Lecture- 2 Introduction Mathematical Modeling Mathematical...
Transcript of Lecture- 2 Introduction Mathematical Modeling Mathematical...
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Automatic Control Systems
Lecture- 2Introduction Mathematical Modeling
Mathematical Modeling of Mechanical Systems
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Lecture Outline
• Introduction to Modeling
– Ways to Study System
– Modeling Classification
• Mathematical Modeling of Mechanical Systems
– Translational Mechanical Systems
– Rotational Mechanical Systems
– Mechanical Linkages
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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.
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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
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Ways to Study a System
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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
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Black Box Model
• When only input and output are known.
• Internal dynamics are either too complex orunknown.
• Easy to Model
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Input Output
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Basic Types of Mechanical Systems
• Translational
– Linear Motion
• Rotational
– Rotational Motion
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Basic Elements of Translational Mechanical Systems
Translational Spring
i)
Translational Mass
ii)
Translational Damper
iii)
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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
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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
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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
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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.
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Common Uses of Dashpots
Door StoppersVehicle Suspension
Bridge SuspensionFlyover Suspension
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Translational Damper
xCF
• Where C is damping coefficient (N/ms-1).
)( 21 xxCF
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Example-1
• Consider the following system (friction is negligible)
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• Free Body Diagram
MF
kf
Mf
k
F
xM
• Where and are force applied by the spring and inertial force respectively.
kf Mf
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Example-1
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• 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
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)()()( skXsXMssF 2
• The transfer function of the system is
kMssF
sX
2
1
)(
)(
• if
12000
1000
Nmk
kgM
2
00102
ssF
sX .
)(
)(
Example-1
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• 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
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Example-2
• Consider the following system
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• Free Body Diagram
k
F
xM
C
MF
kf
Mf
Cf
CMk fffF
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Example-3
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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
)(
)(
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Example-3
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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
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Example-4
• Consider the following system
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• Mechanical Network
k
F
2x
M
1x B
↑ M
k
BF
1x 2x
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Example-4
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• Mechanical Network
↑ M
k
BF
1x 2x
)( 21 xxkF
At node 1x
At node 2x
22120 xBxMxxk )(
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Example-5
• Find the transfer function X2(s)/F(s) of the following system.
1M 2M
k
B
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Example-6
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k
)(tf
2x
1M4B3B
2M
1x
1B2B
↑ M1k 1B)(tf
1x 2x3B
2B M24B
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Example-7
• Find the transfer function of the mechanical translationalsystem given in Figure-1.
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Free Body Diagram
Figure-1
M
)(tf
kf
Mf
Bf
BMk ffftf )(kBsMssF
sX
2
1
)(
)(
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Example-8
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• Restaurant plate dispenser
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Example-9
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• 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
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Example-10
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1k
)(tu
3x
1M
4B3B
2M
2x
2B 5B
2k 3k
1x
1B
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Basic Elements of Rotational Mechanical Systems
Rotational Spring
)( 21 kT
21
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Basic Elements of Rotational Mechanical Systems
Rotational Damper
21
)( 21 CT
T
C
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Basic Elements of Rotational Mechanical Systems
Moment of Inertia
JT
TJ
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Example-11
1
T 1J
1k1B
2k
2J
2 3
↑ J1
1k
T
1 31B
J2
2
2k
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Example-12
↑ J1
1k
1BT
1 32B
3B J24B
2
1
T 1J
1k
3B
2B4B
1B
2J
2 3
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Example-13
1T
1J
1k
2B 2J
22k
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Example-14
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Gear
• Gear is a toothed machine part, suchas a wheel or cylinder, that mesheswith another toothed part totransmit motion or to change speedor direction.
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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.
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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.
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Why Gearing is necessary?
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• 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.
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Gear Trains
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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
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Example of Gear Trains• A most commonly used example of gear trains is the gears of
an automobile.
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Mathematical Modeling of Gear Trains
• Gears increase or descrease angular velocity (whilesimultaneously decreasing or increasing torque, suchthat energy is conserved).
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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
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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.
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B1
B2
N1
N2
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Mathematical Modelling of Gear Trains
• For Gear-1
• For Gear-2
• Since
• therefore
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B1
B2
N1
N2
2211 NN
11111 TBJa Eq (1)
LTBJT 22222 Eq (2)
12
12
N
N Eq (3)
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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)
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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
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Mathematical Modelling of Gear Trains
• After simplification
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)( 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
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Mathematical Modelling of Gear Trains
• For three gears connected together
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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
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Example-15
• Drive Jeq and Beq and relation between appliedtorque τa and load torque TL for three gearsconnected together.
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J1 J2 J3
1
3
2
τa
1N
2N
3N
1B2B
3B
LT