Download Mathematical Modelling of Mechanical and Electrical Systems

Mathematical Modelling of
Mechanical and Electrical
Systems
Prepared By:
Div-ME02
NAME
Enrolment No.
Antala Viral J
130010119004
Bhanderi Sagar S
130010119009
Bhingaradiya Summit
140013119002
Mathematical Modeling
Mathematical modeling seeks to gain an
understanding of science through the use
of mathematical models on HP computers.
Mathematical Models
Think how systems behave with time when
subject to some disturbances.
 In order to understand the behavior of systems,
mathematical models are required.
 Mathematical models are equations which
describe the relationship between the input and
output of a system.
 The basis for any mathematical model is provided
by the fundamental physical laws that govern the
behavior of the system.

Mathematical Modeling
Is often used in place of experiments
when experiments are too large, too
expensive, too dangerous, or too time
consuming.
 Can be useful in “what if” studies; e.g. to
investigate the use of pathogens (viruses,
bacteria) to control an insect population.
 Is a modern tool for scientific investigation.

Building Blocks

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Systems can be made up from a range of building
blocks.
Each building block is considered to have a single
property or function.
Example: an electric circuit system which is made
up from blocks which represent the behaviour of
resistance, capacitance, and inductor, respectively.
By combining these building blocks a variety of
electrical circuit systems can be built up and the
overall input-output relationship can be obtained.
A system built in this way is called a lumped
parameter system.
Mechanical Systems
Translational
• Mass
• Spring
• Damper/dashpot
Rotational
• Inertia
• Damper
• Spring
Mechanical System Building
Blocks

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Basic building block: spring, dashpots, and masses.
Springs represent the stiffness of a system
Dashpots/Damper represent the forces opposing
motion, for example frictional or damping effects.
Masses represent the inertia or resistance to
acceleration.
Mechanical systems does not have to be really made
up of springs, dashpots, and masses but have the
properties of stiffness, damping, and inertia.
All these building blocks may be considered to have a
force as an input and displacement as an output.
Mass



The mass exhibits the property that the bigger the mass the
greater the force required to give it a specific acceleration.
The relationship between the force F and acceleration a is
Newton’s second law as shown below.
Energy is needed to stretch the spring, accelerate the mass and
move the piston in the dashpot. In the case of spring and mass
we can get the energy back but with the dashpot we cannot.
Force
Mass
Acceleration
dv
d 2x
F  ma  m
m
2
dt
dt
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
dx(t ) 1 dFa (t )
v(t ) 

dt
k s dt
t
Fa (t )  k s  v(t )dt
t0
x(t)
Fa(t)
Dashpot/Damper
The dashpot block represents the types of forces
experienced when pushing an object through a fluid or
move an object against frictional forces. The faster the
object is pushed the greater becomes the opposing
forces.
 The dashpot which represents these damping forces that
slow down moving objects consists of a piston moving in
a closed cylinder.
 Movement of the piston requires the fluid on one side of
the piston to flow through or past the piston. This flow
produces a resistive force. The damping or resistive force
is proportional to the velocity v of the piston: F = cv or F
= c dv/dt.

Rotational Damper, Bm (N-m-sec/rad)
Appied torque Ta (t ) (N - m)
Angular velocity  (t ) (rad/sec)
Angular displacement  (t ) (rad)
 (t)  (t)
Ta (t )  Bm (t )
1
 (t ) 
Ta (t )
Bm
d (t )
Ta (t )  Bm (t )  Bm
dt
1 t
 (t ) 
Ta (t )dt

Bm t 0
Fa(t)
Bm
Rotational Spring, ks (N-m-sec/rad)
Appied torque Ta (t ) (N - m)
Angular velocity  (t ) (rad/sec)
Angular displacement  (t ) (rad)
Ta (t )  Bm (t )
1
 (t )  Ta (t )
ks
d (t ) 1 dTa (t )
 (t ) 

dt
k s dt
t
Ta (t )  k s   (t )dt
t0
 (t)  (t)
Fa(t)
ks
Mechanical Building Blocks
Building Block
Spring
Damper
Mass
Spring
Damper
Moment of
inertia
Equation
Translational
F = kx
F = c dx/dt
F = m d2x/dt2
Rotational
T = k
T = c d/dt
T = J d2/dt2
Energy representation
E = 0.5 F2/k
P = cv2
E = 0.5 mv2
E = 0.5 T2/k
P = c2
P = 0.5 J2
Building Mechanical Blocks
Output, displacement

Mass
Mathematical model of a machine
mounted on the ground
d 2x
dx
m
 c  kx  F
2
dt
dt
Ground
Input, force
Building Mechanical Blocks
Moment of inertia

Torque
Torsional resistance
Mathematical model of a rotating
a mass
d 2
d
J
c
 k  T
2
dt
dt
Block model
Shaft
Physical situation
Example 1

Consider the following system
k
F
x
M
C
• Free Body
Diagram
fk
F
M
fC
fM
F  f k  f M  fC
16
Example 1
Differential equation of the system is:
F  Mx  Cx  k x
Taking the Laplace Transform of both
sides and ignoring
Initial conditions we get
F ( s )  Ms 2 X ( s )  CsX ( s )  kX ( s )
X (s)
1

F(s)
Ms 2  Cs  k
17
Example 1
X (s)
1

F(s)
Ms 2  Cs  k
Pole-Zero Map
• if
2
1.5
M  1000 kg
1
C  1000 N / ms 1
Imaginary Axis
k  2000 Nm
1
0.5
0
-0.5
-1
X (s)
0.001
 2
F(s)
s  s  1000
-1.5
-2
-1
-0.5
0
0.5
1
Real Axis
18
Example 2

Find the transfer function of the mechanical
translational system given in Figure-1.
Free Body Diagram
fk
Figure-1
fB
M
f (t )
f (t )  f k  f M  f B
fM
X (s)
1

F(s)
Ms 2  Bs  k
19
Automobile Suspension
20
Automobile Suspension
mxo  b( xo  xi )  k ( xo  xi )  0
(eq .1)
mxo  bxo  kxo  bxi  kxi
eq. 2
Taking Laplace Transform of the equation (2)
ms2 X o ( s )  bsX o ( s )  kXo ( s )  bsX i ( s )  kXi ( s )
X o (s)
bs  k

X i ( s ) ms 2  bs  k
21
Train Suspension
22
Train Suspension
23
Electrical System Building
Blocks

The basic building blocks of electrical systems are
resistance, inductance and capacitance.
2
Resistor : v  iR; P  i R
1
1 2
Inductor : i   vdt; E  Li
L
2
dv
1 2
Capacitor : i  C ; E  Cv
dt
2
Resistance, R (ohm)
Appied voltage v(t )
i(t)
Current i (t )
v(t )  Ri (t )
1
i (t )  v(t )
R
v(t)
R
Inductance, L (H)
Appied voltage v(t )
Current i (t )
di(t )
v(t )  L
dt
t
1
i (t )   v(t )dt
L t0
i(t)
v(t)
L
Capacitance, C (F)
Appied voltage v(t )
Current i (t )
1 t
v(t )   i (t )dt
C t0
dv(t )
i (t )  C
dt
i(t)
v(t)
C
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
s2  
R
1
 R 
 


2L
LC
 2L 
The Transfer Function of Linear
Systems
V2( s )
R2
R2
V1( s )
R
R1  R2
R2

R
 max

V2( s )
ks  1( s )   2( s )
V2( s )
ks  error( s )
ks
Vbattery
 max

Mixed Systems
• Most systems in mechatronics are of the mixed
type, e.g., electromechanical, hydro mechanical, etc
• Each subsystem within a mixed system can be
modeled as single discipline system first
• Power transformation among various subsystems
are used to integrate them into the entire system
• Overall mathematical model may be assembled
into a system of equations, or a transfer function
Electro-Mechanical Example
Power Transformation:
Ra
Torque-Current:
Voltage-Speed:
Tmotor  Kt i a
eb  K bω
u
La
ia
where Kt: torque constant, Kb: velocity constant For an ideal motor
Kt  K b
Combing previous equations results in the following mathematical
model:
 di a
 Ra i a  K b   u
La
 dt

  Bω - K t i a  0
Jω
B
dc

Transfer Function of
Electromechanical Example
Taking Laplace transform of the system’s differential
equations with zero initial conditions gives:
La s  Ra I a (s )  K b (s )  U (s )

u
Js  B (s) - Kt I a (s )  0
Ra
La
B
ia
Kt
Eliminating Ia yields the input-output transfer function
Kt
Ω(s)

U(s) La Js 2  JRa  BLa   BR a  K t K b

Reference
Modern Control Theory by Katsuhiko Ogata, Pearson
Edu. Int.
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