Electromechanical System Dynamics, energy Conversion, and ...rhabash/ELG4112LN03.pdf · Electrical...
Transcript of Electromechanical System Dynamics, energy Conversion, and ...rhabash/ELG4112LN03.pdf · Electrical...
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Electromechanical System Dynamics, energy Conversion, and Electromechanical Analogies
Modeling of Dynamic SystemsModeling 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)
Another approach utilizes the notation of energy to model the dynamic system (Lagrange model).
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Mathematical Modeling and System DynamicsNewtonian 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 in the coordinate system used should be studied.
∑ = aF m
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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.
Fcoulomb :Force
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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 (N-m); J is the moment of inertial about its center of mass (kg/m2); and α is the angular acceleration of the body (rad/s2).
αjM =
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The Lagrange Equations of Motion• Although Newton’s laws of motion form the fundamental foundation for the study
of mechanical systems, they can be straightforwardly used to derive the dynamics of electromechanical motion devices because electromagnetic and circuitry transients behavior must be considered. This means, the circuit dynamics must be incorporated to find augmented models.
• This can be performed by integrating torsional-mechanical dynamics and sensor/actuator circuitry equations, which can be derived using Kirchhoff’s laws.
• Lagrange concept allows one to integrate the dynamics of mechanical and electrical components. It employs the scalar concept rather the vector concept used in Newton’s law of motion to analyze much wider range of systems than F=ma.
• With Lagrange dynamics, focus is on the entire system rather than individual components.
• Γ, D, Π are the total kinetic, dissipation, and potential energies of the system. qiand Qi are the generalized coordinates and the generalized applied forces (input).
ii
ii
i
Qdqd
qd
dDdqd
qd
ddtd
=Π
++Γ
−
Γ
..
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Electrical and Mechanical Counterparts
ResistorRi2
Damper / Friction0.5 Bv2
Dissipative
Capacitor0.5 Cv2
Gravity: mghSpring: 0.5 kx2
Potential
Inductor0.5 Li2
Mass / Inertia0.5 mv2 / 0.5 jω2
Kinetic
ElectricalMechanicalEnergy
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Mathematical Model for a Simple Pendulum
( )θ
θ
cos1 :isenergy potential The21
21 :is bob pendulum theofenergy kinetic The
2.2
−==Π
==Γ
mglmgh
lmmv
y0
x
y
x
mgmg cosθ
l
θ
Ta, θ
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Electrical Conversion
InputElectrical Energy
OutputMechanical
Energy
CouplingElectromagnetic
Field
Irreversible Energy Conversion
Energy Losses
Energy Transfer in Electromechanical Systems
flux. theis where;di WWhere
),(),( :is nt,displacemeangular andcurrent of
function a as torque,neticelectromag themotion, rotationalFor
ic ψψ
θθ
θ
∫=
=didW
iT ce
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Electromechanical Analogies
• From Newton’s law or using Lagrange equations of motions, the second-order differential equations of translational-dynamics and torsional-dynamics are found as
dynamics) (Torsional )(
dynamics) onal(Translati )(
2
2
2
2
tTkdtdB
dtdj
tFxkdtdxB
dtxdm
asm
asv
=++
=++
θθθ
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For a series RLC circuit, find the characteristic equation and define the analytical relationships between the characteristic roots and circuitry parameters.
LCLR
LRs
LCLR
LRs
LCs
LRs
dtdv
Li
LCdtdi
LR
dtid a
122
122
are roots sticcharacteri The
01
11
2
2
2
1
2
2
2
−
+−=
−
−−=
=++
=++
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Resistance, R (ohm)
v(t) R
i(t)
)(1)(
)()()(Current
)( voltageAppied
tvR
ti
tRitvti
tv
=
=
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Inductance, L (H)
v(t) L
i(t)
∫=
=
t
tdttv
Lti
dttdiLtv
titv
0
)(1)(
)()(
)(Current )( voltageAppied
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Capacitance, C (F)
v(t) C
i(t)
dttdvCti
dttiC
tv
titv
t
t
)()(
)(1)(
)(Current )( voltageAppied
0
=
= ∫
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Translational Damper, Bv (N-sec)
Fa(t)
x(t)
∫=
==
=
=
t
ta
v
mma
am
ma
dttFB
tx
dttdxBtvBtF
tFB
tv
tvBtFtxtvtF
0
)(1)(
)()()(
)(1)(
)()((m) )(position Linear (m/sec) )(ocity Linear vel
Newtonin )( force Appied a
Bm
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Translational Spring, k (N)
Fa(t)
x(t)
∫=
==
=
=
t
tsa
a
s
as
sa
dttvktF
dttdF
kdttdxtv
tFk
tx
txktFtxtvtF
0
)()(
)(1)()(
)(1)(
)()((m) )(position Linear (m/sec) )(ocity Linear vel
Newtonin )( force Appied a
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Rotational Damper, Bm (N-m-sec/rad)
Fa(t)
θ (t)
∫=
==
=
=
t
ta
m
mma
am
ma
dttTB
t
dttdBtBtT
tTB
t
tBtTt
ttT
0
a
)(1)(
)()()(
)(1)(
)()((rad) )(nt displacemeAngular
(rad/sec) )(locity Angular vem)-(N )( torqueAppied
θ
θω
ω
ωθ
ω
ω (t)
Bm
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Rotational Spring, ks (N-m-sec/rad)
Fa(t)
θ (t)
∫=
==
=
=
t
tsa
a
s
as
ma
dttktT
dttdT
kdttdt
tTk
t
tBtTt
ttT
0
)()(
)(1)()(
)(1)(
)()((rad) )(nt displacemeAngular
(rad/sec) )(locity Angular vem)-(N )( torqueAppied a
ω
θω
θ
θθ
ω
ω (t)
ks
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Mass Grounded, m (kg)
Fa(t)
x (t)
∫=
==
t
ta
a
dttFm
tv
dttxdm
dtdvmtF
txtvtT
0
)(1)(
)()(
(m) )(position Linear (m/sec) )(ocity Linear vel
m)-(N )( torqueAppied
2
2
a
v (t)
m
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Mass Grounded, m (kg)
Fa(t)
θ (t)
∫=
==
t
ta
a
dttTJ
t
dttdJ
dtdJtT
tt
tT
0
2
2
a
)(1)(
)()(
(rad) )( nt displacemeAngular (rad/sec) )( locity Angular ve
m)-(N )( torqueAppied
ω
θω
θω
ω (t)
m
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Steady-State Analysis• State: The state of a dynamic system is the smallest set of variables
(called state variables) so that the knowledge of these variables at t= t0, together with the knowledge of the input for t ≥ t0, determines the behavior of the system for any time t ≥ t0.
• State Variables: The state variables of a dynamic system are the variables making up the smallest set of variables that determine the state of the dynamic system.
• State Vector: If n state variables are needed to describe the behavior of a given system, then the n state variables can be considered the n components of a vector x. Such vector is called a state vector.
• State Space: The n-dimensional space whose coordinates axes consist of the x1 axis, x2 axis, .., xn axis, where x1, x2, .., xn are state variables, is called a state space.
• State-Space Equations: In state-space analysis we are concerned with three types of variables that are involved in the modeling of dynamic system: input variables, output variables, and state variables.
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State Variables of a Dynamic System
Dynamic SystemState x(t)
u(t) Input y(t) Output
x(0) initial condition
dynamics thedescribing equations theand inputs, excitation thestate,present given the
system, a of response future thedescribe variablesstate The
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Electrical Example: An RLC Circuit
C
L
Ru(t)
iL
iC
vC
( ) ( )
Lc
c
cL
LC
itudtdv
Ci
txtxCvLi
tixtvx
−+==
+=
==
)(
junction at the KCL USEnetwork theofenergy
initial total theis )( and )(2/12/1
)( );(
0201
2221
ξ
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The State Differential Equation
Equation) alDifferenti (StateBu Axx.
+=
Equation)(Output Du Cxy +=
+
=
++++++=
++++++=
++++++=
mnmn
m
nnnnn
n
n
n
mnmnnnnnnn
mmnn
mmnn
u
u
bb
bb
x
xx
aaa
aaaaaa
x
xx
dtd
ububxaxaxax
ububxaxaxax
ububxaxaxax
.....
................
.
. . .
.
......
......
......
1
1
1112
1
21
22221
11211
2
1
112211
.212122221212
.111112121111
.
State Vector
matrix nsmission direct tra :D matrix;Output :Cmatrixinput :B matrix; State :A
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The Output Equation
Equation) alDifferenti (StateBu Axx.
+=
Equation)(Output Du Hxy +=
x
x
xx
hhh
hhhhhh
y
yy
y
xaxaxayxhxhxhyxhxhxhy
nbnbb
n
n
b
nbnbbb
nn
nn
H.
. . .
.
...
......
2
1
21
22221
11211
2
1
2211
22221212
12121111
=
=
=
+++=+++=+++=
matrix nsmission direct tra :D matrix;Output :Hmatrixinput :B matrix; State :A
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Example 1: Consider the given series RLC circuit. Derive the differential equations that map the circuitry dynamics.
V(t)
R
L
C
i(t)
( ))(1
1
)(
tvRivLdt
di
iCdt
dv
tvRivdtdiL
idtdv
C
c
c
c
c
+−−=
=
+−−=
=
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Example 2: Using the state-space concept, find the state-space model and analyze the transient dynamics of the series RLC circuit.
BuAxvL
iv
LR
L
C
dtdidtdv
dtdxdtdx
dtdx
tvtitxtvtx
tvRivLdt
di
iCdt
tdv
ac
c
c
c
+=
+
−=
=
=
==
+−−=
=
10
- 1
1 0
control theis )(states) theare (These )()( );()(
))((1
1)(
2
1
21
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Continue with Values..• Assume R = 2 ohm, L = 0.1 H,
and C = 0.5 F, find the following coefficients.
• The initial conditions are assumed to be vc(t0)=vc0=15 V; and I (t0) = i0 = 5 A.
• Let the voltage across the capacitor be the output; y(t)= vc(t). The output equation will be
• The expanded output equation in y
• The circuit response depends on the value of v (t)
=
=
100
B and 20- 10-2 0
A
=
=
515
20
100 x
xx
[ ] [ ]0 1H ;H 0 1 ==
= x
iv
y c
[ ] [ ] DuHxviv
y ac +=+
= 0 0 1