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:
ƒ Use the standard equation of motion (Newton’s Law)
for mechanical systems.
ƒ Use circuits theorems (Ohm’s law and Kirchhoff’s laws:
KCL and KVL).
ƒ Today’s approach utilizes the notation of energy to
model the dynamic system (Lagrange model).
3
•
•
•
•
•
•
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?
5
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
Newton approach requires that we find
accelerations in all three directions,
equate F = ma, solve for the constraint
forces and then eliminate
these to reduce the problem
to " characteristic size".
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.
7
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 α is the angular
acceleration of the body
(rad/s2).
M = jα
8
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 vector indicating a position of the material point in space)
x 
r =  y 
z 
Vector F(r, t) represents a force field, which may be calculated by taking
into account interactions with other particles, or interactions with electromagnetic
waves, or gravitational fields.
The second law of Newton is an idealisation, of course, even if one was to neglect
quantum and relativistic effects. There is no justification why only a second time derivative
of r should appear in that equation. Indeed if energy is dissipated in the system usually
first time derivatives will appear in the equation too. If a material point loses energy due to EM
radiation, third time derivatives will come up.
9
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.
10
Electrical and Mechanical Counterparts
“Energy”
Energy
Mechanical
Electrical
Kinetic
(Active)
Ke
Mass / Inertia
0.5 mv2 / 0.5 jω2
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
11
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 kinetic energy; V is the potential energy)
1 2
mv ; V = mgh
2
d  ∂L  ∂L ∂P
−
+
= Qi
Lagrange's Equation : 
dt  ∂q&i  ∂qi ∂q& i
Ke =
P is power function (half rate at which energy is dissipated); Qi are generalized external inputs
(forces) acting on the system If there are three generalized coordinates, there will be three equations.
Note that the above equation is a second - order differential equation
12
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, θ, φ); 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.
13
Cont..
• Use a coordinate transformation to convert between sets
of generalized coordinates (x = r sin θ cos φ ; y = r sin θ
sin φ ; z = r cos θ ).
• 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.
14
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 
15
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
−
=0
Lagrang' s Equation : 

dt  ∂q& i  ∂qi
d  ∂L 
∂L
∂L

 = m&x&;
= mx&;
= − kx
Do the derivatives :


dt  ∂q& i 
∂q& i
∂qi
d  ∂L  ∂L
−
Combine all together : 
= m&x& + kx = 0

dt  ∂q& i  ∂qi
x
16
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 have the generalized coordinate q = x, and thus with the applied force Q = f , we write
the Lagrange equation :
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
∂x 2
2
d
= (mx& 2 ) − (− Kx − mg ) + ( Bx& )
dt
17
= 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 have the generalized coordinate q (charge), and with the applied force Q = u , we have
u=
d  ∂L  ∂L ∂P
  −
+
&
dt  ∂q  ∂q ∂q&
d ∂ 1 2 1 2
1 2
∂ 1
∂ 1
( ( Lq& −
q )) − ( Lq& 2 −
q ) + ( Rq& 2 )
dt ∂q& 2
2C
2C
∂q 2
∂q& 2
Q
Q
di
d
= ( Lq& ) + + Rq& = Lq&& + + Rq& = L + vc + Ri
C
dt
dt
C
i = q& and q = Cvc for a capacitor. This is just KVL equation
=
18
Electromechanical System: Capacitor Microphone
About them see: http://www.soundonsound.com/sos/feb98/articles/capacitor.html
This system has two degrees of freedom
(electrical and mechanical : charge q and
displacement x from equilibrium)
1 2 1 2
1 2 1 2
Lq& + mx& ; V =
q + Kx
2
2
2C
2

εA  ε is the dielectric constant of the air (F/m),


C=
xo − x  A is the area of the plate, xo - x is the plate separation 
Ke =
1 2 1 2
1 2
(xo − x ) q + Kx ; P = Rq& + Bx&
V=
2
2
2
2ε A
1
1
1
(xo − x )q 2 − 1 Kx 2
L = Lq& 2 + mx& 2 −
2
2
2εA
2
1
2
19
2
∂P
∂L q
∂L
= Bx&
=
− Kx;
= mx&;
∂x&
∂x 2εA
∂x&
∂L
∂L (xo − x )q ∂P
= Lq&;
=
= Rq&
;
∂q&
∂q
∂q&
εA
Then we obtain the two Lagrange equations
q2
m&x& + Bx& + Kx −
= f
2ε A
1
( xo − x ) q = v
Lq&& + Rq& +
εA
20
Robotic Example
θ 
q =   Generalized coordinates (θ angular position; r radial length; both vary)
r 
τ 
Q =   Applicable forces to each component; τ is the torque; f is the force
f
1
1
J = mr 2 ; K e = Jθ& 2 + mr& 2 ; V = mgr sin (θ )
2
2
1
1
The power dissipation : P = B1θ& 2 + B2 r& 2
2
2
1
1
L = K e − V = Jθ& 2 + mr& 2 − mgr sin (θ )
2
2
 ∂P 
 ∂L 
 ∂L 
 ∂P  ∂θ&   B1θ& 
∂L  ∂θ   Jθ&  mr 2θ&  ∂L  ∂θ  − mgr cos(θ )
= =
= = =
= =
;
;

∂q&  ∂L  mr&  mr&  ∂q  ∂L  mrθ& 2 − mg sin(θ ) ∂q&  ∂P   B2 r& 
 ∂r& 
 ∂r& 
 ∂r 
21
The Lagrange equation becomes
d  ∂L  ∂L ∂P
+
Q =   −
dt  ∂q&  ∂q ∂q&
mr 2θ&& + 2mrr&θ& − mgr cos(θ )
  B1θ& 
Q=
 −  &2
+

m&r&

 mrθ − mg sin(θ )  B2 r& 
mr 2 0  θ&&  B1 2mrθ&  θ& mgr cos(θ ) τ 
= 
  + 

  + 

m  &r&  − mrθ& B2  r&  mg sin(θ )   f 
0
M (q )q&& + V (q, q& ) + G (q ) = Q
M (q ) is the inertia matrix; V (q, q& ) is the Coriolis/centripetal vector
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
Assume q1 and q 2 as the independent generalized coordinates, where q1 is the electric
charge in the first loop and q 2 is the electric charge in the second loop.
The generalized force applied to the system is denoted as Q1
i1
i2
; q 2 = ; Q1 = U a (t ).
s
s
The total magnetic energy (kinetic energy) is :
We should know that : i1 = q&1 ; i2 = q& 2 ; q1 =
Ke =
1
1
1
L1q&12 + L12 (q&1 − q& 2 ) 2 + L2 q& 22
2
2
2
23
∂K e
∂K e
= (L1 + L12 )q&1 − L12 q& 2
= 0;
∂q&1
∂q1
∂K e
∂K e
= 0;
= (L2 + L12 )q& 2 − L12 q&1
∂q& 2
∂q2
Use the equation for the total electric energy (potential energy)
q
q
∂V
1 q12 1 q 22 ∂V
+
= 1 and
= 2
V=
;
∂q 2 C 2
2 C1 2 C 2 ∂q1 C1
The total heat energy dissipated : P =
∂P
∂P
1
1
= R1q&1 and
= R2 q& 2
R1q&12 + R2 q& 22 ;
∂q&1
∂q& 2
2
2
∂K e ∂P ∂V
∂K e ∂P ∂V
d ∂K e
d ∂K e
+
+
=0
+
+
= Q1 ;
(
)−
(
)−
&
&
&
&
dt ∂q1
dt ∂q 2
∂q 2 ∂q 2 ∂q 2
∂q1 ∂q1 ∂q1
( 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 −
− R1q&1 + L12 q&&2 + U a ; q&&2 =
− R2 q& 2 
q&&1 =
( 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 q 2 as the independent generalized coordinates :
ia = q&1 ; i L = q& 2 ; u a (t ) = Q1
∂K e
1 2 ∂K e
d  ∂K e 

 = 0
&
= 0;
= 0;
K e = Lq 2 ;
2
∂q1
∂q&1
dt  ∂q&1 
∂K e
∂K e
d  ∂K e

= Lq& 2 ;
= 0;
dt  ∂q& 2
∂q 2
∂q& 2

 = Lq&&2

25
2
1 (q1 − q 2 )
The total potential energy is : V =
2
C
− q1 + q 2
∂V q1 − q 2
∂V
=
=
and
∂q 2
∂q1
C
C
The total dissipated energy is : P =
∂P
= Rq&1 and
∂q&1
d  ∂K e

dt  ∂q&1
1 2 1
Rq&1 + RL q& 22
2
2
∂P
= RL q& 2
∂q& 2
 ∂K e ∂P ∂V
d  ∂K e  ∂K e ∂P ∂V
 −
 −

=0
+
+
+
+
= Q1 ;
&
&
&
dt  ∂q 2  ∂q 2 ∂q 2 ∂q 2
 ∂q1 ∂q1 ∂q1
− q + q2
q −q
=0
Rq&1 + 1 2 = u a ; Lq&&2 + RL q& 2 + 1
C
C
q1 − q 2 
1  − q + q2
1

&q1 =  1
&
&
&
+ u a ; q 2 =  − RL q 2 +

C 
R
C
L

By using Kirchhoff' s law, we get
du c 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
θr
Load
TL
ir
ωr, Te
Rotor
Te : electromagnetic torque
TL : Load torque
ur
Stator
Rr
us
Lr
Ls
is
ir
q1 = ; q 2 = ; q3 = θ r ;
s
s
q&1 = is ; q& 2 = ir ; q& 3 = ω 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
+
∂q1 ∂q&1 ∂q1
dt ∂q&1
∂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 q1q 2 + Lr q 2 + Jq& 3
2
2
2
Ns Nr
Ns Nr
Mutual inductance : Lsr (θ r ) =
; LM = Lsr max =
ℜ m (θ r )
ℜ m (90 0 )
K ee =
Lsr (θ r ) = LM cos θ 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 following partial derivatives result : e = 0; e = Ls q&1 + LM q& 2 cos q3
∂q&1
∂q1
Ke =
∂K e
∂K e
∂K e
∂K e
&
&
&
&
= Jq& 3
= − LM q1q2 sin q3 ;
= LM q1 cos q3 + Lr q 2 ;
= 0;
∂q& 3
∂q3
∂q& 2
∂q 2
29
We have only a mechanical potential energy: Spring with a constant ks
The potential energy of the spring with constant k s : V =
1
k s q32
2
∂V
∂V
∂V
= 0;
= 0;
= k s q3
∂q1
∂q 2
∂q3
The total heat energy dissipated is expressed as : 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 original values, we have three differential equations for servo - system
dis
di
+ LM cos θ r r − LM i r sin θ r
dt
dt
di
di
Lr r + LM cos θ r s − LM i s sin θ r
dt
dt
Ls
J
d 2θ r
dt 2
+ LM is ir sin θ r + Bm
dθ r
+ Rs i s = u s
dt
dθ r
+ Rr i r = u r
dt
dθ r
+ k sθ r = −TL
dt
30
dθ r
= ω).
dt
Also, using stator current and rotor current, angular velocity, and position as state variables
The last equation should be written in terms of rotor angular velocity (
dis
1
=
dt
Ls Lr − L2M cos 2 θ r
1 2


 − Rs Lr is − LM isω r sin 2θ r + Rr LM ir cos θ r + Lr LM ω r sin θ r + Lr u s − LM cos θ r u r 
2


dir
1 2
1
1


R
L
i
L
L
i
R
L
i
L
i
L
u
L
u
ω
θ
ω
sin
2
θ
cos
θ
sin
=
−
−
−
−
−
+

s M s
s M s r
r
r s r
M r r
M
r s
s r
dt
2
2
Ls Lr − L2M cos 2 θ r 

dω r 1
= (− LM is ir sin θ r − Bmω r − k sθ r − TL )
dt
J
dθ r
= ωr
dt
dω
1
Considering the third equation : r = (− LM is ir sin θ r − Bmω r − k sθ r − TL )
dt
J
We can obtain the expression for the electromagnetic torque Te developed :
Te = − LM is ir sin θ r
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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