204 spring 2011.pdf

2.04A Class Project: the Tower

with Active Damping

2.004 Spring 13 Lecture 09 Wednesday, Feb. 27 1

Problem Wind loading of skyscrapers causes tall building sway.

Upper floor occupants suffer from motion sickness when the building sways in the wind since people are sensitive to accelerations as small as 0.05 m/s2 (0.005 g).

Too much building sway can also lead to long-term structural damage.

The Hancock Tower in Boston had a problem with falling windows. (The Hancock Tower now has two passively controlled 300 ton sliding masses on the 58th floor.)

+ = Courtesy of Rob Pongsajapan on flickr. CC-BY

&DUWRRQVVRXUFHVXQNQRZQ$OOULJKWVUHVHUYHG7KLVFRQWHQWLVH[FOXGHGIURPRXU&UHDWLYH&RPPRQVOLFHQVH)RUPRUHLQIRUPDWLRQVHHKWWSRFZPLWHGXKHOSIDTIDLUXVH


Simplified Building Model

We can model a tall building as a singledegree of freedom lumped-parameter system.


Passive Vibration Damping

Sliding Mass, m2

Spring

Damper

Wind force, One way to stabilize these Fw(t) tall builds from swaying too much during earthquakes or from high winds is to install enormous pendulum weights. When the building sways sideways the pendulum doesn't want to move (inertia) and exerts a pull in the opposite direction.

Passive Damper (Tuned Mass Damper)

Ground

2.04A Spring 13 Lecture 16&17 March 12&14 (Tue-Thu) 4

Skyscrapers Burj Khalifa Skyscraper became the tallest building (828 meters or 2,717 feet) in the world when it was officially opened on 4th January 2010.

TIME Diagram by Joe Lertola Time, Inc. All rights reserved. This content is excluded from ourCreative Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.


Burj Khalifa (http://www.burjkhalifa.ae/)

Burj Khalifa

Taipei 101

Courtesy of Natalie of designedbynatalie.com. CC-BY-NC

Courtesy of ADTeasdale on flickr. CC-BY


Taipei 101 (http://www.taipei-101.com.tw)

Courtesy of Jirka Matousek on flickr. CC-BY

Courtesy of Stefan Tan. Used with permission.


The Tuned Mass Damper in Taipei 101

The passive wind damper with a diameter of 5.5 meters and weighting 660 metric tons, is also the largest in the world now.

Taipei Financial Center Corp. All rights reserved. This contentis excluded from our Creative Commons license. For moreinformation, see http://ocw.mit.edu/help/faq-fair-use/.

Courtesy of Daniel M. Shih. Used with permission.


Active Damper Design

Sliding Mass, m2 controller sensor

Wind force, Fw(t)

The actuator is commanded by a control system, which requires sensor feedback

Spring

Damper

actuator


Experimental System

air-bearingsvoice-coils

wire spring

accelerometer tall building


Lecture 09 Wednesday, Feb. 27 2.004 Spring 13 11

Available Impulse Response Data (Course Lockers\2.004\Labs\Tower Data)

Building Response Damper Response

B1 and K1 B2 and K2

Compare your transfer function impulse response to this one.

Open Loop System Response


Estimating Parameters (Building)

= ln x1 x2

= 1

N 1 ln x1 xN

Logarithmic decrement method

Damped period

Determine after 36.8% amplitude decay

2.004 Spring 13 Lecture 09 Wednesday, Feb. 27

13

Estimating Parameters (Damper)


MIT OpenCourseWarehttp://ocw.mit.edu

2.04A Systems and Controls Spring 2013

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPT OF MECHANICAL ENGINEERING

2.004 Dynamics and Control II

Laboratory Note: Description of the Experimental Rotational Plant1

INTRODUCTION

In the first series of 2.004 laboratory experiments you will be designing and testing various schemes for controlling the velocity and position of a rotational flywheel. This handout introduces you to the experimental plant. You should study the handout before the first lab session. We will be referring to this handout throughout the term. DESCRIPTION OF HARDWARE The rotary inertia (flywheel): Figure 1 is a photograph of the rotational flywheel that you will be using. It consists of a large metal (copper) flywheel which rides in high quality ball bearings. The lower bearing is an angular contact type of bearing. These bearings are designed to taking thrust loads, and in this case the lower bearing supports the full weight of the flywheel. The upper bearing is a conventional ball bearing. The bearings have very low friction: if you set this flywheel rotating it will typically take several minutes to come to a stop.

Figure 1. The copper flywheel and bearings.

Magnet in place for eddy current damping

Magnet in storage location

Copper Flywheel

1 The hardware described here was developed by Prof. E Sachs. These notes are based upon notes prepared by Prof. Sachs.

Upper bearing is recessed in this hole

In the experiments we will add linear (viscous) damping to the system using a phenomenon knows as eddy-current damping. When a permanent magnet (see Fig. 1above) is moved across the surface of an electrical conductor it induces internal currents in the material (eddy currents), which in turn create a magnetic field that opposes the original field (re-read Lenzs law from 8.02). The result is a force on the magnet proportional to the relative velocity of magnet and metal, in a direction which opposes the motion. It therefore requires work to maintain motion between the magnet and the conductor. Energy is dissipated in the conductor as heat (i2R losses), and the effect is that the eddy-current forces appear as a dissipative frictional force.

Our flywheel is made of copper, for several reasons:

1) Copper is highly electrically conductive, and therefore large eddy-currents can be induced in it, 2) Copper is non-magnetic which is important because ferromagnetic forces would overwhelm the eddy-current forces, and, 3) Copper has a high physical density with the result that our flywheel is massive (has a large moment of inertia). You will be provided with several magnets, which may be placed under the rim of the flywheel to achieve a range of damping coefficients. The dimensions of the copper flywheel are shown below In Fig. 2. All dimensions are in inches. The density of copper is 8,230 kg/m3.

Figure 3. Dimensions of the copper flywheel.

Attached to the shaft, under the flywheel, is an optical rotary encoder, shown in

Fig. 3. This consists of a transparent plastic disk, scribed with 1,024 radial lines. As the disk rotates, these lines interrupt a light beam, and are recorded electronically as a sequence of pulses. Specialized electronics (US Digital ETACH and EDAC2 modules) are used to compute either the angular velocity of the shaft or its angular position. Because there is no mechanical contact, the encoder adds no friction to the system. You will use these signals from these modules in your control system designs.

Figure 2. The rotary encoder for measuring angular velocity and position of the flywheel. The Gear Train: The flywheel will be driven by a dc electric motor through a step-down gear train, as shown in Fig. 4. A large gear wheel, with 180 teeth, is attached to the flywheel shaft, and is driven by a smaller gear with 44 teeth from the electric motor shaft.

Encoder

Figure 4: Spur gear drive train.

Flywheel drive-shaft with larger gear (180 teeth)

Motor drive-shaft with smaller gear (44 teeth)

The Electric Motor: The system is driven by a permanent magnet dc motor (Maxon 148867). This is a high-quality low-friction 150w, 24 v. dc permanent magnet motor. A specification sheet for the motor is attached. A flexible shaft coupling is used to connect the motor to the drive-shaft to minimize the effects of any shaft misalignment.

Fig. 5 Complete experimental set-up with dc motor.

We will be covering the operation of a dc motor in class, but for now we can simply assume that the torque produced by a motor is directly proportional to the current flowing in the windings. Therefore we can define the torque applied to the flywheel by specifying the motor current. The complete rotational set-up is shown in Fig. 5. UPDATED PLANT All the experimental rotational plants were updated in the summer of 2010 to improve the mechanical connection between the flywheel and its drive shaft by adding an aluminum hub on top of the copper flywheel. The updated plant is shown in Fig. 6. This aluminum hub adds about 0.94 lbs of mass to the flywheel, which is not significant compared to the weight of the copper. In addition, the height between the lower and the upper plate was increased to accommodate the added height of the flywheel. Two pieces of plexiglass were also added to the long sides of the upper plate as a safety measure.

Electric motor

Encoder output

Electric cable

Flexible coupling

Fig. 6 Updated plant.


2.04A Systems and ControlsSpring 2013


2.04A System Dynamics and Control

Spring 2013

G. Barbastathis

02/05/2013 2.004 System Dynamics and Control Spring 2013

1

System Input Output y(t) Mechanical,

electrical, etc. elements

x(t)

May be scalar or vector May be scalar or vector Usually can be defined by Usually the user specifies

the user a desired form It is subject to noise (or errors) Sometimes it can be measured,

others it can not

Described by ordinary differential equations (ODEs) operating on the input (dynamics) If the ODEs operate on the output as well, then we refer to it as a feedback system Can be linear or nonlinear

The purpose of control is to ensure that the output waveform resembles the waveform desired by the user, despite the systems dynamics and disturbances by noise

Usually, control requires feedback

2.004 System Dynamics and Control Spring 2013 02/05/2013

2

Example1: Hard Disk Drive (HDD)

Speed Control Head-Disk Tracking


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Courtesy of Robert Scholten. Used with permission.

3

Hard Disk Drives


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4

Example 2: The Segway

http://www.segway.com/

Speed Control Stability Control


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5

Example 3: Manufacturing Automation

http://www.youtube.com/watch?v=iwIzPjS5L6w


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6

System Classifications

Single vs Multiple Inputs / Single vs Multiple Outputs SISO (single input - single output)

SIMO MISO MIMO

Feed-forward vs feedback Linear vs nonlinear


7

Feedforward vs feedback

Feedforward: acts without taking the output into account Example: your dishwasher does not measure the

cleanliness of plates during its operation Feedback: the output is specified by taking

the input into account (somehow)

x(t)y(t)

Plant

Controller

+


8

HDD Control System

Michael L. Workman, PhD thesis, Stanford University, 1987.

Karman Tam et al, US Patent 5,412,809


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9

2.04A Learning Objectives

Learn the process of modeling linear time-invariant (LTI) dynamical systems in dual domains: in the time domain using ordinary differential equations and in the Laplace domain (s-domain).

Understand the behavior of LTI systems qualitatively and quantitatively, both in the transient and steady-state regimes, and appreciate how it impacts the performance of electro-mechanical systems.

Introduce feedback control and understand, using the s-domain primarily, how feedback impacts transient and steady-state performance.

Learn how to design proportional, proportional-integral, proportional-derivative, and proportional-integral-derivative feedback control systems meeting specific system performance requirements.

Introduce qualitatively the frequency response of LTI systems and how it relates to the transient and steady-state system performance.


10

What you need

8.01 and 8.02 basic behavior of mechanical and electrical elements

18.03 Linear ordinary differential equations (ODEs) and systems of ODEs Laplace transforms

2.003/2.03

from a physical description of system, derive the set of ODEs that describe it

We will review these here as necessary; but please refer back to your materials from these classes, anticipating the topics that we cover


11


Lab Rules - IMPORTANT!!

You must stay within the designated 2.04A/2.004 lab space

No working on other classes (e.g. your 2.007 project) allowed in the machine shop

2.678

2.672

ENTER

3-062N

12


Lab: Equipment Overview

Oscilloscope

Function Generator

DC Power Supply

Power Amp

Rotational Plant

Data Acquisition Interface Panel

PC

Data Acquisition Hardware:National Instruments PCI-6221

Tachometer

Please read the equipment description before before Thu PM lab!

13


Dynamic System

Input OutputTime VaryingFrequency Dependent

Time VaryingFrequency Dependent

Mechanical, Electrical, Fluid, Thermal

Linear System Theory

Control System Design

Modeling

Real Physical SystemMulti-Domain Ideal ElementsNetwork Representations

Idealized Representation: Lumped Model

State Variables and LODEs

Resulting Equations of Motion Laplace Transforms

Transfer Function RepresentationsPole Zero Analysis

Frequency Response Analysis

Standard Inputs and Prediction of ResponsesTransient ResponseBode Diagrams

System Design Changes;Feedback ControlSteady State ErrorsInput Tracking

Response Shaping

Desired Input - Output Response

Framework for system control

14


Example: Car Suspension System modeling

System dynamics (as is)

System Control

Model: ordinary differential equation (ODE)Hardware or other mathematical representation

Model

Response

Desiredresponse

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15


Physical realization of systems

Mechanical Electrical Fluid Thermal Electromechanical Mechano-fluid Electro-thermal Electromechanicalfluidthermal

16


Complex Interconnected Systems?

Combine Mechanical, Electrical Fluid and Thermal

Common Modeling Method Linear, Lumped Parameter

Circuit-Like Analysis:

Common Analytical Tools Linear System Theory

Powerful Design Tools Feedback Control

2.12, 2.14, 2.151

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17


Dynamic System

Input Output

Mechanical, Electrical, Fluid, Thermal

Control System Design

Linear System Theory

Modeling

Real Physical SystemMulti-Domain Ideal ElementsNetwork Representations

Idealized Representation: Lumped Model

State Variables and LODEs

Resulting Equations of Motion LaPlace Transforms

Transfer Function RepresentationsPole Zero Analysis

Frequency Response Analysis

Standard Inputs and Prediction of ResponsesTransient ResponseBode Diagrams

System Design Changes;Feedback ControlSteady State ErrorsInput Tracking

Response Shaping

Desired Input - Output Response

A simple modeling example

18

y1(t) x1(t)

ax1(t)ay1(t) y1(t) + y2(t) x1(t) + x2(t)

ay1(t) + by2(t)

a, b : constant scalar or tensors


Linear Systems Suppose

The system is linear iff

and

Corollary

x2(t) y2(t)

ax1(t) + bx2(t)

19


ModelingHardware

Idealized Lumped Parameter Model

mathematical representation

oror

m2m1

k

b

y(t)

time domain equation(s) of motion

X(s)Y (s) =

2ns +n2s2 + 2ns +n2

Laplace domain transfer function

x(t)

x+ a2x + a1x = b2y + b1yODE

x1x2

=a11 a12a21 a22

x1x2

+b1b2

y(t)

y(t) = c1 c2x1x2

State Space

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20

=) md2x

dt2+Kx+ fvx = Ku+ fvu=) m

d2x

dt2= Kx+Ku fvx+ fvu

Car suspension model Mass spring viscous damper system

Model

Force balance

Free body diagram

FK0

Mg

Free body diagram (no motion)

FK0

Mg

force due to spring in equilibrium

force because spring changes length during motion

force due to viscous damping

System ODE : 8

md2x

= FK + Fvdt2 =) md

2x = Kx + Ku fvx+ fvu

FK = -K(x - u) dt2 linear differential >: = -fv(x - u)

equation)

Fv

d2xEquation of motion: =) m + Kx + fvx = Ku + fvudt2


General Linear Time-Invariant (LTI) system

dnx dn1x dnx dx dmy dm1y d1yan La1+ an1 + bm1dt n dt n1 + an 2 dt n dt + a0 x = bm dtm dtm1 + Lb1 dt1 + b0u

nth-order Linear Ordinary Differential Equation (ODE) with constant coefficients (time-invariant)

general solution:

x(t) = xhomogeneous(t) + xforced(t)

homogeneous solution: y=0 (no forcing term) forced solution: a guess solution for the system behavior when y(t)0


Homogeneous and forced solutions

Homogeneous solution:

x(t) = C0 + C1es1t + C2es2 t + L+ Cnesnt where in general si = i + j (complex number)

s

Im(s)

Re(s)Forced solution: sometimes difficult to guess but for specific forces of interest, quite easy.

For example, if y(t)=constant, then yforced=constant as well (but a different constant!)


Commonly used input functions

Step function Ramp (aka Heaviside) function

0, t < 0; 0, t < 0; ramp(t) = t, t > 0.step(t) = 1, t > 0.

= t step(t)

Impulse Sinusoidal (aka delta-function function (t) or Dirac function) 0, t < 0;

f(t) = sin(!t), t 0.

= sin(!t) step(t) t [sec]


1st order system Mv + bv = f(t)

mass viscous force damping

Impulse response: equivalent to setting an initial condition v(t=0)

v(t = 0) = v0

t/ v(t) = v0e , t > 0 M

time constant = b

Step response: f(t) is the step function (or Heaviside function)

0, t < 0;f(t) = F0 step(t) = F0, t > 0.

F0 v(t) = 1 e t/

, t > 0

b v(t = 0) = 0

02/07/2013 2.004 System Dynamics and Control Spring 2013 5

0.368 e 1

one time constant

1st order system: step response

steady state (final value)

0.632 1 e 1

Nise Figure 4.3

2.004 Fall 07 Lecture 06 Monday, Sept. 17

Figure John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

How is this different than the car suspension system?



2nd order system: step response


Mechanical system components: translation

Nise Table 2.4


Table John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

11/28/13 2.004 Spring 13'

Rotational Plant

Copper flywheel

Gear set

**Read the Description of the Experimental Rotational Plant**

Flywheel

Encoder output

DC motor

Gear set

Flex coupling

Power cable

J 3 102 kg m21


Mechanical system components: rotation

Nise Table 2.5 Table John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

2

02/07/2013

Mechanical system components: rotation: gears

Nise Figure 2.27, 2.28 Figure John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

2.004 System Dynamics and Control Spring 20133


Gear transformations

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4


Rotational mechanical system: example

Nise Figure 2.30a-b


5

2/13/2013 2.004 Spring 13'

Damping Model

Viscous Damping

Coulomb damping results from the sliddry surfaces. The friction generated by thmotion of the two surfaces is a source of edissipation. It is opposite to the direction and is independent of surface area, displaposition, and velocity

ing of two e relative nergy

of motion cement or

Coulomb Damping

Static

Kinetic

6


Coulomb friction

velocity friction velocityfriction

Coulomb friction is in opposite direction to the velthe magnitude of the friction force is independent magnitude of the velocity

ocity;of the

Example: Block sliding on a rough surface

7


Viscous friction

velocity friction velocityfriction

Viscous friction is in opposite direction to the velocthe magnitude of the friction force is proportional tmagnitude of the velocity

ity;o the

8

11/28/13 2.004 Spring 13'

Eddy Currents Viscous friction

Eddy currents are generated when there is relative motion between a conducting object and a magnetic field. The rotating currents in the conducting object are due to electrons experiencing a Lorentz force that is perpendicular to their motion and the magnetic field (F=qvB).

The Lorentz force results in current in the radial direction on the flywheel; these currents, since the wheel is turning, result in an opposing magnetic field and a force resisting the motion.

The Eddy current and the resisting force are both proportional to the velocity |v|; therefore, they resist motion in a way that is exactly equivalent to viscous friction.

9

In-class experiment

Familiarize with the laboratory equipment and software tools

Study the frictional characteristics of the motor, gear train, and bearings in the flywheel system

Explore the effect of damping on the flywheel system

10

11/28/12 2.004 Spring 12'

Optical Encoders

Regular phase

Quadrature phase

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11

US Digital Optical Encoder E6S-2048-187

Quick, simple assembly and disassembly

Rugged screw-together housing

Positive finger-latching connector

Accepts .010" axial shaft play

Tracks from 0 to 100,000 cycles/sec

64 - 2500 CPR | 256 to 10,000 PPR

2 channel quadrature TTL squarewave outputs

Optional index (3rd channel)

-40 to +100C operating temperature

Fits shaft diameters from 2mm to 1"

11/28/12 2.004 Spring 12'

Max speed = 2,400 rpm

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12

!(t) =V0(t)

Kt

You will monitor the angular velocity profile using the ETach2 electronic

tachometer that is attached to the rotary encoder on the flywheel shaft.

It produces an analog voltage vo proportional to the shaft speed ! ! Vo = Kt!

where the tachometer constant Kt = 0.016 volts/rpm

Converting Tachometer reading

13

Spin the flywheel by hand, and record the angular velocity decay !(t), using the computer-based

Chart Recorder(VI),(Remember to convert the Chart Recorder output to angular velocity.)

Repeat the same procedure with one and two magnets (damper) on the flywheel

Generate clearly labeled plots and indicate, for each case, which kind of friction dominate the

damping behavior. For viscous damping, also compute the time constant . (J 3 102 kg m2)

Extra credit: Can you determine the damping coecient for each viscous case and torque due to

Coulomb friction (Assume the eect of Coulomb damping is significantly smaller than the eect

of viscous damping)

Hand in your results at the end of class.

Procedure

Save your data so that it is readily available later(online or on USB.)

You will need them for Problem Set 1.

14

11/28/13 2.004 Spring 13'

Put a meaningful title Label each axis (with proper unit) Label each data source

>> mydata;>> plot(t_mydata,v_mydata);>> title(Lab 1 (no mag));>> xlabel(Time (s));>> ylabel(Voltage(V));>> legend(Velocity,Linear);>> grid;>>

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Making A Good Plot with MATLAB

MATLAB commands:

15

For pure Coulomb damping: J !(t) = Text(t) Tc(t)

(The external torque is 0 for our experiment, Tc(t) is relatively constant Tc)

Solving this ODE =) !(t) = !0 TcJ t

For pure viscous damping: J !(t) + b!(t) = Text =) !(t) + bJ!(t) = 0

Solving this ODE =) !(t) = !0e( bJ )t

Solving the equations of motion

16

2/13/2013 2.004 Spring 13'

Fitting an Exponential Function

Estimate from the time at which the response has decayed to:

Slope method

17

2/13/2013 2.004 Spring 13'

Fitting an Exponential Function

Log linear fit:

18

Goals for today

Review of translational dynamical variables: position, velocity

Review of rotational dynamical variables: angle, angular velocity

Electrical dynamical variables: charge, current, voltage

Basic electrical components Resistance Capacitance Inductance

DC Motor: an electro-mechanical element basic physics & modeling equation of motion transfer function

Experiment: step and ramp response of the flywheel driven by the DC motor open loop (no feedback)

2.04A Spring 13 Lecture 06 Thursday, Feb. 14 1

Impedances: translational mechanical

Nise Table 2.4


!"##$#!#%&

Impedances: rotational mechanical

Nise Table 2.5


!"##$#!#%&

Electrical dynamical variables: charge, current, voltage


Electrical resistance


Capacitance /1


Capacitance /2


Inductance


Summary: passive electrical elements; Sources

2.04A Spring 13 Lecture 06 Thursday, Feb. 14

9

!"##$#!#%&

Combining electrical elements: networks

Nise Figure 2.6

Network analysis relies on two physical principles

Kirchhoff Current Law (KCL) Kirchhoff Voltage Law (KVL)

charge conservation energy conservation


!"##$#!#%&

Impedances in series and in parallel


!"##$#!#%&

The voltage divider


!"##$#!#%&

Example: the RC circuit


!"##$#!#%&

Interpretation of the RC step response

+

+

+ +

+

+

+

+++

+

+ +

++ + + +

+

++ +

+

+

+

+

+


!"##$#!#%&

Example: RLC circuit with voltage source

Nise Figure 2.3

Nise Figure 2.4


!"##$#!#%&

Quick summary of electrical systems


!"##$#!#%&

Power dissipation in electrical systems


!"##$#!#%&

DC Motor as a system

Transducer:

Converts energy from one domain (electrical) to another (mechanical)


!"##$#!#%&

Physical laws applicable to the DC motor

Lorentz law: Faraday law:magnetic field applies force to a current moving in a magnetic field results (Lorentz force) in potential (back EMF)


!"##$#!#%&

DC motor: principle and simplified equations of motion

multiple windings N: continuity of torque


!"##$#!#%&

DC motor: equations of motion in matrix form

multiple windings N: continuity of torque


!"##$#!#%&

DC motor: why is Km=Kv?


!"##$#!#%&

DC motor with mechanical load and realistic electrical properties (R, L)

inductance dissipation (due to windings) (resistance of windings)

dissipation load (viscous friction (inertia)

in motor bearings)


!"##$#!#%&


inductance dissipation (due to windings) (resistance of windings)


in motor bearings)


!"##$#!#%&


inductance (due to windings)

dissipation (resistance of windings)


in motor bearings)


!"##$#!#%&

Lab 06: Running the flywheel with DC motor open loop

Observe motor behavior under different driving voltages.

Examine transient response of a DC motor

Important: always be ready to turn off the break of power amplifier when motor is spinning too fast!!

Experimental Setup: Open-loop

Signal Generator

Power

Amplifier CH1 DC motor-flywheel

system

ETACH2 ACH2 PC Data-

Acquisition

ACH 1

2.004 Spring 13 Lecture 06 Thursday, Feb. 14 26

Procedure

Make sure all devices are powered off; connect function generator, power amplifier, flywheel system and PC data acquisition system as shown in the previous slide.

Add one magnet to the flywheel damper, open Chart Recorder to record ACH1 and ACH2; turn on function generator, power amplifier and start experiment.

Obtain system response for a ramp function with Freq: 0.2 Hz, Amp: 0.5 V, offset: 0 V. Repeat experiment using a sine function with same parameters. Record your response. Referring to materials we learned from last lecture, comment on the behavior of DC motor- flywheel system.

Use a DC signal with 0.2 V offset. Start experiment and record DC motor transient response data. Convert voltage signal to motor speed (you will need to make use of gear ratio). Generate appropriate plots of motor speed & amplified current V.S time. Compute mechanical power of the DC motor.

Set your function generator to generate a square function(SQUA), set frequency to 0.04 Hz, amplitude to 0.200V and offset to 0.100V. Collect a full period of flywheel response and function generator signal. (You can take a screen shots of the plot in Chart Recorder)


n1=

n2 180

44

Some Hints

Gear Ratio:

Gear 2

F

Gear 1

Unit Conversion: Power Conservation:

Ka = 2.0 A/V, Km = 0.0292 N-m

Pmechanical(t) = T (t) (t)= Km i(t) (

/A

t)


Lab assignment p.1

Comment on how todays experiments involving step input are interpreted differently than we did in Lab 05.

When the DC motor is driven by a step function, how many poles/ zeros do we need to consider, and where are they? How do the magnets (eddy brakes) influence their locations?

2.04A Spring 13 Lab assignment 06 Thursday, Feb. 14 - PLEASE RETURN TO THE T.A. 29

Lab assignment p.2

When the DC motor is driven by a ramp function, how many poles/ zeros do we need to consider, and where are they? How do the magnets (eddy brakes) influence their locations?

Comment on the qualitative and quantitative agreement of your derivations with experiment. Attach sheets with your experimental results.

2.04A Spring 13 Lab assignment 06 Thursday, Feb. 14 - PLEASE RETURN TO THE T.A. 30

!

"#$

%&'"!&()*

Lecture 07 Thursday, Feb. 212.004 Spring 13

Todays plan

1st order system response: review the role of zeros

2nd order system response: example: DC motor with inductance response classifications:

overdamped underdamped undamped

Linearization from pendulum equation to the harmonic oscillator

1


Review: step response of 1st order systems

Nise Figure 4.3

steady state(final value)

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2

J !(t) + b!(t) = f(t) f(t) = F0 step(t)

!1 =F0b


Steady-state

Note that as t, the exponential decays away, and the step response tends to 1, i.e. the value of the driving force.

More generally, if we consider a flywheel-like viscously damped system with equation of motion expressed as 1st order linear time-invariant ODE

and step excitation

!(t) =F0b

1 et/

, t > 0

=

J

b

[where f(t) and F0 have units of torque,]the step response is

the angular velocity as t tends to

This long-term value is known as the systems steady state In systems where the output physically represents a velocity, the steady

state is also known as terminal velocity.

3

limt!1g(t) = lims!0sG(s).


Steady state in the Laplace domain: the final value thm.

It turns out that we can predict the steady state of a system directly in the Laplace domain by using the following property known, for obvious reasons, as the final value theorem:

which holds generally if g(t) and G(s) form a Laplace transform pair.

(s)

F (s)=

1

Js+ b In the case of the flywheel, the transfer function is

F0F (s) =F

s) 0(s) =

s (Js+ b)For the step response we have

lims!0s(s) =F0b

It is easy to verify that

in agreement with the result for the steady state of this system in the previous page

4

H(s) =s+ z

D(s)= s 1

D(s)+ z 1

D(s)


A word on zeros

Transfer functionwith a zero

Derivative Amplificationoperator (gain)

H0(s) =1

s+ p

Example: compare the step response of the two systems below:

H(s) =s+ z

s+ p

5

Example:

p = 0.5;z = 1.0

U(s) =1

sF0(s) F0(s) =

1

s

1

s+ p=

1

p

1

s 1

s+ p

) f0(t) = 1

p

1 ept , t > 0.

F (s) F (s) = sF0(s) + zF0(s)

) f(t) = ddt

f0(t) + zf0(t) = ept +

z

p

1 ept , t > 0.


Step response without zero vs with zero

H0(s) =1

s+ p

Please verify yourselves thispartial-fraction

expansion!

H(s) =s+ z

s+ p

U(s) =1

s

2 0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t [sec]

f 0 [a

.u.]

2 0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

t [sec]

f [a.

u.]

withoutzero

withzero

j

-0.5

j

-0.5-1

6


Zero on the right-hand side?

2 0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t [sec]

f 0 [a

.u.]

2 0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

t [sec]

f [a.

u.]

withoutzero(lhs)

withzero(lhs)

j

-0.5

j

-0.5-1

2 0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t [sec]

f [a.

u.]

2 0 2 4 6 8 101

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

t [sec]

f [a.

u.]

j

-0.5 0.25

j

-0.5 1.0

Example:

p = 0.5;

7


The general 2nd order system


8



Nise Figure 4.10 John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

9



Nise Figure 4.11 John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

10


The underdamped 2nd order system


11




12



forced response,sets steady state

Nise Figure 4.10


13


Transients in the underdamped 2nd order system

Nise Figure 4.14


14


Transient qualities from pole location in the s-plane

Nise Figure 4.19


15


Linearization This technique can be used to approximate a non-linear system


16


Linearizing systems: the pendulum


17

Lecture 08 Thursday, Feb. 21 LAB2.004 Spring 13

Observing 2nd order system behavior

Today we will change the flywheel output so it becomes a 2nd order system employ a feedback scheme such that we can tune the 2nd order

system response from overdamped to undamped

You need to know 1st order systems Laplace transforms

We will cover (here and in Lecture 07) the basics of 2nd order system behavior

You do not need to know yet how feedback systems work (we will cover in detail in the second

part of the class)

1

r(t) !(t)J !(t) + b!(t) = r(t)

(s)

R(s)=

1

Js+ b

(s)R(s)

(t) =

Z t0!(t0)dt0

(s) = (s)s

1

(s)

R(s)=

1

s(Js+ b)


Flywheel: switching to angular position as output /1

We modeled the flywheel as a SISO system with the torque r(t) as input and the angular velocity (t) as output. Schematically,

Time domain (ODE) Laplace domain (TF)

Now let us consider instead the angular displacement (t) as the

output. This is related to (t) via an integral:

In the Laplace domain, the relationship becomes

Using the TF for (s) we obtain

2

r(t) R(s)


Flywheel: switching to angular position as output /2

Schematically,

Time domain (ODE) Laplace domain (TF)

Some important observations: the diagrams in this and the previous page correspond to the same physical system; however, the physical quantity that we choose to represent the output is different;

the ODE for (t) can be obtained by inverse-Laplace transforming the TF (s)/R(s) [recall that multiplying by s in the Laplace domain is equivalent to taking a derivative in the time domain] or by deriving the equation of motion in the time domain using Newtons law for rotational systems. The results are of course consistent (check that!)

J (t) + b(t) = r(t)(s)

R(s)=

1

s(Js+ b)

(t) (s)

3

The transfer function(s) 1

=R(s) s(Js+ b)

has two poles: one at b/J (same as the 1storder system we had consideredearlier, when angular velocity was the output) and one at the origin. The pole atthe origin results from the integrator, i.e. the integral operation that convertsangular velocity !(t) to angular position (t).

se look more like the standard 2ndorder system, jThis is physically implemented by connecting theositive inlet of the amplifier and the output of the tive inlet and specifying a feedback gain K in the b/J 0t going into the details of the derivation (we willllowing weeks) the transfer function of the system

Low gain (overdamped)(s) K

=R(s) Js2 + bs+K jto observe experimentally that at low gain, i.e.

To make the flywheel responwe can introduce feedback.function generator to the pposition meter to the negacomputer interface. Withoudo that extensively in the fowith feedback becomes

The purpose of the lab issmall values of K, the response is overdamped; whereas at high gain, i.e. forlarge values of K, the response becomes underdamped. b/J 0We will verify these observations with mathematical rigor in the following weeks.For now, please consider K as a knob which you can turn to tune the response High gain (underdampfrom overdamped to underdamped. More details will come, we promise!

:open-loop poles


Flywheel with feedback: a tunable 2nd order system

b/J 0

j

(no feedback or K=0)

ed)

4


Lab procedure

First, please read carefully the previous three introductory pages, and make sure you are comfortable with the flywheel as a 2nd order system (now that weve defined the angular position as output) and the role of gain K in tuning the transfer function.

Derive the roots of the polynomial Js2+bs+K=0 and verify that for small values of K they are both real, whereas for large values of K they form a complex pair (i.e. both have the same real part, and the imaginary parts have equal magnitude but opposite signs).

With help from your TA, verify all the connections in your system, especially where the inputs and outputs are applied. Identify the definition of K in your digital interface.

Now use a step function as input to the flywheel, and try different values of K. Record at least two overdamped responses (for two low values of K) and two underdamped responses (for two high values of K) and the values of K for which you observed them.

For the underdamped responses, estimate the values of damping ratio and damped oscillation frequency d (you will find definitions of these quantities in Lecture 07.)

Hand in printouts of your results (or email in digital form) to the TA at the end of the lab.

5

Todays goal

Introduce root locus Close loop transfer function &characteristic equation Root locus with an example

Rules for sketching root locus

Observation of Root Locus with MATLAB s graphical user interface


Close-loop transfer function and characteristic equation Consider a unity feedback control loop

John Wiley & Sons. All rights reserved. This content is excluded from our Creative

N (s)G(s) =

D(s)

Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.

Closed-loop transfer function:

KG(s) KN(s)Gclosed(s) = =

1 +KG(s) D(s) +KN(s)

The closed-loop characteristic equation:

1 +KG(s) = 0

More generally: D(s) +KN(s) = 0


What is root locus

Root locus is all values of s that satises the system characteristic equation:

1 + KG(s) = 0 or more generally: D(s) + KN(s) = 0

as the loop gain K varies from 0 to 1

Example: 0.3162

G(s) = s + 2



Cranking up the gain JType 0 system (no disturbance)

Type 1 system with disturbance



Cranking up the gain LType 1 system (no disturbance)



Cranking up the gain: poles and step response

Root Locus


Root Locus for non-unity negative feedback systems

Closed loop TF:

Open loop TF: KG(s)

Nise Figure 8.1

Caveat: K>0 (i.e., negative

feedback)

Condition for closed-loop pole: denominator of closed-loop TF must equal zero:


Nise Figure 8.10

Breakaway point Root Locus terminology

Real-axis segment

Breakaway point

Asymptote angle

Asymptote real-axis intercept

Break-in point

Nise Figure 8.12

Nise Figure 8.25 Nise Figure 8.19

RL imaginary axis intercept

Departure/Arrival anglesBranches



8

Root-locus sketching rules

Rule 1: # branches = # poles Rule 2: always symmetrical about the real axis Rule 3: real-axis segments are to the left of an odd number of real-

axis finite poles/zeros




Example

Rule 4: RL begins at poles, ends at zeros



Poles and zeros at infinity


Root Locus sketching rules

Rule 5: Asymptotes: angles and real-axis intercept

Nise Figure 8.12




Rule 6: Real axis break-in and breakaway points

maxK for this real axis segment

minK for this real axis segment

Nise Figure 8.13




Lecture 11 Thursday, Feb. 28 2.04A Spring 13

Rule 6: Real axis break-in and breakaway points

Nise Figure 8.13

maxK for this real axis segment

minK for this real axis segment


14


Rule 7: Imaginary axis crossings

system response contains undamped terms at this point






2.04A Spring 13 Lecture 11 Thursday, Feb. 28

system response contains undamped terms at this point

16

Root Locus sketching rules summary

Rule 1: # branches = # poles Rule 2: symmetrical about the real axis Rule 3: real-axis segments are to the left of an odd number of real-

axis finite poles/zeros Rule 4: RL begins at poles, ends at zeros Rule 5: Asymptotes: angles, real-axis intercept

Rule 6: Real-axis break-in and breakaway points

Rule 7: Imaginary axis crossings (transition to instability)


Practice 1: Sketch the Root Locus


Nise Figure P8.2


Practice 2: Are these Root Loci valid? If not, correct them

Nise Figure P8.1 John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.


In-class Experiment 4

What Is Root Locus Design?

A common technique involving iterating on a design by manipulating the compensator gain, poles, and zeros in the root locus diagram.

As system parameter k varies over a range of values, the root locus diagram shows the trajectories of the closed-loop poles of the feedback system.

SISO Design Tool in MATLAB: A graphical-user interface that allows the user to tune control

parameters from root locus design and system response simulation.

12/03/13 2.004 Spring 12' 20

System Modeling

Km/NKa Js + B G(s)

Kt 21

System Parameters

22

Procedures

In MATLAB workspace, construct necessary system data (transfer functions) based on the system model

Graphically tune the control parameters of the following general forms.

C(s) = a1, C(s) = a2 + b2s, C(s) = a3 + b3/s

At the end of the class, turn in your result parameters, root locus plots and system response.

12/03/13 2.004 Spring 13' 23

---------------

Useful Matlab Commands

>> B=0.014;J=0.03;N=44/180;ka=2;km=0.0292;kt=0.016; >> G=tf([ka*km/N],[J,B]);

Transfer function:Setup transfer function

0.2389

0.03 s + 0.0872

12/18/12 2.004 Spring 12' 24

Tips 1

In the command window type in sisotool tool open the SISO Design Tool Interface.

Select appropriate control architecture

Courtesy of The MathWorks, Inc. Used with permission. MATLAB and Simulink are registered trademarks of TheMathWorks, Inc. See www.mathworks.com/trademarks for a list of additional trademarks. Other product orbrand names may be trademarks or registered trademarks of their respective holders.

Enter or import system system data (G, H) from workspace. 12/03/13 2.004 Spring 12' 25

Tips 2

Under Compensator Editor tab, create general form ofKp Kis2 + s +Kd Kdcontroller model (e.g : Kds + Kp + Ki/s = Kd )s


12/03/13 2.004 Spring 12'

26

Tips 3

Select design plots you want to use and click on show design

plot under Graphical Tuning.

You can drag/add/remove poles & zeros in this graphical root locus design window. Simulation result is instantaneous.


12/03/13 2.004 Spring 12'

27

Tip4

Select Step for Plot 1, check closed loop r to y. Show analysis plot under Analysis Plot tab generates a real-time step response of your system. (You can also look at other plots)


28


2.04A Systems and Controls Spring 2013


Lecture 12 Tuesday, March 52.04A Spring 13

Todays goal

Root Locus examples and how to apply the rules single pole single pole with one zero two real poles two real poles with one zero three real poles three real poles with one zero

Extracting useful information from the Root Locus transient response parameters limit gain for stability

1

X(s)Y (s)

X(s)

Y (s)

OL

=1

s+ 2

X(s)

Y (s)

CL

=K

s+ 2 +K

As K varies from 0 to 1 . . .

K = 0K =1


Root Locus definition

Root Locus is the locus on the complex plane of closed-loop poles as the feedback gain is varied from 0 to .

2

+

K G(s) =

1

s+ 2G(s) =

1

s+ 2

Y (s) X(s)

2

j!Open-Looppole

2

j!Open-LooppoleRoot Locus



Rule 1: # branches = # poles Rule 2: always symmetric with respect to the real axis Rule 3: real-axis segments are to the left of an odd number of real-

axis finite poles/zeros

3

8 7 6 5 4 3 2 1 0 10.4

0.3

0.2

0.1

0

0.1

0.2

0.3

0.4Root Locus

m (seconds1)

jt (s

econ

ds1

)

X(s)

Y (s)

OL

=1

s+ 2

6 5 4 3 2 1 0 10.2

0.15

0.1

0.05

0

0.05

0.1

0.15

0.2Root Locus

m (seconds1)

jt (s

econ

ds1

)

X(s)

Y (s)

OL

=s+ 5

s+ 2

1 branch

to the left of the single pole

1 branch

to the left of the pole

nothing to the leftof one zero +

one pole

X(s)

Y (s)

CL

=K (s+ 5)

(K + 1) s+ (5K + 2)closedloop

pole

= 5K + 2

K + 1! 5, as K !1

X(s)

Y (s)

CL

=K

s+ (K + 2)closedloop

pole

= (K + 2)! 1, as K !1



Rule 4: begins at poles, ends at zeros

4

6 5 4 3 2 1 0 10.2

0.15

0.1

0.05

0

0.05

0.1

0.15

0.2Root Locus

m (seconds1)

jt (s

econ

ds1

)

X(s)

Y (s)

OL

=s+ 5

s+ 2

1 branch

to the left of the pole

nothing to the leftof one zero +

one pole

K = 0K =1

8 7 6 5 4 3 2 1 0 10.4

0.3

0.2

0.1

0

0.1

0.2

0.3

0.4Root Locus

m (seconds1)

jt (s

econ

ds1

)

X(s)

Y (s)

OL

=1

s+ 2

1 branch

to the left of the single pole

K = 0K =1

We say that this TF has a zero at infinity

3.5 3 2.5 2 1.5 1 0.5 0 0.52

1.5

1

0.5

0

0.5

1

1.5

2Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 3)

a = /2a = 2



Rule 5: Real-axis intercept and angle of asymptote

5

Two branches (Rule 1) Symmetric (Rule 2)On real axis to the left of the

first pole at -1 (Rule 3) Two zeros at infinity (Rule 4)

8 7 6 5 4 3 2 1 0 15

4

3

2

1

0

1

2

3

4

5Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)

Three branches (Rule 1) Symmetric (Rule 2)On real axis to the left of the first pole

at -1 and the third pole at -3 (Rule 3) Three zeros at infinity (Rule 4)

fia =

Pnite polesP finite zeros

#finite poles#P finite zerosa =

(2m+ 1)

#finite poles#P finite zeros

a = 2 a = /3

SolveXn

1

b zn =Xq

1

b pq



Rule 6: Real axis breakaway and break-in points b

6

3.5 3 2.5 2 1.5 1 0.5 0 0.52

1.5

1

0.5

0

0.5

1

1.5

2Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 2)

a = /2a = 2

Two branches (Rule 1) Symmetric (Rule 2)On real axis to the left of the

first pole at -1 (Rule 3) Two zeros at infinity (Rule 4)

8 7 6 5 4 3 2 1 0 15

4

3

2

1

0

1

2

3

4

5Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)



a = 2 a = /3

b = 2b 1.423

Solve KG (j!x) = 1




7

8 7 6 5 4 3 2 1 0 15

4

3

2

1

0

1

2

3

4

5Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)



a = 2 a = /3

b 1.423

!x 3.32K = 60


What else is the Root Locus telling us

Gain = product of distances to the poles

8

8 7 6 5 4 3 2 1 0 15

4

3

2

1

0

1

2

3

4

5Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)

p 12p 15p 20

p11

K = 60

3.5 3 2.5 2 1.5 1 0.5 0 0.52

1.5

1

0.5

0

0.5

1

1.5

2Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =1

(s+ 1) (s+ 2)

p2p 2

K = 2


The zeros are pulling the Root Locus

Because of Rule 4 Therefore, adding a zero makes the response

faster stable

9

16 14 12 10 8 6 4 2 0 23

2

1

0

1

2

3Root Locus

m (seconds1)

jt (s

econ

ds1

)

6 5 4 3 2 1 0 115

10

5

0

5

10

15Root Locus

m (seconds1)

jt (s

econ

ds1

)

G(s) =s+ 5

(s+ 1) (s+ 3)G(s) =

s+ 5

(s+ 1) (s+ 2) (s+ 3)


Practice 1: Sketch the Root Locus

Nise Figure P8.2

10



Practice 2:Are these Root Loci valid? If not, correct them

Nise Figure P8.1

11


Classification of feedback compensators

compensator plant C(s) +

K(s + z

Gp(s)

R(s)

Proportional-Derivative (PD) speeds up response improves stability (RL moves to the left) may worsen steady state error noisy


K Gp(s) C(s) R(s)

plantcompensator +

K s + z s

Gp(s) C(s) R(s)

plantcompensator +

K(s + z

+ K2 s

Gp(s) C(s)

R(s) plantcompensator

+

Proportional (P) Proportional-Integral (PI)

Proportional-Integral-Derivative (PID)

can meet one transient specification (e.g. rise time) other specifications (e.g. overshoot) unmet steady state error

other specifications (e.g. overshoot) unmet decreases stability (RL moves to the right)

can meet one specification (e.g. rise time) zero steady-state error

complete transient specification (rise time & overshoot) no steady-state error

2.004 Spring 13 Lecture 13 Tuesday, Mar. 5, 2013 1

Compensator rules of thumb Nise Figure 9.1Nise Figure 9.7

Integral action eliminates steady-state error; but, by itself, the integrator slows down the response;

therefore, a zero (derivative action) speeds the response back up to match the response speed of the uncompensated system

Derivative action speeds up the transient response; it may also improve the steady-state error; but differentiation is a noisy process

(we will deal with this later in two ways: the lead compensator and the PID controller)



Example



Evaluating different PD controllers Nise Figure 9.15Gc(s) = K(s + 3)

Gc(s) = K

K=23.7 %OS=23.2 Tr=0.688 sec

Gc

K=51.4 %OS=25.4 Tr=0.197 sec

Gc(s) = K(s + 4)

(proportional)

(s) = K(s + 2) (pole-zero

cancellation)

K=35.3 %OS=27.5 Tr=0.236 sec

K=20.86 %OS=27.2 Tr=0.305 sec



Example: speeding up the response

The closed-loop poles shown here are with proportional control, designed for



z


135

%O

S=constant

PD

2 0

j

0.3162 s (s + 2)

K(s + z

)

pc

pc+




PD

K(s + z

) 0.3162 s (s + 2)




PD

PD compensator

K(s + z

) 0.3162 s (s + 2)

zero

closed-loop poles

K=6.325 gives %OS=6.7; Ts=1.86sec John Wiley & Sons. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.


Experiments

Closed-loop position control with derivative control action:

Experiment #1: P and PD Control of Position

Experiment #2: Compare your results with a Simulink Simulation

Deliverables:

Properly annotated plots showing your results

Comments and discussions on your observations and results

3/5/2013 2.004 Spring 13' 11

Precision Position Control Example: Scanning Electron Microscope (SEM) In Semiconductor Fabrication Process

300 mm diameter wafer 45 nm feature size

Control Console

Vacuum Chamber

Diameter of a human hair 0.1 mm = 1e5 nm

200,000x magnification

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12

Position Sensing Using Encoder and EDAC2

Reset Button

Encoder Types: Incremental (Relative): Only the relative

position of the shaft is known. No absolute 0 position.

Absolute: Unique code for each shaft position (e.g., by adding a reference input to an incremental encoder).

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13

P, PI and PD Controllers

P Controller

Encoder Kpos

(t)

R(s)

R(s)


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14

Plant Transfer Functions

Now we are dealing with POSITION CONTROL, DO NOT use the velocity transfer function.

(s) KaKm/N =TF voltage to velocity: Vc(s) Jeqs + Beq

(s) 1position to velocity: = (s) s

(s) 1 (s) KaKm/NTF voltage to position: Gp(s) = = = Vc(s) s Vc(s) s (Jeqs +Beq)

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Comparison of Closed-Loop Transfer Functions

Let

(s) Js2 +Bs +KKpKpos

pKKP Control: GCL(s) = = R(s)

Common form s 2 + 2!ns + !2 n

(s)PD Control: GCL(s) = R(s)

= K (Kp +Kds)

Js2 + (B +KdKKpos) s +KKpKpos

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System Parameters

J 0.03 N-m2

B 0.014 N-m-s/rad

Ka = 2.0 A/V

Km 0.0292 N-m/A V s )( 1rev Kt = (0.016 )(60 ) = 0.153 V/(rad/s) rev/min min 2rad

44 N = = 0.244180

Ke 1.5 V/rad 17

2nd Order System Poles


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18

Control Action Comparison

P improve speed but with steady-state error D improve stability but sensitive to noise I improve steady state error but with less stability, overshoot, longer transient,

integrator windup (we will discuss PI and PID control next week)

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Procedure EX1 Connect the computer-based as before except we need to use the EDAC2

instead of ETACH2. The set-up is very similar. Install one magnet. Important: ALWAYS RESET EDAC2 BEFORE EACH TRIAL (The flywheel could go out of control, be ready to stop the loop at any time.)

Before starting the experiment, spin the flywheel so that the position mark on the flywheel is observable and define an initial position.

Set the function generator to output a DC signal with 1.0 V offset. Set Kp to 2, 3, 4 and start experiment. Record a transient response for each case.

Spin the wheel back to its initial position. Set Kp to 2 and Kd to 1. Repeat the experiment. Try two other (reasonable) combinations of Kp and Kd control parameters. Compare P and PD control results.

Select your favorite combination of PD control parameters, run experiment as before after the plant reaches steady state, continuously change your DC offset by pressing the up or down button; observe controlling of the plant.

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Procedure EXP2

Define a Matlab SISO model to represent the flywheel with voltage as input and position as output. Simulate a P control with Kp = 2.

Change your controller design to PD. Choose 2 combinations of Kp and Kd values from EXP1 and run simulation.

Compare your simulation with results from EXP1 and comment on agreement or discrepancy between theory and experiment.

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Lecture 14 Thursday, March 72.04A Spring 13

Improving the steady-state error: simple integratorIntegrator as a Compensator:

Eliminates the steady-state error,since it increases the system Type;

however, our desirable closed-looppole A is no longer on the

root locus;

this is because the new pole at s=0changes the total angular

contributions to A so that the180 condition is no longer

satisfied.

This means that our desirabletransient response characteristics

that would have been guaranteed by Aare no longer available L

Nise Figure 9.3

1



Improving the steady-state error: PI controllerIdeal Integral Compensator

(or Proportional-Integral Compensator):

Includes a zero on the negative realaxis but close to the integrators pole

at the origin. The zero

has approximately the sameangular contribution to A as theintegrators pole at the origin;therefore, the two cancel out;

moreover, it contributes the samemagnitude to the pole at A, so

A is reached with the same feedbackgain K.

The net effect is that we have fixedthe steady-state error without

affecting the transient response JNise Figure 9.3

2


PI compensator TF: K1 +K2s

=K1

s+ K2

K1

s

=K(s+ z)

s

where K K1, z K2K1


Implementing the PI controller

Nise Figure 9.8

Another implementation is the lag compensator, which you can learn aboutin more advanced classes (e.g., 2.14.)

3



Example (Nise 9.1)

Nise Figure 9.4

4



Steady-state and transients with the PI controller

Nise Figure 9.7

Steady-state error=0FIXED!

Transient:UNCHANGED!

Nise Figure 9.5

UNCOMPENSATED(Proportional controller)

Nise Figure 9.6

COMPENSATED(PI controller)

5


Lecture 14 Thursday, March 72.04A Spring 13 6

In-class experiments

Elimination of steady-state error through the use of integral (I), and proportional plus integral (PI) control. Experiment #1: Pure Integral Control & PI Control Experiment #2: Compare your results with a Simulink Simulation

Hand in: Properly annotated plots showing your results. Comments and discussions on your observations and results. (How do the

P, I, and PI control actions look and feel like?)


(Another) Limitation of P-control Steady-state error Kp is limited by the saturation limit of the system Large Kp may amplify noise and disturbances and lead to instability


P Control Steady-Sate Errors

In real world a set-point profile is often more complex than a simple step input.


PI Controller

P Controller


2.04A Spring 13 Lecture 14 Thursday, March 7 10

P, I, and PI Comparison

P Control: steady-state error I Control: overshoot, longer transient, integrator windup

Set-point

Vss = lims!0 =

A

SGcl(s) = A

KpKKtJs+B +KpKKt

Gcl(s) =Vt(s)

R(s)=

(Kps+Ki)KKtJs2 + (B +KpKKt)s+KiKKt

Vss = lims!0

A

sGcl(s)

= A

Vt(s)

R(s)=

(Kps+Ki)KKtJs2 + (B +KpKKt)s+KiKKt

= A

K = KaKm

1

N


Comparison of Closed-Loop Transfer Functions

Let

P Control:

PI Control: A real zeroComplex pole pair

2.04A Spring 13 Lecture 14 Thursday, March 7 12

Procedures

EXP1: Connect the computer-based controller same as Lab 5, install one magnet Make function generator to produce a DC with offset = 1.0 V. Set Kp = 0, Ki = 0.5

and 0.2. Make sure power amplifier break is turned on and record system response for 4 to 5 seconds for each case.

Set Kp = 2, and Ki = 1 and record your response data. Compute steady-state error. Compare this result with pure integral control and pure proportional control from Lab 5. Change Kp and Ki both to 3, repeat experiment. Discuss your results, pay attention to the motion of flywheel when the in square wave drops down to 0. What effect does a integrator have on system performance?

EXP2: Modify the controller design in your Simulink model from Lab 5 to have a

form of ; set KKp +Kis p to 2 and Ki to 1 and run simulation for 5 seconds. Comments and discussions on your observations and results. (How do the P, I,

and PI control actions look and feel like?)

2.04A Spring 13 Lecture 14 Thursday, March 7

System Parameters


2.04A Class Project

Tall Building Active Damping

2.04A Spring 13 Lecture 16&17 March 12&14 (T ue-Thu) 1

Problem Wind loading of skyscrapers causes tall building sway.

Upper floor occupants suffer from motion sickness when the building sways in the wind since people are sensitive to accelerations as small as 0.05 m/s2 (0.005 g).

Too much building sway can also lead to long-term structural damage.

The Hancock Tower in Boston had a problem with falling windows. (The Hancock Tower thnow has two passively controlled 300 ton sliding masses on the 58 floor.)

+ = Courtesy of Rob Pongsajapan on flickr. CC-BY

Cartoons sources unknown. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.


Simplified Building Model

We can model a tall building as a singledegree of freedom lumped-parameter system.

2.04A Spring 13 Lecture 16&17 March 12&14 (T ue-Thu) 3

Passive Vibration Damping

Sliding Mass, m2

Spring

Damper

Wind force, One way to stabilize these Fw(t)

tall builds from swaying toomuch during earthquakes or from high winds is to install enormous pendulum weights. When the building sways sideways the pendulum doesn't want to move (inertia) and exerts a pull in the opposite direction.

Passive Damper (Tuned Mass Damper)

Ground

Lecture 16&17 March 12&14 (T ue-Thu) 4 2.04A Spring 13

Taipei 101 (http://www.taipei-101.com.tw)

Courtesy of Jirka Matousek on flickr. CC-BY Courtesy of Stefan Tan. Used with permission.


Taipei 101

The passive wind damper with a diameter of 5.5 meters and weighting 660 metric tons, is also the largest in the world

Documents

204 spring 2011.pdf