/a), is defined as the time for e (1/a), is defined as the time for e-at-at to to decay to 37%of its initial value decay to 37%of its initial value or or the time it takes the time it takes for step response to reach 63% of its final value. for step response to reach 63% of its final value.

3.3 First Order System.3.3 First Order System.

1)()()(

sK

sRsCsG

111)(

sB

sA

sK

ssC

teBAtc ')(

atetc 1)(

4

Step responseStep response,,

ateKKtc 21)(

)()(

assKsC

Cont’d…Cont’d…

Figure 3.2: (a) First Order Response to a Unit Step.

5

The poleThe pole of the transfer function is at of the transfer function is at –a–a, the , the farther the pole from the imaginary axis, the farther the pole from the imaginary axis, the faster the transient response.faster the transient response.

Rise time (TRise time (Trr), ), the time the response to go the time the response to go from the 0.1 to 0.9 of its final value. from the 0.1 to 0.9 of its final value. TTrr=2.2/a=2.2/a..

Settling time (TSettling time (Tss)),, time range when the time range when the response to reach and stay within 2% of its response to reach and stay within 2% of its final value. Let c(t) = 0.98 then the final value. Let c(t) = 0.98 then the TTss.=4/a..=4/a.


6

3.4 Second Order System.3.4 Second Order System. The The transfer function,transfer function,

For Impulse response,For Impulse response, Where,Where,

Standard Form,Standard Form,

Where Where KK is the dc gain, is the dc gain, is the damping ratio, is the damping ratio, nn is the undammped natural is the undammped natural

frequency.frequency.

012

0

)()(

bsbsa

sRsC

2

2

1

1)(

ss

sC

22

2

2)()(

nn

n

ss

KsRsC

02 22 nnss

122,1 nns

WhereWhere

7

Example of 2Example of 2ndnd order system order system responses.responses.

Figure 3.3: Second Order System, pole plots and Step Figure 3.3: Second Order System, pole plots and Step Response.Response.


8

General 2General 2ndnd Order System. Order System.

Natural Frequency (Natural Frequency (nn),),

Damping Ratio (Damping Ratio ()),,Example 3.1Example 3.1: Find the Natural Frequency (: Find the Natural Frequency (nn) and ) and Damping Ratio (Damping Ratio ()),,

Solution:Solution:nn = 6 and = 6 and =0.35.=0.35.

)()(

)()()(

2 basssbsC

sGsRsC

n

n

n

n

nn

n

ab

ab

sssG

2

2

2)(

2

22

2


)362.4(36)( 2

ss

sG

9

From previous page,From previous page,

Figure 3.4: Left; Plot for an underdamped 2Figure 3.4: Left; Plot for an underdamped 2ndnd Order System. Order System. Right; Step Response for 2Right; Step Response for 2ndnd Order System Damping Cases. Order System Damping Cases.

22

2

2 2)(

nn

n

ssbassbsG

)()(

)()()(

2 basssbsC

sGsRsC

)cos()( 21 teKKtc t


1kos

10


Figure 3.5: Second Order Response as a Function of Damping Figure 3.5: Second Order Response as a Function of Damping Ratio.Ratio.


11

The transfer function,The transfer function, Poles at the origin from the Poles at the origin from the unit stepunit step and and

two real poles from the two real poles from the systemsystem.. Constant force response and force Constant force response and force

response.response.

3.4.1 Over Damped Response.3.4.1 Over Damped Response. ))146.1)(854.7(

9)99(

9)( 2

sssssssC

tt eKeKKtc 146.13

854.721)(

12

Example 3.1:Example 3.1: Over Damped Response. Over Damped Response.Find the step response of the system.

Solution:Expand the partial fraction.

Take the inverse Laplace Transform.

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Under Damped transfer function,Under Damped transfer function,

When 0 < When 0 < < 1 The transfer function is,< 1 The transfer function is, The Pole position is,The Pole position is,

3.4.2 Under Damped 3.4.2 Under Damped Response.Response.

dndn

n

jsjsK

sRsC

2

)()(

21 nd

dddd

nn

n

jsK

jsK

sssG

*

22

2

2)(

)92(9)( 2

sss

sC

14


Figure 3.6: Second Order Response as a Function of Damping Figure 3.6: Second Order Response as a Function of Damping Ratio.Ratio.


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Performance Measures.Performance Measures.

21

n

pT Peak Time

Overshoot

nsT

4

Settling Time

Figure 3.7: (Top) The 2nd Order Underdamped

Response Specification. (bottom) Percent overshoot

versus damping factor


%1001

11

%100%

21

e

cctc

M pp

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Performance Measures.Performance Measures.

Poles positionPoles position

Figure 3.8: Lines of constant peak time, Tp , settling time, Ts , and percent overshoot, %OS Note: Ts2 < Ts1 ; Tp2 < Tp1; %OS1 < %OS2

dn

pT

21

dnsT

44

21100%

eOS

dddd

nn

n

jsK

jsK

sssG

*

22

2

2)(


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Pole Placement.Pole Placement.

Figure 3.9: Step responses of second-order underdamped systems as poles move: (a) with constant real part; (b) with constant imaginary part; (c) with constant damping ratio


POd

sd TPO

sd T

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The transfer function,The transfer function,3.4.3 Critically Damped.3.4.3 Critically Damped.

22

)()(

n

n

ssRsC

19

Example 3.3:Example 3.3: Critically Damped Critically Damped Response.Response.Find the step response of the system.Solution:Expand the partial fraction.

20

Dominant Pole.Dominant Pole. The formula that describing %OS, tThe formula that describing %OS, tss, t, tp p

were were derived only for system with two derived only for system with two complex complex poles poles and and no zeros.no zeros.

A system with more than two poles or A system with more than two poles or zeros can be zeros can be approximated as a second approximated as a second order system that has just order system that has just two complex two complex dominant poles.dominant poles.

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Dominant Pole.Dominant Pole.

Figure 3.11: Component responses of a three-pole system: (a) pole plot; (b) component responses: non-dominant pole is

near dominant second-order pair (Case I), far from the pair (Case II), and at infinity (Case III).


Cannot be represented as second order system

Will approach second order system

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Cont’d…Cont’d…Effect of adding a zero to a two-pole systemEffect of adding a zero to a two-pole system The closer is the zero to dominant poles, The closer is the zero to dominant poles, the greater its the greater its effect on transient response.effect on transient response. As the zero move away from dominant As the zero move away from dominant poles, the poles, the response approaches that of the two response approaches that of the two pole system.pole system. Starting at poles

1±j2.828, then consecutively add zeros at -3, -5, -10.

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(i) Stable system. Natural response

approaches zero. Poles in LHP.

(ii) Unstable system. Natural response

grows. Poles in RHP.

(iii) Marginally stable system.

Natural response neither grows/approaches zero.

Poles on j axis.

3.5 Stability.3.5 Stability.

Figure 3.12: Closed-loop poles and response: a. stable system;

b. unstable system

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3.6 Routh-Hurwitz Stability 3.6 Routh-Hurwitz Stability Criteria.Criteria.

What is Routh-Hurwitz Criterion (RHC)?What is Routh-Hurwitz Criterion (RHC)? Through the RHC method we can tell how Through the RHC method we can tell how

many close-loop system many close-loop system polespoles are in the left are in the left half plane, in the right half-plane and on half plane, in the right half-plane and on the jthe j-axis. We can find the number of poles -axis. We can find the number of poles in each section of the s-plane, but in each section of the s-plane, but cannotcannot find their coordinatefind their coordinate. .

The number of roots of the polynomial that The number of roots of the polynomial that are in the right half-plane is equal to the are in the right half-plane is equal to the number of changes in the first column.number of changes in the first column.

The RHC method requires The RHC method requires two stepstwo steps;;(1) Generate the data table called Routh (1) Generate the data table called Routh table.table.(2) Interpret the Routh table to tell number (2) Interpret the Routh table to tell number of close loop system poles in the left half of close loop system poles in the left half plane, in the right half-plane and on the jplane, in the right half-plane and on the j--axis. axis.

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The Close-Loop Transfer function.The Close-Loop Transfer function.

Initial layout for the Routh-Hurwitz Table.Initial layout for the Routh-Hurwitz Table.

Completed Completed Routh TableRouth Table..


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Example 3.4:Example 3.4: Routh-Hurwitz. Routh-Hurwitz.Make a Routh table from the system shown below.

Solution:Find the equivalent close loop system.Figure (b) above.

Interpretation:There are two sign changes in the first column.1 -72 103The system is unstable, two poles exist in the right half plane.

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Example 3.5:Example 3.5: Routh-Hurwitz. Routh-Hurwitz.

The number of RHP poles = The number of The number of RHP poles = The number of SIGN CHANGES of COL 1SIGN CHANGES of COL 1

TWOTWO sign changes: sign changes: RHP Poles =2RHP Poles =2

P(s) = s3 + 10s2 + 31s + 1030

s3 1 31 0 s2 10 1030 0 s1

b1 = 11031311

= - 72 b2 = 1

0101

= 0 b3 = 1

0101

= 0

s0 c1 = 72

0721031

=

103

c2 = 7207201

= 0 c3 = 72

07201

= 0

)595.87068.1)(595.87068.1)(4136.13()(10303110)( 23

jsjsssPssssP

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Solution:Solution:

Two sign changes: 2 RHP (UNSTABLE)Two sign changes: 2 RHP (UNSTABLE) Poles: 2 LHP and 2 RHPPoles: 2 LHP and 2 RHP

2006116200)( 234

ssss

sT

s 4 1 1 1 2 0 0

s 3 6 1 6 1 0

s 2 1 0 1 2 0 0 2 0 0

s 1 - 1 9 0 0

s 0 2 0 0 0

)54.227.1)(54.227.1)(54.227.4)(54.227.4()(2006116)( 234

jsjsjsjssPsssssP

29


Assume Assume is small POSITIVE : TWO sign is small POSITIVE : TWO sign changeschanges

Poles: 2 RHP, 3 LHPPoles: 2 RHP, 3 LHP

3563210)( 2345

sssss

sT

s 5 1 3 5

s 4 2 6 3

s 3 0 7 /2 0

s 2

76 3 0

s 1

141264942 2

0 0

s 0 3 0 0

)7.051.)(7.051.)(5.134.)(5.134.)(66.1()(35632)( 2345

jsjsjsjsssPssssssP

Solution:Solution:

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Solution:Solution:

Assume Assume is small is small positivepositive: : TwoTwo sign sign changeschanges

Poles: 2 RHP, 3 LHPPoles: 2 RHP, 3 LHP

1232321)( 2345

sssss

sT

1232321)( 2345

sssss

sT

s 5 2 2 2

s 4 3 3 1

s 3 0 4 / 3 0

s 2

43 1 0

s 1

12931612 2

0 0

s 0 1 0 0

)5.041.)(5.041.)(89.33.)(89.33.)(33.1()(123232)( 2345

jsjsjsjsssPssssssP

31

568426710)( 2345

sssss

sT

s 5 1 6 8

s 4 7 1 4 2 6 5 6 8

s 3 0 4 1 0 1 2 3 0 0 0

s 2 3 8 0

s 1 1 / 3 0 0

s 0 8 0 0


Solution:Solution:

NO sign changes: No RHP (STABLE)NO sign changes: No RHP (STABLE) Row of ZEROS indicate existence of Row of ZEROS indicate existence of

complex poles & Symmetric Equationscomplex poles & Symmetric Equations Poles: 1 LHP and 4 on jw axisPoles: 1 LHP and 4 on jw axis

0124)(86)(

3

24

ssds

sdPsssP

)414.1)(414.1)(2)(2)(7()(5684267)( 2345

jsjsjsjsssPssssssP

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When a purely even or odd polynomial is a When a purely even or odd polynomial is a factor of the original polynomial.factor of the original polynomial.

Even polynomial only have roots that are Even polynomial only have roots that are symmetrical about the origin.symmetrical about the origin.

The symmetry can occur under 3 The symmetry can occur under 3 conditions:conditions:

1. The roots are symmetrical and real.2. The roots are symmetrical and imaginary.3. The roots are quadrantal.

Entire row is zeroEntire row is zero

2

2

1 1

3

3

j

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Solution:Solution:

2 sign changes: 2 RHP (symmetric)2 sign changes: 2 RHP (symmetric) Poles: 2 RHP, 4 LHP and 2 on jPoles: 2 RHP, 4 LHP and 2 on j axis axis

128192128964824103128)( 2345678

ssssssss

sT

s 8 1 1 0 4 8 1 2 8 1 2 8

s 7 3 1 2 4 8 9 6 3 2 1 9 2 6 4

s 6 2 1 1 6 8 6 4 3 2 1 2 8 6 4

s 5 0 6 3 0 3 2 1 6 0 6 4 3 2 0 0 0

s 4 8 / 3 1 6 4 / 3 8 6 4 2 4

s 3 - 8 - 1 - 4 0 - 5

s 2 3 1 2 4 8

s 1 3

s 0 8

064326)(64328)(

35

246

sssds

sdPssssP

)73.11)(73.11)(73.11)(73.11)(2)(2)(1)(2()(128192128964824103)( 2345678

jsjsjsjsjsjssssPsssssssssP

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Main use is to determine the position of the Main use is to determine the position of the poles, which in turns can determine the poles, which in turns can determine the stability of the response.stability of the response.

Use of Routh Hurwitz Use of Routh Hurwitz CriteriaCriteria

KssssK

sRsC

21)()(

2

ExampleA closed-loop transfer function is given by

Determine the range for K for the system to be always stable and its oscillating frequency before it becomes unstable.

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SolutionSolution:: Characteristic equation is:Characteristic equation is:

Expand the equationExpand the equation

Form the Routh’s arrayForm the Routh’s array

0212 Kssss

0233 234 Kssss

s4 1 3 Ks3 3 2s2 K

s1

s0 K

37

329

7914

373314 KK

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SolutionSolution:: For no sign changeFor no sign change Referring to row 4Referring to row 4

which gives,which gives, and row 5,and row 5, Hence its range,Hence its range,

Oscillating frequency, Oscillating frequency,

07914 K

914K

0K

9140 K091437 2

srad32

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Steady stateSteady state

)()()(1

1)( sRsHsG

sE

From the diagram

Consider

And

Use the final value theorem and define steady state error, ess that is given by

)(sG

)(sH

Y(s)R(s) E(s)

+ -

B(s)

1...11

1...11)(21

21

sTsTsTs

sTsTsTKsGbjbb

naiaa

1...11

1...11)(

21

21

sTsTsTsTsTsT

sHylyy

xkxx

)(lim)(lim0

ssEteest

ss

38

Unit stepUnit step Unit step input,

From

Steady state error,

We define step error coefficient,

Thus, the steady state error is

By knowing the type of open-loop transfer function,

we can know step error coefficient and thus the steady state error

ssR 1)(

)()()(1

1)( sRsHsG

sE

ssHsGse

sss

1)()(1lim

0

)()(lim0

sHsGKs

s

Sss K

e

1

1

)()( sHsG

)()(lim0

sHsGKs

s

1...11

1...11lim21

21

0

sTsTsTs

sTsTsTKbjbb

naiaa

s

1...11

1...11

21

21

sTsTsTsTsTsT

ylyy

xkxx

39

For open-loop transfer function of type 0:



,KK s Kess

1

1

,sK 01

1

sse

,sK 01

1

sse

Unit stepUnit step

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Example:A first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain,determine the steady state error for a unit step input and the final value of the output.

Solution:The block diagram of the system is

As we are looking for a steady state error for a step input, we need to know,

Knowing the open-loop transfer function, then

And steady state error of,

Its final value is,

519

500

s

limsHsGlimKsss

sK

61

511

11

S

ss Ke

65611 ssy

Unit stepUnit step

195s

Y(s) R(s) + -

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Unit RampUnit Ramp

ttr )( , while its Laplace form is 21)(s

sR As in the above section, we know that

)()()(1

1)( sRsHsG

sE

Thus, its steady state error is

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1)()(1

1lim

ssHsGse

sss

Define ramp error coefficient, ;K r )()(lim0

sHssGKs

r

Which the steady state error as

rss K

e 1

Just like for the unit step input we can conclude the steady state error for a unit ramp through the type

of the open-loop transfer function of the system.

For open-loop transfer function of type 0: 0rK

For open-loop transfer function of type 1: KKr

For open-loop transfer function of type 2: rK

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Example:

A missile positioning system is shown.

(i) Find its closed-loop transfer function )()(

ss

i

m

(ii) Determine its undamped natural frequency and its damping ratio if 310K(iii) Determine the steady state error, if the input is a unit ramp.

(iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii).

)14.0(01.0ss

K mi

Compensator DC motor +

-

Unit RampUnit Ramp

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Solution:(a) By Mason rule, the closed-loop transfer function is

KssK

KssK

ssK

ssK

ss

i

m

025.05.2025.0

01.04.001.0

)14.0(01.0.1

)14.0(01.0.

)()(

22

KssK

KssK

025.05.2025.0

01.04.001.0

2

2

(b) If 310K

,

255.225

)()(

2

ssss

i

m

Comparing with a standard second order transfer function

22

2

2)()(

nn

n

i

m

ssK

ss

Unit RampUnit Ramp

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Comparing

252 nThus undamped natural frequency

5n rad.s-1

and

5.22 n

damping ratio of 25.0

(c) To determine the ramp error coefficient, we must obtain its open-loop transfer function

)14.0(01.0.)(

ss

KsGo

As it is a type 1, the system will have a finite ramp error coefficient, putting 310K

)14.0(10)(

sssGo

10)14.0(

10.lim)(lim00

ss

sssGKs

os

r

Hence steady state error of 1.01

rss K

e

Unit RampUnit Ramp

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Unit ParabolaUnit Parabola

Its time function 2)( ttr ,while its Laplace 32)(s

sR ,thus its steady state error is

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21 s)s(H)s(G

slimesss

Define parabolic error coefficient, paK

)s(H)s(GslimKspa

2

0

Similarly we can determine its steady state error by knowing the type of the open-loop transfer function

For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2:

0paK

0paK

KK pa

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In summary we can make a table of the steady state error for the above input

sK11

rK1

paK2

Unit step Unit ramp Unit parabolic

Type 0

Type 1 0

Type 2 0 0

1 Time Response. CHAPTER 3. 2 3.2 Poles and Zeros and System Response. Figure 3.1: (a) System...

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Transcript of 1 Time Response. CHAPTER 3. 2 3.2 Poles and Zeros and System Response. Figure 3.1: (a) System...