Transcript Document

Dynamic Responses of Systems
The most important function of a model devised for measurement or
control systems is to be able to predict what the output will be for a
particular input.
We are not only interested with a static situation but wanted to see
how the output will change with time when there is a change of input
or when the input changes with time.
To do so, we need to form equations which will indicate how the
system output will vary with time when the input is varying with time.
This can be done by the use of differential equations.
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Examples of Dynamic Systems
• An example of a first order
system is water flowing out of
a tank. q1 is the forcing input.
• Another
example,
is
a
thermometer being placed in a
hot liquid at some temperature.
The rate at which the reading
of the thermometer changes
with time. TL is the forcing
input.
dh
RA
 pgh  q1
dt
dT
RC
 T  TL
dt
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Electromechanical System Dynamics, energy
Conversion, and Electromechanical Analogies
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)
 Another approach utilizes the notation of energy to model the
dynamic system (Lagrange model).
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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 in the
coordinate system used should
be studied.
 F  ma
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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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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.
, D,  are the total kinetic, dissipation, and potential energies of the system. qi
and Qi are the generalized coordinates and the generalized applied forces
(input).

d  d

dt  .
 d qi

 d dD d
 . 
 Qi

dqi
 dqi d q
i

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Electrical and Mechanical Counterparts
Energy
Mechanical
Electrical
Kinetic
Mass / Inertia
0.5 mv2 / 0.5 j2
Inductor
0.5 Li2
Potential
Gravity: mgh
Spring: 0.5 kx2
Capacitor
0.5 Cv2
Dissipative
Damper / Friction
0.5 Bv2
Resistor
Ri2
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Mathematical Model for a Simple Pendulum
1 2 1  
The kinetic energy of the pendulum bob is :   mv  m l  
2
2  
The potential energy is :   mgh  mgl 1  cos 
.
2
y
Ta, 
0
l

mg
x
x
y
mg cos
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Electrical Conversion
Input
Electrical Energy
Output
Mechanical
Energy
Coupling
Irreversible
Electromagnetic Energy Conversion
Field
Energy Losses
Energy Transfer in Electromechanical Systems
For rotational motion, the electromag netic torque, as a function
dWc (i,  )
of current and angular displaceme nt, is : Te (i, ) 
d
Where Wc  di; where  is the flux.
i
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Electromechanical Analogies
•
From Newton’s law or using Lagrange equations of motions, the secondorder differential equations of translational-dynamics and torsionaldynamics are found as
m
j
d 2x
dt 2
d 2
dt 2
 Bv
dx
 k s x  Fa (t ) (Translational dynamics)
dt
 Bm
d
 k s  Ta (t ) (Torsional dynamics)
dt
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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.
d 2i
R di
1
1 dva


i
2
L dt LC
L dt
dt
R
1
2
s  s
0
L
LC
The characteristic roots are
2
s1  
R
1
 R 
 


2L
LC
 2L 
2
R
1
 R 
s2  
 
 
2L
LC
 2L 
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Translational Damper, Bv (N-sec)
Appied force Fa (t ) in Newton
Linear vel ocity v(t ) (m/sec)
Linear position x(t ) (m)
Fa (t )  Bm v(t )
1
v(t ) 
Fa (t )
Bm
dx(t )
Fa (t )  Bm v(t )  Bm
dt
1 t
x(t ) 
Fa (t )dt

Bv t0
x(t)
Fa(t)
Bm
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Translational Spring, k (N)
Appied force Fa (t ) in Newton
Linear vel ocity v(t ) (m/sec)
Linear position x(t ) (m)
Fa (t )  k s x(t )
1
x(t ) 
Fa (t )
ks
x(t)
Fa(t)
dx(t ) 1 dFa (t )
v(t ) 

dt
k s dt
t
Fa (t )  k s  v(t )dt
t0
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Rotational Damper, Bm (N-m-sec/rad)
Appied torque Ta (t ) (N - m)
Angular velocity  (t ) (rad/sec)
Angular displaceme nt  (t ) (rad)
Ta (t )  Bm (t )
1
 (t ) 
Ta (t )
Bm
 (t)  (t)
Fa(t)
Bm
d (t )
Ta (t )  Bm (t )  Bm
dt
1 t
 (t ) 
Ta (t )dt

Bm t 0
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Rotational Spring, ks (N-m-sec/rad)
Appied torque Ta (t ) (N - m)
Angular velocity  (t ) (rad/sec)
Angular displaceme nt  (t ) (rad)
Ta (t )  Bm (t )
1
 (t )  Ta (t )
ks
d (t ) 1 dTa (t )
 (t ) 

dt
k s dt
 (t)  (t)
Fa(t)
ks
t
Ta (t )  k s   (t )dt
t0
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Mass Grounded, m (kg)
Appied torque Ta (t ) (N - m)
Linear vel ocity v(t ) (m/sec)
Linear position x(t ) (m)
dv
d 2 x(t )
Fa (t )  m
m
dt
dt 2
x (t) v (t)
Fa(t)
m
1 t
v(t )   Fa (t )dt
m t0
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Mass Grounded, m (kg)
Appied torque Ta (t ) (N - m)
Angular velocity  (t ) (rad/sec)
Angular displaceme nt  (t ) (rad)
d
d 2 (t )
Ta (t )  J
J
dt
dt 2
1 t
 (t )   Ta (t )dt
J t0
 (t)  (t)
Fa(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
x(0) initial condition
y(t) Output
u(t) Input
Dynamic System
State x(t)
The state variables describe the future response of a system,
given the present state, the excitation inputs,
and the equations describing the dynamics
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Electrical Example: An RLC Circuit
x1  vC (t ); x2  i L (t )
  1 / 2 Li L2  1 / 2 Cvc2
L
iL
x1 (t 0 ) and x2 (t 0 ) is the total initial
energy of the network
USE KCL at the junction
ic  C
iC
u(t) vC
C
R
dvc
 u (t )  i L
dt
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The State Differential Equation
.
x1  a11 x1  a12 x2  ...  a1n xn  b11u1  ...  b1m u m
.
x 2  a 21 x1  a 22 x2  ...  a 2 n xn  b21u1  ...  b2 m u m
.
x n  a n1 x1  a n 2 x2  ...  a nn xn  bn1u1  ...  bnmu m
State Vector
 x1  a11 a12 a1n
  
d  x2  a 21 a 22 a 2 n

.


. .
dt .
  
 xn  a n1 a n 2 a nn
  x1 
   b11....b1m  u1 
  x2   .......... ..  . 
 
 .  
   bn1....bnm  u m 
  xn 
.
x  Ax  Bu (State Differenti al Equation) A : State matrix; B : input matrix
C : Output matrix; D : direct transmission matrix
y  Cx  Du (Output Equation)
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The Output Equation
y1  h11 x1  h12 x2  ...  h1n xn
y 2  h21 x1  h22 x2  ...  h2 n xn
yb  ab1 x1  ab 2 x2  ...  abn xn
.
 y1  h11 h12 h1n
  
y 2  h21 h22 h2 n

y

.  .
. .
  
 yb  hb1 hb 2 hbn
  x1 
 
  x 2   Hx
 . 
 
  xn 
x  Ax  Bu (State Differenti al Equation) A : State matrix; B : input matrix
H : Output matrix; D : direct tra nsmission matrix
y  Hx  Du (Output Equation)
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Example 1: Consider the given series RLC circuit. Derive the
differential equations that map the circuitry dynamics.
dvc
C
i
dt
di
L  vc  Ri  v(t )
dt
dvc 1
 i
dt
C
di 1
  vc  Ri  v(t ) 
dt L
R
i(t)
L
V(t)
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.
dv(t ) 1
 i
dt
C
di 1
 (vc  Ri  v(t ))
dt L
x1 (t )  vc (t ); x2 (t )  i (t ) (These are the states)
v(t ) is the control
 dx1   dvc  
0





dx
dt
dt



dt  dx2   di   1
 dt   dt   L
1
0 

C v c   
    1 va  Ax  Bu
R  i   
 L 

L
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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)
2
0
0 
A
and B   

- 10 - 20 
10 
 x10  15 
x0      
 x20  5 
vc 
y  1 0    Hx; H  1 0
i 
v c 
y  1 0    0va  Hx  Du
i 
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