course outline - university of sydneyweb.aeromech.usyd.edu.au/amme3500/course_documents...3 dr....

1

Amme 3500 : System Dynamics and Control

System Response

Dr. Stefan B. Williams

Slide 2 Dr. Stefan B. Williams Amme 3500 : Introduction

Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the s-plane 4 22 Mar Block Diagrams Assign 1 Due 5 29 Mar Feedback System Characteristics 6 5 Apr Root Locus Assign 2 Due 7 12 Apr Root Locus 2 8 19 Apr Bode Plots No Tutorials 26 Apr BREAK 9 3 May Bode Plots 2 Assign 3 Due 10 10 May State Space Modeling 11 17 May State Space Design Techniques 12 24 May Advanced Control Topics 13 31 May Review Assign 4 Due 14 Spare

Slide 3 Dr. Stefan B. Williams Amme 3500 : System Response

Engineering as Design

•  Engineering Design is the systematic, intelligent generation and evaluation of specifications for products whose form and function achieve stated objectives and satisfy specified constraints

•  The purpose of design is to derive from a set of specifications a description of a product sufficient for its realization


Problem Identification •  Design constraints must be identified

– Restrictions or limitations on a behaviour or a value or some other aspect of a designed object’s performance

•  Requirement analysis –  allows us to verify that we have reached a solution –  a description of the requirements that characterise a

solution –  the criteria to be used to verify that the requirements

have been met –  produces an understanding of the nature and scope of

the remaining activities

2


Response Vs. Pole Location •  As we saw previously,

a qualitative understanding of the effect of poles and zeros on the system response can help us to quickly estimate performance

Re{s}

Im{s}

stable unstable


Response Vs. Pole Location

•  We would now like to gain a deeper insight into the system performance as a function of pole and zero locations

•  This will help us to describe the system performance quantitatively

•  We will also see how this allows us to design a system to meet certain design constraints


General First Order System

•  A first order system without zeros can be described by:

•  The resulting impulse response is

•  When –  σ > 0, pole is located at σ < 0,

exponential decays = stable –  σ < 0, pole is at σ > 0,

exponential grows = unstable

1( )( )

H ss σ

=+

( ) th t e σ−=

( )1( ) 1 ty t e σ

σ−= −

Step Response


General First Order System •  We define the time

constant of the system as

•  This is the time taken for the system to decay to 1/e or 37% of its initial value or rise to 63% of step response

1τσ

=

3



•  The Rise Time Tr is defined as the time for the waveform to go from 0.1 to 0.9 of its final value

0.9 0.1

2.31 0.11

2.2

RT t t

σ σ

σ

= −

= −

=

0.9

0.1

( ) 10.9 10.1 1

t

t

t

y t eee

σ

σ

σ

−

−

−

= −

= −

= −



•  The Settling Time Ts is defined as the time for the waveform to reach and stay within a certain percentage of its final value

•  Values of 1%, 2% and 5% are often used

0.98 14 (2%)

sT

S

e

T

σ

σ

−= −

=

4.6 (1%)

3 (5%)

S

S

T

T

σ

σ

=

=


Example •  A system has a

transfer function

•  The time constant is: –  1/σ = 0.02

•  Settling time (2%) is: –  4/σ = 0.08

•  Rise Time is: –  2.2/σ = 0.044

50( )50

G ss

=+

Step Response

Time (sec)

Ampl

itude

0 0.02 0.04 0.06 0.08 0.1 0.120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


General Second Order System

•  Many systems of interest are of higher order

•  For a first order system we have a single, real pole

•  Second order systems are quite common and are generally written in the following standard form

2

2 2( )2

n

n n

G s Cs s

ωςω ω∞=

+ +

4


Natural Frequency and Damping Ratio

•  The Natural Frequency, ωn, of a second-order system is the frequency of oscillation of the system without damping

•  The Damping Ratio, , of a second-order is the ratio of the exponential decay frequency and the natural frequency

!


General Second Order System

•  For a second order system, we can have four distinct combinations of real and imaginary poles – Two real poles -σ1, -σ2

– Two identical real -σ1

– Two complex poles -σ ± jωd

– Two imaginary ± jωd



•  How does this relate to the pole location in terms of their real and imaginary parts?

•  Complex poles are always in complex conjugate pairs so

ds jσ ω= − ±

22

( ) ( )( )

( )d d

d

d s s j s j

s

σ ω σ ω

σ ω

= + − + +

= + +



•  By multiplying out and comparing the denominator of the two forms we find that

21n

d n

σ ζω

ω ω ζ

=

= −

2 22 2( ) 2d n ns s sσ ω ξω ω+ + = + +

5



•  We can now relate the natural frequency and damping ratio to the s-plane

21n

d n

σ ζω

ω ω ζ

=

= −

( )1sinθ ζ−=

q

Slide 18

Natural Response of Second Order System

Dr. Stefan B. Williams Amme 3500 : System Response

0 2 2

2( )2

n

n n

sC s ys s

ςωςω ω+=

+ +

•  We’d like to find the natural response of a general 2nd order system

•  For the case of zero initial velocity we have

˙ ̇ y + 2!"n ˙ y + "n2y = 0

s2Y (s) # sy0 # ˙ y 0 + 2!"n sY(s) # y0( ) + "n2Y (s) = 0

Y (s) =s + 2!"n( )y0 + ˙ y 0s2 + 2!"ns + "n

2



•  We’d like to find the natural response of a general 2nd order system

•  After completing squares

•  Take Inverse Laplace and simplify

2

2

0 2 2 2

0 2 2

( ) 11

( )( ) (1 )

( )

( )

n n

n n

dd

d

sC s y

s

sy

s

ζζω ω ζζ

ζω ω ζσσ ωω

σ ω

+ + −−

=+ + −

+ +=

+ +

0 2 2

2( )2

n

n n

sC s ys s

ςωςω ω+=

+ +

0( ) cos sintd d

d

y t y e t tσ σω ωω

− ⎛ ⎞= +⎜ ⎟

⎝ ⎠



•  We can use this to estimate the system parameters

•  By displacing the system and measuring its response we can identify the peaks and estimate natural frequency and damping ratio

•  This is known as the log decrement

2

0

2

0

0

02

0

2

2ln

1 ln21

1 ln2

nd

nd

n

d

n

n

nn

d

n

nn

n

ny y e

y y e

y ny

yn y

yn y

πζωω

πζωω

πω

πζωω

ζωπω ζ

ζπ

−

−

⎛ ⎞=⎜ ⎟

⎝ ⎠

=

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟− ⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

;

0( ) cos sintd d

d

y t y e t tσ σω ωω

− ⎛ ⎞= +⎜ ⎟

⎝ ⎠

6


A Familiar Mechanical Example

M!!y (t ) + Kd !y (t ) + Ky (t ) = f (t )

•  Earlier, we considered this mechanical system

•  Now we can consider the natural and forced response

M

f(t)

y(t)

M s 2Y (s ) ! sy (0) ! !y (0)( )+ Kd sY (s ) ! y (0)( )+ KY (s ) = F (s )Slide 22 Dr. Stefan B. Williams Amme 3500 : System Response

Example: Natural Response

•  The natural response of the system can be found by assuming no input force

•  Taking inverse Laplace transform will give us the system response

Ms2 + Kd s + K( )Y (s) = Ms + Kd( )y(0) + M˙ y (0)

Y (s) =Ms + Kd( )y(0) + M˙ y (0)

Ms2 + Kd s + K( )



•  Suppose we wish to find the natural response of the system to an initial displacement of 0.5m

•  Assume the mass of the system is 1kg and the systems constant kd is 2 Nsec/m and k is 20 N/m



•  The natural response of the system can be found by assuming no input force

0 2 22

2

2( ) (0)2

2 ,

d

n

d n n

dn n

KssMY s y y

K K s ss sM M

K KM M

ςωςω ω

ζω ω

⎛ ⎞+⎜ ⎟ +⎝ ⎠= =+ +⎛ ⎞+ +⎜ ⎟⎝ ⎠

= =

7



•  Substituting values, we find Impulse Response

Time (sec)

Ampl

itude

0 1 2 3 4 5 6-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5 System: h

Time (sec): 0 Amplitude: 0.5

System: h Time (sec): 2.88 Amplitude: 0.0279

2!"n = 2,"n2 = 20,

"n = 20,! =120,,"d = 19

y(t) = y0e#$t cos"d t +

$"d

sin"d t% & '

( ) *

= 0.5e# t cos 19t +119sin 19t%

& ' ( ) *


Example: Natural Response •  We can work backwards from the response to

estimate the parameters 22.88

0.69 4.36

n dcyclessradHzs

ω ω =

= =

;

Impulse Response

Time (sec)

Ampl

itude

0 1 2 3 4 5 6-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5 System: h

Time (sec): 0 Amplitude: 0.5

System: h Time (sec): 2.88 Amplitude: 0.0279

01 ln2

1 0.5ln4 0.02790.230

n

yn y

ζπ

π

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠=

;


Step Response of Second Order System

•  We’d like to find the step response of the standard form

•  After some partial fraction expansion and completing squares

•  Take Inverse Laplace and simplify

2

2

2 2 2

2

2 2

( ) 111( )

( ) (1 )

( )11

( )

n n

n n

d

d

sC s

s s

s

s s

ζζω ω ζζ

ζω ω ζζσ ωζ

σ ω

+ + −−

= −+ + −

+ +−

= −+ +

( )2

2 2( )

2n

n n

C ss s s

ωςω ω

=+ +

( ) 1 cos sintd d

d

c t e t tσ σω ωω

− ⎛ ⎞= − +⎜ ⎟

⎝ ⎠Slide 28 Dr. Stefan B. Williams Amme 3500 : System Response

Step Response vs. Damping Ratio

•  Damping ratio determines the characteristics of the system response

8



( )

1 1

1 2

1 2

1 2

( ) cos( )

( ) cos

( )

( )

d

n

td

t t

t t

c t A t

c t Ae t

c t K e K te

c t K e K e

σ

σ σ

σ σ

ω φ

ω φ−

− −

− −

= −

= −

= +

= +



•  The damping ratio describes the degree of damping in the system

•  For an underdamped system, rising damping will lower the overshoot of the system


Time Domain Specifications •  Specifications for a control system design often

involve requirements associated with the time response of the system –  Peak Time, tp, is the time taken to reach maximum

overshoot –  Overshoot, Mp, is the maximum amount the system

overshoots its final value –  Settling time, ts, is the time for system transients to

decay –  Rise time, tr, is the time it takes for the system to

reach the vicinity of its new set point


Time Domain Specifications •  Rise time, settling

time and peak time yield information about the speed of response of the transient response

•  This can help a designer determine if the speed and nature of the response is appropriate

9


Peak Time

•  Peak time is found by differentiating and finding the first zero crossing after t=0

•  After some manipulation we find tp occurs when sinwdt=0

L[ !c (t )] = sC (s )

=!n2

s 2 + 2!"n +!n2

21p

n

d

t πω ζπω

=−

=


Percent Overshoot •  Substituting the peak

time back into the step response yields

•  Notice this is only a function of the damping ratio. We can invert to find the required damping ratio for a particular overshoot

21

( ) 1 cos sin

1

,0 1

d

d

pd

p

c t e

e

M e

πσω

πσω

πζζ

σπ πω

ζ

−

−

−−

⎛ ⎞= − +⎜ ⎟

⎝ ⎠

= +

= ≤ <

2 2

ln( )

ln ( )p

p

M

Mζ

π

−=

+


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ζ

Mp,

%

Overshoot vs. Damping Ratio for second order system

Percent Overshoot

•  We can plot the relationship between percent overshoot and damping ratio

•  Two frequently used specifications are –  Mp=5% for ζ=0.7 –  Mp=16% for ζ=0.5

5%

16%


Settling Time

•  The settling time can be derived in a similar manner as for a first order system

•  The settling time is essentially decided by the transient exponential

0.024

4

n st

sn

e

so t

ζω

ζω

σ

− =

=

=

10


Rise Time •  Rise time is not as straightforward

to compute for this case •  A precise analytical relationship

does not exist •  Instead we use an approximation

as suggested in Franklin (Section 3.4.1).

•  Nise suggests an alternative approach based on a graph of normalised rise time

1.8r

n

tω

≅


Time Domain Specifications

•  Lines of constant peak time,Tp , settling time, Ts , and percent overshoot, %OS and rise time

•  Note: Ts2 < Ts1 ; Tp2 < Tp1;

%OS1 < %OS2

wn


Step response and Pole Location

•  Step responses of second-order underdamped systems as poles move: a. with constant real part σ; b. with constant imaginary part jω; c. with constant damping ratio ζ


Example : Transforming Specifications

•  Find allowable region in the s-plane for the poles of a transfer function to meet the requirements

tr ≤ 0.6 sec Mp ≤ 10% ts ≤ 3 sec

0.6ζ ≥

4 1.53

σ ≥ =

1.8 3.0 / secnr

radt

ω ≥ =

sin-1z

s

wn

Im(s)

Re(s)

11


Example : Transient Response through Component Design

•  Given the system shown here, find J and D to yield 20% overshoot and a settling time of 2 seconds for a step input torque T(t)



•  First the differential equation

•  Find the transfer function

•  The second order properties are

J !!! + D !! + K! = T

2

1/( ) JG s D Ks sJ J

=+ +

2n nK DandJ J

ω ζω= =



•  From problem statement

•  Hence

•  For a 20% overshoot, ζ = 0.456 so

•  From problem statement K = 5 so J = 0.26kgm2 and D = 1.04 Nms/rad

42sn

Tζω

= =

44 22 n

D JandJ K

ζω

= = =

0.052JK

=


Additional Poles and Zeros

•  The preceding developing holds for a second order system with no zeros

•  What happens to the system response if we add additional poles or zeros?

•  This depends on their location in the s-plane

12


Additional Poles

•  Consider an additional pole at a

•  By partial fraction expansion and completing squares we have the step response

( )( )2

2 2( )

2n

n n

H ss s

ωα ζω ω

=+ + +

( ) ( )2 2

( )( ) n d

n d

B s CA DT ss ss

ζω ωαζω ω

+ += + +++ +

( )( ) ( )

cos sinnt td d

c t Au te B t C t Deζω αω ω− −

=

+ + +


Additional Poles


Additional Poles

•  As the pole location is moved to the left in the LHP, the effect on the response becomes less and the transient dies out more quickly


Additional Poles

•  How much further from the dominant poles does the third pole have to be for its effect on the second-order response to be negligible?

•  This depends on the accuracy required by the design

•  We often assume the exponential decay is negligible after five time constants

•  The accuracy of this assumption should always be verified

13


Zeros

•  Consider a zero at β •  For large β, the zero

acts as a scaling factor

•  For smaller β, the derivative increases the speed of response and overshoot

( )( )

2

2 2( )

2n

zn n

sH s

sβ ωζω ω+

=+ +

2 2

2 2 2 22 2( ) ( )

n n

n n n n

ss ssH s H s

ω βωζω ω ζω ω

β

= ++ + + +

= +Step Response

Time (sec)

Ampl

itude

0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

No zero

Zero at -3

Zero at -5

Zero at -10


Zeros

•  If we compare the normalized step response, we find that the derivative term increases the responsiveness of the system


Zeros •  We have seen that a pole

in the RHP makes a system unstable

•  A zero in the RHP causes the response to be depressed

•  If the derivative term is larger than the scaled response, this can lead to non-minimum phase behaviour


Effects of Pole-Zero Patterns

•  For a second order system with no finite zeros, the transient response parameters are approximated by – Rise time :

– Overshoot :

– Settling Time (2%) :

5%, 0.716%, 0.520%, 0.45

pMζζζ

=⎧⎪≅ =⎨⎪ =⎩

1.8r

n

tω

≅

4st σ≅

14


Effects of Pole-Zero Patterns •  An additional pole in the left half-plane (LHP) will

increase the rise time significantly if the extra pole is within a factor of 5 of the real part of the complex poles

•  A zero in the LHP will increase the overshoot if the zero is within a factor of 4 of the real part of the complex poles

•  A zero in the RHP will depress the overshoot (and may cause the step response to be non-minimum phase)


Conclusions

•  We have looked at the characteristics of first and second order systems

•  We have derived specifications that describe the response of the system to a step input

•  We have also looked at the effect of extra poles and zeros on the system response


Further Reading

•  Nise – Sections 4.1-4.8

•  Franklin & Powell – Section 3.3-3.5

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