lecture- 2 introduction mathematical modeling mathematical...
Post on 24-Mar-2020
40 Views
Preview:
TRANSCRIPT
Automatic Control Systems
Lecture- 2Introduction Mathematical Modeling
Mathematical Modeling of Mechanical Systems
1
Lecture Outline
• Introduction to Modeling
– Ways to Study System
– Modeling Classification
• Mathematical Modeling of Mechanical Systems
– Translational Mechanical Systems
– Rotational Mechanical Systems
– Mechanical Linkages
2
Model
• A model is a simplified representation or
abstraction of reality.
• Reality is generally too complex to copy exactly.
• Much of the complexity is actually irrelevant in
problem solving.
3
What is Mathematical Model?
A set of mathematical equations (e.g., differential eqs.) thatdescribes the input-output behavior of a system.
What is a model used for?
• Simulation
• Prediction/Forecasting
• Prognostics/Diagnostics
• Design/Performance Evaluation
• Control System Design
Ways to Study a System
5
System
Experiment with actual System
Experiment with a model of the System
Physical Model Mathematical Model
Analytical Solution
Simulation
Frequency Domain Time Domain Hybrid Domain
Black Box Model
• When only input and output are known.
• Internal dynamics are either too complex orunknown.
• Easy to Model
6
Input Output
Basic Types of Mechanical Systems
• Translational
– Linear Motion
• Rotational
– Rotational Motion
7
Basic Elements of Translational Mechanical Systems
Translational Spring
i)
Translational Mass
ii)
Translational Damper
iii)
Translational Spring
i)
Circuit Symbols
Translational Spring• A translational spring is a mechanical element that
can be deformed by an external force such that thedeformation is directly proportional to the forceapplied to it.
Translational Spring
Translational Spring• If F is the applied force
• Then is the deformation if
• Or is the deformation.
• The equation of motion is given as
• Where is stiffness of spring expressed in N/m
2x1x
02 x1x
)( 21 xx
)( 21 xxkF
k
F
F
Translational Mass
Translational Mass
ii)
• Translational Mass is an inertiaelement.
• A mechanical system withoutmass does not exist.
• If a force F is applied to a massand it is displaced to x metersthen the relation b/w force anddisplacements is given byNewton’s law.
M)(tF
)(tx
xMF
Translational Damper
Translational Damper
iii)
• Damper opposes the rate ofchange of motion.
• All the materials exhibit theproperty of damping to someextent.
• If damping in the system is notenough then extra elements (e.g.Dashpot) are added to increasedamping.
Common Uses of Dashpots
Door StoppersVehicle Suspension
Bridge SuspensionFlyover Suspension
Translational Damper
xCF
• Where C is damping coefficient (N/ms-1).
)( 21 xxCF
Example-1
• Consider the following system (friction is negligible)
15
• Free Body Diagram
MF
kf
Mf
k
F
xM
• Where and are force applied by the spring and inertial force respectively.
kf Mf
Example-1
16
• Then the differential equation of the system is:
xMkxF
• Taking the Laplace Transform of both sides and ignoring initial conditions we get
MF
kf
Mf
Mk ffF
)()()( skXsXMssF 2
17
)()()( skXsXMssF 2
• The transfer function of the system is
kMssF
sX
2
1
)(
)(
• if
12000
1000
Nmk
kgM
2
00102
ssF
sX .
)(
)(
Example-1
18
• The pole-zero map of the system is
2
00102
ssF
sX .
)(
)(
Example-2
-1 -0.5 0 0.5 1
0
𝑗 2
Pole-Zero Map
Real Axis
Ima
gin
ary
Axis
−𝑗 2
Example-2
• Consider the following system
19
• Free Body Diagram
k
F
xM
C
MF
kf
Mf
Cf
CMk fffF
Example-3
20
Differential equation of the system is:
kxxCxMF
Taking the Laplace Transform of both sides and ignoring Initial conditions we get
)()()()( skXsCsXsXMssF 2
kCsMssF
sX
2
1
)(
)(
Example-3
21
kCsMssF
sX
2
1
)(
)(
• if
1
1
1000
2000
1000
msNC
Nmk
kgM
/
1000
00102
sssF
sX .
)(
)(-1 -0.5 0 0.5 1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Pole-Zero Map
Real Axis
Imagin
ary
Axis
Example-4
• Consider the following system
22
• Mechanical Network
k
F
2x
M
1x B
↑ M
k
BF
1x 2x
Example-4
23
• Mechanical Network
↑ M
k
BF
1x 2x
)( 21 xxkF
At node 1x
At node 2x
22120 xBxMxxk )(
Example-5
• Find the transfer function X2(s)/F(s) of the following system.
1M 2M
k
B
Example-6
25
k
)(tf
2x
1M4B3B
2M
1x
1B2B
↑ M1k 1B)(tf
1x 2x3B
2B M24B
Example-7
• Find the transfer function of the mechanical translationalsystem given in Figure-1.
26
Free Body Diagram
Figure-1
M
)(tf
kf
Mf
Bf
BMk ffftf )(kBsMssF
sX
2
1
)(
)(
Example-8
27
• Restaurant plate dispenser
Example-9
28
• Find the transfer function X2(s)/F(s) of the following system.
Free Body Diagram
M1
1kf
1Mf
Bf
M2
)(tF
1kf
2Mf
Bf2kf
2k
BMkk fffftF 221
)(
BMk fff 11
0
Example-10
29
1k
)(tu
3x
1M
4B3B
2M
2x
2B 5B
2k 3k
1x
1B
Basic Elements of Rotational Mechanical Systems
Rotational Spring
)( 21 kT
21
Basic Elements of Rotational Mechanical Systems
Rotational Damper
21
)( 21 CT
T
C
Basic Elements of Rotational Mechanical Systems
Moment of Inertia
JT
TJ
Example-11
1
T 1J
1k1B
2k
2J
2 3
↑ J1
1k
T
1 31B
J2
2
2k
Example-12
↑ J1
1k
1BT
1 32B
3B J24B
2
1
T 1J
1k
3B
2B4B
1B
2J
2 3
Example-13
1T
1J
1k
2B 2J
22k
Example-14
Gear
• Gear is a toothed machine part, suchas a wheel or cylinder, that mesheswith another toothed part totransmit motion or to change speedor direction.
37
Fundamental Properties• The two gears turn in opposite directions: one clockwise and
the other counterclockwise.
• Two gears revolve at different speeds when number of teethon each gear are different.
Gearing Up and Down
• Gearing up is able to convert torque tovelocity.
• The more velocity gained, the more torquesacrifice.
• The ratio is exactly the same: if you get threetimes your original angular velocity, youreduce the resulting torque to one third.
• This conversion is symmetric: we can alsoconvert velocity to torque at the same ratio.
• The price of the conversion is power loss dueto friction.
Why Gearing is necessary?
40
• A typical DC motor operates at speeds that are far too
high to be useful, and at torques that are far too low.
• Gear reduction is the standard method by which a
motor is made useful.
Gear Trains
41
Gear Ratio• You can calculate the gear ratio by using
the number of teeth of the driverdivided by the number of teeth of thefollower.
• We gear up when we increase velocityand decrease torque.Ratio: 3:1
• We gear down when we increase torqueand reduce velocity.Ratio: 1:3
Follower
Driver
𝐺𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑖𝑛𝑝𝑢𝑡 𝑔𝑒𝑎𝑟
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑜𝑢𝑝𝑢𝑡 𝑔𝑒𝑎𝑟=
𝐼𝑛𝑝𝑢𝑡 𝑇𝑜𝑟𝑞𝑢𝑒
𝑂𝑢𝑝𝑢𝑡 𝑇𝑜𝑟𝑞𝑢𝑒=𝑂𝑢𝑡𝑝𝑢𝑡 𝑆𝑝𝑒𝑒𝑑
𝐼𝑛𝑝𝑢𝑡 𝑆𝑝𝑒𝑒𝑑
Example of Gear Trains• A most commonly used example of gear trains is the gears of
an automobile.
43
Mathematical Modeling of Gear Trains
• Gears increase or descrease angular velocity (whilesimultaneously decreasing or increasing torque, suchthat energy is conserved).
44
2211 NN
1N Number of Teeth of Driving Gear
1 Angular Movement of Driving Gear
2N Number of Teeth of Following Gear
2 Angular Movement of Following Gear
Energy of Driving Gear = Energy of Following Gear
Mathematical Modelling of Gear Trains
• In the system below, a torque, τa, is applied to gear 1 (withnumber of teeth N1, moment of inertia J1 and a rotational frictionB1).
• It, in turn, is connected to gear 2 (with number of teeth N2,moment of inertia J2 and a rotational friction B2).
• The angle θ1 is defined positive clockwise, θ2 is defined positiveclockwise. The torque acts in the direction of θ1.
• Assume that TL is the load torque applied by the load connectedto Gear-2.
45
B1
B2
N1
N2
Mathematical Modelling of Gear Trains
• For Gear-1
• For Gear-2
• Since
• therefore
46
B1
B2
N1
N2
2211 NN
11111 TBJa Eq (1)
LTBJT 22222 Eq (2)
12
12
N
N Eq (3)
Mathematical Modelling of Gear Trains
• Gear Ratio is calculated as
• Put this value in eq (1)
• Put T2 from eq (2)
• Substitute θ2 from eq (3)
47
B1
B2
N1
N2
22
11
1
2
1
2 TN
NT
N
N
T
T
22
11111 T
N
NBJa
)( La TBJN
NBJ 2222
2
11111
)( La TN
N
N
NB
N
NJ
N
NBJ
2
12
2
121
2
12
2
11111
Mathematical Modelling of Gear Trains
• After simplification
48
)( La TN
N
N
NB
N
NJ
N
NBJ
2
12
2
121
2
12
2
11111
La TN
NB
N
NBJ
N
NJ
2
112
2
2
11112
2
2
111
La TN
NB
N
NBJ
N
NJ
2
112
2
2
1112
2
2
11
2
2
2
11 J
N
NJJeq
2
2
2
11 B
N
NBBeq
Leqeqa TN
NBJ
2
111
Mathematical Modelling of Gear Trains
• For three gears connected together
49
3
2
4
3
2
2
12
2
2
11 J
N
N
N
NJ
N
NJJeq
3
2
4
3
2
2
12
2
2
11 B
N
N
N
NB
N
NBBeq
Example-15
• Drive Jeq and Beq and relation between appliedtorque τa and load torque TL for three gearsconnected together.
50
J1 J2 J3
1
3
2
τa
1N
2N
3N
1B2B
3B
LT
top related