Modeling of Dynamic Systems

  Modeling of Dynamic Systems

  Ref: Control System Engineering Norman Nise : Chapters 2 & 3

  Chapter objectives :

  Review the Laplace transform Learn how to find a mathematical model, called a transfer

  function

  Learn how to find a mathematical model, called a state

  variable representation How to convert between transfer function and state space models How to linearize a non-linear system

  Modeling in Frequency Domain Laplace Transforms Review

  1. Standard notation in dynamics and control (shorthand notation)

  2. Converts mathematics to algebraic operations

  3. Advantageous for block diagram analysis Laplace transforms can be used in process control for:

  1. Solution of differential equations (linear)

  2. Analysis of linear control systems (frequency response)

  3. Prediction of transient response for different inputs

  ∞

• st _-

  L(f (t)) = f (t)e dt Examples Examples _{∞ ∞}

• _{-st − st} a a a

  L(a)= ae dt = − e = − − = _{∞ ∞} s s s _{∞ -bt -bt -st -(b+s)t − ( b + s)t}

  1

  1 = = =

  L(e )= e e dt e dt -e _∞ b+s s+b df df

• _-st ′ = = =

  L(f ) L e dt

  sL(f) − f(0)

  dt dt Usually define f(0) = 0 (e.g., the error) Table Laplace Transforms for Various Time-Domain Functions _{f(t) F(s)}

  f(t) F(s)

• = =
• =

  4

  , ( )

  ( )

  1

  5

  2

  5

  s y t s s

  4

  − +

  =

  From LT Table, ( )

  0.8

  0.5 0.5 (3-37)

  t y t e

  −

  s Y s s s

  5

  = +

  First, take L of both sides ( )

  f(t) F(s) LT Example 1

  Solve the ODE,

  ( )

  5

  4

  2 1 (3-26)

  dy y y dt

  ( )

( )

  2 (3-34)

  2

  5

  1 4 sY s Y s

  s

  − + = Rearrange,

  ( ) ( )

  5

• Take L
• 1
• L

  Partial fraction expansion

  1. Case 1. Roots of the denominator of F(s) are real and

  distinct

  2. Case 2. Roots of the denominator of F(s) are real and

  repeated

  3. Case 3. Roots of the denominator of F(s) are complx or

  imaginary _{3 2} LT Example 2

  d y d y dy

• =
₃

  6 ₂

  11 6 y

  4

  dt dt dt

  ′ ′′

  y( ) = y ( )= y ( ) = system at rest (s.s.)

  To find transient response for u(t) = unit step at t > 0

  1. Take Laplace Transform (L.T.)

  2. Factor, use partial fraction decomposition 3. Take inverse L.T.

  Step 1 Take L.T. (note zero initial conditions)

  4

  3

  2

• s Y(s)+ s Y(s)+

  6 11 sY(s) 6 ( ) Y s =

  s

  4 Y(s) = Rearranging, _{3 2}

• ( s

  6 s 11 s 6) s

  Step 2a. Factor denominator of Y(s) _{3 2}

• + + s(s

  6 s 11 s+ 6 )=s(s+ 1 )(s+ 2 )(s+ 3 )

  Step 2b. Use partial fraction decomposition

  4 _{1 2 3 4}

• = +

  s(s + )(s + )(s + )

  1

  2 3 s s 1 s 2 s

  3 Multiply by s, set s = 0

  4 _{2 3 4} = + + + ₁ s

  (s + )(s + )(s + )

  1

  2 3 s _{s = s =} 1 s 2 s

  3

  4

  2 = = ₁ 1 2 3 ⋅ ⋅

3 For a , multiply by (s+1), set s = -1 (same procedure

  2

  for a a )

  3,

  4

  2 = − = = − ₂ 2 , ₃ 2 , ₄

  3

  2

  2 2 2/3

  Y(s)=

  − −

  Step 3. Take inverse of L.T. ) +

  (

  s s s

s

  3 + + 1 +

  2

  3

  2 _{− t − 2 t −}

  2 _{3 t} − + −

  y(t)=

  2 e 2 e e

  3

  3

  2 (check original ODE)

  → ∞ → = =

  t y(t) t 0 (0) y 0.

  3 You can use this method on any order of ODE, limited only by factoring of denominator polynomial (characteristic equation) Practice Matlab ….. a description of a system in term of

  Mathematical model

  equations

  Types of model

  Models of dynamics system can be of many kinds, including the following: Mental, intuitive or verbal models

  Graphs and tables Mathematical models

  Constructing a model: Mathematical modeling or first principle modeling.

  Process or system identification.

  One example of dynamic system automobile

  The procedure of deriving models :

• Understand the physics and interaction of the elements
• Construct a simplified diagrammatic rep of the system
• Apply element and interconnection laws
• Draw the FBD
• Identify or define the inputs, outputs and state variables
• Establish the system equations
• Obtain the desired form of the system model
• If the model is non-linear, determine the equilibrium conditions and obtain a linearized model.

1. Mechanical Systems translational & rotational

  Translational mechanical systems can have only horizontal or vertical motion

a. State variables

  Variables of trans mechanical systems are :

  x , displacement in meters (m) v , velocity in meters per second (m/s)

  2 a , acceleration in meter per second square (m/s ) f , force in Newton (N) w , energy in Joule (J) p , power in watts (W) all variables are functions of time

b. The element laws include in trans systems are mass, friction and stiffness.

  They relate the external force to the acceleration, velocity and displacement associated with the element.

  Mass Friction when two bodies slide over each other there is a frictional

  force f between them that is a function of the relative velocity between the sliding surface

  Stiffness when a mechanical element is subjected to a force f and goes through a change in length x, it can be characterized by a stiffness element.

c. Interconnection Laws D’Alembert’s law is developed from Newton’s law for translational system.

  For a constant mass :

  dv ( f ) = M _{ext i i} dt

  where the summation over the index i includes all the external forces (f ) _{ext i} acting on the body.

  dv f − M =

  ( ) _{ext i i} dt The sum of the forces is zero provided -Mdv/dt is thought of as an additional forces this fictitious force is called inertial force or D’Alembert force.

  = D’Alembert’s law _i f _i

  The law of reaction force accompanying any force of one element on

  another, there is a reaction force on the first element of equal magnitude and opposite direction (Newton’s third law : reaction forces)

  The law for displacements if the ends of two elements are connected, those ends are forced to move with the same displacement and velocity.

d. Obtaining the system model FBD

  Example 1 Example 2

  Example 3 Example 4

  Series and parallel combination b b k k ^{1 2}

  1

  2

b =

k = eq
• eq

  b b

• ¹
² k k

  1

  2

• b b b

  =

• =

  k k k eq

  1 2 eq

  1

  2 Rotational mechanical systems are modeled using the same techniques as those for translational mechanical systems

  a. State variables

  Variables of trans mechanical systems are : , angular displacement in radians (rad)

  , velocity in radians per second (rad/s)

  2 , angular acceleration in radians per second squared (rad/s ) , torque in Newton-meters (N.m) all variables are functions of time

  b. The element laws include in rotational systems are moment of inertia, friction, stiffness, levers and gears. _{2 2} Moment of Inertia J in kilogram-meters (kg.m ) ₂

  where r is the distance from the axis of reference and

  J = r dm dm is the mass of the small element.

  The net torque applied about the fixed axis of rotation is given by

  d = where J is the angular momentum of a body.

  τ ( J ω )

  dt

  Parallel axis theorem states ₂

  J = J ma

• where a is the distance between the parallel axes and J

  is the moment of inertia about the principal axis

  Friction A rotational friction element is one for which there is an algebraic

  relationship between the torque and the relative angular velocity between two surfaces.

  Rotational viscous friction arises when two rotating bodies are separated by a film of oil.

  τ = b ∆ ω = b ω − ω

  ( ) _{2 1} Rotational devices characterized by viscous friction Stiffness is usually associated with a torsional spring, such as the mainspring of a clock, or with a relatively thin shaft.

  For a linear torsional spring or flexible shaft, where k is the stiffness constant with units of newton-

  τ θ

  = k ∆ ₂ - meters (N.m) and = _{1 2} - (a) Rot stiffness el with one end fixed. (b) Rot stiffness el with = ₁ Potential energy is stored in a twisted element and for a linear spring or shaft is given by

  1 ₂ θ

  W = k

  2

  The Lever an ideal lever is assumed to be a rigid bar pivoted at a point and having no mass, no friction, no momentum and no stored energy.

  Let be the angular displacement of the lever from the horizontal position.

  The lever

  For small displacements

  d d d ^{2 2}

  2 v = v f = f x = x ^{2 1 2 1}

  2

  1 d d _{1 1} d

1 Gears an ideal gear is assumed to have no moment of inertia, no friction, no stored energy, and a perfect meshing of the teeth.

  N is a gear ratio A pair of gears r n _{2 2} N = =

  where r and n denote the radius and number of teeth

  r n _{1 1 1} and are the angular displacements for the ₂ gears

  ω θ r r _{1 2 1 2}

  ; ; θ = θ = = N = = N

  r r _{1 1 2 2}

  θ r ω _{2 1 2} r _{1 1} and are the angular ₂ displacements for the gears

  Ideal gears (a) ref position, (b) after rotation

  τ ω τ ω _{1 1} + = _{2 2} FBD for a pair of ideal gears

c. Interconnection Laws

  D’Alembert’s law for a body with constant moment of inertia rotating about

  a fixed axis, τ − ω = _i ( ) J _{ext i} where the summation over i includes all the torques acting on the body, and the term –J can be considered an inertial torque. _i τ = D’Alembert’s law _i the torque –J is directed opposite to the positive sense of , and

  The law of reaction force for bodies that are rotating about the same axis,

  any torque exerted by one element on another is accompanied by a reaction torque of equal magnitude and opposite direction on the first element.

  The law for angular displacements Rot sys to illustrate the law of ang disp

  The reference marks on the rims are at the top of the two disks when no torque is applied. The net angular displacement for the shaft K with respect to its ₂ unstressed condition is . - _{2 1 i} ( ∆ θ ) = around any closed path _i

d. Obtaining the system model FBD

  (a) Rotational system, (b) & (c) Its corresponding FBD