Course Outline - University of Sydneyweb.aeromech.usyd.edu.au/AMME3500/Course_documents/...General...
Transcript of Course Outline - University of Sydneyweb.aeromech.usyd.edu.au/AMME3500/Course_documents/...General...
1
Amme 3500/5501 : System Dynamics & Control
Review
Dr. Stefan B. Williams
Slide 2 Dr. Stefan B. Williams Amme 3500 : Introduction
Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the s-plane 4 22 Mar Block Diagrams Assign 1 Due 5 29 Mar Feedback System Characteristics 6 5 Apr Root Locus Assign 2 Due 7 12 Apr Root Locus 2 8 19 Apr Bode Plots No Tutorials 26 Apr BREAK 9 3 May Bode Plots 2 10 10 May State Space Modeling Assign 3 Due 11 17 May State Space Design Techniques 12 24 May Advanced Control Topics 13 31 May Review Assign 4 Due 14 Spare
Slide 3 Dr. Stefan B. Williams Amme 3500 : Review
Assessment
• 4 Assignments (40%) – Assignment 1 : Due Week 4 (5%) – Assignment 2 : Due Week 6 (10%) – Assignment 3 : Due Week 9 (10%) – Assignment 4 : Due Week 13 (15%)
• Final Exam (60%)* * Note you are expected to pass the exam to pass this course
Slide 4 Dr. Stefan B. Williams Amme 3500 : Review
Course Objectives
• To introduce the methods used for the analysis and design of linear time invariant feedback control systems
2
Slide 5 Dr. Stefan B. Williams Amme 3500 : Review
What is a Control System?
• An interconnection of components forming a system configuration that will provide a desired system response
• The study of control provides us with a process for analysing and understanding the behaviour of a system given some input
• It also introduces methods for achieving the desired system response
Slide 6 Dr. Stefan B. Williams Amme 3500 : Review
What sorts of Systems? • Many systems can be considered using control
system techniques – Vehicle Control and Design (land, sea, air, space) –
understanding how the system responds to external disturbances
– Biomedical (cardiac system, dialysis machine) – design and control of the system with relation to human interaction
– Stock Market - understanding the relationship between interest rates and inflation where one drives the other
– Biological - determining the relationship between the outcome of a natural phenomenon when the system is disturbed
Slide 7 Dr. Stefan B. Williams Amme 3500 : Review
A General Control System • Many control systems can be
characterised by these components
Sensor
Actuator Process Control
Reference r(t)
Output y(t)
- +
Error e(t)
Control Signal
u(t)
Plant
Disturbance
Sensor Noise
Feedback
Slide 8 Dr. Stefan B. Williams Amme 3500 : Review
Components of a Block Diagram
• A block diagram is made up of signals, systems, summing junctions and pickoff points
3
Slide 9 Dr. Stefan B. Williams Amme 3500 : Review
Mathematical Modelling
• A system model is one or more equations that describe the relationship between the system variables – often the input(s) and output(s) of the system
• For physical systems, these equations are derived from study of the physical properties of the system such as mechanics, fluids, electrical, thermodynamics, etc.
Slide 10 Dr. Stefan B. Williams Amme 3500 : Review
Translation Mechanical Elements • Force-velocity,
force-displacement, and impedance translational relationships for springs, viscous dampers, and mass
Slide 11 Dr. Stefan B. Williams Amme 3500 : Review
Rotational Mechanical Elements • Torque-angular
velocity, torque-angular displacement, and impedance rotational relationships for springs, viscous dampers, and inertia
Slide 12 Dr. Stefan B. Williams Amme 3500 : Review
Electrical Elements Voltage-current, voltage-charge, and
impedance relationships for capacitors, resistors, and inductors
4
Slide 13 Dr. Stefan B. Williams Amme 3500 : Review
Comparing Solution Methods
• Starting with an impulse response, h(t), and an input, u(t), find y(t)
u(t), h(t)
U(s), H(s)
convolution y(t)
L L-1
Multiplication, algebraic manipulation
Y(s)
Slide 14 Dr. Stefan B. Williams Amme 3500 : Review
Table of Laplace Transforms
• We normally use tables of Laplace transforms rather than solving the Laplace equations directly
• This greatly simplifies the transformation process
Slide 15 Dr. Stefan B. Williams Amme 3500 : Review
Laplace Transform Theorems
• The Laplace Transform is a linear transformation between functions in the t domain and s domain
Slide 16 Dr. Stefan B. Williams Amme 3500 : Review
General Second Order System
• Second order systems are quite common and are generally written in the following standard form
• Many systems of interest are of higher order
2
2 2( )2
n
n n
G s Cs s
ωςω ω∞=
+ +
5
Slide 17 Dr. Stefan B. Williams Amme 3500 : Review
Natural Frequency and Damping Ratio
• We can relate the natural frequency and damping ratio to the s-plane
21n
d n
σ ζω
ω ω ζ
=
= −
( )1sinθ ζ−=
θ
Slide 18 Dr. Stefan B. Williams Amme 3500 : Review
Step Response vs. Damping Ratio
• Damping ratio determines the characteristics of the system response
Slide 19 Dr. Stefan B. Williams Amme 3500 : Review
Time Domain Specifications • Rise time, settling
time and peak time yield information about the speed of response of the transient response
• This can help a designer determine if the speed and nature of the response is appropriate
Slide 20 Dr. Stefan B. Williams Amme 3500 : Review
Effects of Pole-Zero Patterns
• For a second order system with no finite zeros, the transient response parameters are approximated by – Rise time :
– Overshoot :
– Settling Time (2%) :
5%, 0.716%, 0.520%, 0.45
pMζζζ
=⎧⎪≅ =⎨⎪ =⎩
1.8r
n
tω
≅
4st σ≅
6
Slide 21 Dr. Stefan B. Williams Amme 3500 : Review
Example : Transforming Specifications
• Find allowable region in the s-plane for the poles of a transfer function to meet the requirements
tr ≤ 0.6 sec Mp ≤ 10% ts ≤ 3 sec
0.6ζ ≥
4 1.53
σ ≥ =
1.8 3.0 / secnr
radt
ω ≥ =
sin-1z
s
wn
Im(s)
Re(s)
Slide 22 Dr. Stefan B. Williams Amme 3500 : Review
Input, system type and steady-state error
• This table shows the relationship between input, system type, error constants and steady-state error
Slide 23 Dr. Stefan B. Williams Amme 3500 : Review
Properties of the Root Locus
• We can easily find the root locations for a second order system
• What about for a general, possibly higher order, control system?
• Poles exist when the characteristic equation (denominator) is zero
( )( )1 ( ) ( )
KG sT sKG s H s
=+
1 ( ) ( ) 0KG s H s+ =
Slide 24 Dr. Stefan B. Williams Amme 3500 : Review
Root Location
• The location of the roots, and hence the nature of the system performance, are a function of the system gain K
• In order to solve for this system performance, we must factor the denominator for specific values of K
• We define the root locus as the path of the closed-loop poles as the system parameter varies from 0 to ∞
7
Slide 25 Dr. Stefan B. Williams Amme 3500 : Review
Root Locus • The CL roots move
from the OL towards the zeros
• Additional poles move towards infinity along well defined asymptotes
Slide 26 Dr. Stefan B. Williams Amme 3500 : Review
Frequency Response • We saw that the steady state response for an
LTI system excited by a sinusoid with unit amplitude and frequency w0 will also exhibit a sinusoidal output of frequency w0 with magnitude M(w0) and a phase f(w0) where
• Notice that both the magnitude and the phase of the response on dependent on the frequency of the input w0
0 0
0 0
( ) ( )( ) ( )
M j G jj G jω ω
φ ω ω==∠
Slide 27 Dr. Stefan B. Williams Amme 3500 : Review
Bode Plot
• We often plot the magnitude and phase of the system response as a function of the input frequency
• The magnitude is normally plotted in dB=20logM(w) vs. log(w)
• The phase is plotted in degrees vs. log(w) • The resulting graph is called the Bode plot
Slide 28 Dr. Stefan B. Williams Amme 3500 : Review
First Order Bode Plot Techniques
• Normalized and scaled Bode plots for a. G(s) = s; b. G(s) = 1/s; c. G(s) = (s + a); d. G(s) = 1/(s + a)
8
Slide 29 Dr. Stefan B. Williams Amme 3500 : Review
PID Controller
• The complete three-term controller is described by
Kp+Ki/s+Kds -
+
R(s) E(s) C(s) G(s)
0( ) ( )
t
p I du t K e K e d K eτ τ= + +∫ &
Slide 30 Dr. Stefan B. Williams Amme 3500 : Review
PD Controller • The ideal derivative compensator effectively
adds a pure differentiator to the forward path of the control system
• This is effectively equivalent to an additional zero
• As you should by now be aware, the location of the open loop poles and zeros affects the root locus and hence the transient response of the closed loop system
( ) ( )cU s K s z= +
Slide 31 Dr. Stefan B. Williams Amme 3500 : Review
Root Locus
Real Axis
Imag
Axis
-15 -10 -5 0-8
-6
-4
-2
0
2
4
6
8
PD Controller Root Locus
Real Axis
Imag
Axis
-15 -10 -5 0-8
-6
-4
-2
0
2
4
6
8Root Locus
Real Axis
Imag
Axis
-15 -10 -5 0-8
-6
-4
-2
0
2
4
6
8
• Consider a simple second order system whose root locus looks like this (roots -1, -2)
• Adding a zero to this system drastically changes the shape of the root locus
• The position of the zero will also change the shape and hence the nature of the transient response
Zero at -3
Zero at -5
Slide 32 Dr. Stefan B. Williams Amme 3500 : Review
Bode Diagram
Frequency (rad/sec)
Phas
e (d
eg)
Mag
nitu
de (d
B)
10-1 100 101 1020
45
90
0
5
10
15
20
25
PD Controller • The Bode plot for a PD
controller looks like this • The stabilizing effect is
seen by the increase in phase at frequencies above the break frequency
• However, the magnitude grows with increasing frequency and will tend to amplify high frequency noise
9
Slide 33 Dr. Stefan B. Williams Amme 3500 : Review
Lead Compensation
• For compensation using passive components, a pole and zero will result
• If the pole position is selected such that it is to the left of the zero, the resulting compensator will behave like an ideal derivative compensator
• The name Lead Compensation reflects the fact that this compensator imparts a phase lead
( )( )( ) c
c
K s zc cs pU s z p+
+= <
Slide 34 Dr. Stefan B. Williams Amme 3500 : Review
Lead Compensation Example
• First find a point in the s-plane that we’d like to have on the root locus
• Place the lead pole at 20 and solve for the position of the zero
3.5 3.5 3j− +
Root Locus
Real Axis
Imag
Axis
-20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0-10
-8
-6
-4
-2
0
2
4
6
8
10
5.420( ) s
sD s ++=
x
θ1=120 θ2=112.4 θ3=20.2 therefore θc=72.6
Slide 35 Dr. Stefan B. Williams Amme 3500 : Review
Bode Diagram
Frequency (rad/sec)
Phas
e (d
eg)
Mag
nitu
de (d
B)
0
5
10
15
20
10-1
100
101
102
0
45
90
Lead Compensation
• The Bode plot for a Lead compensator looks like this
• The frequency of the phase increase can be designed to meet a particular phase margin requirement
• The high frequency magnitude is now limited
Slide 36 Dr. Stefan B. Williams Amme 3500 : Review
PI Controller
• If we rewrite the transfer function for the integral compensator we find
• This is simply a pole at the origin and a zero at some other position to be selected based on our design requirements – normally close to the origin to minimize the angular contribution of the compensator
( ) cs zU s Ks+=
10
Slide 37 Dr. Stefan B. Williams Amme 3500 : Review
PI Controller Example
Slide 38 Dr. Stefan B. Williams Amme 3500 : Review
Bode Diagram
Frequency (rad/sec)
Phas
e (d
eg)
Mag
nitu
de (d
B)
0
5
10
15
20
25
10-1 100 101-90
-45
0
PI Controller • The Bode plot for a PI
controller looks like this • The break frequency is
usually located at a frequency substantially lower than the crossover frequency to minimize the effect on the phase margin
Slide 39 Dr. Stefan B. Williams Amme 3500 : Review
Lag Compensation • As with the lead compensation, using passive
components results in a pole and zero
• If the pole position is selected such that it is to the right of the zero near the origin, the resulting compensator will behave like an ideal integral compensator although it will not increase the system type
• The name Lag Compensation reflects the fact that this compensator imparts a phase lag
( )( )( ) ,c
c
K s zc cs pU s z p+
+= >
Slide 40 Dr. Stefan B. Williams Amme 3500 : Review
Lag Controller Example
11
Slide 41 Dr. Stefan B. Williams Amme 3500 : Review
Bode Diagram
Frequency (rad/sec)
Phas
e (d
eg)
Mag
nitu
de (d
B)
0
5
10
15
20
10-1 100 101 102
-60
-30
0
Lag Compensation • The Bode plot for a Lag
compensator looks like this
• This compensator effectively raises the magnitude for low frequencies
• The effect of the phase lag can be minimized by careful selection of the centre frequency
Slide 42 Dr. Stefan B. Williams Amme 3500 : Review
State-space strategies - “Modern Control”
• System equations are described in matrix form. • Most fundamental are the state variables: A set of
parameters that completely describes the current state of the system. – For example position and velocity of a moving body
• Using measurements of the outputs, the system state can be computed - observers
• Using the system state, control strategies can be devised to achieve desired performance. – Pole placement – Optimal controllers
• Attractive for multi-input multi-output (MIMO) systems
Slide 43 Dr. Stefan Williams Amme 3500 : State Space
State Space Modelling
• Using the state space approach, we represent a system by a set of n first-order differential equations:
• The output of the system is expressed as: !x = Ax +Bu
[ ]= +y Cx Dux - state vector y - output vector u - input vector
A - state matrix B - input matrix C - output matrix D - feedthrough or feedforward matrix (often zero)
Slide 44
State Space Modelling
• We can draw a block diagram describing the general State Space Model
Dr. Stefan Williams Amme 3500 : State Space
!x = Ax +Bu [ ]= +y Cx Du
˙ x
x
12
Slide 45 Dr. Stefan Williams Amme 3500 : State Space
State Space Control • We can represent a
general state space system as a Block Diagram.
• If we feedback the state variables, we end up with n controllable parameters.
• State feedback with the control input =-Kx +r.
* N.S. Nise (2004) “Control Systems Engineering” Wiley & Sons
Slide 46 Dr. Stefan Williams Amme 3500 : State Space
State Space Control
• We can then control the pole locations by finding appropriate values for K
• This allows us to select the position of all the closed loop system roots during our design.
• There are a number of methods for selecting and designing controllers in state space, including pole placement and optimal control methods via the Linear Quadratic Regulator algorithm.
Slide 47
State Space Control • Setting u=-Kx+r yields
• Rearranging the state equation and taking LT yields
• Select values of K so that the eigenvalues (root locations) of (A-BK) are at a particular location
Dr. Stefan Williams Amme 3500 : State Space
˙ x (t) = Ax(t) + B(!Kx(t) + r)= A ! BK( )x(t) + Br
y(t) = Cx(t)
sX(s) = A ! BK( )X(s) + BR(s)sI! A ! BK( )( )X(s) = BR(s)
X(s)R(s)
= sI! A ! BK( )( )!1B
Slide 48
Advanced Topics
• We have really only scratched the surface of the study of control – Non-linear systems – Switching Control – Adaptive Control – MIMO systems
Dr. Stefan B. Williams Amme 3500 : Review
13
Slide 49 Dr. Stefan B. Williams Amme 3500 : Review
What about the Exam???
• You should be familiar with the concepts reviewed in this lecture – Modelling of dynamic systems – Specification of second order systems – Root Locus – Bode Plots – Design and properties of PID (and variants),
Lead and Lag controllers – State Space Modelling and Design
Slide 50 Dr. Stefan B. Williams Amme 3500 : Review
What about the Exam???
• You will not be required to find roots of high order systems
• You will be provided with a selected set of equations you may require for solving the problems
• In order to prepare I would suggest that you – Review the assignment questions, making sure you
understand the material covered this semester – Look over previous years’ exams
Slide 51
How do they do that?
Dr. Stefan B. Williams Amme 3500 : Review Slide 52 Dr. Stefan B. Williams Amme 3500 : Review
Conclusions
• In order to understand system performance, we must be able to model these systems
• The study of control provides us with a process for analysing, understanding and deisgning for the behaviour of a system