1
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
Feedback Control Systems
First half: Professor POO Aun Neow
Second half: Professor V Subramaniam
ME2142/TM3142 Feedback Control Systems2
Introduction and Basic Concepts
ME2142/TM3142 Feedback Control Systems3
• 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?
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ME2142/TM3142 Feedback Control Systems4
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)
ME2142/TM3142 Feedback Control Systems5
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
ME2142/TM3142 Feedback Control Systems6
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
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ME2142/TM3142 Feedback Control Systems7
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
ME2142/TM3142 Feedback Control Systems8
Some example of control System?Some example of control System?
ME2142/TM3142 Feedback Control Systems9
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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ME2142/TM3142 Feedback Control Systems10
Example: Open-loop vs Closed LoopExample: Open-loop vs Closed Loop
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
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
ME2142/TM3142 Feedback Control Systems11
Example: Open-loop vs Closed LoopExample: Open-loop vs Closed Loop
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
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
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
ME2142/TM3142 Feedback Control Systems12
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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ME2142/TM3142 Feedback Control Systems13
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))
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))
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)
ME2142/TM3142 Feedback Control Systems14
Linear and Non-linear SystemsLinear and Non-linear Systems
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))
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))
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.
ME2142/TM3142 Feedback Control Systems15
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.
Linear and Non-linear SystemsLinear and Non-linear Systems
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= + ≠ = +
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ME2142/TM3142 Feedback Control Systems16
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∆ = ∆
ME2142/TM3142 Feedback Control Systems17
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
ME2142/TM3142 Feedback Control Systems18
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
2
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);
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ME2142/TM3142 Feedback Control Systems1
Review of Laplace TransformsReview of Laplace Transforms
ME2141/ME2142E
Feedback Control Systems
ME2141/ME2142E
Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
Complex variableComplex variable
ωσ js +=
Real part Imaginary part
1−=j
Re
Im
σ
ω s
s-plane
Mathematical preliminariesMathematical preliminaries
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
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:
ME2142/TM3142 Feedback Control Systems5
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
ME2142/TM3142 Feedback Control Systems6
Laplace TransformsLaplace Transforms
End of L1End of L1
3
ME2142/TM3142 Feedback Control Systems7
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 ⋅+⋅=
ME2142/TM3142 Feedback Control Systems8
Properties of Laplace TransformsProperties of Laplace Transforms
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
ME2142/TM3142 Feedback Control Systems9
Properties of Laplace TransformsProperties of Laplace Transforms
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
ME2142/TM3142 Feedback Control Systems10
Properties of Laplace TransformsProperties of Laplace Transforms
Final-Value Theorem:
Initial-Value Theorem:
)(lim)(lim0
ssFtfst →∞→
=
)(lim)0( ssFfs ∞→
=+
ME2142/TM3142 Feedback Control Systems11
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
ME2142/TM3142 Feedback Control Systems12
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
ME2142/TM3142 Feedback Control Systems13
Using Table and Properties Using Table and Properties
ExampleExampleFind the inverse of
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=
+
ME2142/TM3142 Feedback Control Systems14
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
ME2142/TM3142 Feedback Control Systems15
Using Partial Fractions Using Partial Fractions
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
ME2142/TM3142 Feedback Control Systems16
Using Partial Fractions Using Partial Fractions
ExampleExampleFind the inverse of
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
ME2142/TM3142 Feedback Control Systems17
Using Partial Fractions Using Partial Fractions
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 )(
)()(
ME2142/TM3142 Feedback Control Systems18
Using Partial Fractions Using Partial Fractions
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
ME2142/TM3142 Feedback Control Systems19
Solving differential equations Solving differential equations
ME2142/TM3142 Feedback Control Systems20
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 =+−+−− &
ME2142/TM3142 Feedback Control Systems21
Solving differential equations Solving differential equations
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
ME2142/TM3142 Feedback Control Systems22
Laplace TransformsLaplace Transforms
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
ME2142/TM3142 Feedback Control Systems1
Modelling of Physical Systems
The Transfer Function
Modelling of Physical Systems
The Transfer Function
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
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
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
Mechanical Systems – Torsional SystemsMechanical Systems – Torsional Systems
ωθ &&& JJT ==
ME2142/TM3142 Feedback Control Systems5
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
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
Mechanical Systems - springsMechanical Systems - springs
)( 21 xxKf −=Important: Note directions and signs
)( 21 θθ −= GT
ME2142/TM3142 Feedback Control Systems6
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
.
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems7
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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.
ME2142/TM3142 Feedback Control Systems8
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
+==
)()()(
ME2142/TM3142 Feedback Control Systems9
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems10
Capacitance
Or
Complex impedance
Capacitance
Or
Complex impedance
eq
C =
Ceq =
dtde
Cdtdq
i ==
)(sECI =
)/(1 sCX c =
cIXsC
IE ==1
e i C
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems11
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems12
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems13
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems14
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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 ==−
ME2142/TM3142 Feedback Control Systems15
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems16
Modelling of Physical Dynamic SystemsModelling of Physical Dynamic Systems
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
ME2142/TM3142 Feedback Control Systems17
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
ME2142/TM3142 Feedback Control Systems18
End
1
ME2142/TM3142 Feedback Control Systems1
System Transient/Time ResponseSystem Transient/Time Response
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
System Characteristic EquationSystem Characteristic Equation
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
ME2142/TM3142 Feedback Control Systems5
System Characteristic EquationSystem Characteristic Equation
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.
ME2142/TM3142 Feedback Control Systems6
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
ME2142/TM3142 Feedback Control Systems7
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.
ME2142/TM3142 Feedback Control Systems8
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.
ME2142/TM3142 Feedback Control Systems9
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
ME2142/TM3142 Feedback Control Systems10
System Response – First-order systemsSystem Response – First-order systems
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
ME2142/TM3142 Feedback Control Systems11
System Response – First-order systemsSystem Response – First-order systems
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.
ME2142/TM3142 Feedback Control Systems12
System Response – First-order systemsSystem Response – First-order systems
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
ME2142/TM3142 Feedback Control Systems13
System Response – First-order systemsSystem Response – First-order systems
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
ME2142/TM3142 Feedback Control Systems14
§ 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 −+−=
ME2142/TM3142 Feedback Control Systems15
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
ME2142/TM3142 Feedback Control Systems16
+
-
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
ME2142/TM3142 Feedback Control Systems17
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.
ME2142/TM3142 Feedback Control Systems18
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
ME2142/TM3142 Feedback Control Systems19
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
ME2142/TM3142 Feedback Control Systems20
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
ME2142/TM3142 Feedback Control Systems21
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
ME2142/TM3142 Feedback Control Systems22
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
ME2142/TM3142 Feedback Control Systems23
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
−− −−=⇔+
Step Response – Second-order systemsStep Response – Second-order systems
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
ME2142/TM3142 Feedback Control Systems24
We use Entry 17 in Dr Chen’s tables. We use Entry 17 in Dr Chen’s tables.
Step Response – Second-order systemsStep Response – Second-order systems
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
ME2142/TM3142 Feedback Control Systems25
Step Response – Second-order systemsStep Response – Second-order systems
Normalized response curves
For fast response,is usually
desirable.7.0≈ζ
If no overshoot is required, is usually used.
1>ζ
ME2142/TM3142 Feedback Control Systems26
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
ME2142/TM3142 Feedback Control Systems27
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
ME2142/TM3142 Feedback Control Systems28
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
=
Measures of transient performanceMeasures of transient performance
ME2142/TM3142 Feedback Control Systems29
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
Measures of transient performanceMeasures of transient performance
ME2142/TM3142 Feedback Control Systems30
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=
Measures of transient performanceMeasures of transient performance
11
ME2142/TM3142 Feedback Control Systems31
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
ME2142/TM3142 Feedback Control Systems1
Steady-State CharacteristicsSteady-State Characteristics
ME2142/ME142E Feedback Control SystemsME2142/ME142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
Steady-State Errors – Static Error ConstantsSteady-State Errors – Static Error Constants
)(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
ME2142/TM3142 Feedback Control Systems5
Steady-State Errors – Static Error ConstantsSteady-State Errors – Static Error Constants
)(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
ME2142/TM3142 Feedback Control Systems6
Steady-State Errors – Static Error ConstantsSteady-State Errors – Static Error Constants
)(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
ME2142/TM3142 Feedback Control Systems7
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.
ME2142/TM3142 Feedback Control Systems8
End
1
ME2142/TM3142 Feedback Control Systems1
Block Diagram AlgebraBlock Diagram Algebra
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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 +=+
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
Blocks in series or cascaded blocks
Block Diagram AlgebraBlock Diagram Algebra
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
ME2142/TM3142 Feedback Control Systems5
Blocks in parallel
Block Diagram AlgebraBlock Diagram Algebra
G1
G2
++ YX
G1 + G2
YX
ME2142/TM3142 Feedback Control Systems6
G+
+ ZX
Y
G+
+ ZX
Y
GX Y
X
GX Y
X
XG
Y
Z
+
+XG
Y
Z
+
+
Block Diagram AlgebraBlock Diagram Algebra
G
G
+
+ ZX
Y
GX Y
1/GX
+
+ Z
Y
XG
1/G
3
ME2142/TM3142 Feedback Control Systems7
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
ME2142/TM3142 Feedback Control Systems8
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
ME2142/TM3142 Feedback Control Systems9
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
ME2142/TM3142 Feedback Control Systems10
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
ME2142/TM3142 Feedback Control Systems11
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.
ME2142/TM3142 Feedback Control Systems12
Block Diagram ManipulationBlock Diagram Manipulation
Example: Determine C(s)/R(s)Example: Determine C(s)/R(s)
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
ME2142/TM3142 Feedback Control Systems13
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
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
ME2142/TM3142 Feedback Control Systems14
Block Diagram ManipulationBlock Diagram Manipulation
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
ME2142/TM3142 Feedback Control Systems15
End
1
ME2142/TM3142 Feedback Control Systems1
Concepts of System StabilityConcepts of System Stability
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
Higher-Order System response
Plot of poles on the s-plane
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
ME2142/TM3142 Feedback Control Systems5
Higher-Order System response
Plot of poles on the s-plane
Higher-Order System response
Plot of poles on the s-plane
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
ME2142/TM3142 Feedback Control Systems6
Some typical responsesSome typical responses
Stable systemsStable systems
3
ME2142/TM3142 Feedback Control Systems7
Some typical responsesSome typical responses
Stable systemsStable systems
ME2142/TM3142 Feedback Control Systems8
Some typical responsesSome typical responses
An Unstable systemsAn Unstable systems
ME2142/TM3142 Feedback Control Systems9
Higher-Order System response
Dominant Poles
Higher-Order System response
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
ME2142/TM3142 Feedback Control Systems10
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.
ME2142/TM3142 Feedback Control Systems11
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.
ME2142/TM3142 Feedback Control Systems12
Routh’s Stability CriterionRouth’s Stability Criterion
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
ME2142/TM3142 Feedback Control Systems13
10
211
3213
43212
75311
6420
.
.
.
fsees
cccsbbbbs
aaaasaaaas
n
n
n
n
L
LL
−
−
−
Routh ArrayRouth Array
Routh’s Stability CriterionRouth’s Stability Criterion
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
−=
ME2142/TM3142 Feedback Control Systems14
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.
Routh’s Stability CriterionRouth’s Stability Criterion
SimilarlySimilarly
Routh ArrayRouth Array
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
−=
ME2142/TM3142 Feedback Control Systems15
Routh’s Stability CriterionRouth’s Stability Criterion
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
Routh ArrayRouth Array
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
ME2142/TM3142 Feedback Control Systems16
Routh’s Stability CriterionRouth’s Stability Criterion
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.
ME2142/TM3142 Feedback Control Systems17
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.
Routh’s Stability CriterionRouth’s Stability Criterion
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 −+
ME2142/TM3142 Feedback Control Systems18
Routh’s Stability CriterionRouth’s Stability Criterion
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
−−
Routh ArrayRouth Array
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
ME2142/TM3142 Feedback Control Systems19
End
1
ME2142/TM3142 Feedback Control Systems1
Root Locus AnalysisRoot Locus Analysis
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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.
ME2142/TM3142 Feedback Control Systems3
Consider the system with
Root Locus AnalysisRoot Locus Analysis
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
ME2142/TM3142 Feedback Control Systems4
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
ME2142/TM3142 Feedback Control Systems5
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 =
ME2142/TM3142 Feedback Control Systems6
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
ME2142/TM3142 Feedback Control Systems7
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
ME2142/TM3142 Feedback Control Systems8
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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.
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
ME2142/TM3142 Feedback Control Systems9
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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.
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.
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
ME2142/TM3142 Feedback Control Systems10
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 =
++
+=
ME2142/TM3142 Feedback Control Systems11
0 1-1
j1
-2-3
j2
-j1
-j2
Re
Im
-p1
-p2
p0-z
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
−++++
=
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.
ME2142/TM3142 Feedback Control Systems12
0 1-1
j1
-2-3
j2
-j1
-j2
Re
Im
-p1
-p2
p0-z
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
−++++
=
)())((
)(
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
ME2142/TM3142 Feedback Control Systems13
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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.
ME2142/TM3142 Feedback Control Systems14
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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
ME2142/TM3142 Feedback Control Systems15
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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
ME2142/TM3142 Feedback Control Systems16
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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
ME2142/TM3142 Feedback Control Systems17
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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.
ME2142/TM3142 Feedback Control Systems18
Manual plotting – Procedure and GuidelinesManual plotting – Procedure and Guidelines
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
ME2142/TM3142 Feedback Control Systems19
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 ±−=
ME2142/TM3142 Feedback Control Systems20
Example Root Locus PlotsExample Root Locus Plots
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
ME2142/TM3142 Feedback Control Systems21
Example Root Locus PlotsExample Root Locus Plots
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
ME2142/TM3142 Feedback Control Systems22
Example Root Locus PlotsExample Root Locus Plots
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 ±−−=
ME2142/TM3142 Feedback Control Systems23
End
1
ME2142/TM3142 Feedback Control Systems1
Root Locus Analysis
Example
Root Locus Analysis
Example
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
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
ME2142/TM3142 Feedback Control Systems3
Root Locus PlottingRoot Locus Plotting
Consider a systemConsider a system K-
+)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
ME2142/TM3142 Feedback Control Systems4
Root Locus PlottingRoot Locus Plotting
Consider a systemConsider a system K-
+)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
ME2142/TM3142 Feedback Control Systems5
Root Locus PlottingRoot Locus Plotting
Consider a systemConsider a system K-
+)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 ±=
ME2142/TM3142 Feedback Control Systems6
Root Locus PlottingRoot Locus Plotting
Consider a systemConsider a system K-
+)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
ME2142/TM3142 Feedback Control Systems7
Root Locus PlottingRoot Locus Plotting
Consider a systemConsider a system K-
+)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
ME2142/TM3142 Feedback Control Systems8
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
ME2142/TM3142 Feedback Control Systems1
Control Actions – System CompensationControl Actions – System Compensation
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
ME2142/TM3142 Feedback Control Systems2
ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems
• 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.
ME2142/TM3142 Feedback Control Systems3
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
ME2142/TM3142 Feedback Control Systems4
Derivative (D) ControlDerivative (D) Control
Commonly Used Control Actions Commonly Used Control Actions
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 +=)(
ME2142/TM3142 Feedback Control Systems5
Proportional-plus-Derivative (PD) ControlProportional-plus-Derivative (PD) Control
Commonly Used Control Actions Commonly Used Control Actions
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)(
ME2142/TM3142 Feedback Control Systems6
Effects of Control Actions
dc motor under speed feedback control
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
ME2142/TM3142 Feedback Control Systems7
Effects of Control Actions
dc motor under speed feedback control
Effects of Control Actions
dc motor under speed feedback control
Effect of Disturbance inputEffect of Disturbance input
With R=0, the block diagram becomes
ME2142/TM3142 Feedback Control Systems8
Effects of Control Actions
dc motor under speed feedback control
Effects of Control Actions
dc motor under speed feedback control
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
ME2142/TM3142 Feedback Control Systems9
Effects of Control Actions
dc motor under speed feedback control
Effects of Control Actions
dc motor under speed feedback control
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
ME2142/TM3142 Feedback Control Systems10
Effects of Control Actions
dc motor under speed feedback control
Effects of Control Actions
dc motor under speed feedback control
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
ME2142/TM3142 Feedback Control Systems11
Effects of Control Actions
dc motor under position feedback control
Effects of Control Actions
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
ME2142/TM3142 Feedback Control Systems12
Effects of Control Actions
dc motor under position feedback control
Effects of Control Actions
dc motor under position feedback control
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
ME2142/TM3142 Feedback Control Systems13
An informative website on PID gain tuning and effects of various control actions.
http://www.expertune.com/tutor.html
ME2142/TM3142 Feedback Control Systems14
End