5. MechatronicsWorkshop KCCDay2 Session1-2.pdf

Mechatronics for the 21st Century System Design and Performance Specifications

Mechatronic System Design Process

START HERE Simplifying Assumptions

System Design Concept Engineering Judgment

Past Experience & Experiments

ld r o W l a e R

Apply Laws of Nature

Concept Physical Model

Expected Component and OpenLoop System Response

Identify Model Parameters

Solve Equations: Analytical & Numerical

Re-evaluate Physical Model Assumptions & Parameters no

Predicted Open-Loop System Response

Agreement? yes

Past Experience & Experiments

Control System Design

no Expected Closed-Loop System Response

Concept Mathematical Model

Design & Simulate

Re-evaluate

Predicted Closed-Loop System Response

Agreement?

iS m u la ti o n W o rl d

yes

Improve System Design: Parameters and/or Configuration / Concept

no

Is predicted response acceptable with respect to specifications?

Improve Control Design: Feedback, Feedforward, Observers, Filters

no

yes

Build and Test Physical System Check That System Meets Specifications Evaluate, Iterate, and Improve As Needed

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Mechatronics Master Class Schedule

Day 1

Day 2

Day 3

Session 1

Mechatronics and Innovation

Modeling & Analysis of Dynamic Physical Systems

High-Performance Mechatronic Motion Systems

Session 2

Human-Centered Design

Session 3

Model-Based Design

Session 4

Mechatronic System Design

Automotive Mechatronics

Control System Design: Feedback, Feedforward, & Observers

Web-Handling Mechatronic Applications Fluid Power Mechatronic Applications

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Modeling Engineering Systems


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Engineering System Investigation Process

Engineering System Investigation Process

START HERE

Physical System System Measurement

Physical Model

Modeling The cornerstone of modern engineering practice !

Parameter Identification

Measurement Analysis

Mathematical Model

Mathematical Analysis Comparison: Predicted vs. Measured

Design Changes

YES

Is The Comparison Adequate ?

NO


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• Apply the steps in the process when: – An actual physical system exists and one desires to understand and predict its behavior. – The physical system is a concept in the design process that needs to be analyzed and evaluated. • After recognizing a need for a new product or service, one generates design concepts by using: past experience (personal and vicarious), awareness of existing hardware, understanding of physical laws, and creativity. • The importance of modeling and analysis in the design process has never been more important. Design concepts can no longer be evaluated by the build-and-test approach because it is too costly and time consuming. Validating the predicted dynamic behavior in this case, when no actual physical system exists, then becomes even more dependent on one's past hardware and experimental experience. Modeling Engineering Systems

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• What is a Physical Model? – In general, a physical model is an imaginary physical system which resembles an actual system in its most significant features, but which is simpler, more ideal, and is thereby more amenable to analytical studies. – There is a hierarchy of physical models of varying complexity possible. – The difference between a physical system and a physical model is analogous to the difference between the actual physical terrain and a map. – Not oversimplified, not overly complicated - a slice of reality. – The very crux of engineering analysis and the hallmark of every successful engineer is the ability to make shrewd and viable approximations which greatly simplify the system and still lead to a rapid, reasonably accurate prediction of its behavior. Modeling Engineering Systems

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• What is a Mathematical Model? – We apply the Laws of Nature to the Physical Model, NOT to the Physical System, to obtain the Mathematical Model. – What Laws of Nature? • Newton’s Laws of Motion • Maxwell’s Equations of Electromagnetism • Laws of Thermodynamics – Once we have the Mathematical Model of the Physical Model, we then solve the equations, either analytically or numerically, or both to get the greatest insight, to predict how the Physical Model behaves. This predicted behavior must then be compared to the actual measured behavior of the Physical System. Modeling Engineering Systems

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Balance: The Key to Success

Computer Simulation Without Experimental Verification Is At Best Questionable, And At Worst Useless! Modeling Engineering Systems

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Balance in Engineering is the Key! • The essential characteristic of an engineer and the key tosets success is a balance between the following of skills: – modeling (physical and mathematical), analysis (closed-form and numerical simulation), and control design (analog and digital) of dynamic physical systems – experimental validation of models and analysis and understanding the key issues in hardware implementation of designs


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• Dynamic Physical System – Any collection of interacting elements for which there are cause-and-effect relationships among the timedependent variables. The present output of the system depends on past inputs.

• Analysis of the Dynamic Behavior of Physical Systems – Cornerstone of modern technology – More than any other field links the engineering disciplines

• Purpose of a Dynamic System Investigation

– Understand & predict the dynamic behavior of a system – Modify and/or control the system, if necessary


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• Essential Features of the Study of Dynamic Systems – Deals with entire operating machines and processes rather than just isolated components. – Treats dynamic behavior of mechanical, electrical, electromechanical, fluid, thermal, chemical, and mixed systems. – Emphasizes the behavioral similarity between systems that differ physically and develops general analysis and design tools useful for all kinds of physical systems. – Sacrifices detail in component descriptions so as to enable understanding of the behavior of complex systems made from many components. Modeling Engineering Systems

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– Uses methods which accommodate component descriptions in terms of experimental measurements, when accurate theory is lacking or is not cost-effective, and develops universal lab test methods for characterizing component behavior. – Serves as a unifying foundation for many practical application areas, e.g., vibrations, measurement systems, control systems, acoustics, vehicle dynamics, etc. – Offers a wide variety of computer software to implement its methods of analysis and design.


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Electro-Dynamic Vibration Exciter Physical System vs. Physical Model


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• This "moving coil" type of device converts an electrical command signal into a mechanical force and/or motion and is very common, e.g., vibration shakers, loudspeakers, linear motors for positioning heads on computer disk memories, and optical mirror scanners. • In all these cases, a current-carrying coil is located in a steady magnetic field provided by permanent magnets in small devices and electrically-excited wound coils in large ones. • Two electromechanical effects are observed in such configurations: – Generator Effect: Motion of the coil through the magnetic field causes a voltage proportional to velocity to be induced into the coil.

– Motor Effect: Passage of current through the coil causes it to experience a magnetic force proportional to the current. Modeling Engineering Systems

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• Flexure Kf is an intentional soft spring (stiff, however, in the radial direction) that serves to guide the axial motion of the coil and table. • Flexure damping Bf is usually intentional, fairly strong, and obtained by laminated construction of the flexure spring, using layers of metal, elastomer, plastic, and so on. • The coupling of the coil to the shaker table would ideally be rigid so that magnetic force is transmitted undistorted to the mechanical load. Thus Kt (generally large) and Bt (quite small) represent parasitic effects rather than intentional spring and damper elements. • R and L are the total circuit resistance and inductance, including contributions from both the shaker coil and the amplifier output circuit. Modeling Engineering Systems

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Mathematical Modeling: Laws of Nature applied to Physical Mode Equations of Motion

M t  x t −= K−f x t +Bf x t− B+t (x c −x t ) K t (x c M c  xc = − Bt−(xc − x t ) −K t+(x c

State-Space Representation

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

Li = ei − Ri − Kx c

0 1 0 ⎡ ⎢ − K− K− − B BK t f t ⎢ f ⎡ x t ⎤ ⎡t MM ⎢ Mt ⎢  ⎥ ⎢ tt xt ⎢ ⎢ ⎢ 0 00 ⎥ ⎢ x c = ⎢ ⎥ ⎢ B ⎢ K ⎢ ⎥ −K ⎢ t t t c x ⎢  ⎢ Mc ⎥ ⎢ Mc M ⎢ ⎢⎣ i ⎥⎦ ⎢⎣c ⎢ 0 0 0 ⎢⎣


x t ) Ki

0 ⎤

0 Bt Mt 1

−B t

Mc

−K L

⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ K ⎥ Mc ⎥ ⎥ −R ⎥ L ⎥⎦

⎡0⎤ ⎢0⎥ x t ⎢ ⎥ ⎢0⎥e xc + ⎢0⎥ xt

x c i

⎢⎢ 1 ⎥⎥ ⎢⎣ L ⎥⎦ K. Craig

Analytical Frequency Response Plots

Typical Parameter Values (SI= Units) L 0.0012 R = 3.0 K = 190 Kt = 8.16E8 Bt = 3850 Kf = 6.3E5 Bf = 1120 Mc = 1.815

resonance due to flexures

parasitic resonance due to coil-table coupling

Mt = 6.12


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Experimental Frequency Response Plot

B&K PM Vibration Exciter Type 4809 Modeling Engineering Systems

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Electro-Pneumatic Transducer

Σ


Σ

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This system can be collapsed into a simplified approximate overall model when numerical values are properly chosen:


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• It is interesting to note here that while the block diagram shows one input for the system, i.e., command voltage ein , there are possible undesired inputs that must also be considered. • For example, the ambient temperature will affect the electric coil resistance, the permanent magnet strength, the leaf-spring stiffness, the damper-oil viscosity, the air density, and the dimensions of the mechanical parts. All these changes will affect the system output pressure po in some way, and the cumulative effects may not be negligible.


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Temperature Feedback Control System: A Larger-Scale Engineering System Desired Temperature (set with RV)

RV

Block Diagram of an Temperature Control System eE

Σ

Bridge Circuit

pM

eM Amplifier

Controller

ElectroPneumatic Transducer

xV Valve

T Chemical Process

RC Actual Temperature (measured with RC)


Thermistor

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• Note that in the block diagram of this system, the detailed operation of the electropneumatic transducer is not made apparent; only its overall input/output relation is included. • The designer of the temperature feedback-control dynamic system would consider the electropneumatic transducer an off-the-shelf component with certain desirable operating characteristics. • The methods of system dynamics are used by both the electropneumatic transducer designer and the designer system. of the larger temperature feedback-control


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Physical Modeling Less Real, Less Complex, More Easily Solved

TruthModel

DesignModel

More Real, More Complex, Less Easily Solved

Hierarchy Of Models Always Ask: Why Am I Modeling? Modeling Engineering Systems

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• Physical System – Define the physical system to be studied, along with the system boundaries, input variables, and output variables.

• Physical System to Physical Model

– In general, a physical model is an imaginary physical system which resembles an actual system in its salient features, but which is simpler, more ideal, and is thereby more amenable to analytical studies. There is a hierarchy of physical models of varying complexity possible.

– Not oversimplified, not overly complicated - a slice of reality.


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– The astuteness with which approximations are made at the onset of an investigation is the very crux of engineering analysis. – The ability to make shrewd and viable approximations which greatly simplify the system and still lead to a rapid, reasonably accurate prediction of its behavior is the hallmark of every successful engineer. – What is the purpose of the model? Develop a set of performance specifications for the model based on the specific purpose of the model. What features must be included? How accurately do they need to be represented?


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– The challenges to physical modeling are formidable: • Dynamic behavior of many physical processes is complex. • Cause and effect relationships are not easily discernible. • Many important variables are not readily identified. • Interactions among the variables are hard to capture.

• Engineering Judgment is the Key!


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– In modeling dynamic systems, we consider matter and energy as being continuously, though not necessarily uniformly, distributed over the space within the system boundaries. – This is the macroscopic or continuum point of view. We consider the system variables as quantities which change continuously from point to point in the system as well as with time and this always leads to a distributed-parameter physical model which results in a partial differential equation mathematical model. – Models in thisatmost general form level. behave most like the real systems the macroscopic


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– Because of the mathematical complexity of these models, engineers find it necessary and desirable to work with less exact models in many cases. – Simpler models which concentrate matter and energy into discrete lumps are called lumped-parameter physical models and lead to ordinary differential equation mathematical models. – An understanding of the difference between distributed-parameter and lumped-parameter models is vital to the intelligent formulation and use of lumped models. – The time-variation of the system parameters can be random or deterministic, and if deterministic, variable or constant. Modeling Engineering Systems

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• Comments on Truth Model vs. Design Model – In modeling dynamic systems, we use engineering judgment and simplifying assumptions to develop a physical model. The complexity of the physical model depends on the particular need, and the intelligent use of simple physical models requires that we have some understanding of what we are missing when we choose the simpler model over the more complex model. – The truth model is the model that includes all the known relevant characteristics of the real system. This model isdesign, often too forinuse in engineering butcomplicated is most useful verifying design changes or control designs prior to hardware implementation. Modeling Engineering Systems

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– The design model captures the important features of the process for which a control system is to be designed or design iterations are to be performed, but omits the details which you believe are not significant. – In practice, you may need a hierarchy of models of varying complexity: • A very detailed truth model for final performance evaluation before hardware implementation • Several less complex truth models for use in evaluating particular effects • One or more design models


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Approximation

Mathematical Simplification

Neglect small effects

Reduces the number and complexity of the equations of motion

Assume the environment is independent of system motions

Reduces the number and complexity of the equations of motion

Replace distributed characteristics with appropriate lumped elements

Leads to ordinary (rather than partial) differential equations

Assume linear relationships

Makes equations linear; allows superposition of solutions

Assume constant parameters

Leads to constant coefficients in the differential equations

Neglect uncertainty and noise

Avoids statistical treatment


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• Neglect Small Effects – Small effects are neglected on a relative basis. In analyzing the motion of an airplane, we are unlikely to consider the effects of solar pressure, the earth's magnetic field, or gravity gradient. To ignore these effects in a space vehicle problem would lead to grossly incorrect results!

• Independent Environment – In analyzing the vibration of an instrument panel in a vehicle, we assume that the vehicle motion is independent of the motion of the instrument panel.


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• Lumped Characteristics – In a lumped-parameter model, system dependent variables are assumed uniform over finite regions of space rather than over infinitesimal elements, as in a distributed-parameter model. Time is the only independent variable and the mathematical model is an ordinary differential equation. – In a distributed-parameter model, time and spatial variables are independent variables and the mathematical model is a partial differential equation. – Note that elements in a lumped-parameter model do not necessarily correspond to separate physical parts of the actual system.


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– A long electrical transmission line has resistance, inductance, and capacitance distributed continuously along its length. These distributed properties are approximated by lumped elements at discrete points along the line. – It is important to note that a lumped-parameter model, which may be valid in low-frequency operations, may not be valid at higher frequencies, since the neglected property of distributed parameters may become important. For example, the mass of a spring may be neglected in low-frequency operations, but it becomes an important property of the system at high frequencies.


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• Linear Relationships – Nearly all physical elements or systems are inherently nonlinear if there are no restrictions at all placed on the allowable values of the inputs, e.g., saturation, dead-zone, square-law nonlinearities. – If the values of the inputs are confined to a sufficiently small range, the srcinal nonlinear model of the system may often be replaced by a linear model whose response closely approximates that of the nonlinear model. – When a linear equation has been solved once, the solution is general, holding for all magnitudes of motion.


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– Linear systems also satisfy conditions for superposition. – The superposition property states that for a system initially at rest with zero energy: (1) multiplying the inputs by any constant multiplies the outputs by the same constant, and (2) the response to several inputs applied simultaneously is the sum of the individual responses to each input applied separately.

• Constant Parameters – Time-varying systems are ones whose characteristics change with time. – Physical problems are simplified by the adoption of a model in which all the physical parameters are constant. Modeling Engineering Systems

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• Neglect Uncertainty and Noise – In real systems we are uncertain, in varying degrees, about values of parameters, about measurements, and about expected inputs and disturbances. Disturbances contain random inputs, called noise, which can influence system behavior. – It is common to neglect such uncertainties and noise and proceed as if all quantities have definite values that are known precisely.


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• Summary – The most realistic physical model of a dynamic system leads to differential equations of motion that are: • nonlinear, partial differentialparameters equations with timevarying and space-varying • These equations are the most difficult to solve. – The simplifying assumptions discussed lead to a physical model of a dynamic system that is less realistic and to equations of motion that are: • linear, ordinary differential equations with constant coefficients • These equations are easier to solve and design with. Modeling Engineering Systems

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Classes of Systems


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Pure and Ideal Elements vs. Real Devices • A pure element refers to an element (spring, damper, inertia, resistor, capacitor, inductor, etc.) which has only the named attribute. • For example, a pure spring element has no inertia or friction and is thus a mathematical model (approximation), not a real device. • The term ideal, as applied to elements, means linear, that is, the input/output relationship of the element is linear, or straight-line. The output is perfectly proportional to the input. • A device can be pure without being ideal and ideal without being pure. Modeling Engineering Systems

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• From a functional engineering viewpoint, nonlinear behavior may often be preferable, even though it leads to difficult equations. • Why do we choose to define and use pure and ideal elements when we know that they do not behave like the real devices used in designing systems? – Once we have defined these pure and ideal elements, we can use these as building blocks to model real devices more accurately. • For example, if a real spring has significant friction and mass, we model it as a combination of pure/ideal spring, mass, and damper elements, which may come quite close in behavior to the real spring.


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Physical Model of a Real Spring


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• There are three basic types of building blocks, two that can store energy and one that dissipates energy. – All Mechanical Systems have three attributes - mass, springiness, and energy dissipation. The basic elements corresponding to these attributes are inertia (mass), damper (friction), and spring (compliance). – All Electrical Systems have three attributes inductance, resistance, and capacitance. The corresponding basic elements are inductor, resistor, and capacitor.


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Mechanical System Elements • It is hard to imagine any engineering system that does not have mechanical components. • Motion and force are concepts used to describe the behavior of engineering systems that employ mechanical components. • The three basic mechanical building block elements are: – Spring (elasticity or compliance) element – Damper (friction or energy dissipation) element – Mass (inertia) element • There are both translational and rotational versions of these basic building blocks. Modeling Engineering Systems

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• These are passive (non-energy producing) devices. • Driving Inputs are force and motion sources which cause elements to respond. • Each of the elements has one of two possible energy behaviors: – stores all the energy supplied to it – dissipates all energy into heat by some kind of “frictional” effect • Spring stores energy as potential energy. • Mass stores energy as kinetic energy. • Damper dissipates energy into heat. • The Dynamic Response of each element is important.


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• Spring Element – A spring is a fundamental mechanical component found intentionally or unintentionally in almost every mechanical system. – Real-world spring is neither pure nor ideal. – Real-world spring has inertia and friction. – Pure spring has only elasticity - it is a mathematical model, not a real device. – Some dynamic operation requires that spring inertia and/or damping not be neglected. – Ideal spring is linear. However, nonlinear behavior may often be preferable and give significant performance advantages. F = K(x 2 − x1 ) Modeling Engineering Systems

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Pure & Ideal Spring Element Differential Work Done f ) dx =( (K x)=dx

s

Total Work Done

∫

K x)d=x

x0

0

(

s

=


K s x 02 2

=

Cs f 02 2

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• Damping Element – A damper is a mechanical component often found in engineering systems. – A pure damper dissipates all the energy supplied to it, i.e., converts the mechanical energy to thermal energy. – Various physical mechanisms, usually associated with some form of friction, can provide this dissipative action, e.g., • Coulomb (dry friction) damping • Material (solid) damping • Viscous (fluid) damping


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– Shown is a typical mechanical viscous damper. If the mass and springiness of the piston and cylinder are small, then the force will be a function of the relative velocity between the piston and the cylinder. – Forces on the two ends of the damper are exactly equal and opposite at all times (just like a spring); pure springs and dampers have no mass or inertia. This is NOT true for real springs and dampers.

F = B(v 2 − v1 )


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Pure & Ideal Damper Element Step Input Force causes instantly (a pure damper has no of inertia) Step dx/dt a and a Ramp of x Modeling Engineering Systems

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• Mass or Inertia Element – All real mechanical components used in engineering systems have mass. – A designer rarely inserts a component for the purpose of adding inertia; the mass or inertia element often represents an undesirable effect which is unavoidable since all materials have mass. – There are some applications in which mass itself serves a useful function, e.g., flywheels – a flywheel is an energy-storage device and can be used as a means of smoothing out speed fluctuations in engines or other machines.


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– Newton’s Law defines the behavior of mass elements and refers basically to an idealized “point mass”:

∑ forces = (mass)(acceleration) – The concept of rigid body is introduced to deal with practical situations. For pure translatory motion, every point in a rigid body has identical motion. – Real physical bodies never display ideal rigid behavior when being accelerated. – The pure and ideal inertia element is a model, not a real object. Real inertias may be impure (have some springiness and friction) but are very close to ideal. F = Ma Modeling Engineering Systems

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Pure & Ideal Inertia Element Real inertias may be impure (have some springiness and friction) but are very close to ideal.

x f

(D) =

θ

1

1 (D) = T JD 2

MD 2

Inertia Element stores energy as kinetic energy:

Mv 2 2

or

J ω2 2


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Cantilever Beam

Inertia, Energy Dissipation, Compliance – Does it have these attributes? Can you model this real device using the pure and ideal building blocks?


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Ideal vs. Real Sources • External driving agencies are physical quantities which pass from the environment, through the interface into the system, and cause the system to respond. • In practical situations, there may be interactions between the environment and the system; however, we often use the concept of ideal source. • An ideal source (force, motion, voltage, current, etc.) is totally unaffected by being coupled to the system it is driving. • For example, a “real” 6-volt battery will not supply 6 volts to a circuit! The circuit will draw some current from the battery and the battery’s voltage will drop. Modeling Engineering Systems

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Physical Model to Mathematical Model • We derive a mathematical model to represent the physical model, i.e., apply the Laws of Nature to the physical model and write down the differential equations of motion. • The goal is a generalized treatment of dynamic systems, including mechanical, electrical, electromechanical, fluid, thermal, chemical, and mixed systems. – Define System: Boundary, Inputs, Outputs – Define Variables: Through and Across Variables – Write System Relations: Dynamic Equilibrium Relations and Compatibility Relations – Write Constitutive Relations: Physical Relations for Each Element – Combine: Generate State Equations Modeling Engineering Systems

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• Define System – A system must be defined before equilibrium and/or compatibility relations can be written. – Unless physical boundaries of a system are clearly specified, any equilibrium and/or compatibility relations we may write are meaningless.

• Define Variables – Physical Variables • Select precise physical variables (velocity, voltage, pressure, flow rate, etc.) with which to describe the instantaneous state of a system, and in terms of which to study its behavior.


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– Through Variables • Through variables (one-point variables) measure the transmission of something through an element, e.g., – electric current through a resistor – fluid flow through a duct – force through a spring – Across Variables • Across variables (two-point) variables measure a difference in state between the ends of an element, e.g., – voltage drop across a resistor – pressure drop between the ends of a duct – difference in velocity between the ends of a damper Modeling Engineering Systems

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– In addition to through and across variables, integrated through variables (e.g., momentum) and integrated across variables (e.g., displacement) are important.

• Write System Relations – Dynamic Equilibrium Relations • Write dynamic equilibrium relations to describe the balance - of forces, of flow rates, of energy - which must exist for the system and its subsystems. • Equilibrium relations are always relations among through variables, e.g., – Kirchhoff’s Current Law (at an electrical node) – continuity of fluid flow – equilibrium of forces meeting at a point Modeling Engineering Systems

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– Compatibility Relations • Write system compatibility relations to describe how motions of the system elements are interrelated because of the way they are interconnected. • These are inter-element or system relations. • Compatibility relations are always relations among across variables, e.g., – Kirchhoff’s Voltage Law around a circuit – pressure drop across all the interconnected stages of a fluid system – geometric compatibility in a mechanical system


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• Write Physical Relations for Each Element – These relations are called constitutive physical relations as they concern only individual elements or constituents of the system. – They are natural physical laws which the individual elements of the system obey, e.g., • mechanical relations between force and motion • electrical relations between current and voltage • electromechanical relations between force and magnetic field • thermodynamic relations between temperature, pressure, etc.


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– They are relations between through and across variables of each individual physical element. – They may be algebraic, differential, integral, linear or nonlinear, constant or time-varying. – They are purely empirical relations observed by experiment and not deduced from any basic principles.

• Combine System Relations and Physical Relations to Generate System Differential Equations


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• Study Dynamic Behavior – Study the dynamic behavior of the mathematical model by solving the differential equations of motion either through mathematical analysis or computer simulation. – Dynamic behavior is a consequence of the system structure - don’t blame the input! – Seek a relationship between physical model structure and behavior. – Develop insight into system behavior.


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• Comparison: Actual vs. Predicted – Compare the predicted dynamic behavior to the measured dynamic behavior from tests on the actual physical system; make physical model corrections, if necessary.

• Make Design Decisions – Make design decisions so that the system will behave as desired: • modify the system (e.g., change the physical parameters of the system, add a sensor, change the actuator or its location) • control the system (e.g., augment the system, typically by adding a dynamic system called a compensator or controller) Modeling Engineering Systems

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Dynamic System Response


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Dynamic System Response • LTI Behavior vs. Non-LTI Behavior • Solution of Linear, Constant-Coefficient, Ordinary Differential Equations – Classical Operator Method – Laplace Transform Method • Laplace Transform Properties • 1st-Order Dynamic System Time and Frequency Response • 2nd-Order Dynamic System Time and Frequency Response • Dynamic System Response Example Problems Dynamic System Response

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LTI Behavior vs. Non-LTI Behavior • Linear Time-Invariant (LTI) Systems – A frequency-domain is limited to describing elements transfer that are function linear and time invariant – severe restrictions! No real-world system meets them! – Linear Time-Invariant System Properties • Homogeneity: If r(t) → c(t), then kr(t) → kc(t) • Superposition: If r (t) → c (t) and r (t) → c (t), then 2 2 r1(t) + r2(t) → c1(t) 1+ c2(t) 1 • Time Invariance: If r(t) → c(t), then r(t-t1) → c(t-t1) Dynamic System Response

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– Comments on the Principle of Superposition • The principle of superposition states that if the system has an input that can be expressed as the sum of signals, then the response of the system can be expressed as the sum of the individual responses to the respective signals. Superposition applies if and only if the system is linear. • Using the principle of superposition, we can solve for the system responses to a set of elementary signals. We are then able to solve for the response to a general signal simply by decomposing the given signal into sum of the elementary components and,aby superposition, concluding that the response to the general signal is sum of the responses to the elementary signals. Dynamic System Response

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• In order for this process to work, the elementary signals need to be sufficiently rich that any reasonable signal can be expressed as a sum of them, and their responses have to be easy to find. The most common candidate for elementary signals for use in linear systems are the impulse and the exponential. • The unit impulse δ(t) is a pulse of zero duration and infinite height. The area under the unit impulse (its strength) is equal to one.

∫

τ+

τ−

δ(t − τ) dt=

1

• The impulse δ(t) is defined by δ(t-τ) = 0 for all t ≠ τ. It has the property that if f(t) is continuous at t = τ, then ∞ Dynamic System Response

∫

−∞

f ( τ) δ(t

−τ) d τ =f (t)

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• The impulse is so short and so intense that no value of f matters except over the short range where the δ occurs. The integral can be viewed as representing the function f as a sum of impulses. • To find the response to an arbitrary input, the principle of superposition tells us that we need only find the response to a unit impulse. • For a linear, time-invariant system, the impulse response, i.e., the response at time t to an impulse applied at time τ, can be expressed as h(t – τ) because the response at time t to an input applied at depends only the difference between thetime timeτ the impulse is on applied and the time we are observing the response. Dynamic System Response

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• For linear, time-invariant systems with input u(t), the superposition integral, called the convolution integral, is: ∞

y(t) =

∞

τ )h(t− τ )τd=

−∞ u(

∫

−∞

∫

−u(t τ τ)h(τ=) d∗

u h

• Here u(t) is the input to the system and h(t-τ) is the impulse response of the system.


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y + ky =δ =u

– Convolution Example

∫

0+

0

−

y dt + k

(t)

∫

0+

0−

= y(0− )

y dt =

∫

0+

0−

δ(t)

⎡⎣ y(0+ ) − y(0− ) ⎤⎦ + k(0) = 1 y(0+ ) = 1

Assume y = Aest

y + ky = 0

+ kAe = 0 s = −k A =1 y(t) = h(t) = e − kt for t > 0 Ase

st

st

y(0 + ) = 1

impulse response y(t) = h(t) = e − kt 1(t) The response to a general input u(t) is

1(t)

0 t<0 1 t≥0

unit step function Dynamic System Response

given by the convolution of this impulse response and the input:

∫ =∫

y(t) =

∞

e − kt 1(t) u(t − ττ ) d

−∞ ∞ − kt

0

e

u(t − τ τ) d

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– A consequence of convolution is the concept of transfer ∞ function. y(t) = h( τ )u(t − τ τ) d Convolution Integral

∫ = ∫ τh( )τ e −∞ ∞ −∞ ∞

s( t −τ )

stτ − s

d

=

Input st

est

∞τ −

s

= ∫−∞τh( ) e =eτ d τe−∞ ∫ τ h( ) e ∞ = est H(s) where =H(s) τ τ ∫−∞ h(

d ) e − sτ d

– Note that both the input and output are exponential time functions and that the output differs from the input only in the amplitude H(s). – The function H(s) is the transfer function from the input to output of the system. It is the ratio of the Laplace transform of the output to the Laplace transform of the input assuming all initial conditions are zero. Dynamic System Response

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– If the input is the unit impulse function δ(t), then y(t) is the unit impulse response. The Laplace transform of δ(t) is 1 and the Laplace transform of y(t) is Y(s), so Y(s) = H(s). – The transfer function H(s) is the Laplace transform of the unit impulse response h(t) where the Laplace transform is defined by:

∫ =∫

H(s) =

∞

−∞ ∞

0

h(t) e − st dt

h(t) e − st dt since h(t) = 0f or t < 0

– Thus, if one wishes to characterize a linear timeinvariant system, one applies a unit impulse, and the resulting response is a description (inverse Laplace transform) of the transform function. Dynamic System Response

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– A common way to use the exponential response of a linear time-invariant system is in finding the frequency response. A – Euler’s relation is: A cos(ωt) = (e jωt + e − jωt ) 2 – Let s = jω and, by superposition, the response to the sum of these two exponentials which make up the cosine signal, is the sum of the responses: A y(t) = ⎡⎣ H(ω j )e+jωt −ωH( j )e − jωt ⎤⎦ 2 – The transfer function H(jω) is a complex number that can be expressed in polar form (magnitude and phase form) as H(jω) = M(ω)ejΦ(ω). A t ) ω +φ ⎤⎦ ω +AM φ cos( t ) y(t) = M ⎡⎣ e j( ωt++φ) e− j(= 2 Dynamic System Response

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– This means that if a system represented by the transfer function H(s) has a sinusoidal input with magnitude A, the output will be sinusoidal at the same frequency with magnitude AM and will be shifted in phase by the angle Φ. – The response of a linear time-invariant system to a sinusoid of frequency ω is a sinusoid with the same frequency and with an amplitude ratio equal to the magnitude of the transfer function evaluated at the input frequency. The phase difference between input and output signals is given by the phase of the transfer function evaluated at the input frequency. The magnitude ratio and phase difference can be computed from the transfer function or measured experimentally. Dynamic System Response

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– Transfer Functions, the basis of classical control theory, require LTI systems, but no real-world system is completely LTI. There are regions of nonlinear operation and often significant parameter variation. – Examples of Linear Behavior • addition, subtraction, scaling by a fixed constant, integration, differentiation, time delay, and sampling – No practical control system is completely linear and most vary over time. – LTI systems can be represented completely in the frequency domain. Non-LTI systems can have frequency response plots, however, these plots change depending on system operating conditions. Dynamic System Response

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• Non-LTI Behavior – Non-LTI behavior is any behavior that violates one or more of the three criteria for an LTI system. – Within nonlinear behavior, an important distinction is whether the variation is slow or fast with respect to the loop dynamics. • Slow Variation – When the variation is slow, the nonlinear behavior may be viewed as a linear system with parameters that vary during operation. – The dynamics can still be characterized effectively with a transfer function. However, the frequency response plots will change at different operating points. Dynamic System Response

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• Fast Response – If the variation of the loop parameter is fast with respect to the loop dynamics, the situation becomes more complicated. Transfer functions cannot be relied upon for analysis. – The definition of fast depends on the system dynamics. • Fast vs. Slow – The line between fast and slow is determined by comparing the rate at which the parameter changes to the bandwidth of the control system. If the parameter variation occurs over a period of time no faster than 10 times the control loop settling time, the effect can be considered slow for most applications. Dynamic System Response

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• Dealing with Nonlinear Behavior – Nonlinear behavior can usually be ignored if the changes in parameter values effect the loop gain by no more than about 25%. A variation this small will be tolerated by systems with reasonable margins of stability. – If a parameter varies more than that, there are at least three courses of action: • Modify the plant • Tune for the worst-case conditions • Compensate for the nonlinearity of the control loop


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• Modify the Plant – Modifying the plant to reduce the parameter variation is the most straightforward solution to nonlinear behavior. It cures the problem without adding complexity to the controller or compromising system performance. – This solution is commonly employed in the sense that components used in control systems are generally better (closer to LTI) than components used in open-loop systems. – Enhancing the LTI behavior of a loop component can increase its cost significantly. Components for control systems are often more expensive than open-loop components. Dynamic System Response

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• Tune for the Worst-Case Conditions – Assuming that the variation from the non-LTI behavior is slow with respect to the control loop, its effect is to change gains in the control loop. In this case, the operating conditions can be varied to produce the worst-case gains while tuning the control system. Doing so will ensure stability for all operating conditions. – Tuning the system for worst-case operating conditions generally implies tuning the proportional gain of the inner loop when the plant gain is at its maximum. This ensures that the inner loop will be stable in all conditions; parameter variation will only lower the loop gain, which will reduce responsiveness but will not cause instability. Dynamic System Response

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– The other loop gains (inner loop integral and the outer loops) should be stabilized when the plant gain is minimized. This is because minimizing the plant gain reduces the inner loop response; this will provide the maximum phase lag to the outer loops and again provides the worst case for stability. – So tune the proportional gain with a high plant gain and tune the other gains with a low plant gain to ensure stability in all conditions. – The penalty for tuning for worst case is the reduction in responsiveness. Consider the proportional gain. Because the proportional term is tuned with the highest plant gain, the loop gain will be reduced at operating points where the plant gain is low. – In general, you should expect to lose responsiveness in proportion to plant variation. Dynamic System Response

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• Compensate in the Control Loop – Compensating for the nonlinear behavior in the controller requires that a gain equal to the inverse of the non-LTI behavior be placed in the loop. – This is called gain scheduling. By using gain scheduling, the impact of the non-LTI behavior is eliminated from the control loop. – Gain scheduling requires that the non-LTI behavior be slow with respect to any transfer functions between the non-LTI component and the scheduled gain. – This is a less onerous requirement than being slow with respect to the loop bandwidth because the loop components are always much faster than the loop itself. Dynamic System Response

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– Gain scheduling assumes that the non-LTI behavior can be predicted to reasonable accuracy (generally ± 25%) based on information available to the controller. This is often the case. – Many times, a dedicated control loop will be placed under the direction ofcontroller a larger system more flexible system can becontroller. used to The accumulate information on a changing gain and then modify gains inside dedicated controllers to affect the compensation. – The chief shortcoming of gain scheduling via the system controller is limited speed. The system controller may be unable to keep up with the fastest plant variations. Still this solution is commonly employed because of the controller’s higher level of flexibility and broader access to information. Dynamic System Response

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Solution of Linear, Constant-Coefficient Ordinary Differential Equations • A basic mathematical model used in many areas of engineering is the linear, ordinary differential equation with constant coefficients: an

dnqo dt

n

+ a n −1

b

d n −1q o dt

++ n −1

+

d mqi b+ m dt m

dq a1= o dt

m −1

d m−1q i dt

++ m −1

a 0qo b+ 

dq i bq 1 dt

0

i

• qo is the output (response) variable of the system • qi is the input (excitation) variable of the system • an and bm are the physical parameters of the system Dynamic System Response

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• Straightforward analytical solutions are available no matter how high the order n of the equation. • Review of the classical operator method for solving linear differential equations with constant coefficients will be useful. When the input qi(t) is specified, the right hand side of the equation becomes a known function of time, f(t). • The classical operator method of solution is a three-step procedure: – Find the complimentary (homogeneous) solutionqoc for the equation with f(t) = 0. – Find the particular solution qop with f(t) present. – Get the complete solution qo = qoc + qop and evaluate the constants of integration by applying known initial conditions. Dynamic System Response

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• Step 1 – To find qoc, rewrite the differential equation using the differential operator notation D = d/dt, treat the equation as if it were algebraic, and write the system characteristic equation as: a n Dn

+ a n −1D n+−1 + + a=1D

a0

0

– Treat this equation as an algebraic equation in the unknown D and solve for the n roots (eigenvalues) s1, s2, ..., sn. Since root finding is a rapid computerized operation, we assume all the roots are available and now we state rules that allow one to immediately write down qoc.


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q oc

– Real, unrepeated root s1:

= c1es t 1

– Real root s2 repeated m times: q oc

= c0 e+s t 2

s2 t c+ 1 te

+c +2 t 2 es t 2



c m t m es2 t

– When the a0 to an in the differential equation are real numbers, any appear alwaysthen come in complex pairs a ± roots ib: that might q oc

= ceat sin(bt + φ)

– For repeated root pairs a ± ib, a ± ib, and so forth, we have: q oc

= c0 eat sin(bt + φ+ 0 )

c1te at sin(bt + φ+

1

)



– The c’s and φ’s are constants of integration whose numerical values cannot be found until the last step. Dynamic System Response

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• Step 2 – The particular solution qop takes into account the "forcing function" f(t) and methods for getting the particular solution depend on the form of f(t). – The method of undetermined coefficients provides a simple method of getting particular solutions for most f(t)'s of practical interest. – To check whether this approach will work, differentiate f(t) over and over. If repeated differentiation ultimately leads to zeros, or else to repetition a finite numberwill of different functions, of then the method work. time


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– The particular solution will then be a sum of terms made up of each different type of function found in f(t) and all its derivatives, each term multiplied by an unknown constant (undetermined coefficient). – If f(t) or one of its derivatives contains a term identical to a term in qoc, the corresponding terms should be multiplied by t. – This particular solution is then substituted into the differential equation making it an identity. Gather like terms on each side, equate their coefficients, and obtain a set of simultaneous algebraic equations that can be solved for all the undetermined coefficients.


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• Step 3 – We now have qoc (with n unknown constants) and qop (with no unknown constants). – The complete solution qo = qoc + qop. – The initial conditions are then applied to find then unknown constants.


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• Certain advanced analysis methods are most easily developed through the use of the Laplace Transform. • A transformation is a technique in which a function is transformed from dependence on one variable to dependence on another variable. Here we will transform relationships specified in the time domain into a new domain wherein the axioms of algebra can be applied rather than the axioms of differential or difference equations. • The transformations used are the Laplace transformation (differential equations) and the Z transformation (difference equations). • The Laplace transformation results in functions of the time variable t being transformed into functions of the frequency-related variable s. Dynamic System Response

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• The Z transformation is a direct outgrowth of the Laplace transformation and the use of a modulated train of impulses to represent a sampled function mathematically. • The Z transformation allows us to apply the frequency-domain analysis and design techniques of continuous control theory to discrete control systems. • One use of the Laplace Transform is as an alternative method for solving linear differential equations with constant coefficients. Although this method will not solve any equations that cannot be solved also by the classical operator method, it presents certain advantages.


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– Separate steps to find the complementary solution, particular solution, and constants of integration are not used. The complete solution, including initial conditions, is obtained at once. – There is never any question about which initial conditions are needed. In the classical operator method, the initial conditions are evaluated at t = 0+, a time just after the input is applied. For some kinds of systems and inputs, these initial conditions are not the same as those before the input is applied, so extra work is required to find them. The Laplace Transform method uses the conditions before the input is applied; these are generally physically known and are often zero, simplifying the work. Dynamic System Response

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– For inputs that cannot be described by a single formula for their entire course, but must be defined over segments of time, the classical operator method requires a piecewise solution with tedious matching of final conditions of one piece with initial conditions of the The Laplace Transform method handles suchnext. discontinuous inputs very neatly. – The Laplace Transform method allows the use of graphical techniques for predicting system performance without actually solving system differential equations. • All theorems and techniques of the Laplace Transform derive from the fundamental definition for the direct Laplace Transform F(s) of the time function f(t): L [ f( t) ] = F(s) =

∫

∞

0


f( t)e − st dt t = >0 s

complex σω=variable + K. Craig

i 32

• This integral cannot be evaluated for all f(t)'s, but when it can, it establishes a unique pair of functions, f(t) in the time domain and its companion F(s) in the s domain. Comprehensive tables of Laplace Transform pairs are available. Signals we can physically generate always have corresponding Laplace transforms. When we use the Laplace Transform to solve differential equations, we must transform entire equations, not just isolatedf(t) functions, so several theorems are necessary for this. • Linearity (Superposition and Amplitude Scaling) Theorem: L [ a1f(1 t) + a 2 f 2]=(t) [


] L[ +a1f( ]1

t) = L +a 2 f 2 (t)

a1F1 (s) a 2 F2 (s)

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• Differentiation Theorem: ⎡ df ⎤ = sF(s) − f (0) L ⎢⎣ dt ⎥⎦

⎡ d 2f ⎤ 2 df L ⎢ 2 ⎥ = s F(s) − sf (0) − (0) dt ⎣ dt ⎦ ⎡dnf ⎤ n df L ⎢ n ⎥ = s F(s) − s n−−1f (0) s−n−−2 (0) dt ⎣ dt ⎦



d n −1f dt

n −1

(

– f(0), (df/dt)(0), etc., are initial values of f(t) and its derivatives evaluated numerically at a time instant before the driving input is applied. • Multiplication by Time: – Multiplication by time corresponds to differentiation in the frequency domain. d Lt[ f (t) ] = − F(s) ds Dynamic System Response

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• Integration Theorem: L ⎡ f (t)dt ⎤ =

⎣∫

L ⎡⎣ f

⎦

(−n)

(t) ⎤⎦ =

)

where f

(−n

F(s) s F(s)

∫∫ (t)

n

s

=

+

f(

−1)

(0)

s n

+∑

f(

−k)

(0)

n − k +1

k =1

s )  f (t)(dt)

n

(

= f and

-0

(t)

f (t)

– Again, the initial values of f(t) and its integrals are evaluated numerically at a time instant before the driving input is applied. • Convolution Theorem: – Convolution in the time domain corresponds to multiplication in the frequency domain. L [ f12(t) ∗ f (t) ] = F1 (s)F2 (s) Dynamic System Response

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• Time Delay Theorem: – The Laplace Transform provides a theorem useful for the dynamic system element called dead time (transport lag) and for dealing efficiently with discontinuous inputs. u(t) = 1.0 ⇒ t > 0 u(t) = 0 ⇒ t < 0

⇒ u(t − a)= 1.0

> u(t − a)= ⇒ 0
L [ f (t − a)u(t − a) ] = e − as F(s) t

a a

• Time Scaling Theorem: – If the time t is scaled by a factor a, then the Laplace transform of the time-scaled signal is Dynamic System Response

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1 a

⎛ ⎝

F⎜ 36

• Final Value Theorem: – If we know Q0(s), q0(∞) can be found quickly without doing the complete inverse transform by use of thefinal lim t →∞ f (t) = lim s→0 sF(s) value theorem. – This is true if the system and input are such that the output approaches a constant value as t approaches ∞ – The DC gain of a system, the steady-state value of the unit-step response, is given by: 1 DC Gain = lim s G(s) = lim G(s) s →0 s s →0 • Initial Value Theorem: – This theorem is helpful for finding the value off(t) just after the input has been applied, i.e., att = 0+. In getting the F(s) needed to apply this theorem, our usual definition of initial conditions as those before the input is applied must be used. lim t →0 f (t) = lim s →∞ sF(s) Dynamic System Response

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• Impulse Function

δ(t) δ(t) = 0⇒ ≠t

p(t)

Approximating function for the unit impulse function

1/b

b

t

∫

+ε

−ε

0

δ(t)dt = 1⇒ε>

δ(t) = lim b→0 p(t) ⎡ du ⎤ =sU(s) L [ δ(t)=] L = ⎢⎣ dt ⎥⎦ =

s

0

1 s

1

– The step function is the integral of the impulse function, or conversely, the impulse function is the derivative of the step function.


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– When we multiply the impulse function by some number, we increase the “strength of the impulse”, but “strength” now means area, not height as it does for “ordinary” functions. • An impulse that has an infinite magnitude and zero duration is mathematical fiction and does not occur in physical systems. • If, however, the magnitude of a pulse input to a system is very large and its duration is very short compared to the system time constants, then we can approximate the pulse input by an impulse function. • The impulse input supplies energy to the system in an infinitesimal time.


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Approximate and Exact Impulse Functions If es =1.0 (unit step function), its derivative is the unit impulse function with a strength (or area) of one unit.

This “non-rigorous” approach does produce the correct result.


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• Inverse Laplace Transformation – A convenient method for obtaining the inverse Laplace transform is to use a table of Laplace transforms. In this case, the Laplace transform must be in a form immediately recognizable in such a table. – If a particular transform F(s) cannot be found in a table, then we may expand it into partial fractions and write F(s) in terms of simple functions ofs for which inverse Laplace transforms are already known. – These methods for finding inverse Laplace transforms are based on the fact that the unique correspondence of a time function and its inverse Laplace transform holds for any continuous time function.


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Time Response & Frequency Response 1st-Order Dynamic System Example: RC Low-Pass Filter

out

in

in


out

Dynamic System Investigation of the RC Low-Pass Filter

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Zero-Order Dynamic System Model

Step Response


Frequency Response

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Validation of a Zero-Order Dynamic System Model


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1st-Order Dynamic System Model τ = time constant K = steady-state gain

t=τ

Slope at t = 0


q o

=

Kq is

τ

e

−

t

τ

q o

= t =0

Kq is

τ

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• How would you determine if an experimentallydetermined step response of a system could be represented by a first-order system step response?

q(o

t − ⎞ ⎛ τ is t) = Kq ⎜ 1 − e ⎟ ⎝ ⎠

q o (t) − Kq is Kq is 1−

q o (t) Kq is

⎡ ⎣

log10 ⎢1 −

=e

= −e −

−

Straight-Line Plot:

⎡ q o (t) ⎤ log10 ⎢1 − vs.t ⎥ ⎣ Kq is ⎦

t

τ

Slope = -0.4343/τ

t

τ

q o (t) ⎤

=−⎥ Kq is ⎦


t

τ

=− 10 e log

0.4343

t

τ K. Craig

47

– This approach gives a more accurate value of τ since the best line through all the data points is used rather than just two points, as in the 63.2% method. Furthermore, if the data points fall nearly on a straight line, we are assured that the instrument is behaving as a first-order type. If the data deviate considerably from a straight line, we know the system is not truly first order and a τ value obtained by the 63.2% method would be quite misleading. – An even stronger verification (or refutation) of first-order dynamic characteristics is available from frequencyresponse testing. If the system is truly first-order, the amplitude ratio follows the typical low- and highfrequency asymptotes (slope 0 and –20 dB/decade) and the phase angle approaches -90° asymptotically. Dynamic System Response

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– If these characteristics are present, the numerical value of τ is found by determining ω (rad/sec) at the breakpoint and using τ = 1/ωbreak. Deviations from the above amplitude and/or phase characteristics indicate non-first-order behavior.


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• What is the relationship between the unit-step response and the unit-ramp response and between the unit-impulse response and the unit-step response? – For a linear time-invariant system, the response to the derivative of an input signal can be obtained by differentiating the response of the system to the srcinal signal. – For a linear time-invariant system, the response to the integral of an input signal can be obtained by integrating the response of the system to the srcinal signal and by determining the integration constants from the zero-output initial condition.


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• Unit-Step Input is the derivative of the Unit-Ramp Input. • Unit-Impulse Input is the derivative of the Unit-Step Input. • Once you know the unit-step response, take the derivative to get the unit-impulse response and integrate to get the unit-ramp response.


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System Frequency Response


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Bode Plotting of 1st-Order Frequency Response

0.01 = −40 dB 0.1 = −20 dB 0.5 = −6dB 1.0 = 0 dB 2.0 = 6 dB 10.0 = 20 dB 100.0 = 40 dB Dynamic System Response

dB = 20 log10 (amplitude ratio) decade = 10 to 1 frequency change octave = 2 to 1 frequency change K. Craig

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Analog Electronics: RC Low-Pass Filter Time Response & Frequency Response

⎡ein ⎤ ⎡ RCs + 1 − R ⎤ ⎡eout ⎤ ⎢ i ⎥ = ⎢ Cs −1 ⎥⎦ ⎢⎣ i out ⎥⎦ ⎣ in ⎦ ⎣ e out ein Dynamic System Response

=

1 RCs + 1

=

1

τs + 1

wheni

out

=0 K. Craig

54

Time Response to Unit Step Input 1 0.9 0.8 0.7 e0.6 d u itl p0.5 m A0.4

R = 15 KΩ C = 0.01 μF

0.3 0.2 0.1 0

012345678 Time (sec)

-4

x 10

Time Constant τ = RC Dynamic System Response

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• Time Constant τ – Time it takes the step response to reach 63% of the steady-state value • Rise Time Tr = 2.2 τ – Time it takes the step response to go from 10% to 90% of the steady-state value • Delay Time Td = 0.69

τ

– Time it takes the step response to reach 50% of the steady-state value


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Frequency Response R = 15 KΩ C = 0.01 μF

Bandwidth = 1/τ

e out ein

=) (iω

K

= iωτ + 1


K ∠0 2 ωτ + ∠2 ( )1t

K



=

−1 ωτ an

∠ − ωτ 2 ωτ ( + )1

2

tan −1 K. Craig

57

• Bandwidth – The bandwidth is the frequency where the amplitude ratio drops by a factor of 0.707 = -3dB of its gain at zero or low-frequency. – For a 1st -order system, the bandwidth is equal to 1/τ. – The larger (smaller) the bandwidth, the faster (slower) the step response. – Bandwidth is a direct measure of system susceptibility to noise, as well as an indicator of the system speed of response.


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Amplitude Ratio = 0.707 = -3 dB

Phase Angle = -45°

Response to Input 1061 Hz Sine Wave 1

Input

0.8

0.6

0.4

0.2 e d u ti l p m a

0

-0.2

Output

-0.4

-0.6

-0.8

-1

0

0.5


1

1.5

2 time (sec)

2.5

3

3.5

4 x 10

-3

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Time Response & Frequency Response 2nd-Order Dynamic System Example: 2-Pole, Low-Pass, Active Filter R

R 4

7

C5 R1

R3

C2

-

R6

-

+

ein

+ eout

Dynamic System Investigation of the Two-Pole, Low-Pass, Active Filter Dynamic System Response

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Physical Model Ideal Transfer Function eout ein

⎛ R⎞⎛7 ⎜ R⎟⎜ ⎝ ⎠⎝6

(s) = s

2

1⎞

⎛ 1 1 ++ + + ⎝⎜ R 3C 2 R1C 2


⎟

R 1R 3C⎠2 C5 1

⎞

R 4C2

⎠⎟

s

1 R 3 R 4 C 2 C5

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2nd-Order

Dynamic System Model

ωn  ζ

a2

d 2q 0 dt

2

1 d 2q 0

ω2n

dt 2

+ a1 +

dq 0 dt

+ a 0q0 = b0qi

2ζ dq 0

ωn

dt

+ q 0 = Kq i

a0 a2

= undamped natural frequency

a1

damping = ratio 2 a 2a 0 b0

K a 0

= steady-state gain

Step Response of a 2nd-Order System


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62

1 d 2q 0

ω2n

dt 2

2ς

+ ζ ωn

dq 0

+ = q0

dt

Step Response of a 2nd-Order System

Kq i

Underdamped

⎡ qo

= Kq is −⎢⎣1

1

−ζωn t

ωsin− ζ+n

1− ζ2 e

−1

2

(

Critically Damped q o

1

− tζ

sin

= Kq is −⎡⎣1+ ω (1

n

⎤

2

<ζ1

t)e

)

−ωn t

⎤⎦=ζ ⎤ ⎥

2

n

= Kq is ⎢ ⎢ ⎢ ⎣


2

ζ −1

1 1

⎡ ζ + ζ 2 − 1 ( −ζ+ ζ −1)ω t e ⎢1 − 2 Over- q damped o

⎥⎦

⎥ ζ − ζ − 1 ( −ζ− ζ −1 )ω t ⎥ + e ⎥ 2 2 ζ −1 ⎦ 1

2

ζ>

2

n

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63

Frequency Response of a 2nd-Order System Qo

K

Qi ( s ) = s 2

2ζ s + +1 2 ωn ωn

Laplace Transfer Function Sinusoidal Transfer Function

Qo

iω

Qi

( )


=

K

⎡ ⎛ ω ⎞2 ⎤ 2 4ζ 2 ω2 ⎢1 − ⎜ ⎟ ⎥ + 2 ωn ⎢⎣ ⎝ ωn ⎠ ⎥⎦

∠ tan −1

2ζ

⎛⎜ ω − ωn ⎝ ωn ω K. Craig

64

Frequency Response of a 2nd-Order System


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65

-40 dB per decade slope

Frequency Response of a 2nd-Order System


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• Some Observations • When a physical system exhibits a natural oscillatory behavior, a 1st-order model (or even a cascade of several 1st-order models) cannot provide the desired response. Theissimplest modeldynamic that does possess that possibility the 2nd-order system model. • This system is very important in control design. – System specifications are often given assuming that the system is 2nd order. – For higher-order systems, we can often use dominant pole techniques to approximate the system with a 2nd-order transfer function.


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• Damping ratio ζ clearly controls oscillation; ζ < 1 is required for oscillatory behavior. • The undamped case (ζ = 0) is not physically realizable (total absence of energy loss effects) but gives us, mathematically, a sustained oscillation at frequency ωn. • Natural oscillations of damped systems are at the damped natural frequency ωd, and not at ωn.

ωd =ωn 1−

• In hardware design, an optimum value of ζ = 0.64 is often used to give maximum response speed without excessive oscillation. • Undamped natural frequency ωn is the major factor in response speed. For a given ζ response speed is directly proportional to ωn. Dynamic System Response

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• Thus, when 2nd-order components are used in feedback system design, large values of ωn (small lags) are desirable since they allow the use of larger loop gain before stability limits are encountered. • For frequency response, a resonant peak occurs for ζ < 0.707. The peak frequency is ωp and the peak amplitude ratio depends only on ζ.

ωω =ζ−p

n

1 2

2

peak=amplitude ratio

K 2ζ 1 − ζ 2

• Bandwidth – The bandwidth is the frequency where the amplitude ratio drops by a factor of 0.707 = -3dB of its gain at zero or low-frequency. Dynamic System Response

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– For a 1st -order system, the bandwidth is equal to 1/τ. – The larger (smaller) the bandwidth, the faster (slower) the step response. – Bandwidth is a direct measure of system susceptibility as well as an indicator of the system speedtoofnoise, response. – For a 2nd-order system: BW =ω n 1− ζ 2

+2

−2ζ 4+ ζ2

4

4

– As ζ varies from 0 to 1, BW varies from 1.55ωn to n. For a value of n. For 0.64 = 0.707, BW = ω ζ ω most design considerations, we assume that the bandwidth of a 2nd-order all pole system can be approximated by ωn.


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G(s)

=

Kω2n

s 2 + 2ς ωn s

+ω2n

= −ςωn ±iωn s1,2 = −σ ± iωd

1 −ς2

s1,2

y ( t ) =−1 e −σt ⎛⎜ω cos +

⎝

tr

≈

1.8

ts

≈

4.6

Mp

ωn ςωn

=e

d

t

ωσ sin ωd

d

t ⎞⎟

⎠

risetime settlingtime

−πς2 1−ς

≈ ⎛⎜−1 ⎝

Location of Poles of Transfer Function

0( ≤ζ<

ζ ⎞ ⎟ 0.6 ⎠

0 ≤ζ (≤

1 overshoot ) 0.6


)

All-Pole 2nd-Order Step Response K. Craig

71

ωn ≥

1.8

σ≥

tr

ζ≥

0.6−(1 M p ) ≤ζ0≤

4.6 ts

0.6

Time-Response Specifications vs. Pole-Location Specifications Dynamic System Response

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• Experimental Determination of ζ and ωn –

ζ and ωn can be obtained in a number of ways from step or frequency-response tests.

– For an underdamped second-order system, the values of ζ and ωn may be found from the relations: −

Mp

T = 2π

ωd

=e

πζ 1−ζ 2

ωd =ω n −1ζ


1

⇒ζ=

2

ω ⇒=

2

⎛ ⎞ π ⎜ ⎟ +1 ⎜ log e ( M p ) ⎟ ⎝ ⎠ n

d ω = 1 − ζ2

2π T 1− ζ

2

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– Logarithmic Decrement δ is the natural logarithm of the ratio of two successive amplitudes.

=δ

⎛ x (t) ⎞ ⎜ x ( t + T=) ⎟⎟ ⎝ ⎠

ln ⎜

ln (ζω =e n

ζω T

)

n 2 π = ζωn 2π = 2πζ = ζω ωd ω n− ζ1 2 − ζ1 2 δ ζ= 2 ( 2π ) + δ 2

δ=

1 n

ln

n

T

Free Response of a 2nd-Order System T=

B1 Bn +1

x) ( t Dynamic System Response

2π

ωd

= Be −ζω ( t sin)ω+ dφt n

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– If several cycles of oscillation appear in the record, it is more accurate to determine the period T as the average of as many distinct cycles as are available rather than from a single cycle. – If a system is strictly linear and second-order, the value of n is immaterial; the same value of ζ will be found for any number of cycles. Thus if ζ is calculated for, say, n = 1, 2, 4, and 6 and different numerical values of ζ are obtained, we know that the system is not following the postulated mathematical model. • For overdamped systems (ζ > 1.0), no oscillations exist, and the determination of ζ and ωn becomes more difficult. Usually it is easier to express the system response in terms of two time constants. Dynamic System Response

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– For the overdamped step response:

⎡ ζ + ζ 2 − 1 ( −ζ+ ζ −1)ω t e ⎢1 − 2 2 ζ −1 ⎢ q o = Kq is ⎢ 2

⎤ ⎥ ⎥ ⎥

2

n

+ ζ − ζ2 − 1 e( 2 ζ −1

⎢ ⎣ qo Kq is

1

t

t

1

2

−ζ− ζ2 − 1 ω

ζ>

t

) ⎥ n

⎦

τ1 − τ τ2 − τ = e − e +1 τ2 − τ1 τ 2− τ 1

– where

τ1




1

(

ζ − ζ −2 ω1

)



n

1

τ2 ζ + ζ − ω2

(

1

)

n

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– To find τ1 and τ2 from a step-function response curve, we may proceed as follows: • Define the percent incomplete response Rpi as:

⎛ q ⎞ R pi  ⎜ 1 − Kqo ⎟100 is ⎠ ⎝ • Plot Rpi on a logarithmic scale versus time t on a linear scale. This curve will approach a straight line for large t if the system is second-order. Extend this line back to t = 0, and note the value P1 where this line intersects the Rpi scale. Now, τ1 is the time at which the straight-line asymptote has the value 0.368P1. Dynamic System Response

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• Now plot on the same graph a new curve which is the difference between the straight-line asymptote and Rpi. If this new curve is not a straight line, the system is not second-order. If it is a straight line, the time at which this line has the value 0.368(P1-100) is numerically equal to τ2. • Frequency-response methods may also be used to find τ1 and τ2.


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Step-Response Test for Overdamped Second-Order Systems


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Frequency-Response Test of Second-Order Systems


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Dynamic System Response Examples


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• Problem Statement – An underdamped 2nd-order system model has the following transfer function: Kω2n G(s) = 2 s + 2ς ωn s +ω2n 2

– Part 1:

s1,2 s1,2

= −ςωn ±iωn = −σ ± iωd

1 −ς

• Using the properties and formulas for 2nd-order systems, discuss the relationships between the step-response-parameters rise time, settling time, and overshoot, and the frequency-responseparameters bandwidth and peak amplitude as the model parameters vary. Use plots as needed in your presentation. Dynamic System Response

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• Suggestion: Pick a base system. Generate 4 families of plots – ωd constant, vary σ – σ constant, vary ωd – ωn constant, vary ζ – ζ constant, vary ωn • Show both time-response and frequencyresponse plots. Include discussion.


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– Part 2: • Investigate the effects on the time (step) response and frequency response of adding a real pole or a real zero to the 2nd-order transfer function. The pole and zero are added separately. In classical deign using root-locus or frequency-response techniques, real poles and zeros are added (lead, lag, lead-lag controllers) to modify system dynamics, and so it is important to have a good understanding of these effects. Use plots as needed in your presentation.


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• Suggestion: Pick a base second-order system. – Add a negative real pole (s + p) to the transfer function and move the pole from the left towards the srcin and describe its effect on the time-response and frequencyresponse plots. – Add a negative real zero (s + z) to the transfer function and move the zero from the left towards the srcin and describe its effect on the time-response and frequencyresponse plots.


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– Part 3: • Now add a positive real zero to your base second-order system and evaluate the step response for the system. Explain your observations. • Physically, what might cause a transfer function to have a right-half plane zero?


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• Problem Solution

ωd2 + σ 2 ω2n = 2 G(s) = 2 2 s + 2ζωn s + ωn s + 2σ + s ω ( +d2 σ – Base System G(s) – Effects of σ:

2

= s2

+ 2s + 2

σ =1 ωd = 1

2

ωn =

) 2

ζ = 0.707

• ωd = 1, σ = [0.5, 1, 5] – Effects of ωd: • σ = 1, ωd = [0.5, 1, 5] – Effects of ωn: • ζ = 0.707, ωn = [0.5√2, √2, 5√2] – Effects of ζ: • ωn = √2, ζ = [0.866, 0.707, 0.5] Dynamic System Response

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Effects of varying σ

ωd = 1, σ = [0.5, 1, 5] σ = 0.5 σ=5

σ=5

As σ increases: ts decreases tr decreases M decreases p BW increases

σ = 0.5

σ=5 σ = 0.5


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Effects of varying ωd

σ = 1, ωd = [0.5, 1, 5] ωd = 0.5 ωd = 5 ωd = 5

As ωd increases: ts is fixed tr decreases M increases p BW increases

ωd = 0.5 ω =5 d

ωd = 0.5


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Effects of varying ωn

ζ = 0.707, ωn = [0.5√2, √2, 5√ ωn = 0.5 ωn = 5

ωn = 5 ωn = 0.5

As ωn increases: ts decreases

ω =5

tr

Mdecreases p is fixed BW increases

n

ωn = 0.5 Dynamic System Response

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ζ =0.5

Effects of varying ζ

ωn = √2, ζ = [0.866, 0.707, 0.5] ζ =0.866

ζ = 0.5

As ζ increases: ts increases

ζ =0.866

r decreases tM p increases BW increases

ζ =0.5 ζ =0.866


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• Effect of an Additional LHP Pole

ωd2 + σ 2 ω2n G(s) = 2 = 2 2 s + 2ζωn s + ωn s + 2σ + s ω ( +d2 σ

2

)

– Base System G (s) =

2 s + 2s + 2 2

σ =1 ωd = 1

ωn = 2 ζ = 0.707

– Additional Pole G(s)

= =

2

( ps + 1) (s 2 + 2s + 2)

p = [0, 0.2, 1, 2]

1

+ 1)s + +2 (2p + 2)s 2 ps 3 + (2p


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Effect of an Additional Pole p = [0, 0.2, 1, 2]

increasing p

As p increases (pole gets closer to the srcin): ts increases tr increases Mp decreases to zero BW decreases


increasing p

increasing p

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• Effect of a LHP Zero

ωd2 + σ 2 ω2n G(s) = 2 = s + 2ζωn s + ω2n s 2 + 2σ + s ω ( +d2 σ

2

)

– Base System G (s) =

2 s 2 + 2s + 2

σ =1 ωd = 1

ωn = 2 ζ = 0.707

– Add a Zero G(s)


=

2(zs + 1) (s 2 + 2s + 2)

z = [0,0 .2, 1, 2]

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Effect of a LHP Zero z = [0,0 .2, 1, 2]

increasing z

As z increases (zero gets closer to the srcin): ts increases tr decreases Mp increases BW increases


increasing z

increasing z

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• Effect of a RHP Zero

ωd2 + σ2 ω2n G(s) = 2 = 2 2 s + 2ζωn s + ωn s + 2σ s+ ω ( +d2 σ

2

)

– Base System 2 G ( s ) = s 2 + 2s + 2

σ =1 ωd = 1

ω = 2 n ζ = 0.707

– Add a RHP Zero G(s)

=

G1 (s) =

2 (s 2 + 2s + 2) 2

+ 2 2 (s + 2s + 2) +(s+ 2 G 2 (s) = 2 − (s + 2s + 2) +(s+2 Dynamic System Response

2s 2s +2)+ 2s 2s +2) +

= =

( 2s + 2 ) 2

(s 2s 2) ( −2s + 2) (s 2

2s 2)

G(s) plus its derivative

G(s) minus its derivative K. Craig

97

G1(s)

G(s)

G2(s)


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98

2s + 2 s 2 + 2s + 2 2 s 2 + 2s + 2

2s s 2 + 2s + 2


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2 s 2 + 2s + 2

−2s + 2 s + 2s + 2 2

−2s s 2 + 2s + 2


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• Problem Statement – Satellites often require attitude control for proper orientation of antennas and sensors with respect to the earth. This is a three-axis, attitude-control system. To gain insight into the three-axis problem we often consider one axis at a time. For the one-axis problem, the plant transfer function is G(s) = 1/s2. This results from the equation of motion:  = T Jθ

– J is the moment of inertia of the satellite about its mass center, T is the control torque applied by the thrusters, and θ is the angle of the satellite axis with respect to an inertial reference frame, which must have no angular acceleration. Dynamic System Response

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– A controller has been designed (see below). What are the time-response and frequency-response performance indicators for this closed-loop system. s+3 G c (s) = 220 s + 21

– Assuming the controller is to be implemented digitally, approximate the time lag from the D/A converter to be: 2/T s+2/T


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– Determine the closed-loop system root locations for sample rates ωs = 5 Hz, 10 Hz, and 20 Hz, where the sample period T = 1/ωs seconds. Plot the unit step responses for the analog system and for each sample rate and compare. State your observations regarding closed-loop stability. How fast do you think one should sample in order to have a reasonably smooth response? – The closed-loop block diagram is shown below: R(s) + -


2/T s + 2/T

Controller

1

C(s)

2

s

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• Problem Solution – Open-Loop Transfer Function

⎛ s + 3⎞⎛ ⎞1 ⎟⎜ ⎟ ⎝ s + 21⎠⎝ ⎠s 2

220 ⎜

– Closed-Loop Transfer Function

⎛ s + 3⎞⎛ ⎞1 ⎟⎜ ⎟ 220(s + 3) ⎝ s + 21⎠⎝ ⎠s 2 = s + 3⎞⎛ ⎞1 (s3 + 21s 2 + 220s + 660) ⎛ 1 + 220 ⎜ s + 21 ⎟⎜ ⎟s 2 ⎝ ⎠⎝ ⎠ 220 ⎜

Closed-Loop Poles: -4.55, -8.23 ± 8.80i Closed-Loop Zero: -3


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Closed-Loop Step Response Mp = 29.4% 2% ts = 0.781 sec tr = 0.113 sec


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Open-Loop Frequency-Response Plot

GM = ∞


PM = 47.9º

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Closed-Loop Frequency-Response Plot

-3dB

BW = 16.5 rad/sec


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• Time Delay τ = T/2

T = 0.2, 0.1, 0.05 τ = 0.1, 0.05, 0.025

1

– Approximation

τs + 1

G c (s) = 220

s+3 s + 21

R(s) +

C R

(s) =

-

2/T s + 2/T

Controller

1 s2

C(s)

220(s + 3)

τs 4+ (21 τ + 1)s + 3 +21s 2 + 220s

660

Closed-Loop Poles: τ = 0.1: -25.78, -0.77 ± 8.31i, -3.67 τ = 0.05: -31.72, -2.65 ± 9.87i, -3.98 τ = 0.025: -46.82, -4.99 ± 10.44i, -4.2 Dynamic System Response

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Step Response τ = 0.1 τ = 0.05 τ = 0.025 τ=0


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5. MechatronicsWorkshop KCCDay2 Session1-2.pdf

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