National University of Singaporeguppy.mpe.nus.edu.sg/anpoo/FBControla/ME2142 handouts.pdf1 ME2142/TM3142 Feedback Control Systems 1 Reference Text “Control Systems Engineering”

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ME2142/TM3142 Feedback Control Systems1

Reference Text “Control Systems Engineering” 5th Editionby Noman S Nise. John Wiley & Sons

ReferenceText: “Modern Control Engineering” by K Ogata. Prentice Hall

Website for this portion: http://guppy.mpe.nus.edu.sg/~mpepooan/FDControl/welcome.htm

Reference Text “Control Systems Engineering” 5th Editionby Noman S Nise. John Wiley & Sons

ReferenceText: “Modern Control Engineering” by K Ogata. Prentice Hall

Website for this portion: http://guppy.mpe.nus.edu.sg/~mpepooan/FDControl/welcome.htm

ME2142/ME2142E

Feedback Control Systems

First half: Professor POO Aun Neow

Second half: Professor V Subramaniam

ME2142/ME2142E


First half: Professor POO Aun Neow

Second half: Professor V Subramaniam


Introduction and Basic Concepts


• A control system is an interconnection of components that will provide a desired system response or output response.

• The study of control systems is the study of dynamic systems. A static system needs no control.

• Examples of controlled outputs: temperature, humidity, position, speed, pressure, direction, liquid level, altitude.

• And also: sugar level in humans, inflation, interest rates.

• A control system is an interconnection of components that will provide a desired system response or output response.

• The study of control systems is the study of dynamic systems. A static system needs no control.

• Examples of controlled outputs: temperature, humidity, position, speed, pressure, direction, liquid level, altitude.

• And also: sugar level in humans, inflation, interest rates.

What is a control System?What is a control System?

2


In order for a system to be controllable, there must be a cause-effect relationship for its components, i.e. there must be some input that can cause changes to the output parameter to be controlled.

In order for a system to be controllable, there must be a cause-effect relationship for its components, i.e. there must be some input that can cause changes to the output parameter to be controlled.

What is a control System?What is a control System?

Process or Plant

(Source of energy)

Controlling/actuatinginput,

Output

(chemical process, machine, industrialprocess, economic process)


Open-loop control SystemOpen-loop control System

In an open-loop control system, no feedback from the output is used to control the system.

Based on how the output is required, or desired, to respond, the controller adjusts the input to the plant to achieve this.

Control will only be accurate if plant is highly predictable and there is no internal or external disturbance. Generally used only when good control performance is not required.

Examples: An electric bread toaster. Temperature control of a simple water heater for the shower.

In an open-loop control system, no feedback from the output is used to control the system.

Based on how the output is required, or desired, to respond, the controller adjusts the input to the plant to achieve this.

Control will only be accurate if plant is highly predictable and there is no internal or external disturbance. Generally used only when good control performance is not required.

Examples: An electric bread toaster. Temperature control of a simple water heater for the shower.

Plantor

Process

Desired outputresponse

OutputController


Closed-loop feedback control SystemClosed-loop feedback control System

The sensor measures the actual value of the output, Y, compares this with the desired value, R, and computes the error, E. Based on this error E, the controller generates the input, U, to the plant so as to bring Y to the desired value R.

The sensor measures the actual value of the output, Y, compares this with the desired value, R, and computes the error, E. Based on this error E, the controller generates the input, U, to the plant so as to bring Y to the desired value R.

PlantU Y

Controller

Sensor

R E

Feedback

+

-

R – Set-point or Reference Input E – ErrorU – Plant input Y – Controlled VariableR – Set-point or Reference Input E – ErrorU – Plant input Y – Controlled Variable

3


Generally used when good control performance is required.Accurate control can be achieved even in the presence of plant variations, and internal or external disturbances because such disturbances will affect the output Y, reflected in the error E, ant thus will cause the plant input U to change so as to correct for these disturbances.Can become unstable. Stability becomes an important consideration.

Generally used when good control performance is required.Accurate control can be achieved even in the presence of plant variations, and internal or external disturbances because such disturbances will affect the output Y, reflected in the error E, ant thus will cause the plant input U to change so as to correct for these disturbances.Can become unstable. Stability becomes an important consideration.

PlantU Y

Controller

Sensor

R E

Feedback

+

-

Closed-loop feedback control SystemClosed-loop feedback control System


Some example of control System?Some example of control System?


Example: Open-loop vs Closed LoopExample: Open-loop vs Closed Loop

A walking manA walking man

Process of Walking:Desired output: a point where you want to be.Controller: the brain

Plant or process: the legs

Process of Walking:Desired output: a point where you want to be.Controller: the brain

Plant or process: the legs

Open-loop control:Walking with your eyes closed.Open-loop control:Walking with your eyes closed.

Closed-loop feedback control:Walking with your eyes open.The eyes sensed the actual output, where you are and where you are heading, computes the error in position and in direction, and issues commands to the plant, meaning the legs, to move in such a way so as to reduce the error.

Closed-loop feedback control:Walking with your eyes open.The eyes sensed the actual output, where you are and where you are heading, computes the error in position and in direction, and issues commands to the plant, meaning the legs, to move in such a way so as to reduce the error.

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Dropping a Bomb:Objective of dropping a bomb from a height is to hit a target below.Desired output: Target belowPlant or process: the bomb with its control fins


Open-loop Control or dumb bombsThe controller, meaning the pilot or bombardier, needs to estimate his own height, velocity, distance to target, wind conditions, and characteristics of bomb to decide when and where to release the bomb. Often, hundreds of bombs are needed to hit a specific target.

Open-loop Control or dumb bombsThe controller, meaning the pilot or bombardier, needs to estimate his own height, velocity, distance to target, wind conditions, and characteristics of bomb to decide when and where to release the bomb. Often, hundreds of bombs are needed to hit a specific target.

Photo courtesy U.S. Air Force





Closed-loop Control or “smart bombs”Sensors are incorporated into the bomb to give feedback on its actual position relative to the target. The “error” information is then used to steer the bomb, using its control fins, to the target. Result: one target only needs one bomb.Sensors: TV, Infrared, laser guided, or GPS.

Closed-loop Control or “smart bombs”Sensors are incorporated into the bomb to give feedback on its actual position relative to the target. The “error” information is then used to steer the bomb, using its control fins, to the target. Result: one target only needs one bomb.Sensors: TV, Infrared, laser guided, or GPS.

See also: http://science.howstuffworks.com/smart-bomb1.htmSee also: http://science.howstuffworks.com/smart-bomb1.htm


Study of control SystemsStudy of control Systems

Given a control system, y = f(r,t)

meaning that y is not only a function of r, but also varies with time t.

If y = f(r) then the system is not a dynamic system but is static.

Given a control system, y = f(r,t)

meaning that y is not only a function of r, but also varies with time t.

If y = f(r) then the system is not a dynamic system but is static.

PlantU Y

Controller

Sensor

R E

Feedback

+

-Study of control systems is the study of the dynamics of the system.The response of the controlled variable Y to any input R depends upon the dynamics of the Plant, Controller, and the Sensor or Feedback.

Study of control systems is the study of the dynamics of the system.The response of the controlled variable Y to any input R depends upon the dynamics of the Plant, Controller, and the Sensor or Feedback.

To mathematically describe the dynamic behavior of the control system and its components, differential equations are used. To mathematically describe the dynamic behavior of the control system and its components, differential equations are used.

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Linear and Non-linear SystemsLinear and Non-linear Systems

A system is linear if it satisfy the properties of superposition and homogeneity/scaling.A system is non-linear if it is not linear.

A system is linear if it satisfy the properties of superposition and homogeneity/scaling.A system is non-linear if it is not linear.

Consider a system which has the responses to any two arbitrary inputs u1(t) and u2(t) as

y1(t) = f(u1(t)) and y2(t) = f(u2(t))


y1(t) = f(u1(t)) and y2(t) = f(u2(t))

Property of Superposition is satisfied if the output for a combined input of u1(t) and u2(t) is

y3 = f(u1(t) + u2(t)) = y1(t) + y2(t)

Property of Superposition is satisfied if the output for a combined input of u1(t) and u2(t) is

y3 = f(u1(t) + u2(t)) = y1(t) + y2(t)

Property of homogeneity is satisfied if

y3 = f(Ku1(t)) = Ky1(t)

Property of homogeneity is satisfied if

y3 = f(Ku1(t)) = Ky1(t)




y1(t) = f(u1(t))) and y2(t) = f(u2(t))


y1(t) = f(u1(t))) and y2(t) = f(u2(t))

A system is linear if the properties of superposition and homogeneity are satisfied.

The above system will be linear if the following is satisfied

y3 = f(K1u1(t) + K2u2(t)) = K1y1(t) + K2y2(t)

A system is linear if the properties of superposition and homogeneity are satisfied.

The above system will be linear if the following is satisfied

y3 = f(K1u1(t) + K2u2(t)) = K1y1(t) + K2y2(t)

In general, real physical systems are non-linear if the operating range is very large, However, if operation is considered only about some operating point, and the range of operation is sufficiently small, most systems can be considered to be linear.

In general, real physical systems are non-linear if the operating range is very large, However, if operation is considered only about some operating point, and the range of operation is sufficiently small, most systems can be considered to be linear.


Which of the following systems are linear?

(i) F = Kx (ii) y = x2 (iii) y = mx + b

Which of the following systems are linear?

(i) F = Kx (ii) y = x2 (iii) y = mx + b

For any constants A and B and any two inputs x1 and x2, For any constants A and B and any two inputs x1 and x2,

(i) F1 = f(x1) = Kx1 and F2 = f(x2) = Kx2

Also, F3 = f(Ax1+Bx2) = K(Ax1+Bx2) = AKx1 +BKx2 = AF1 + BF2

Thus properties of superposition and homogeneity is met. Thus linear.

(i) F1 = f(x1) = Kx1 and F2 = f(x2) = Kx2

Also, F3 = f(Ax1+Bx2) = K(Ax1+Bx2) = AKx1 +BKx2 = AF1 + BF2

Thus properties of superposition and homogeneity is met. Thus linear.

(ii)And (not homogenous) Also, (superposition violated)Thus system is not linear or non-linear.

(ii)And (not homogenous) Also, (superposition violated)Thus system is not linear or non-linear.


Examples:Examples:

2( )y f x x= =2( ) ( )f Ax Ax Ay= ≠

2 2 21 2 1 2 1 2 1 2( ) ( ) ( ) ( )f x x x x f x f x x x+ = + ≠ + = +

(iii)And (not homogenous) System is not linear. Can be shown that superposition also violated.

(iii)And (not homogenous) System is not linear. Can be shown that superposition also violated.

( )y f x mx b= = +( ) ( ) ( ) ( )f Ax m Ax b Af x A m x b= + ≠ = +

6


Linear Approximation of SystemsLinear Approximation of Systems

Any non-linear system can be linearised about some operating point and can be considered to be linear within a small operating region about that point

Any non-linear system can be linearised about some operating point and can be considered to be linear within a small operating region about that point

For a any function with

We can use the Taylor Series expansion about some operating point, x0, and have

For small variations about the operating point, second and higher-order terms in can be neglected. Then

or which is linear.

For a any function with

We can use the Taylor Series expansion about some operating point, x0, and have

For small variations about the operating point, second and higher-order terms in can be neglected. Then

or which is linear.

( )y f x=

0 0

220 0

0 2

( ) ( )( ) ...

1! 2!x x x x

x x x xdf d fy f x

dx dx= =

− −= + + +

0( )x x−

0 0( )y f x=

0 0( )y y m x x= + − y m x∆ = ∆


Linear Approximation of SystemsLinear Approximation of Systems

Example:Example:

θ

For the pendulum shown in the figure, the restoring torque due to gravity is given by

Derive the linearised equation about the operating point .

For the pendulum shown in the figure, the restoring torque due to gravity is given by

Derive the linearised equation about the operating point .

sinT MgL θ=

0θ =

Solution:

Since with ,

Solution:

Since with ,

0 0 00

sin ( ) cos(0)( )dT T MgL MgLd θ

θ θ θ θ θθ =

− = − = −

0 0T =

T MgLθ∴ =00 =θ


End

ME2142/ME2142E Feedback Control Systems

Installing OCTAVE in WINDOWS

And

Using OCTAVE for Control Systems Analysis

January 2009

©Department of Mechanical Engineering, and Bachelor of Technology Programme National University of Singapore

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1. Introduction to GNU Octave

GNU Octave (http://www.gnu.org/software/octave/) is a high-level language mostly compatible with MATLAB®(http://www.mathworks.com/). It is primarily intended for numerical computations, for solving common numerical linear algebra problems, manipulating polynomials, and integrating ordinary differential and differential-algebraic equations. It also has facilities for displaying the results of computation in various forms of graphs which makes it really nice. The really nice thing about GNU Octave is that it is a freely redistributable software. You may redistribute it and/or modify it under the terms of the GNU General Public License (GPL http://www.gnu.org/copyleft/gpl.html) as published by the Free Software Foundation (http://www.gnu.org/). On-line documentation on the usage of Octave can be found at http://www.octave.org/doc/index.html http://www.gnu.org/software/octave/doc/interpreter/.

2. OCTAVE for Control Systems Study

Similar to MATLAB, OCTAVE also has extensive tools for various functions including a “control systems toolbox” and it is these that we will primarily be interested in in this first introductory course on control systems. 3. Downloading, Installing and Using OCTAVE

Information on downloading, installing and using OCTAVE can all be found from OCTAVE website at

http://www.gnu.org/software/octave/about.html.

A copy of the latest pre-built installers for OCTAVE for both Windows and Mac OS X can be found via the main OCTAVE site or at

http://octave.sourceforge.net/

For those using Windows, do the following to install a copy of Octave 3.0.3:

1. download and run the installer file octave-3.0.3-setup.exe. Downloading could take a few to tens of minutes depending upon your connection speed.

2. During installation, click “next” at the following screens

3

4

For the next screen, choose “Gnuplot” and click “next”.

Click “next on the subsequent screens and “finish” at the last screen to complete the installation.

3. Test that the installation is running properly by launching OCTAVE

START>All Programs>GNU Octave>Octave to open an Octave window.

Note that you can also view Octave documentations.

4. In the Octave window, at the command line, type (underlined below)

Octave-3.0.3.exe:1> sys=tf(1,[1,2,4]); Octave-3.0.3.exe:2> step(sys);

The figure below, which was copied to clipboard and pasted here, should then appear of a step response for an underdamped 2nd-order system.

5. Congratulations, you have successfully installed Octave Version 3.0.3.

5

6. There are a few demonstration programs for control in the Octave suite:

DEMOControl: is a demo/tutorial program for various control toolbox functions.

AnalDemo: State-Space Analysis demo.

FRDemo: Frequency Response demo.

ModDemo: Model Manipulation demo.

RLDemo: Root Locus demo. You can run these by typing the demo/tutorial name at the command line.

4 Some Notes on Using Octave

4.1 Control System Specification: The control system under study can be specified in several ways.

(a) By using vectors to represent the coefficients of the numerator and denominator polynomials of the closed-loop transfer function. For example, to specify the system with the closed-loop transfer function

? ? ?? ? ? ? ? ? ?

The following commands can be used:

Octave-3.0.3.exe:3> num=[1 3]; Octave-3.0.3.exe:4> den=[1 2 4]; Octave-3.0.3.exe:5> sys1=tf(num,den);

or, alternatively, the following:

Octave-3.0.3.exe:6> sys2=tf([1 3],[1 2 4]);

in which [1 3] specifies the polynomial ?? ? ? ? and [1 2 4] specifies the polynomial ?? ? ?? ? ? ? ?.

To display the data structure of the system, the “sysout” command can be used as: Octave-3.0.3.exe:7> sysout(sys1); Octave-3.0.3.exe:8> sysout(sys2);

(b) By using vectors to specify the zeros and poles of the closed-loop system: For example, to define the system with the closed-loop transfer function ? ?? ? ? ???? ? ? ??? ? ? ?

we can use the commands: octave-3.0.3.exe:9> num=[-3]; octave-3.0.3.exe:10> den=[0 -2 -6]; octave-3.0.3.exe:11> k=5; octave-3.0.3.exe:12> sys3=zp(num,den,k);

or,alternatively, the following:

octave-3.0.3.exe:13> sys4=zp([-3],[0 -2 -6],5); To display the data structure of the system, the sysout command can be used as:

octave-3.0.3.exe:14> sysout(sys3); octave-3.0.3.exe:15> sysout(sys4);

4.2 Transient Responses: The unit impulse response and the unit step response of a system can be easily obtained by using the “impulse” function and the “step” function respectively. These functions produce a plot of the transient response. For example:

octave-3.0.3.exe:16> impulse(sys1);

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octave-3.0.3.exe:17> step(sys4);

4.3 Root Locus Plots: Root locus plots can easily be obtained using the function

“rlocus(sys,[inc, min_k, max_k]),

where “sys” is the open-loop transfer for which the root locus plot is desired, “inc” is the increment in the gain parameter, and “min_k” and “max_k” are the minimum and maximum values of the gain parameter for the plot. For example, if the root locus plot is desired of the system named sys4 specified in Section 4.1 above, the following command can be used:

octave-3.0.3.exe:18> rlocus(sys4);

or octave-3.0.3.exe:19> rlocus(sys4,0.02,0,5);

1


Review of Laplace TransformsReview of Laplace Transforms

ME2141/ME2142E


ME2141/ME2142E



Complex variableComplex variable

ωσ js +=

Real part Imaginary part

1−=j

Re

Im

σ

ω s

s-plane

Mathematical preliminariesMathematical preliminaries


Complex functionComplex function

Complex function

Complex conjugate

Complex function

Complex conjugate

Real part Imaginary part

yx jGGsG +=)(Im

G(s)-plane

xG

yG )(sG

θ

G22yx GGG +=

= −

x

y

G

G1tanθ

θjeGsG =)(

yx jGGsG −=)(

2


Euler’s TheoremEuler’s Theorem

θθθ sincos je j +=

θθθ sincos je j −=−Its Complex conjugates

Euler’s Theorem

( )θθθ jj eej

−−=21

sin

( )θθθ jj ee −+=21

cosSome useful formulas:


Laplace TransformsLaplace Transforms

A mathematical tool that transforms difficult differential equations into simple algebra problems where solutions can be easily obtained. A mathematical tool that transforms difficult differential equations into simple algebra problems where solutions can be easily obtained.

Definition:

Inverse Transform:

0for0)( <= ttf

Normally Tables of Laplace Transform pairs are used for taking the Laplace Transfrom and the Inverse Transforms.

∫∞

−=≡0

)()( dtesFtf stL

L -1 ∫∞+

∞−

=≡jc

jc

st dsesFj

tfsF )(21)()(π



End of L1End of L1

3


From Table L and L

Then L

From Table L and L

Then L

Properties of Laplace TransformsProperties of Laplace Transforms

Linearity:

ExampleExample

2

1s

t = 4

2)2sin( 2 +=

st

4

103)2sin(53 22 ++=+

sstt

⋅=⋅+⋅ atgbtfa )()(L ⋅+btf )( LL )(tg

)()( sGbsFa ⋅+⋅=

L -1 L -1L -1 ⋅=⋅+⋅ asGbsFa )()( ⋅+ bsF )( )(sG

)()( tgbtfa ⋅+⋅=



Translation:

If , then

Variable transform:

ExampleExampleFrom Table L

Then L and L

From Table L

Then L and L

4

3 6s

t =

4

23 6)2(

set

s−

=−

4

23 6

)2(se

ts−

=− 4

32

)2(6

−=

ste t

<≥−

=atatatf

tg0

)()( )()( sFetg as=L

)()( asFtfea t −=L

)(1

)(as

Fa

atf =L



Derivatives:

stf =)(' )0()()0()( fssFftf −=−L LL L )0()0()()0()0()( '2' fsfsFsfsftf −−=−− 2'' )( stf =

.

.

.

)0()0()0()()( )1('21 −−− −−−−= nnnnn ffsfssfstf LL

)()(' ttfsF −=

)()( 2'' tftsF =

)()1()( tftsF nnn −=

.

.

.

L -1

L -1

L -1

4



Final-Value Theorem:

Initial-Value Theorem:

)(lim)(lim0

ssFtfst →∞→

=

)(lim)0( ssFfs ∞→

=+


Finding the Inverse Finding the Inverse

1. Using L-1

2. Using the Table of Transform Pairs

3. Using properties

4. Using Partial Fraction Expansion(We do not normally use Approach 1. We use a combination of 2 to 4.)

1. Using L-1

2. Using the Table of Transform Pairs

3. Using properties

4. Using Partial Fraction Expansion(We do not normally use Approach 1. We use a combination of 2 to 4.)

∫∞+

∞−=

jc

jc

st dsesFj

sF )(21

)(π


Using Table and Properties Using Table and Properties

ExampleExampleFind the inverse of

Solution:

From Table L-1

L-1 5L-1

L-1 2L-1

Thus L-1

ts

=

2

1

=

+ 2)3(5

stte

s3

2 5)3(

1 −=

+

168

)3(51)( 222 +

++

+=sss

sF

=

+168

2s)4sin(2

44

22 ts

=

+

)4sin(25)( 3 ttetsF t ++= −

Lookup from TableLookup from Table

Lookup from Table orUse Translation propertyLookup from Table orUse Translation property

Lookup from TableLookup from Table

5


Using Table and Properties Using Table and Properties


Solution:

We write

522)( 2

8

++=

−

ssesF

s

228

2

8

2)1(2

522

)(++

=++

= −−

se

sse

sF ts

)2sin(2)1(

222

tes

t−=

++

8for0

8for)8(2sin)( )8(

<=≥−= −−

tttesF t

otherwise0andfor)()( =≥−=− atatfsFe as

Using translation propertyUsing translation property

From Table L-1

Thus L-1

Also Property L-1

Thus L-1

)2sin(2

222 t

s=

+


Using Partial Fractions Using Partial Fractions

Frequently

N(s) and D(s) being polynomials in s

Example

or Eqn(1-1)

are the zeros and poles of F(s) respectively. They can either be real or complex. If they are complex, they always occur in conjugate pairs .

)()(

)(sDsN

sF =

nm pppzzz −−−−−− ,,,and,,, 2121 KK

mnasasasabsbsbsbsF

nn

nn

mm

mm >

++++++++=

−−

−− with)(

011

1

011

1

LL

)())(()())((

)(21

21

n

m

pspspszszszsK

sF++++++

=LL



If Eqn(1-1) has distinct poles, then F(s) can always be expanded into a sum of partial fractions:

Eqn(1-2)

where ak are constants.

To determine the value of ak, multiply both sides of Eqn(1-2) by

(s+pk) and let s = -pk.

Eqn(1-3)

n

n

psa

psa

psa

sDsNsF

+++

++

+≡= L

2

2

1

1

)()()(

kpskk sD

sNpsa

−=

+=

)()(

)(

6




Solution:

We let

then

Thus and

)2)(1(3

)(++

+=

sss

sF

21)2)(1(3

)( 21

++

+≡

+++

=sa

sa

sss

sF

22)(1(

)3()1(

11 =

++

++=

−=ssss

sa

12)(1(

)3()2(2

2 −=

++

++=−=s

ssssa

21

12

)(+

−+

=ss

sF 0for2)( 2 ≥−= −− teetf tt



If Eqn(1-1) has multiple poles, each multiple pole pr of order qwill be equivalent to partial fractions of the form:

where bk are constants.

qr

q

rr ps

b

psb

psb

)()()( 221

+++

++

+L

If Eqn(1-1) has denominators of the form each of these will be equivalent to partial fractions of the form:

where a, b, c and d are constants.

)( 2 bass ++

LLLL

+++

++≡

++=

bassdcs

basssN

sF22 )(

)()(



Given , find f(t).

Solution:

Let

Multiplying both sides by (s+1)3

Letting we have

Comparing terms in s2, we have

Comparing constant terms, we have giving

Thus and

ExampleExample3

2

)1(32)(

+++=

ssssF

322

12 )1()1(32 bsbsbss ++++=++

1−=s 23 =b

11 =b

33

221

3

2

)1()1()1()1(32)(

++

++

+≡

+++=

sb

sb

sb

ssssF

3213 bbb ++= 02 =b

3)1(2

11

)(+

++

=ss

sF tt etetf −− += 2)(

7


Solving differential equations Solving differential equations



ExampleExampleSolve for y(t) given

Solution:

Transforming

or

giving

Multiplying both sides by s and letting s=0, we get

0)0(,0)0(,352 ===++ yyyyy &&&&

ssYsssYssYsYs

3)()52()(5)(2)( 22 =++=++

)52(3)( 2 ++

=sss

sY522 ++

++≡sscbs

sa

53

=a

[ ] [ ]s

sYyssYysysYs 3)(5)0()(2)0()0()(2 =+−+−− &



ExampleExample

We have

Multiplying both sides by s and letting s=0, we get

Multiplying both sides by the denominators

we have

Comparing terms in s,

Comparing terms in s2,

Thus

And

)52(3)( 2 ++

=sss

sY522 ++

++≡sscbs

sa

53

=a

)52( 2 ++ sss

)()52(3 22 csbsssa ++++=

5/65/60 −=⇒+= cc

5/35/30 −=⇒+= bb

522

53

53

)( 2 +++

−=ss

ss

sY

0for2sin1032cos

53

53)( ≥−−= −− ttetety tt

2222 2)1(

2103

2)1(

153

53

++−

+++−=

ss

ss

8



END

Table of Laplace Transform Pairs

ME2142/ME2142E

January 2008

Department of Mechanical Engineering

&

Bachelor of Technology Programme

National University of Singapore

Laplace Transform Pairs

)(tf )(sF

1 Unit impulse )(tδ 1

2 Unit step )(tu s1

3 t 2

1s

4 )!1(

1

−

−

nt n

, )3, ,2 ,1( K=n ns

1

5 nt , )3, ,2 ,1( K=n 1

!+ns

n

6 ate− as +

1

7 atte− 2)(

1as +

8 atn etn

−−

−1

)!1(1

, )3, ,2 ,1( K=n nas )(

1+

9 atnet −, )3, ,2 ,1( K=n 1)(

!++ nas

n

10 tωsin 22 ωω+s

11 tωcos 22 ω+ss

12 tωsinh 22 ωω−s

13 tωcosh 22 ω−ss

14 ( )atea

−−11

)(1

ass +

15 ( )btat eeab

−− −−1

))((1

bsas ++

16 ( )atbt aebeab

−− −−1

))(( bsass

++

17

−

−+ −− )(

11

1 btat aebebaab ))((

1bsass ++

18 ( )atat ateea

−− −−11

2

2)(1

ass +

19 ( )ateata

−+− 11

2 )(

12 ass +

20 te at ωsin− 22)( ω

ω++ as

21 te at ωcos− 22)( ω++

+as

as

22 te ntn n 2

21sin

1ζω

ζ

ω ζω −−

− , 1<ζ 22

2

2 nn

n

ss ωζωω

++

23 ( )φζωζ

ζω −−−

− − te ntn 2

21sin

1

1

ζζ

φ2

1 1tan

−= − , 1<ζ

22 2 nnsss

ωζω ++

24 ( )φζω

ζζω +−

−− − te n

tn 2

21sin

1

11

ζζ

φ2

1 1tan

−= − , 1<ζ

)2( 22

2

nn

n

sss ωζωω

++

25 tωcos1− )( 22

2

ωω+ss

26 tt ωω sin− )( 222

3

ωω

+ss

27 ttt ωωω cossin − 222

3

)(2

ωω

+s

28 tt ωω

sin21

222 )( ω+s

s

29 tt ωcos 222

22

)( ωω

+−

ss

30 ( )tt 2121

22

coscos1

ωωωω

−− , )( 2

22

1 ωω ≠ )()( 22

221

2 ωω ++ sss

31 ( )ttt ωωωω

cossin21

+

222

2

)( ω+ss

1


Modelling of Physical Systems

The Transfer Function

Modelling of Physical Systems

The Transfer Function

ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems


Differential EquationsDifferential Equations

Differential equation is linear if coefficients are constants or functions only of time t.

Linear time-invariant system: if coefficients are constants.

Linear time-varying system: if coefficients are functions of time.

Differential equation is linear if coefficients are constants or functions only of time t.

Linear time-invariant system: if coefficients are constants.

Linear time-varying system: if coefficients are functions of time.

PlantU Y

In the plant shown, the input u affects the response of the output y.In general, the dynamics of this response can be described by a differential equation of the form

In the plant shown, the input u affects the response of the output y.In general, the dynamics of this response can be described by a differential equation of the form

ubdtdub

dtudb

dtudbya

dtdya

dtyda

dtyda

m

m

m

m

n

n

n

n 01

1

101

1

1 ++++=++++−

−

−

− LL


Newton’s Law

f is applied force, nm is mass in Kg

x is displacement in m.

Newton’s Law

f is applied force, nm is mass in Kg

x is displacement in m.

mf

x

Mechanical Systems – Translational SystemsMechanical Systems – Translational Systems

Mechanical Systems – Fundamental LawMechanical Systems – Fundamental Law

Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems

xmmaf &&==

2


T is applied torque, n-mJ is moment of inertia in Kg-m2

is displacement in radiansis the angular speed in rad/s

T is applied torque, n-mJ is moment of inertia in Kg-m2

is displacement in radiansis the angular speed in rad/s

JT

ωθ

θω


Mechanical Systems – Torsional SystemsMechanical Systems – Torsional Systems

ωθ &&& JJT ==


Rotational:

T are external torques applied on the torsional spring, n-m

G is torsional spring constant, n-m/rad

Rotational:

T are external torques applied on the torsional spring, n-m

G is torsional spring constant, n-m/rad

1θ

2θ

Translational:

f is tensile force in spring, nK is spring constant, n/m

Translational:

f is tensile force in spring, nK is spring constant, n/m

f

x1x2

f

K


Mechanical Systems - springsMechanical Systems - springs

)( 21 xxKf −=Important: Note directions and signs

)( 21 θθ −= GT


Translational:

f is tensile force in dashpot, n

b is coefficient of damping, n-s/m

Translational:

f is tensile force in dashpot, n

b is coefficient of damping, n-s/m

f

x1x2

f

.

b

.

f

x1x2

f

.

b

.


Mechanical Systems – dampers or dashpotsMechanical Systems – dampers or dashpots

)( 21 xxbf && −=

Rotational:

T is torque in torsional damper, n-mb is coefficient of torsional damping,

n-m-s/rad

Rotational:

T is torque in torsional damper, n-mb is coefficient of torsional damping,

n-m-s/rad

2θ&1θ&

)( 21 θθ && −= bT

3



Example Example

Since m = 0, givesm af =∑ 0=− ds ff

Since and

Thus

Or

ybf d &= )( yxKf s −=

0)( =−− ybyxK &

KxKyyb =+&

xy

b KA

Derive the differential equation relating the output displacement y to the input displacement x.

Derive the differential equation relating the output displacement y to the input displacement x.

Free-body diagram at point A,A fsf d

Note: Direction of fs and fd shown assumes they are tensile.

Note: Direction of fs and fd shown assumes they are tensile.


The transfer function of a linear time invariant system is defined as the ratio of the Laplace transform of the output (response) to the Laplace transform of the input (actuating signal), under the assumption that all initial conditions are zero.

The transfer function of a linear time invariant system is defined as the ratio of the Laplace transform of the output (response) to the Laplace transform of the input (actuating signal), under the assumption that all initial conditions are zero.

The Transfer FunctionThe Transfer Function

Previous ExampleAssuming zero conditions and taking Laplace transforms of both sides we have

Transfer Function

This is a first-order system.

Previous ExampleAssuming zero conditions and taking Laplace transforms of both sides we have

Transfer Function

This is a first-order system.

KxKyyb =+&

)()()( sKXsKYsbsY =+

KbsK

sXsYsG

+==

)()()(



Example Example

Free-Body diagram

givesmaf =∑ ods xmff &&=+

m

fs fd

xo

m

K b

xi

xo

)()()()()(2 sKXsbsXsKXsbsXsXms iiooo +=++

ooioi xmxxbxxK &&&& =−+− )()(

iiooo KxxbKxxbxm +=++ &&&&

Thus

Or

And

KbsmsKbs

sXsX

sGi

o

+++==

2)()(

)(Transfer Function . This is a second-order system.

For the spring-mass-damper system shown on the right, derive the transfer function between the output xo and the input xi.

For the spring-mass-damper system shown on the right, derive the transfer function between the output xo and the input xi.

Note: fs and fd assumed to be tensile.

4


Capacitance

Or

Complex impedance

Capacitance

Or

Complex impedance

eq

C =

Ceq =

dtde

Cdtdq

i ==

)(sECI =

)/(1 sCX c =

cIXsC

IE ==1

e i C


Electrical ElementsElectrical Elements Resistance

Units of R: ohms ( )

Resistance

Units of R: ohms ( )

iRe =

Re

i =

Ω

e i R

Inductance

Units of L: Henrys (H)

Or

Inductance

Units of L: Henrys (H)

Or

dtdi

Le =

∫=t

teL

i0

d1

)( sLIIXE L ==

e i L



Electrical Circuits- Kirchhoff’s LawsElectrical Circuits- Kirchhoff’s Laws

Current Law:

The sum of currents entering a node is equal to that leaving it.

Current Law:

The sum of currents entering a node is equal to that leaving it.

0=∑ i

Voltage Law:

The sum algebraic sum of voltage drops around a closed loop is zero.

Voltage Law:

The sum algebraic sum of voltage drops around a closed loop is zero.

0=∑ e



Electrical Circuits- ExamplesElectrical Circuits- Examples

RC circuit: Derive the transfer function for the circuit shown,

and

giving

This is a first-order transfer function.

RC circuit: Derive the transfer function for the circuit shown,

and

giving

This is a first-order transfer function.

ci IXIRE +=

co IXE =

)/(1

)/(1

sCR

sC

XR

X

E

E

c

c

i

o

+=

+=

11

+=

RCs

eii C

R

eo

5



Electrical Circuits- ExamplesElectrical Circuits- Examples

RLC circuit:

and

giving

This is a second-order transfer function.

RLC circuit:

and

giving

This is a second-order transfer function.

cLi IXIXIRE ++=

co IXE =

)/(1)/(1

sCsLRsC

XXRX

EE

cL

c

i

o

++=

++=

11

2 ++=

RCsLCs

ei

i C

R

eo

L



Operational Amplifier – Properties of an ideal Op AmpOperational Amplifier – Properties of an ideal Op Amp

Gain A is normally very large so that compared withother values, is assumed small, equal to zero.Gain A is normally very large so that compared withother values, is assumed small, equal to zero.

)( 12 vvAvo −=)( 12 vv −

The input impedance of the Op Amp is usually very high (assumed infinity) so that the currents i1 and i2 are very small, assumed zero.

The input impedance of the Op Amp is usually very high (assumed infinity) so that the currents i1 and i2 are very small, assumed zero.

Two basic equation governing the operation of the Op Amp

and

Two basic equation governing the operation of the Op Amp

and 0,0 21 == ii2112 or0)( vvvv ==−



Operational Amplifier – ExampleOperational Amplifier – Example

For the Op Amp, assume i1=0 and vs=v+=0.For the Op Amp, assume i1=0 and vs=v+=0.

-

+

vi i1= 0voZi

Z f

i i

if

S

Then orThen or0=+ fi ii 0=+f

o

i

i

Zv

Zv

ThereforeThereforei

i

fo v

ZZ

v −=i

f

i

o

Z

Z

sVsV

−=)()(

6



Operational Amplifier – ExampleOperational Amplifier – Example

-

+

vi i1= 0voZi

Z f

i i

if

S

ii

fo v

ZZ

v −=

For the following

sCRZ ff1+=

+−≡

+−=−=

s

KK

CsRR

R

R

Z

v

v ip

ii

f

i

f

i

o 1


Permanent Magnet DC Motor Driving a LoadPermanent Magnet DC Motor Driving a Load

For the dc motor, the back emf is proportional to speed and is given by where is the voltage constant. The torque produced is proportional to armature current and is given by where is the torque constant.

For the dc motor, the back emf is proportional to speed and is given by where is the voltage constant. The torque produced is proportional to armature current and is given by where is the torque constant.

ωeK

eK

iKT t= tK

Relevant equations:Relevant equations: ωeaa KdtdiLiRe ++=

iKT t= ωω

bdtd

JT +=

e i

RaLa

ωeK J

bT

ω

Note: By considering power in = power out, can show that Ke=KtNote: By considering power in = power out, can show that Ke=Kt


End

1


System Transient/Time ResponseSystem Transient/Time Response



System responseSystem response

The magnitude of the transient response decreases with time and ultimately vanishes leaving only the steady-state response. It is always associated with the component

0with >− ae at

The system response comprises two parts, transient and steady-state.

Transient Response Steady -State Response

Steady -State Error


System Characteristic EquationSystem Characteristic Equation

Consider the system with the closed-loop transfer function, Gc(s) as shown

Gc(s)System

Input OutputR(s) C(s)

The system’s characteristic equation is given by

0)( =sDc

with

where Nc(s) and Dc(s) are polynomials of s.

)()(

)()()(

sDsN

sGsRsC

c

cc ==

Note that the characteristic equation is a property of the system and is not dependent on the input.

2



ExamplesExamplesSpring-mass-damper (Slide 9: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

Spring-mass-damper (Slide 9: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

KbsmsKbs

sXsX

sGi

o

+++

==2)(

)()(

02 =++ Kbsm s

R-C circuit (Slide 12: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

R-C circuit (Slide 12: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

11

+=

RCsEE

i

o

01 =+RCs

Closed-loop feedback system (Slide 8: Block Diagram Algebra)

Transfer Function

Characteristic Eqn:

Closed-loop feedback system (Slide 8: Block Diagram Algebra)

Transfer Function

Characteristic Eqn:

GHG

RC

+=

1

01 =+ GH



The roots of this equation are the closed-loop poles and they determine the transient response of the system.

Characteristic equation

0)( =sDc

)()(

)()()(

sDsN

sGsRsC

c

cc ==

Note that if all the roots, pr, are negative, then the transient response will eventually die away as t increases.

Each root, p, of this equation will contribute a term in the time response of the system. Or

LL ++= p tAetc )(

p te

But if any of the roots is positive, then the transient response will grow without bounds as time increases. The system is then said to be unstable.


Given a dynamic system:Given a dynamic system:

System ResponseSystem Response

We use§ Standard test inputs to excite system and observe

response§ Classify systems with similar characteristics and

identify their performance characteristics with system parameters.

§ How do we specify the characteristics of the response required?

§ How do we compare it with another system?

§ How do we know whether it’s response will adequately meet our needs?

§ How will we know how it will respond to different inputs?

3


2) Ramp input2) Ramp input

000)(

<=≥=

ttAttr

2)(

sAsR =

At

t = 0

r(t)

t

System Response –Test signalsSystem Response –Test signals

1) Step input1) Step input

000)(

<=≥=

ttAtr

sAsR =)(

A

t = 0

r(t)

t

When A = 1, we have a unit step input.Used to study response to sudden changes in input.

When A = 1, we have a unit ramp input.Used to study response to gradual changes in input.


4) Sinusoidal input4) Sinusoidal input

000sin)(

<=≥=

tttAtr ω

t

r(t)

t = 0

3) Impulse input

is the unit-impulse function or Dirac delta function

3) Impulse input

is the unit-impulse function or Dirac delta function

)0()( δAtr =

AsR =)(

A

t = 0

r(t)

t

System Response –Test signalsSystem Response –Test signals

When A = 1, we have a unit impulse input.Used to study response to sudden shocks or impacts.

Used for frequency response analysis.Important method. Will be discuss in the second half of course.

Using test signals (1) to (3) are often known as time response or transient response analysis while using test signal (4) is known as frequency response.


System Response – First-order systemsSystem Response – First-order systems

A first-order system can be written in the standard form

T is known as the time constant and determines the speed of response.

A first-order system can be written in the standard form

T is known as the time constant and determines the speed of response.

1)(

)(

+=

Ts

K

sR

sC

ExamplesExamples

Spring-damper system (Slide 7 of Modelling of Physical Systems)

.

Spring-damper system (Slide 7 of Modelling of Physical Systems)

. KbsK

sXsY

+=

)()(

KbT

Ts=

+= with

11

RC circuit (Slide 12 of Modelling of Physical Systems)

.

RC circuit (Slide 12 of Modelling of Physical Systems)

.RCT

TsRCsEE

i

o =+

=+

= with1

11

1

If the transfer function are the same, then the response y(t) and eo(t)will be the same for the same inputs in x(t) and ei(t) ..

4



with

.

with

.

1)(

)(

+=

Ts

K

sR

sC

Response to a unit step inputResponse to a unit step input

ssR

1)( =

Thus )(1

)( sRTs

KsC

+=

sTsTK 1

)/1(/

+=

TsB

sA

/1++≡

Multiplying both sides by s and letting s=0 gives A = K

Multiplying both sides by (s+1/T) and letting s = -1/T gives B = -K

ThereforeTs

KsKsC

/1)(

+−=

Using tables

)1()( // TtTt eKKeKtc −− −=−= 0for ≥t



Response to a unit step inputResponse to a unit step input)1()( /Ttetc −−=For K = 1

Note: The smaller the time constant T, the faster the response.The shape is always the same.



with

.

with

.

1)(

)(

+=

Ts

K

sR

sC

Response to a unit ramp inputResponse to a unit ramp input

Thus )(1

)( sRTs

KsC

+=

2

1)(

ssR =

2

1

)/1(

/

sTs

TK

+=

)/1(2 TsKT

sKT

sK

++−≡

For K = 1,

)1()( / TteTttc −−−=

with the error e(t) = r(t) – c(t)

Using tables

0for ≥t)()( / TtTeTtKtc −+−=

)1( / TteT −−=

r(t)

t = 0

r(t)

t

ess=Tc(t)

r(t)

t = 0

r(t)

t

ess=Tc(t)

5



with

.

with

.

1)(

)(

+=

Ts

K

sR

sC

Response to a unit impulse inputResponse to a unit impulse input

1)( =sR

ThusTs

TKTs

KsC

/1/

1)(

+=

+=

Or TteTK

tc /)( −=

For K = 1,

TteT

tc /1)( −=

r(t)

t = 0

1

t

r(t)

t = 0

1

t


§ The transient response all contains the term which is determined by the root of the characteristic equation and the parameter T.

§ The transient response all contains the term which is determined by the root of the characteristic equation and the parameter T.

Tte /−

Response toResponse to

Unit Impulse

TteTK

tc /1 )( −=

System Response – Linear time-invariant systems

System Response – Linear time-invariant systems

1)()(

+=

TsK

sRsCPropertiesProperties Characteristic EquationCharacteristic Equation

TsTs 101 −=⇒=+

§ Note that the unit step is the derivative of the unit ramp, and the unit impulse is the derivative of the unit step.

§ Note that similarly, c2(t) is the derivative of c3(t) and c1(t) is the derivative of c2(t) .

§ For linear time-invariant systems, the response to the derivative of an input can be obtained by taking the derivative of the response to the input.

Unit Step

)1()( /2

TteKtc −−=

Unit Ramp

)()( /3

TtTeTtKtc −+−=


Block DiagramBlock Diagram

Permanent Magnet DC MotorPermanent Magnet DC Motor

e i

Ra La

ωeK J

bT

ω

e i

Ra La

ωeK J

bT

ω

The Permanent Magnet DC motor.The Permanent Magnet DC motor.

Governing equationsGoverning equations

ωeaa Kdtdi

LiRe ++=

iKT t=

ωω

bdt

dJT +=

IsLRKE aae )( +=Ω−IKT t=

Ω+= )( bJsT

+

-

E I

aa RsL +1

ΩeK

tKT

bJs +1 Ω

eK

+

-

E

ΩeK

Ω

eK

))(( bJsRsLK

aa

t

++

6


+

-

E

ΩeK

Ω

eK

))(( bJsRsLK

aa

t

+++

-

E

ΩeK

Ω

eK

))(( bJsRsLK

aa

t

++

The Permanent Magnet DC motor.The Permanent Magnet DC motor.

Commonly

bJ

RL

a

a <<

La can then be neglected

Block diagram then becomesBlock diagram then becomes

E +

-

Ω

eK

bJsRK at

+/E +

-

Ω

eK

bJsRK at

+/

GHG

E +=

Ω1

)(

/1

)(/

bJs

RKKbJs

RK

aet

at

++

+=

aet

at

RKKbJsRK

//

++=

1+=

sK

τ

aet

at

RKKbRK

K/

/with

+=

aet RKKbJ

/+=τ

Permanent Magnet DC MotorPermanent Magnet DC Motor


aet RKKbJ

/+=τ

Speed Control of the DC MotorSpeed Control of the DC Motor

1+sK

τ

E Ω

The response to a unit step input is first order with a time constant of

t = 0 t

K

Ω

τ

With speed feedbackWith speed feedback

1+sK

τ

E ΩError+

-

VKc

Controller

11

1

++

+=Ω

sKK

sKK

V c

c

τ

τKKs

KK

c

c

++=

1τ 1''+

=sK

τ

KK c+=

1'with ττ

KKKK

Kc

c

+=

1'

The resultant system is still first-order but the time constant is now much smaller, thus a much faster response.


ExamplesExamples

RLC circuit (see Modelling of Physical Systems)

.

RLC circuit (see Modelling of Physical Systems)

.1

1)()(

2 ++=

RCsLCssEsE

i

o

System Response – Second-order systemsSystem Response – Second-order systems

A second-order system will be of the form

with a, b, c, d and e being constants.

A second-order system will be of the form

with a, b, c, d and e being constants.cbsasesd

sXsY

+++=

2)()(

Spring-mass-damper

KbsmsKbs

sXsX

i

o

+++

=2)(

)(

Standard Form:

.

Standard Form:

. 22

2

21

2

221

2)(

12)()(

nn

n

nn

ssKsK

ssKsK

sRsC

ωζωω

ωζ

ω++

+=++

+=

7


Closed-Loop Position Feedback System(Servomechanism)

Closed-Loop Position Feedback System(Servomechanism)

V Ω

1+sK

τ s1 θ

Gc

controller

E+

-

R

With Gc being a proportional gain Kp

θE+

-

R)1( +ss

KK p

τ

withτ

ωKKp

n =2

τζω 12 =n

τζ

KK p

121

=

natural frequencydamping ratioζ

ω n

KKss

KK

GHG

R p

p

++=

+=

Θ21 τ

In standard format

22

2

2 nn

n

ssR ωζωω

++=

Θ


Examples: Determine the value of gain K for the closed-loop system to have anundamped natural frequency of 4. What will then be the damping factor?

Examples: Determine the value of gain K for the closed-loop system to have anundamped natural frequency of 4. What will then be the damping factor?

System Response – Second-order systemsSystem Response – Second-order systems

2825

2 ++ ss

010282 2 =+++ Kss

02514 222 =++≡+++ nn ssKss ωςω

164)51( 22 ==+= Knω 3=K

Characteristic equation: 01 =+GH 0282

)2(51

2=

+++

ssK

42 =nςω 5.0=ς


Consider Consider 22

2

2 nn

n

ssR ωζω

ω

++=Θ

Time Response – Second-order systemsTime Response – Second-order systems

For , the roots are equal and the system is said to be critically damped.For , the roots are equal and the system is said to be critically damped.

1=ζ np ω−=2,1

The roots of the characteristic equation are

122,1 −±−= ζωζω nnp

For , the roots are both real and unequal and the system is said to be overdamped.For , the roots are both real and unequal and the system is said to be overdamped.

1>ζ 122,1 −±−= ζωζω nnp

For , the roots are a pair of complex conjugates10 << ζ

where is called the damped natural frequency and the response is underdamped.

dn jp ωζω ±−=2,1

21 ζωω −= nd

8


Step Response – Second-order systemsStep Response – Second-order systems

Therefore

withwiths

sR1

)( =22

2

2 nn

n

ssR ωζω

ω

++=Θ

sss nn

n 1)2( 22

2

ωζω

ω

++=Θ

This represents a decaying oscillatory response depending upon with a frequency of oscillation of This represents a decaying oscillatory response depending upon with a frequency of oscillation of

Underdamped ResponseUnderdamped Response 10 << ζ

ζdω

From tables (Entry 24 in Tables), we haveFrom tables (Entry 24 in Tables), we have

)sin(1

1)(2

φωζ

ζω

+−

−=−

tetc d

tn

20 πφ <<0≥t

−= −

ζζ

φ2

1 1tan

21 ζωω −= nd


From tables (Entry 18 in Dr Chen’s tables with ), we haveFrom tables (Entry 18 in Dr Chen’s tables with ), we havena ω=

)1(1

)()(

122

atat ateea

tcass

−− −−=⇔+


We have We have

sss nn

n 1)2( 22

2

ωζω

ω

++=Θ

Critically damped ResponseCritically damped Response 1=ζ

ss n

1)( 2

2

ωω

+=

This represents a non-oscillatory response with an exponentially decaying transient component and a zero steady-state error. The speed of decay of the transient component depends upon the parameter .

This represents a non-oscillatory response with an exponentially decaying transient component and a zero steady-state error. The speed of decay of the transient component depends upon the parameter .nω

tn

t nn teetc ωω ω −− −−= 1)( 0≥tgiving for


We use Entry 17 in Dr Chen’s tables. We use Entry 17 in Dr Chen’s tables.


sss nn

n 1)2( 22

2

ωζω

ω

++=Θ

Overdamped ResponseOverdamped Response 1>ζ

sss nnnn

n

)1)(1( 22

2

−−+−++=

ζωζωζωζω

ω

0≥t

bta t eCeC −− ++= 211

for , C1 and C2 being constants.

The response is non-oscillatory, starts initially with and exponentially rises to . The response is non-oscillatory, starts initially with and exponentially rises to .

0)0( =c1)( =∞c

If , then and the first exponential term will decay much faster than the second. The pole can then be neglected and the system behaves like a first-order system.

If , then and the first exponential term will decay much faster than the second. The pole can then be neglected and the system behaves like a first-order system.

ba >>1>>ζ)( as +

))((

2

bsassn

++=

ω12 −+= ζωζω nna 12 −−= ζωζω nnbwith and

2nab ω=We have

−

−+= −− )

2

(1

1)( btat aebebaab

tcnω

so that

9



Normalized response curves

For fast response,is usually

desirable.7.0≈ζ

If no overshoot is required, is usually used.

1>ζ


Transient Response SpecificationsTransient Response Specifications

Maximum (percent) overshoot:

%100)(

)()(×

∞∞−

=c

ctcM p

p

Delay time

Rise time:10% - 90%, or5% - 95%, or0% - 100%

Peak time

Settling time: time to reach and stay within specified limits, usually 2% or 5%.

Five measures of transient performance – based on 2nd-order underdamped responseFive measures of transient performance – based on 2nd-order underdamped response


Measures of transient performanceMeasures of transient performance

We haveWe have

)sin(1

1)(2

φωζ

ζ ω

+−

−=−

te

tc d

tn

−= −

ζζ

φ2

1 1tan

1)( =rtc giving 0)sin( =+φω rd t or 0=+ φω rd t

Thus φω −=rd t

−−= −

ζζ 2

1 1tan

−−

= −

ζζ 2

1 1tan

Rise TimeRise Time rt

drt ω

βπ −=giving

21 ζωω −= nd

10


We haveWe have

)sin(1

1)(2

φωζ

ζ ω

+−

−=−

te

tc d

tn

−= −

ζζ

φ2

1 1tan

Peak TimePeak Time pt

21 ζωω −= nd

01

)(sin)(

2=

−= −

=

pn

p

tnpd

tt

etdt

tcd ζω

ζ

ωω

giving 0sin =pd tω or K,3,2,,0 πππω =pdt

Therefore for the first peak. d

ptωπ

=



We haveWe have

)sin(1

1)(2

φωζ

ζ ω

+−

−=−

te

tc d

tn

−= −

ζζ

φ2

1 1tan

Maximum OvershootMaximum Overshoot

21 ζωω −= nd

pM

1)( −= pp tcM

])/(sin[1 2

)/(

φωπωζ

ωπζω

+−

=−

dd

dne

)sin(1 2

)1/( 2

φπζ

πζζ

+−

=−−e

As 21)sin( ζφπ −−=+

Therefore πζζ )1/( 2−−= eM p



We haveWe have

)sin(1

1)(2

φωζ

ζ ω

+−

−=−

te

tc d

tn

−= −

ζζ

φ2

1 1tan

Settling TimeSettling Time

21 ζωω −= nd

st

The curves gives the

envelope curves of the transient response. )1/(1 2ζζω −± − tne

is found to be approximately

where “time constant”

st

Tt s 4=Tt s 3=

(2% criterion)

(5% criterion)

n

Tζω

1=


11


End

A Note on neglecting aL On Slide 16 in the presentation on System Response, it was mentioned

that the inductance aL can be neglected. Below are some notes on this.

Consider the transfer function

))((

'bJsRsL

KV aa ++

=ω

We can re-write this as

)1)(1( 21 ++

=ss

KV ττω

where bR

KKa

'= , aa RL /1 =τ and bJ /2 =τ .

Consider the response to a unit step input, V(s) = 1/s.

Then sss

Ks)1)(1(

)(21 ++

=Ωττ

From the Laplace Transform Tables (Chen’s Table entry 17), we write

sss

Ks)/1)(/1(

)/()(21

21

ττττ++

=Ω

the inverse transform as

⎥⎦

⎤⎢⎣

⎡−

−+= −− )11(

/1/111

)/1)(/1()/()( 21 /

1

/

22121

21 ττ

ττττττττ

ϖ tt eeKt

⎥⎦

⎤⎢⎣

⎡−

−+= −− )(11 21 /

2/

112

ττ ττττ

tt eeK

For simplicity, consider the case when K=1. Consider also that 11 =τ and with 21 ττ << with 10/1/ 21 =ττ .

Then )10(911)( 10/tt eet −− −+=ω (1)

If, because 21 ττ << , we neglect 1τ , then we have

ssssss

Ks)10/1(

10/1)110(

1)1(

)(2 +

=+

=+

=Ωτ

From Entry 14 in Dr Chen’s Table, we have

10/10/ 1)1(10/110/1)( tt eet −− −=−=ω (2)

The two responses represented by Equation (1) and (2) are plotted in the figure below.

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time, sec

The two curves shown above, plotted using Equations (1) and (2) are approximately similar, meaning that we can neglect 1τ and the response will still be reasonably accurate with the simpler transfer function.

(2)

(1)

A Note on Derivation of Peak Time, Tp (Slide 27 in System Response)

The output response is

)sin(1

1)(2

φωζ

ζω

+−

−=−

tetc d

tn

(1)

with 21 ζωω −= nd and ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −= −

ζζ

φ2

1 1tan .

Differentiating Eqn (1) by parts, we have

( )dd

t

d

tn tete

dttcd nn

ωφωζ

φωζ

ζω ζωζω

)cos(1

)sin(1

)()(22

+−

−+−

−−=

−−

[ ])cos()sin(1 2

φωωφωζωζ

ζω

+−+−

=−

ttedddn

tn

( ) ( )[ ]φωφωωφωφωζωζ

ζω

sin)sin(cos)cos(sin)cos(cos)sin(1 2

ttttedddddn

tn

−−+−

=−

Noting that ζφ =cos and 21sin ζφ −= , we have

[ ])cos()cossin()sin()sincos(1

)(2

ttetc ddnddn

tn

ωφωφζωωφωφζωζ

ζω

−++−

=−

&

( )[ ])cos()11()sin()1(1

2222

2tte

dnndnn

tn

ωζωζζζωωζωωζζ

ζω

−−−+−+−

=−

[ ])sin(1 2

tedn

tn

ωωζ

ζω

−=

−

Letting 0)( =tc& at ptt = , we thus have

[ ] 0)sin(1 2

=−

−

pdn

t

te pn

ωωζ

ζω

.

Since 0≠− pnte ζω for ∞<pt ,

we have 0)sin( =pd tω giving K,3,2,,0)sin( πππω =pd t

Choosing the first peak gives

d

ptωπ

=

1


Steady-State CharacteristicsSteady-State Characteristics



Consider the unity-feedback systemConsider the unity-feedback system

System TypeSystem Type

RG

C

-

+ E

With … (2-1)

The parameter N associated with the term SN in the denominator represents the “Type” of the system. Example: Type 0 if N=0, Type 1 if N=1 and so on. A free “s” term in the denominator represents an integration. The higher the type number, the better the steady-state accuracy of the closed-loop control system.However, the higher the system type, the greater the problem with system stability.

)1()1)(1(

)1()1)(1()(

21 +++

+++=

sTsTsTs

sTsTsTKsG

pN

mba

L

L


The same unity-feedback systemThe same unity-feedback system

Steady-State Errors – Static Error ConstantsSteady-State Errors – Static Error Constants

RG

C

-

+ E RCR

RE −

=RC

−= 1

GG+

−=1

1G

GG+

−+=

1)1(

GRE

+=

11

Error Transfer function

Thus )()(1

1)( sR

sGsE

+=

and the steady-state error is )(lim teet

ss∞→

=

)(lim0

ssEs→

=

)(1)(

lim0 sG

ssRe

sss +=

→

2



)(1)(lim

0 sGssRe

sss +

=→

For a unit-step inputFor a unit-step input ssR

1)( = and

ssGs

esss

1)(1

lim0 +

=→ )0(1

1G+

=

Static Position Error Constant, Kp is defined as

)0()(lim0

GsGKsp ==

→and

pss K

e+

=1

1

with ,)1()1)(1(

)1()1)(1()(

21 +++

+++=

sTsTsTs

sTsTsTKsG

pN

mba

L

L

for Type 0 systems KK p =

for Type 1 or higher systems K

ess +=

11

∞=pK 0=sse



)(1)(lim

0 sGssRe

sss +

=→

For a unit-ramp inputFor a unit-ramp input2

1)(

ssR = and

20

1)(1

limssG

se

sss +

=→ )(

1lim

0 ssGs→=

Static Velocity Error Constant, Kv is defined as

and )(lim0

ssGKsv →

=v

ss Ke

1=

with ,)1()1)(1(

)1()1)(1()(

21 +++

+++=

sTsTsTs

sTsTsTKsG

pN

mba

L

L

for Type 0 systems

for Type 1 systems

0=sse

0=vK

for Type 2 or higher systems

KKv =K

ess

1=

∞=s se

∞=vK



)(1)(lim

0 sGssRe

sss +

=→

and

Static Acceleration Error Constant, Ka is defined as

and

with ,)1()1)(1(

)1()1)(1()(

21 +++

+++=

sTsTsTs

sTsTsTKsG

pN

mba

L

L

for Type 0 and Type 1 systems

for Type 2 systems

0=ssefor Type 3 or higher systems

Ke

ss

1=

∞=s se

For a unit-acceleration inputFor a unit-acceleration input

,0for2

)(2

≥= tt

tr3

1)(

ssR =

30

1)(1

limssG

ses

ss +=

→ )(1

lim20 sGss→

=

)(lim 2

0sGsK

sa

→=

ass K

e1

=

0=aK

KK a =

∞=aK

3


Steady-State Errors Steady-State Errors

Summaryof steady-State errors

Summaryof steady-State errors

Step Input 1=r

Ramp Input tr =

Accel. Input 2/2tr =

Type 0 system K+1

1 ∞ ∞

Type 1 system 0 K1 ∞

Type 2 system 0 0 K1

§ Type 0 systems have finite steady-state errors for step inputs and cannot follow ramp inputs.

§ Type 1 systems have zero steady-state errors for step inputs, finite errors for ramp inputs, and cannot follow acceleration inputs.

§ Type 2 systems are needed to follow ramp inputs with zero steady-state errors.

§ In general, the higher the static gain of the open-loop transfer function, G(s), the smaller the steady-state errors. However, higher gains normally lead to stability problems.


End

1


Block Diagram AlgebraBlock Diagram Algebra



The Transfer Function Block

Block Diagram RepresentationBlock Diagram Representation

A block diagram is a graphical tool can help us to visualize the model of a system and evaluate the mathematical relationships between their elements, using their transfer functions.

A block diagram is a graphical tool can help us to visualize the model of a system and evaluate the mathematical relationships between their elements, using their transfer functions.

G(s)System

Input OutputR(s) C(s)

)()()(

sRsCsG =

The transfer function G(s) is § defined only for a linear time-invariant system and not

for nonlinear systems.§ Is a property of the system and is independent of the

input to the system.

§ Commutative 1221 GGGG =§ Associative

1221 GGGG +=+


The Summing Point

Block Diagram ElementsBlock Diagram Elements

§ Any number of inputs. Only one output

Signed inputs

X

-Z

Y+

+ + X + Y - Z

output

2


Blocks in series or cascaded blocks


When manipulating block diagrams, the original relationships, or equations, relating the various variables must remain the same.When manipulating block diagrams, the original relationships, or equations, relating the various variables must remain the same.

§ When blocks are connected in series, there must be no loading effect.

G1 G2

X Y ZG1G2

ZX


Blocks in parallel


G1

G2

++ YX

G1 + G2

YX


G+

+ ZX

Y

G+

+ ZX

Y

GX Y

X

GX Y

X

XG

Y

Z

+

+XG

Y

Z

+

+


G

G

+

+ ZX

Y

GX Y

1/GX

+

+ Z

Y

XG

1/G

3


Closed-Loop Feedback SystemClosed-Loop Feedback System

is called the open-loop transfer functionGHEB

=

is called the feedforward transfer functionGEC

=

R is called the reference inputC is the output or controlled variableB is the feedbackE = (R – B) is the error

R G C

-

+ E

HB


Closed-Loop Feedback SystemClosed-Loop Feedback System

C = GE= G(R – B)= G(R – HC)

C(1 + GH) = GR

GHG

RC

+=

1

is the closed-loop transfer functionRC

Also and GC

E =GHR

CGR

E+

==1

11

is called the error transfer functionRE

R G C

-

+ E

HB


Closed-Loop Control Feedback SystemClosed-Loop Control Feedback System

is the open-loop transfer functionHGGEB

pc=

is the feedforward transfer functionpc GGEC

=

G c is the controller transfer functionGp is the plant transfer functionM is the manipulated variableD is the external disturbance

RGp

C

-

+ E

HB

Gc+ +

D

M

4


HGG

GG

GHG

RC

pc

pc

+=

+=

11

Assuming R = 0, we can re-draw

HGG

G

GHG

DC

cp

p

+=

+=

11

Closed-Loop Control Feedback SystemClosed-Loop Control Feedback System

R GpC

-

+ E

HB

Gc+ +

D

M

DGp

C

-

+

GcH


Block Diagram ManipulationBlock Diagram Manipulation

Example: Determine C(s)/R(s)Example: Determine C(s)/R(s)

R G-

+

H

F ++

-

+

I

++

ED

C

aDa

When manipulating blocks, must ensure C(s) does not change, so that C(s)/R(s)

remains same.




b = Fa-Fc+Da

Assume names of signals as shown

We wish to move this signal to before block F

b = Fa-Fc+(D/F)aF

To move signal b to after Block G

5




R G-

+

H

F ++

-

+

I

++

ED

C

aDa

bEb

R G-

+

H

F-

+

I

++

ED/F

C+

a(D/F)a

Gb

R-

+

H

FG-

+

I

+ +

E/G

C1+D/F

Gb

Eb

R-

+

H

FG-

+

I

1+E/G C1+D/F

C = Gb + Eb

C = Gb + (E/G)Gb



ExampleExample

R-

+

H

FG-

+

I

1+E/G C1+D/F R

-

+

I

1+E/G C1+D/F

FGH1FG

+

R-

+

I

C

+

+

+

GE

FGHFG

FD 1

11

CRI

GE

FGHFG

FD

GE

FGHFG

FD

+

+

++

+

+

+

11

11

11

1

GHG

+1


End

1


Concepts of System StabilityConcepts of System Stability



Higher-Order System responseHigher-Order System response

Consider the closed loop transfer functionConsider the closed loop transfer function

)()(

)()()(

sDsN

sGsRsC

c

cc ==

mnasasasabsbsbsb

nn

nn

mm

mm >

++++++++

=−

−

−− with

011

1

011

1

LL

The system is said to be a higher-order system for n > 2.There will be n poles of Gc(s), or roots of the denominator Dc(s).

Gc(s) can thus be written as

)()()(

sRsCsGc =

)2()2)(2)(())((

)())((222

22222

1112

21

21

rrrq

m

sssssspspsps

zszszsK

ωωζωωζωωζ +++++++++

+++=

LL

L

)2( nrq =+whereReal RootsReal Roots

Complex rootsComplex roots∑ ∑= = ++

−+++

+=

q

j

r

k kkk

kkkkk

j

j

scsb

psa

1 122

2

21)(

ωωζζωζω


Higher-Order System responseHigher-Order System response

For a unit step input, we can re-write C(s) in terms of partial fractions as

∑ ∑= = ++

−+++

++=

q

j

r

k kkk

kkkkk

j

j

scsb

psa

sasC

1 122

2

21)(

)(ωωζ

ζωζω

in which we assume that all the poles are distinct, i.e. not repeated.

The time response will then be, by using the Inverse Laplace Transform

∑∑=

−− +−

−++=

r

kkkk

j

ttp

j teeaatckk

j

1

2

2)1sin(

1)( φωζ

ζ

ωζ

where

−= −

k

kk ζ

ζφ

21 1

tan

For a stableresponse, the poles must all have negative real parts.The response of a stable higher-order system thus comprises a sum of a number of decaying exponential curves and decaying damped sinusoidal curves.

2


Higher-Order System response

Plot of poles on the s-plane



Real Poles and their effect on the responseReal Poles and their effect on the response

Re

Im

1p− 2p−

Real pole on the S-Plane.

Each real pole will contribute a term into the response.

tpj

jea −

The more negative the pole, or the farther away to the left from the Imaginary axis it is, the more rapidlythe exponential term decays to zero.

In general, if two poles are such that , then the response

caused by is dominant and that for can be neglected without loss of accuracy.

21 5 pp >

2p

1p






Complex conjugate poles and their effect on the responseComplex conjugate poles and their effect on the response

Each complex pair contributes a decaying damped sinusoidal term to the response.

Complex conjugate pole on the S-Plane.

Lines ofconstant ζ

kζω−

kk ωζ 21−

Im

Re

kφ The more negative the real part , or the farther away to the left the poles are from the Imaginary axis, the more rapidly the term decays to zero.

kζω−

The angle the poles make with the Real Axis determines the damping ratio, the greater the angle, the less the damping ratio.

kφ

−= −

k

kk ζ

ζφ

21

1tan


Some typical responsesSome typical responses

Stable systemsStable systems

3



Stable systemsStable systems



An Unstable systemsAn Unstable systems



Dominant Poles


Dominant Poles

Complex conjugate poles and their effect on the responseComplex conjugate poles and their effect on the response

The relative dominance of closed-loop poles is determined by how far they are from the Imaginary Axis, assuming that there are no zeros nearby. (Zeros affect the relative magnitude of the constant terms associated with the poles, the closer they are the more the effect.)

Usually the response will be adjusted such that one pair of complex conjugate poles will be closer to the Imaginary Axis relative to all the other poles and the response caused by this pair dominates the overall response.This pair is called the dominant closed-loop poles.

Closed-loops poles on the S-Plane.

4


Concepts of System StabilityConcepts of System Stability

§ A system (linear or non-linear) is said to be BIBO (bounded input, bounded output) stable if, for every bounded input, the output is bounded for all time.

§ An LTI (Linear Time-Invariant) system must have all poles in the left-half of the s-plane (negative real parts) for it to be stable.

§ In other words, the roots of the characteristic equation must all have negative real parts.

§ If a pole, or poles, lie on the imaginary axis, the system is critically, or limitedly, stable.

§ If a linear system is unstable, even in the absence of any input, the output will grow without bounds and becomes infinitely large as time goes to infinity.


Routh’s Stability CriterionRouth’s Stability Criterion

§ A system is stable if all the roots of the system’s characteristic

equation have negative real parts.0)()(1 =+ sHsG

§ The problem is if the characteristic equation is of an order higher than two, it is not easy to find the roots. (Of course, there are computer programs, e.g. MATLAB or OCTAVE, that helps with this.)

§ Fortunately, there is a simple criterion, known as Routh’s Stability Criterion (sometimes also known as the Routh-Hurwitz Stability Criterion), which enables us to find out the number of roots of the characteristic equation that lie on the right-half of the s-plane, i.e. have positive real parts, without having to factor the characteristic polynomial.



ProcedureProcedure

1) Form the characteristic equation

00 012

21

10 >=++⋅⋅⋅+++ −−− aaSaSaSaSa nn

nnn 00 012

21

10 >=++⋅⋅⋅+++ −−− aaSaSaSaSa nn

nnn

We assume that ; i.e. any zero roots have been removed.0≠na

0)10536( 232 =+++ ssss010536 2345 =+++ ssss

010536 23 =+++ sss

Example: or

use the equation

2) If any of the coefficients is negative or zero, the system is not stable.

3) If all the coefficients are positive, there is still no guarantee that all the roots have negative real parts. We then form the Routh Arrayand use the Routh Criterion to determine the number of roots with positive real parts.

5


10

211

3213

43212

75311

6420

.

.

.

fsees

cccsbbbbs

aaaasaaaas

n

n

n

n

L

LL

−

−

−

Routh ArrayRouth Array


Characteristic equation 00 012

21

10>=++⋅⋅⋅+++

−−− aaSaSaSaSa nn

nnn 00 012

21

10>=++⋅⋅⋅+++

−−− aaSaSaSaSa nn

nnn

1

50412 a

aaaab

−=

1

50412 a

aaaab

−=

1

30211 a

aaaab

−=

1

70613 a

aaaab

−=

1

70613 a

aaaab

−=


10

211

3213

43212

75311

6420

.

.

.

fs

ees

cccs

bbbbsaaaas

aaaas

n

n

n

n

L

L

L

−

−

−

In developing the array, an entire row can be multiplied by a positive number to simplify the process without affecting the result.

In developing the array, an entire row can be multiplied by a positive number to simplify the process without affecting the result.


SimilarlySimilarly


1

21311 b

baabc

−=

1

21311 b

baabc

−=

1

31512 b

baabc

−=

1

31512 b

baabc

−=

1

41713 b

baabc

−=

1

41713 b

baabc

−=

Routh’s Criterion states that the number of roots with positive real parts is equal to the number of changes in sign of the coefficients in the first column of the array.

Routh’s Criterion states that the number of roots with positive real parts is equal to the number of changes in sign of the coefficients in the first column of the array.

1

21211 e

eddef

−=

1

21211 e

eddef

−=



ExampleExample

0322

13

0 =+++ asasasa

If , then there is no sign change and there is no roots with positive real parts.If , then there is no sign change and there is no roots with positive real parts.

03021 >− aaaa

If , then there are two sign changes. Therefore there are two roots with positive real parts.If , then there are two sign changes. Therefore there are two roots with positive real parts.

03021 <− aaaa 03021 <− aaaa


0

1

312

203

s

s

aasaas

1

3021

1

aa

aaaa −

3a

Determine the conditions for the following equation to have only roots with negative real parts.Determine the conditions for the following equation to have only roots with negative real parts.

6



Special case 1Special case 1

§ If the sign of the coefficient in the row above is the same as that below (as in this case), then there are a pair of imaginary roots.

§ If the sign of the coefficient in the row above is the same as that below (as in this case), then there are a pair of imaginary roots.

A zero occurs in the first column of any row while the remaining terms are not zero, or there is no remaining term.Solution: The zero term is replaced by a small positive number and the array is processed accordingly.

ExampleExample

033 23 =+++ sss

0

1

2

3

3311

ss

ss

Routh ArrayRouth Arrayε→0

3

§ If the sign of the coefficient in the row above is different from that below, there is one sign change indicating one root with positive real part.


For such cases, form an auxiliary polynomial with the coefficients of the row above the all-zero row and using the coefficients of the derivative of this polynomial to replace the all-zero row.


Special case 2Special case 2

If all the coefficients, or the only one coefficient, in a derived row are zero, it means that there are roots of equal magnitude located symmetrically about the origin. Example: The characteristic equation have factors such as or .

))(( σσ −+ ss))(( ωω jsjs −+



Special case 2 – ExampleSpecial case 2 – Example

Note that because not all the coefficients are positive, this indicates that there is at least one root with a positive real part.

0502548242 2345 =−−+++ sssss

00

5048225241

3

4

5

s

ss

−−


There is one change in sign in the first column – one root with +ve real part.

Use this as auxiliary polynomial

50482)( 24 −+= sssP

sssP 968)( 3 +=&

New Routh ArrayNew Routh Array

5007.112

5024968

5048225241

0

1

2

3

4

5

−

−

−−

ss

ssss

7


End

1


Root Locus AnalysisRoot Locus Analysis




The transient response, and stability, of the closed-loop system is determined by the values of the roots of the characteristic equation or, in other words, the location of the closed-loop poles on the s=plane.The open-loop transfer function can be written in the form

where K is an adjustable gain, the z’s and p’s are the zeros and poles of the open-loop transfer function.

0)()(1 =+ sHsG

Consider the closed-loop system RG

C

-

+ E

HB

)())(()())((

)()(21

21

n

m

pspspszszszs

KsHsG++++++

=LL

As the gain K changes, the values of the closed-loop poles will change and thus the transient response, and stability.The root locus plot is a plot of the loci of the closed-loop poles on the s-plane as the gain K varies from 0 to infinity.


Consider the system with


ExampleExample

Root Locus Plot

)2)(1()()(

++=

sssK

sHsG

Poles vs KPoles vs KK P 1 P2 P3 0 0 -1 -2

0.1 -0.054 -0.90 -2.05 0.2 -0.12 -0.79 -2.09 0.4 -0.42+j0.09 -0.42- j0.09 -2.16 0.7 -0.38+j0.41 -0.38- j0.41 -2.25 1 -0.34+j0.56 -0.34- j0.56 -2.32 2 -0.24+j0.86 -0.24- j0.86 -2.52 4 -0.10+j1.9 -0.10-j1.9 -2.80 7 0.04+j1.50 0.04- j1.50 -3.09

10 0.15+j1.73 0.15- j1.73 -3.31

K=0

K=0.2

K=0.4

K=1

K=7

K=1

K=7

3 poles, thus 3 loci.

Characteristic Eqn: 0)2)(1( =+++ Ksss

2


The root loci are plotted either

§ Manually, or

§ Using a computer program such as OCTAVE(easy if you have the program and knows how to use it.)

The root loci are plotted either

§ Manually, or

§ Using a computer program such as OCTAVE(easy if you have the program and knows how to use it.)

Plotting the Root LociPlotting the Root Loci


Manual plotting – Root Locus ConceptsManual plotting – Root Locus Concepts

The Characteristic Equation is first written in the form

where K is a constant gain.

The Characteristic Equation is first written in the form

where K is a constant gain.

0)(1)()(1 =+=+ sKFsHsG

The roots of the characteristic equation are the values of s which satisfy the equation

for °=−= 18011)( nsKF K,5,3,1 ±±±=n

Or when °= 180)( nsF K,5,3,1 ±±±=n

The magnitude condition is satisfied by having

1)( =sKF giving )(

1sF

K =


Manual plotting – Root Locus ConceptsManual plotting – Root Locus Concepts

Consider

Determining the phase angle for F(s)Determining the phase angle for F(s)

))()()(()(

)()()(4321

1

pspspspszsK

sKFsHsG++++

+==

Then

where

0 Re

Im

-p2

-z1

s

s

(s+p2)

(s+z1)

2θ

1φ

0 Re

Im

-p2

-z1

s

s

(s+p2)

(s+z1)

2θ

1φ

°=−−−−= 180)( 43211 nsF θθθθφ K,5,3,1 ±±±=n

Note that °±≡ 360nθθ

K,3,2,1=n

)( 11 zs +=φ

)( 11 ps +=θ

)( 22 ps +=θ

)( 33 ps +=θ

)( 44 ps +=θ

3


Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines

1) Locate the poles and zeros of the open-loop transfer, G(s)H(s), function on the s plane.

1) Locate the poles and zeros of the open-loop transfer, G(s)H(s), function on the s plane.

2) There are as many loci, or branches, as poles of the G(s)H(s).

2) There are as many loci, or branches, as poles of the G(s)H(s).

3) Each branch starts from a pole of G(s)H(s) and ends in a zero. If there are no zeros in the finite region, then the zeros are at infinity.

3) Each branch starts from a pole of G(s)H(s) and ends in a zero. If there are no zeros in the finite region, then the zeros are at infinity.

Reason: 0)()( =+ sKNsD

When K=0, and roots are roots of D(s).0)( =sD

0)()(

1)()(1 =+=+sDsN

KsHsG

When , and roots are roots of N(s).1>>K 0)( =sN



4) The loci exist on the real axis only to the left of an odd number of poles and/or zeros. Complex poles and/or zeros have no effect because, for a point on the real axis, the angles involved are equal and opposite.


0 1- 1

j1

- 2-3

j2

-j1

-j2

Re

Im

- p2

-p 1

- p3

-z1

1θ

2θ

3θs

Consider a test point, s, on the real axis as shown.

Every pair of complex conjugate poles (or zeros) will contribute a pair of angles, and such that

They can thus be ignored.

Every pair of complex conjugate poles (or zeros) will contribute a pair of angles, and such that

They can thus be ignored.

1θ 2θ°=+ 36021 θθ





0 1- 1

j1

- 2-3

j2

-j1

-j2

Re

Im

- p2

-p 1

- p3

-z1

1θ

2θ

3θs

Consider a test point, s, on the real axis as shown.

Each real pole or zero to the left of point s does not contribute to the angle sum and thus can be ignored.

Each real pole or zero to the left of point s does not contribute to the angle sum and thus can be ignored.

°± 180

°180nK,5,3,1 ±±±=n

Each pole/zero to the right of point s contributes an angle of . An odd number of them will thus contribute a total of where

4


Consider a test point, s, on the realaxis to the left of the pole at s=0.

Locus exists to left of odd No. of zeros/polesLocus exists to left of odd No. of zeros/poles

Example: Consider the characteristic equationExample: Consider the characteristic equation

)22)(22(

)3(jsjss

sKGH

−++++

=

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

-p1

-p2

p0-z

One zero at s=-z=-3;

One zero at s=0;

Complex conjugate poles at s=-p1=-2-j2, and s=-p2=-2+j2Complex conjugate poles at

s=-p1=-2-j2, and s=-p2=-2+j2

)(

))((

)(

21

sKFpspss

zsK =

++

+=


0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

-p1

-p2

p0-z



)22)(22(

)3(jsjss

sKGH

−++++

=

Consider a test point, S, on the realaxis to the left of ONE pole at s=0.

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

1θ

2θ

3θ

φ s

ss+p1

-p1

-p2

s+p2

s+z

This test point will be part of theroot locus if °= 180)( nsF

)(

))((

)(

21

sKFpspss

zsK =

++

+=

Or °=−−− 180321 nθθθφ

Note that ,and

°=+ 032 θϑ °= 0φ °= 1801θ

Point S forms part of the root locus.


0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

-p1

-p2

p0-z



)22)(22(

)3(jsjss

sKGH

−++++

=

)())((

)(

21

sKFpspss

zsK =

++

+=

Point S is not part of of the rootlocus. Note S is to the left of aneven No. of zero/pole

Consider a test point, S, on the realaxis to the left of the zero at s=-3.

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

1θ

2θ

3θ

φs

s

s+p1

-p1

-p2

s+p2

s+z

Note that ,

and

Thus

°=+ 032 θϑ °== 1801θφ °= 0)(sF

5



0 Re

Im

- p1

- p2

- p3

- z1

θ

θ

θθ

6) Loci which terminates at infinity approach asymptotes in doing so.For a test point S at infinity, each pole/zero contributes an equal phase angle.Thusfor

Z/P=No. of zeros/poles

°=− 180)( nPZ θ

K,5,3,1 ±±±=n

PZn

−°=∴ 180θ

5) Because complex roots must occur in conjugate pairs, i.e. symmetrical about the real axis, the root-locus plot is symmetrical about the real axis.

5) Because complex roots must occur in conjugate pairs, i.e. symmetrical about the real axis, the root-locus plot is symmetrical about the real axis.



ZP

zp m

i

n

ia −

−= ∑∑ 11σ

7) All the asymptotes start from a point on the real axis with coordinate

Example: For

)2)(1(

1)()(

++=

ssssHsG

Then

13

03−=

−−=aσ

30180180

−°

=−

°=

nPZ

nθ °°°= 300,180,60

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im



8) Break-in and breakaway points (on real axis)

At a break-in point, value of K increases as the loci moves onto the real axis and away from the break-in point.At a breakaway point, values of K increases along the real axis from both sides and reach maximum at the breakaway point.

Breakaway pt.

Break-in pt.

σ=sAlong the real axis, . Thus, at the break-in or breakaway points,

provided K>0 and exists on the root loci.

0==σsds

Kd

σ

6



9) Imaginary Axis Crossing

Two approaches:Im Axis Crossing

a) Use Routh Criteria to determine the value of K at which the system is critically stable. This is indicated by a value of zero in the first column but with no sign change in the first column of the Routh Array.

b) Since the roots are on the imaginary axis, by letting in the characteristic equation and solve for and K. This is done by equating both the real and imaginary parts of the characteristic equation to zero.

ωjs =

ω



10) Angle of Departure from complex poles and Angle of Arrival from complex zeros.

Angle of Departure

This is done by taking a test point very close to the complex pole, or zero, and applying the angular criteria.



10) Angle of Departure from complex poles and Angle of Arrival from complex zeros.

0 Re

Im

-p1

-p 2

-z1

2θ

1φ

1θ

0 Re

Im

-p1

-p 2

-z1

2θ

1φ

1θ

Example, for diagram on right,

°±=+− 180)( 211 θθφ

This is done by taking a test point very close to the complex pole, or zero, and applying the angular criteria.

7


Example Root Locus PlotsExample Root Locus Plots

MATLAB Program

>> gh=tf([1],[1 2 2 0])

Transfer function:1

-----------------s^3 + 2 s^2 + 2 s

>> rlocus(gh)

Poles at

11,0 js ±−=



MATLAB Program

>> h=zpk([-3 -4],[0 -1],[1])

Zero/pole/gain:(s+3) (s+4)-----------s (s+1)

>> rlocus(h)>>

Poles at s=0, -1

Zeros at s=-3, -4



MATLAB Program

>> g1=tf([1],[1 2 5])Transfer function:

1-------------s^2 + 2 s + 5>> g2=zpk([-2],[0 -1],[1])Zero/pole/gain:(s+2)-------s (s+1)>> gh=g1*g2Zero/pole/gain:

(s+2)----------------------s (s+1) (s^2 + 2s + 5)

>> rlocus(gh)Poles at

Zero at

Poles at

Zero at

21,1,0 js ±−−=

2−=s

8



MATLAB Program

>> g1=tf([1],[1 2 5])

Transfer function:1

-------------s^2 + 2 s + 5>> g2=zpk([],[0 -4],[1])Zero/pole/gain:

1-------s (s+4)>> gh=g1*g2Zero/pole/gain:

1----------------------s (s+4) (s^2 + 2s + 5)>> rlocus(gh)

Poles atPoles at 21,4,0 js ±−−=


End

1


Root Locus Analysis

Example

Root Locus Analysis

Example



Root Locus PlottingRoot Locus Plotting

Consider a systemConsider a system K-

+)2)(1(

1++ sss

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im1) Locate poles and zeros1) Locate poles and zeros

2) 3 poles gives 3 branches2) 3 poles gives 3 branches

3) Each branch starts from a pole and ends at a zero. 3 zeros at infinity.

3) Each branch starts from a pole and ends at a zero. 3 zeros at infinity.

)2)(1()10) (10)(10(

)2)(1(1

+++++

=++ sss

ssssss




+)2)(1(

1++ sss

4) On the real axis, loci exist only to the left of an odd number of poles/zeros.

4) On the real axis, loci exist only to the left of an odd number of poles/zeros.

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

5) Root-locus plot will be symmetrical about the real axis.

5) Root-locus plot will be symmetrical about the real axis.

6) Asymptotes are at angles6) Asymptotes are at angles

°±°±°±=−

°240or180or60

30180n

7) Asymptotes start from a point on the real axis with

7) Asymptotes start from a point on the real axis with

13

03−=

−−=aσ

2




+)2)(1(

1++ sss

8) Breakqway point:8) Breakqway point:

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

We have 0)2)(1(

1 =++

+sss

K

0)2)(1( =+++∴ Ksss

giving 5774.1or4226.0 −−=s

Since there is no loci at s=-1.5774 (corresponding to K=-8.23<0), the breakaway point is at s=-0.4226.

Breakaway pt.S=-0.4226

)23( 23 sssK ++−=∴

0263 2 =−−−= ssdsKd




+)2)(1(

1++ sss

9) Imaginary Axis Crossing (a):9) Imaginary Axis Crossing (a):

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

We have

giving

Crossing atS=1.414

023 23 =+++ Ksss

Ks

Ks

Kss

0

1

2

3

36

321

−

6=K

Auxiliary Polynomial

03 2 =+ Ks 2js ±=




+)2)(1(

1++ sss

9) Imaginary Axis Crossing (b):9) Imaginary Axis Crossing (b):

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

0 1-1

j1

-2-3

j2

-j1

-j2

Re

Im

We have Crossing atS=1.414

023 23 =+++ Ksssωjs =Letting

we have 02)(3)( 23 =+++ Kjjj ωωω

or 0)2()3( 32 =−+− ωωω jK

02 3 =− ωωThus giving 2js ±=

and giving03 2 =− ωK 6=K

3




+)2)(1(

1++ sss

MATLAB Program

clear allh=zpk([],[0 -1 -2],[1])rlocus(h)end

MATLAB Program

clear allh=zpk([],[0 -1 -2],[1])rlocus(h)end


End

Graphic Approach to determining the value of K from the Root Locus Consider the characteristic equation

0)(1)()(1 =+=+ sKFsHsG (1)

If F(s) is written in factors of zeros and poles, then we have

0

)())(()())((

121

21 =++++++

+n

m

pspspszszszs

KLL

(2)

From Equation (1), we have

°−=−= 180/11)(sKF (3)

meaning a magnitude of 1 and a phase angle of °− 180 Equation (3) is equivalent to the following two conditions being met since K has no phase angle.

i) Phase angle of F(s) is -180°

ii) Magnitude of K is given by )(

1sF

K = . (4)

The root locus contains all the roots of the characteristic equation as the parameter K varies from 0 to infinity. This means that any point on the root locus will have values of s which satisfy Equations (1), (2), (3) and (4). We can, therefore, make use of Equation (4) to determine the value of K if we can find the value of

)(sF for any point on the root locus plot. For this, we can write

n

m

pspsps

zszszssF

++⋅+

++⋅+=

L

L

21

21)(

For any point s on the s-plane, the vector, rs + , where rs −= is a zero or pole, is a vector starting from the point rs −= to the point s on the s-plane. Refer to Slide 6 of the lecture presentation on Root Locus Analysis for an illustration. Once the vector is known on the s-plane, its length can be determined geometrically.

Example:

Consider the characteristic equation given by 0)100)(1(

)10(101 =

+++

+ssssK v (5)

The root locus plot is shown in the figure on the next page. Here there is one zero at s=-10 and three poles at s=0, s=-1, and s=-100. These are pointed to by blue arrows in the figure. Suppose we wish to find values of Kv which makes the damping ratio for the complex pair of poles be 0.6. A line for 6.0=ζ is drawn and where this cuts the root locus plot, at three points, will be the three solutions meeting this condition. To determine the value of Kv, take for example the solution at the point P. If an accurate plot is available, we can read the value of s as 105.7 js ±−= . The vectors for the single zero and the three poles, 100+s , s , 1+s and 100+s are shown and pointed to by red arrows. (Refer to Slide 6 of Lecture Presentation for an illustration of the vector.) The value of Kv at the point P can then be obtained from obtaining the lengths of these vectors and will be given by

1010

1001

+⋅

+⋅+⋅=

s

sssK v and will work out to be 135=vK .

Note that there are three solutions given by the condition that the damping ratio for the complex pair of poles being 0.6. These three solutions are

i) 71.054.0,9.99;8.0 3,21 jssK v ±−=−==

ii) 105.7,9.85;135 3,21 jssK v ±−=−== iii) 5.596.44,7.11;648 3,21 jssK v ±−=−==

Note that only for cases (i) and (ii) above will the pair of complex poles be dominant since the pole at

1ss = is then much further to the left of the imaginary axis. For case (iii), the pole at 1ss = will dominate the response since it is comparatively much closer to the imaginary axis than the pair of complex poles. The resulting response will then not be oscillatory but will be like a first order response.

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0-40

-30

-20

-10

0

10

20

30

40Root Locus

Real Axis

Imag

inar

y A

xis

6.0=ζ

100+s

10+s

1+s

s

0=s

1−=s10−=s100−=s

3

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0-40

-30

-20

-10

0

10

20

30

40Root Locus

Real Axis

Imag

inar

y A

xis

1


Control Actions – System CompensationControl Actions – System Compensation




• Discusses the “control actions” typically used for the controller.

• What do you do if the system’s output does not meet with your requirements? • – Apply compensation to the system, or

system compensation, to get it to response closer to what you want.

• Discusses the “control actions” typically used for the controller.

• What do you do if the system’s output does not meet with your requirements? • – Apply compensation to the system, or

system compensation, to get it to response closer to what you want.


Proportional (P) ControlProportional (P) Control

Commonly Used Control Actions Commonly Used Control Actions

R GpC

-

+ EGc

+ +

D

M

controller plant

eKm p= pc KsG =)( Kp is proportional gain

Integral (I) ControlIntegral (I) Control

This is the simplest form of control action and is used on its own or in conjunction with other control actions.

∫= dteKm i sK

sG ic =)( Ki is integral gain

Integral control reduces, or eliminates, steady-state errors. It also increase the order of the system and make it more prone to instability.

2


Derivative (D) ControlDerivative (D) Control


Kd is derivative gain

Proportional-plus-Integral (PI) ControlProportional-plus-Integral (PI) Control

§ Derivative control action is akin to “anticipatory” control action.§ It tends to increase the damping in, and stability of, the system. § Although is does not directly affect the steady-state error, it

allows higher proportional gains to be used, thereby reducing steady-state errors.

§ It is never used on its own but always with proportional control.

Addition of integral control action to proportional control reduces steady-state errors but also tend to make the system more oscillatory.

eKm d &= sKsG dc =)(

∫+= dteKeKmip s

KKsG i

pc +=)(


Proportional-plus-Derivative (PD) ControlProportional-plus-Derivative (PD) Control


Proportional-plus-Integral-plus Derivative (PID) ControlProportional-plus-Integral-plus Derivative (PID) Control

Addition of derivative control action adds anticipatory action and tend to increase system damping and stability.

Because three gains are involved, tuning, or adjusting, of the gains is not an easy task.

eKeKm dp &+= sKKsG dpc +=)(

eKdteKeKm dip &++= ∫ sKsK

KsG di

pc ++=)(

Sometimes Gc(s) is written in the formTi is Integral Time

Td is Derivative or Rate Time

++= sT

sTKsG d

ipc

11)(


Effects of Control Actions

dc motor under speed feedback control



D

R C

-

+ EGc

+ +T1+sT

K

m

m

C is output speedR is reference inputT is motor torqueD is disturbance/load torque

Proportional (P) Control with Gc(s)=KpProportional (P) Control with Gc(s)=Kp

11

1

++

+=

sTKK

sT

KK

RC

m

mp

m

mp

mpm

mp

KKsT

KK

++=

1 1+=

T sK

with

mp

mp

KK

KKK

+=

1

mp

m

KKT

T+

=1

§ The resulting system is still first-order and will always be stable.

§ The resulting time constant, T, is much smaller, thus giving a much faster speed of response.

Example Example

)1/(1 mK+ )1/(1 mpKK+§ Steady-state error to unit step is reduced from to but is still non-zero since the system is Type 0.

3






Effect of Disturbance inputEffect of Disturbance input

With R=0, the block diagram becomes






D

R C

-

+ EGc

+ +T1+sT

K

m

m

With R=0, the block diagram becomes

A step disturbance thus causes the output c(t) to change.Steady-state error, r – c, is non-zero. Therefore speed regulation is affected.

Proportional (P) ControlEffect of Disturbance inputProportional (P) ControlEffect of Disturbance input

11

1

++

+=

sT

KKsT

K

DC

m

mp

m

m

D C

-

+

Kp

1+sTK

m

m

-M

mpm

m

KKsTK++

=1

For a unit step disturbance ssD /1)( =

sKKsTK

sCmpm

m 11

)(++

=mp

m

sss KK

KssCc

+==

→ 1)(lim

0






D

R C

-

+ EGc

+ +T1+sT

K

m

m

PI ControlPI Controls

KKsG ipc +=)(

With PI control, the forward transfer function becomes

sKsK ip +

=

Becomes a Type 1 system: steady-state error to step input will be zero.

.)1(

)(+

+sTs

KKsK

m

mip

mipm

mip

KKsKsTs

KKsK

RC

)()1(

)(

+++

+=Also,

m

mi

m

mp

mmip

TKK

sT

KKs

TKKsK

++

+

+=

1/)(

222

21

2 nn

n

sssK

ωζωω

+++

=

m

mp

TKK

K =1m

min T

KK=ω

mmi

mp

TKK

KK

2

1+=ζwith

Increasing Kp increases damping. Increasing Ki increases natural frequency and reduces damping.

4






D

R C

-

+ EGc

+ +T1+sT

K

m

m

PI Control – effect on disturbancePI Control – effect on disturbance

sKKsG i

pc +=)(

sKsK ip +

=

With

D C

-

+1+sT

K

m

m

-M

Gc

sKsK

sTK

sTKDC

ip

m

m

mm

)()1(

1

)1/(+

++

+=)()1( ipmm

m

KsKKssTsK

+++=

sKsKKssTsK

sCipmm

m 1)()1(

)(+++

=

ssD /1)( =

0)(lim0

==→

ssCcsss



dc motor under position feedback control



C is output positionR is reference inputT is motor torqueE is position error

Proportional (P) Control with Gc(s)=KpProportional (P) Control with Gc(s)=Kp

§ The resulting system is a Type 1 second-order system.§ Steady-state position error to a step input is zero.

R C

-

+ EGc

T)1( +sTs

K

m

m

Characteristic Equation

0)1(

1 =+

+sTs

KK

m

mp02 =++ mpm KKssT 02 22 =++ nn ss ωζω

With andm

mpn T

KK=ω

mmp TKK21

=ζ

Increasing Kp increases and thus speed of response but reduces damping.

nω






R C

-

+ EGc

T)1( +sTs

K

m

m

Characteristic Equation

Thus andm

mpn T

KK=ω

Open-loop transfer is

PD Control withPD Control with

sKKsG dpc +=)(

)1(

)()(

+

+=

sTs

sKKKsG

m

dpmOL

Closed-loop system remains a Type 1 system with zero steady-state error to step inputs.

0)(1 =+ sGOL 0)()1( =+++ sKKKsTs dpmm 012 =+

++

m

mp

m

dm

T

KKs

TKK

s

m

dmn T

KK+= 12ζωmmp

dm

TKK

KK

2

1 +=ζ

Addition of derivative control allows damping to be increased without affecting system type and natural frequency.

5


An informative website on PID gain tuning and effects of various control actions.

http://www.expertune.com/tutor.html


End

Documents

National University of Singaporeguppy.mpe.nus.edu.sg/anpoo/FBControla/ME2142 handouts.pdf1 ME2142/TM3142 Feedback Control Systems 1 Reference Text “Control Systems Engineering”