Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Lumped Parameter Systems
Outline of Today’s Lecture
 Review
 Engineering Modeling Procedure
 State Space Models
 Lumped Parameter Systems
 DC Armature control motor
 Balance Systems
Models
SENSE
REAL WORLD
OBSERVATIONS
FORMULATE
TEST
EXPLANATION/
PREDICTION
INTERPRET
MATHEMATICAL
MODEL
Engineering Modeling Procedure
 Understand the problem
 What are the factors and relevant relationships?
 What assumptions can be made?
 What equilibrium conditions exist?
 What should the result look like?
 Draw and label an engineering sketch
 Free body diagram
 Hydraulic schematic
 Electrical schematic
 Write the equilibrium equations (usually differential or difference)
 Newton 2nd Law
 Kirchoff Laws for current and voltages
 Flow continuity laws
 Solve the equations for the desired result
 Check the validity of the results
Modeling is an Iterative Process
Understand
the Problem
Can you
formulate a
model?
Sketch
YES
Mathematical
Model
NO
NO
YES
NO
YES
Use the
Model
Do the
results
represent
reality?
Can you
solve the
model?
Validate
the Results
Solve
the Model
Modeling Terms
 System: a functional group of interrelated things
 State: A condition (which may or may not be physical) of the system




regarding form, structure, location, thermodynamics or composition
State vector: a collection of variables that fully describe the object
over time
Input: an external object provide to the system
Output: a dependent variable (often a state) from within the system
that can be measured or quantified
Dynamics: a chance process of the state variables over time
State Space Formulation
Continuous Models
 Let x be a vector formed of the state variables
x  {x1(t ), x2 (t ),...}T
 The number of components of the state vector is called the order
 Formulate the system as
dx
 Ax  Bu
dt
y  Cx  Du
State Transition Equation
Output Equation
 The matrices A, B, C and D have constant elements
 The matrix A is the called the State Dynamics Matrix
 The matrix B is called the Input or Control Matrix
 The matrix C is called the Output or Sensor Matrix
 The matrix D is called the Pass Through or Direct term
State Space Formulation
Discrete Models
 Let x be a vector formed of the state variables
x  {x1(t ), x2 (t ),...}T
 The number of components of the state vector is called the order
 Formulate the system as
h  Time Step Size, often assumed to be 1
x(t  h)  h  Ax(t )  Bu(t ) 
y (t )  Cx(t )  Du(t )
State Transition Equation
Output Equation
 The matrices A, B, C and D have constant elements
 The matrix A is the called the State Dynamics Matrix
 The matrix B is called the Input or Control Matrix
 The matrix C is called the Output or Sensor Matrix
 The matrix D is called the Pass Through or Direct term
State Space Formulation
 Procedure:
 Develop the equations of equilibrium
 Put the equilibrium equations in the form of the highest
derivative equal the remainder of the terms
 Make a choice of states, the input and the outputs
 Write the equilibrium equations in terms of the state variables
 Construct the dynamics, the input, the output and the pass
through matrices
 Write the state space formulation
Distributed vs. Lumped
Parameters
 Distributed parameter
 Analysis is at the material
element level
 Partial differential equations
describe the transfer of force
from the constitutive equations
 FEM/BEM often used
 Lumped parameter
 Analysis is at the component level
 Component properties are self
contained and complete
 ODE/Diff E based on linking
component parameters
 Equations solved analytically or
numerically
Distributed vs. Lumped
Parameters
 Distributed parameter systems
 physically better descriptions
 more accurate results when done correctly
 Lumped parameter systems
 simpler
 quicker results
?
 Both can be used in building controls
 Lumped parameter descriptions are appropriate when the
property being examined is of much greater magnitude than the
added accuracy that would be gained using a distributed parameter
model
Lumped Parameter Variables
From Richard C. Dorf, Modern Control Systems, 6 ed.
Mechanical Systems
What are the noises from wheel speed?
Determine the number of equations
need form the number of inertial
coordinates (qe,qd,qa,and qw)
and their linkages
Equilibrium Equations Needed:
1. Engine to clutch
T  J cq d  bc q e  q d


2. Clutch to transmission
J cq d  kd q d  N tf q a   0
3. Transmission to wheel
J tq a  kd  N tf q a  q d   ka (q a  q w )  0
4. Wheel to ground
J wqw  btqw  ka (qw  qa )  ktqw  0
Mechanical Systems
What are the noises from wheel speed?






  bcq d  T  bcq e  kd N tf q a  kd qd
J cq d  kd q d  N tf q a   0  J cq d  kd N tf q a  kd q d 
T  J cq d  bc q e  q d  J cq d  T  bc q e  q d
J tq a  kd  N tf q a  q d   ka (q a  q w )  0  J tq a  kd q d   kd N tf  ka qa  kaq w
J wq w  btq w  ka (q w  q a )  ktq w  0  J wq w  btq w  kaq a   ka  kt q w
Inputs are T and q . The Output is q w

State variables are q d ,q a ,q a ,q w ,q w
 kd
 b
c
q d  
q   0

a
d    kd
q a  
dt    J t
q w   0
q w  

 0

kd N tf
0
0
1
0
0
ka
Jt
0
0
0
ka
Jw
0 
bc
0


kd N tf  ka
Jt
q d 
q 
 a
y  0 0 0 0 1 q a 
 
q w 
q w 
k a  kt
Jw

0 
 q    1
d
0     bc
 q a  
 0
0  q a   
 
0
 q w  
 0
1 
q w  
b 
 0
 t 
J w 

1

0
T q e 

0 

0

0
Lumped Parameter Model of an
Armature Controlled DC Motor
dia

Voltage
Loop:
L
 Ra ia  Vb  Va
a

dt

 Back Voltage: V  K dq

b
b
dt

 Motor Torque: T  Kia

2
 Rotations NSL: J d q  b dq  T

dt
dt
Ra ia K b dq Va
 dia




 dt
La
La dt La
 2
 d q  Kia  b dq
 dt
J
J dt
 dq 
State vector  ia ,
  ia , 
 dt 
Input is Va Output is 
What is the speed?
Assume the friction term is f  bq
 Ra

d ia   La

dt    K
 J
i 
y  0 1  a 
 
Kb 
1
La  ia   
    La Va
b     
 
 0 
J 

Lumped Parameter Model of an
Armature Controlled DC Motor
Assume the friction term is f  bq
dia

 Voltage Loop: La dt  Ra ia  Vb  Va

 Back Voltage: V  K dq

b
b
dt

 Motor Torque: T  Kia

2
 Rotations NSL: J d q  b dq  T

dt
dt
Ra ia K b dq Va
 dia
 dt   L  L dt  L
a
a
a
 2
 d q  Kia  b dq
 dt
J
J dt
 dq 
State vector  ia ,
  ia , 
 dt 
Input is Va Output is 
What is the speed?
 Ra

d ia   La

dt    K
 J
Kb 
1
La  ia   
    La Va
b     
 
 0 
J 
Note how the mechanical
and the electrical domains
were put together here:
1) KVL for the electrical
2) NSL for the mechanical
3) Relationship or coupling
equation between the two

i 
y  0 1  a 
 
Is this a good model
for motor angle?
In a controls problem,
sometimes called
Mechatronics, this is often
necessary
Lumped Parameter Model of an
Armature Controlled DC Motor
dia

 Voltage Loop: La dt  Ra ia  Vb  Va

 Back Voltage: V  K dq

b
b
dt

 Motor Torque: T  Kia

2
 Rotations NSL: J d q  b dq  T

dt
dt
Ra ia K b dq Va
 dia
 dt   L  L dt  L
a
a
a
 2
 d q  Kia  b dq
 dt
J
J dt
 dq 
State vector  ia ,
, q   ia ,  ,q 
 dt 
Input is Va Output is q
What is the motor angle?
Same process,
different question,
different formulation
 Ra
 L
ia   a
d    K
 
dt    J
q  
 0



ia 
y  0 0 1  
 
q 
Kb
La
b
J
1


0
 ia   1

0     0 Va
   
 q   0 
0


Lumped Parameter Model of an
Armature Controlled DC Motor
 Voltage Loop: Ra ia  Vb  Va

dq
 Back Voltage: Vb  K b

dt

 Motor Torque: T  Kia

d 2q
dq
Rotations
NSL:
J
b
T

dt
dt

K b dq Va

i



 a
Ra dt Ra
 b KK b  dq KVa
d 2q

  

 2

dt
J
JR
dt
JRa
Ki
d
q
b
d
q
a 


 a 
 dt
J
J dt
What is the motor angle?
If the inductance La is small such
that it can be neglected, then
another simpler formulation is
 dq 
State vector   ,q   ,q 
 dt 
Input is Va Output is q
 bRa  KK b
d    
JRa

dt q  
1

 
y  0 1  
q 

 K 
0   
 JR V
 q   a  a
0
 0 
Balance Systems
A large number of control problems are called balance systems
where an object must be maintained in technically an
unstable position
Balance Dynamics
General Dynamics Equation form is
M (q, q, q)  C  q, q   B(q, q, q, u)
External Forcing terms
Energy Dissipating (Rayleigh) Terms
Energy Conserving Terms
This equation is usually nonlinear
Example: Inverted Pendulum
d2
NSL in p direction: Mp  m 2 ( p  l sin q )  bp  F 
dt


Mp  m p  l (  sin qq 2  cosqq )  bp  F
 M  m  p  ml sin qq 2  ml cosqq  bp  F
NSL about pivot:  J  ml 2 q  mlp cos q  mgl sin q  q  0
0
 M  m ml cos q   p  0 ml sin q   0   b 0   p  
 F 
 ml cos q J  ml 2  q   0
 q 2    0   q    mgl sin q    0 
0

  
  
  
  
Where b is the viscous friction at the wheels and  is the viscous friction in the pin
Clearly Nonlinear
Example: Inverted Pendulum
Assuming q and q are small, then sin q  q , cos q  1 and qq 2  0
without the friction terms,
0
 M  m ml cos q   p   0 ml sin q   0  
 F 


 ml cos q J  ml 2  q   0
 q 2   mgl sin q    0 
0

  
  
  
ml   p   0   0   F 
M  m

2 
 ml
 q    0 
J

ml
q

mgl

  
   
 M  m  p  mlq



2
J

ml
q

mlp

mgl
q

0




F
2

F  mlq   J  ml q  mglq


M

m
ml





  J  ml 2 q  mglq 
ml
q

( g ( M  m )q  F )
p
2

J
(
M

m
)

Mml
ml

F  mlq
p
 M  m
F   M  m p
ml


(  p  gq )
q
2

ml
J

ml

ml

1
ml
2 2
2

p

(

gm
l
q

(
J

ml
)F )
q
(

p

g
q
)
2

J ( M  m )  Mml
J  ml 2

F   M  m p
Example: Inverted Pendulum

1
m 2 l 2 gq  F  J  ml 2 
2
J ( M  m )  Mml
ml
q
 gq ( M  m)  F 
J ( M  m )  Mml 2
p
0
 p  
  0
d  p 

dt q  0
  
q  0


1
0
m2l 2 g
0 
J ( M  m )  Mml 2
0
0
mlg ( M  m )
0
J ( M  m )  Mml 2
 p
 p
y   0 0 1 0  
q 
 
q 

0


0

  p 
2
J

ml




0  
  p   J ( M  m )  Mml 2 

F

1  q  
0





 q 
ml
0


 J ( M  m )  Mml 2 

Summary
 Lumped Parameter vs. Distributed Parameter Systems
 Distributed parameter systems:
 Material element level
 Partial differential equations describe the transfer of force from the
constitutive equations
 Lumped Parameter Systems
 Component level
 Component properties are self contained and complete with ODE/Diff E based
on linking component parameters for equilibrium equations
 Mechanical system equations
 Electric Motor
 Balance systems
 Next Class: Matlab and Simulink