Transcript Slide 1

Lagrange Equations
Use kinetic and potential energy to solve for motion!
References
http://widget.ecn.purdue.edu/~me563/Lectures/EOMs/Lagrange/In_Focus/page.html
System Modeling: The Lagrange Equations (Robert A. Paz: Klipsch School of Electrical and
Computer Engineering)
Electromechanical Systems, Electric Machines, and Applied Mechatronics by Sergy E. Lyshevski,
CRC, 1999.
Lagrange’s Equations, Massachusetts Institute of Technology @How, Deyst 2003 (Based on
notes by Blair 2002)
1
We use Newton's laws to describe the motions of objects. It works well
if the objects are undergoing constant acceleration but they can
become extremely difficult with varying accelerations.
For such problems, we will find it easier to express the solutions with
the concepts of kinetic energy.
2
Modeling of Dynamic Systems
Modeling of dynamic systems may be done in several
ways:
 Use the standard equation of motion (Newton’s Law)
for mechanical systems.
 Use circuits theorems (Ohm’s law and Kirchhoff’s laws:
KCL and KVL).
 Today’s approach utilizes the notation of energy to
model the dynamic system (Lagrange model).
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•
•
•
•
•
•
Joseph-Louise Lagrange: 1736-1813.
Born in Italy and lived in Berlin and Paris.
Studied to be a lawyer.
Contemporary of Euler, Bernoulli, D’Alembert, Laplace, and Newton.
He was interested in math.
Contribution:
– Calculus of variations.
– Calculus of probabilities.
– Integration of differential equations
– Number theory.
4
Equations of Motion: Lagrange Equations
• There are different methods to derive the dynamic equations of a
dynamic system. As final result, all of them provide sets of equivalent
equations, but their mathematical description differs with respect to
their eligibility for computation and their ability to give insights into the
underlying mechanical problem.
• Lagrangian method, depends on energy balances. The resulting
equations can be calculated in closed form and allow an appropriate
system analysis for most system applications.
• Why Lagrange:
–
–
–
–
–
Scalar not vector.
Eliminate solving for constraint forces (what holds the system together)
Avoid finding acceleration.
Uses extensively in robotics and many other fields.
Newton’s Law is good for simple systems but what about real systems?
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Mathematical Modeling and System Dynamics
Newtonian Mechanics: Translational Motion
• The equations of motion of
mechanical systems can be
found using Newton’s second
law of motion. F is the vector
sum of all forces applied to the
body; a is the vector of
acceleration of the body with
respect to an inertial reference
frame; and m is the mass of
the body.
• To apply Newton’s law, the
free-body diagram (FBD) in the
coordinate system used should
be studied.
 F  ma
Newtonapproachrequiresthat wefind
accelerations in all three directions,
equateF  ma, solvefor the constraint
forces and then eliminate
theseto reducethe problem
to " characteristicsize".
6
Force : Fcoulomb
Translational Motion in Electromechanical Systems
• Consideration of friction is essential for understanding the operation
of electromechanical systems.
• Friction is a very complex nonlinear phenomenon and is very difficult
to model friction.
• The classical Coulomb friction is a retarding frictional force (for
translational motion) or torque (for rotational motion) that changes its
sign with the reversal of the direction of motion, and the amplitude of
the frictional force or torque are constant.
• Viscous friction is a retarding force or torque that is a linear function
of linear or angular velocity.
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Newtonian Mechanics: Translational Motion
• For one-dimensional rotational
systems, Newton’s second law
of motion is expressed as the
following equation. M is the
sum of all moments about the
center of mass of a body (Nm); J is the moment of inertial
about its center of mass
(kg/m2); and  is the angular
acceleration of the body
(rad/s2).
M  j
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Newton’s Second Law
The movement of a classical material point is described by the second law of Newton:
m
d 2 r (t )
dt
2
 F (r , t ) (r is a vectorindicatinga positionof the material pointin space)
x 
r   y 
z 
Vector F(r, t) representsa force field, which may be calculatedby taking
intoaccountinteractions with otherparticles,or interactions with electromagnetic
waves,or gravitational fields.
The secondlaw of Newtonis an idealisation, of course,even if one was to neglect
quantumand relativistic effects. There is no justification whyonlya secondtime derivative
of r shouldappearin thatequation.Indeedif energyis dissipatedin thesystemusually
first time derivatives willappearin theequation oo.
t If a material pointlosesenergydue to EM
radiation,third time derivatives willcome up.
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Energy in Mechanical and Electrical Systems
• In the Lagrangian approach, energy is the key issue. Accordingly,
we look at various forms of energy for electrical and mechanical
systems.
• For objects in motion, we have kinetic energy Ke which is always a
scalar quantity and not a vector.
• The potential energy of a mass m at a height h in a gravitational
field with constant g is given in the next table. Only differences in
potential energy are meaningful. For mechanical systems with
springs, compressed a distance x, and a spring constant k, the
potential energy is also given in the next table.
• We also have dissipated energy P in the system. For mechanical
system, energy is usually dissipated in sliding friction. In electrical
systems, energy is dissipated in resistors.
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Electrical and Mechanical Counterparts
“Energy”
Energy
Mechanical
Electrical
Kinetic
(Active)
Ke
Mass / Inertia
0.5 mv2 / 0.5 j2
Inductor
1 2 1 2
Li  Lq
2
2
Potential
V
Gravity: mgh
Spring: 0.5 kx2
Capacitor
0.5 Cv2 = q2/2C
Dissipative
P
Damper / Friction
0.5 Bv2
Resistor
1 2 1 2
Ri  Rq
2
2
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Lagrangian
The principle of Lagrange’s equation is based on a quantity called
“Lagrangian” which states the following: For a dynamic system in which
a work of all forces is accounted for in the Lagrangian, an admissible
motion between specific configurations of the system at time t1 and t2
in a natural motion if , and only if, the energy of the system remains
constant.
The Lagrangian is a quantity that describes the balance between no
dissipative energies.
L  K e  V ( K e is the kineticenergy;V is the potentialenergy)
1 2
mv ; V  mgh
2
d  L  L P

Lagrange's Equation: 

 Qi
dt  qi  qi qi
Ke 
P is powerfunction(half rate at whichenergyis dissipated); Qi are generalized externalinputs
(forces) actingon thesystemIf there are three generalized coordinates, there willbe three equations.
Notethat theaboveequationis a second- order differential equation
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Generalized Coordinates
• In order to introduce the Lagrange equation, it is important to first
consider the degrees of freedom (DOF = number of coordinatesnumber of constraints) of a system. Assume a particle in a space:
number of coordinates = 3 (x, y, z or r, , ); number of constrants
= 0; DOF = 3 - 0 = 3.
• These are the number of independent quantities that must be
specified if the state of the system is to be uniquely defined. These
are generally state variables of the system, but not all of them.
• For mechanical systems: masses or inertias will serve as generalized
coordinates.
• For electrical systems: electrical charges may also serve as
appropriate coordinates.
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Cont..
• Use a coordinate transformation to convert between sets
of generalized coordinates (x = r sin  cos  ; y = r sin 
sin  ; z = r cos  ).
• Let a set of q1, q2,.., qn of independent variables be
identified, from which the position of all elements of the
system can be determined. These variables are called
generalized coordinates, and their time derivatives are
generalized velocities. The system is said to have n
degrees of freedom since it is characterized by the n
generalized coordinates.
• Use the word generalized, frees us from abiding to any
coordinate system so we can chose whatever parameter
that is convenient to describe the dynamics of the
system.
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For a large class of problems, Lagrange equations can be written in
standard matrix form
 L   L 
 P 
 q   q 
 q   f 
 1  1
 1   1
 .

.
 . 
d .

 - 
  
 
dt .
 .

.
 . 
 L   L 
 P   f n 

 



 q n   q n 
 q n 
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Example of Linear Spring Mass System and Frictionless
Table: The Steps
k
m
1 2 1 2
Lagrangian: L  K e  V  mx  kx
2
2
d  L  L

Lagrang's Equation: 
0

dt  q i  qi
L
d  L 
L

  mx;
Do the derivatives :
 mx;
 kx


q i
dt  q i 
qi
d  L  L

Combineall together: 
 mx  kx  0

dt  q i  qi
x
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Mechanical Example: Mass-Spring Damper
1 2
mx
2
1
V  Kx 2  mgh  x 
2
1
1
L  K e  V  mx 2  Kx 2  mgh  x 
2
2
1
P  Bx 2
2
Ke 
We havethe generalized coordinateq  x, and thuswith theappliedforce Q  f , we write
the Lagrangeequation:
d  L  L P

 
dt  x  x x
d  1
1
 ( ( mx 2  Kx 2  mg(h  x)))
dt x 2
2
 1
1
 1
 ( mx 2  Kx 2  mg(h  x))  ( Bx 2 )
x 2
2
x 2
d
 (mx 2 )  ( Kx  mg)  ( Bx )
dt
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 mx  Kx  mg  Bx
f 
Electrical Example: RLC Circuit
1 2
Lq
2
1 2
V
q
2C
Ke 
L  Ke V 
P
1 2 1 2
Lq 
q
2
2C
1 2
Rq
2
We havethe generalized coordinateq (charge),and with theappliedforce Q  u, we have
u
d  L  L P
  


dt  q  q q
d  1 2 1 2
 1
1 2
 1
( ( Lq 
q ))  ( Lq 2 
q )  ( Rq 2 )
dt q 2
2C
q 2
2C
q 2
d
Q
Q
di
 ( Lq )   Rq  Lq   Rq  L  vc  Ri
dt
C
C
dt
i  q and q  Cvc for a capacitor.This is just KVL equation

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Electromechanical System: Capacitor Microphone
About them see: http://www.soundonsound.com/sos/feb98/articles/capacitor.html
This systemhas two degreesof freedom
(electrical and mechanical: chargeq and
displacement x from equilibrium)
1 2 1 2
1 2 1 2
Lq  mx ; V 
q  Kx
2
2
2C
2

A   is the dielectricconstantof the air (F/m),


C
xo  x  A is the area of the plate,xo - x is the plateseparation
Ke 
1 2
1 2 1 2
xo  x  q  Kx ; P  Rq  Bx
V
2 A
2
2
2
1 2 1 2
1
1 2
2


xo  x q  Kx
L  Lq  mx 
2
2
2A
2
1
2
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L
L q
P
 mx;

 Kx;
 Bx
x
x 2A
x
L
L xo  x q P
 Lq;

;
 Rq
q
q
A
q
Then we obtain thetwo Lagrangeequations
2
q2
mx  Bx  Kx 
 f
2 A
1
 xo  x  q  v
Lq  Rq 
A
20
Robotic Example
 
q    Generalized coordinates (θ angularposition;r radiallength;both vary)
r 
 
Q    Applicableforces to each component; is the torque; f is the force
f
1 2 1 2
J  mr ; V  mgr sin 
2
2
1
1
The powerdissipatio
n : P  B1 2  B2 r 2
2
2
1
1
L  K e  V  J 2  mr 2  mgrsin 
2
2
 P 
 L 
 L 
 P     B1 
L     J  mr 2 L     mgr cos 
  
 
 
;
;

2



q  L  mr  mr  q  L  mr  mg sin( ) q  P   B2 r 
 r 
 r 
 r 
J  mr 2 ; K e 
21
The Lagrange equation becomes
d  L  L P
Q    

dt  q  q q
mr 2  2mrr  mgr cos 
  B1 
Q
   2


mr

 mr  mg sin( )  B2 r 
mr 2 0    B1 2mrθ   mgr cos( )  
 
   

   

m  r   mrθ B 2  r  mg sin( )   f 
0
M (q )q  V (q, q )  G (q)  Q
M (q) is the inertiamatrix; V (q, q ) is the Coriolis/centripetalvector
G (q) is the gravity vector;Q is the input vector
22
Example: Two Mesh Electric Circuit
R1
Ua(t)
C1
L1
q1
L2
L12
C2
q2
R2
Assumeq1 and q 2 as the independent generalized coordinates, where q1 is the electric
chargein thefirst loopand q 2 is the electricchargein thesecondloop.
The generalized force appliedto the systemis denotedas Q1
i1
i2
; q 2  ; Q1  U a (t ).
s
s
The totalmagneticenergy(kineticenergy)is :
We shouldknow that : i1  q1 ; i2  q 2 ; q1 
1
1
1
2
2



K e  L1q1  L12 (q1  q 2 )  L2 q 22
2
2
2
23
K e
K e
 0;
 L1  L12 q1  L12 q 2
q1
q1
K e
K e
 0;
 L2  L12 q 2  L12 q1
q2
q 2
Use the equationfor the totalelectricenergy(potentialenergy)
q
q
1 q12 1 q22 V
V
V

;
 1 and
 2
2 C1 2 C 2 q1 C1
q2 C 2
The totalheat energydissipated: P 
1
1
P
P
R1q12  R2 q 22 ;
 R1q1 and
 R2 q 2
2
2
q1
q 2
K e P V
K e P V
d K e
d K e
(
)


 Q1 ;
(
)


0




dt q1
q1 q1 q1
dt q2
q2 q 2 q 2
( L1  L12 )q1  L12 q2  R1q1 
q1
q2





 U a ; - L12 q1  ( L2  L12 )q 2  R2 q 2 
0
C1
C2
 q1



q2
1
1
 
 L12 q1 
q1 
 R1q1  L12 q2  U a ; q2 
 R2 q 2 
( L1  L12 )  C1
( L2  L12 ) 
C2


24
Another Example
ia(t)
iL(t)
R
Ua(t)
q1
L
C
uc
q2
uL
RL
Use q1 and q2 as the independent generalized coordinates :
ia  q1 ; iL  q 2 ; u a (t )  Q1
K e
1 2 K e
d  K e 

  0

K e  Lq2 ;
 0;
 0;
2
q1
q1
dt  q1 
K e
K e
d  K e 

  Lq2
 0;
 Lq 2 ;
q2
dt  q 2 
q 2
25
2
1 q1  q 2 
The totalpotentialenergyis : V 
2
C
 q1  q 2
V q1  q 2
V

and

q1
C
q 2
C
The totaldissipatedenergyis : P 
1 2 1
Rq1  RL q 22
2
2
P
P
 Rq1 and
 RL q 2
q1
q 2
d  K e

dt  q1
 K e P V
d  K e  K e P V
 

 


 Q1 ;


0



dt  q 2  q 2 q 2 q 2
 q1 q1 q1
q1  q 2
 q1  q 2




Rq1 
 u a ; L q2  RL q 2 
0
C
C
q q 
1   q  q2
1

q1   1
 u a ; q2    RL q 2  1 2 
R
C
L
C 

By usingKirchhoff's law,we get
duc 1  u c
u (t )  di
1
 
 i L  a ; L  u c  RL i L 
dt
C R
R  dt
L
26
Directly-Driven Servo-System
r
Load
TL
ir
r, Te
Rotor
Te : electromagnetictorque
TL : Load torque
ur
Stator
Rr
us
Lr
Ls
is
ir
q1  ; q2  ; q3   r ;
s
s
q1  is ; q 2  ir ; q 3   r ;
Q1  u s ; Q2  u r ; Q3  TL
is
Rs
27
The Lagrange equations are expressed in terms of each independent
coordinate
K e P V
d K e
(
)


 Q1
dt q1
q1 q1 q1
K e P V
d K e
(
)


 Q2
dt q 2
q 2 q 2 q 2
K e P V
d K e
(
)


 Q3
dt q 3
q3 q 3 q3
28
The total kinetic energy is the sum of the total electrical (magnetic) and
mechanical (moment of inertia) energies
1
1
1
Ls q12  Lsr q1q 2  Lr q 22 (Electrical); K em  Jq 32 (Mechanical)
2
2
2
1
1
1 2
2
2




K e  K ee  K em  Ls q1  Lsr q1q2  Lr q2  Jq 3
2
2
2
Ns Nr
Ns Nr
Mutualinductance: Lsr ( r ) 
; LM  Lsr max 
 m ( r )
 m (900 )
K ee 
Lsr ( r )  LM cos r  LM cos q3 ( LM is magnetizing reluctance)
1
1
1
Ls q12  LM q1q 2 cos q3  Lr q 22  Jq 32
2
2
2
K
K
The followingpartialderivatives result: e  0; e  Ls q1  LM q 2 cos q3
q1
q1
Ke 
K e
K e
K e
K e




 0;
 LM q1 cos q3  Lr q 2 ;
  LM q1q 2 sin q3 ;
 Jq 3
q2
q 2
q3
q 3
29
We have only a mechanical potential energy: Spring with a constant ks
The potentialenergyof the springwith constantk s : V 
1
k s q32
2
V
V
V
 0;
 0;
 k s q3
q1
q 2
q3
The totalheat energydissipatedis expressedas : P  PE  PM
1
1
1
Rs q12  Rr q 22 ; PM  Bm q 32
2
2
2
1
1
1
P  Rs q12  Rr q 22  Bm q 32
2
2
2
P
P
P
 Rs q1 ;
 Rr q 2 ; and
 Bm q 3



q1
q 2
q3
PE 
Substituting the originalvalues,we havethree differential equationsfor servo- system
di
di
d
Ls s  LM cos r r  LM i r sin r r  Rs is  u s
dt
dt
dt
di
di
d
Lr r  LM cos r s  LM i s sin r r  Rr ir  u r
dt
dt
dt
d 2 r
d r
J

L
i
i
sin


B
 k s r  TL
M s r
r
m
2
dt
30
dt
dθ r
 ω).
dt
Also,usingstatorcurrent and rotorcurrent,angular velocity,and positionas statevariables
The last equationshouldbe writtenin terms of rotorangular velocity(
dis
1

dt Ls Lr  L2M cos2  r
1 2


  Rs Lr is  LM is r sin 2 r  Rr LM ir cos r  Lr LM  r sin r  Lr u s  LM cos r u r 
2


dir
1
1
1 2




R
L
i

L
L
i

sin


R
L
i

L
i

sin
2


L
cos

u

L
u

s M s
s M s r
r
r s r
M r r
M
r s
s r
dt Ls Lr  L2M cos2  r 
2
2

d r 1
 ( LM is ir sin r  Bm r  k s r  TL )
dt
J
d r
 r
dt
d
1
Considering the third equation: r  ( LM is ir sin r  Bm r  k s r  TL )
dt
J
We can obtain theexpressionfor the electromagnetictorqueTe developed:
Te   LM is ir sin r
31
More Application
Application of Lagrange equations of motion in the modeling of twophase induction motor and generator.
Application of Lagrange equations of motion in the modeling of
permanent-magnet synchronous machines.
Transducers
32