3/24/2015
ME 461 Control Engineering Semester: February 2015 Dr. Sumon Saha Assistant Professor Department of Mechanical Engineering Bangladesh University of Engineering and Technology Tuesday, March 24, 2015
Control Engineering
Today’s topic
Mathematical Modelling & Transfer Function
Control Course Syllabus Engineering Introduction to control systems and their representation by different different equations equations and Laplace Laplace transforms Block diagrams and transfer functions Analog computer solution of system equations System response, control action and system types t ypes Frequency response System analysis System compensation Analogues to control systems Hydraulic and pneumatic control systems Elements of electro-mechanical controls Introduction to digital computer control
Control Engineering
Transfer function
• What is Transfer Function? an algebr algebraic aic expres expressio sion n for the dynamic relati relation on betwee between n the the inpu inputt and outpu utputt of the process model. It is defined so as to be indepe independe ndent nt of the initial initial conditio conditions ns and of the particular choice of forcing function. – Properties: It can be derive derived d only for a linear differential equation model because Laplac Laplace e transf transform orm can be applied only to linear equations. If model is non-linear, then it must be linearized first. – Advantage: Advantage: It is easy to interpret and use in calculating output responses for particular input changes.
3/24/2015
Control Engineering
Transfer function
Control Engineering
Transfer function
Control Engineering
Modeling Mechanical System (Translational)
A transfer function (TF) relates one input and one output: u (t ) U ( s )
→ system →
y (t ) Y ( s )
Transfer function of a system, G(s), is defined as the ratio of the Laplace Transform (LT) of the output variable, Y(s), to the LT of the input variable, U(s), with all the initial conditions are assumed to be zero. Y ( s) G(s) =
where:
U ( s)
Y ( s) = L[ y (t )
] U (s ) = L[ u (t ) ]
Control Engineering
Properties of Transfer Functions
• Additive property – Y (s ) = G 1(s )U 1(s )+ G 2 (s )U 2 (s )
• Multiplicative property – Y 2 (s ) = G 1(s )G 2 (s )U (s )
• ODE equivalence Transfer function Equivalent ODE
Y ( s )
= G (s) =
b1 s + b0
a2 s 2 + a1 s + a0 dy du a2 + a1 + a0 y = b1 + b0 u 2 dt dt dt U ( s ) d 2 y
Viscous friction produces a restraining force proportional to velocity (similar to damping effect of a damper or dashpot).
3/24/2015
Control Engineering
Building Mechanical System
Control Engineering
Building Mechanical System
What is the transfer function for this system? b (dx/dt )
y (‘+’ve)
x (‘+’ve)
m (d 2 x/dt 2 )
m
F
kx Free body diagram
2
F = M
d y dt 2
Free body diagram Mathematical model of system Spring-mass-damper system with an input. Find Y (s )/ X( s )?
↓
∑ F = 0 ⇒M
M s 2Y ( s ) − sy ( 0 ) − y ′ ( 0) + c sY ( s ) − y ( 0) + kY ( s ) = kX ( s) ⇒ Ms 2 + cs + k Y ( s ) = kX ( s )
Control Engineering
⇒
2
d y dt 2
+c
Y ( s) X ( s )
=
Mathematical system the
Spring-mass-damper system with an input. Find X(s)/F(s)?
dy
+ ky = kx dt k Ms + cs + k 2
Example: Transfer Function
2
→
d x
∑ F = 0 ⇒m dt
m s 2 X ( s ) − sx ( 0 ) − x′ ( 0) + b sX ( s ) − x ( 0) + kX ( s ) = F ( s )
⇒ ms + bs + k X ( s ) = F ( s ) 2
Control Engineering
model
⇒
2
+b
X ( s ) F ( s)
of
the
dx
+ kx = F dt
=
1 ms + bs + k 2
Example: Transfer Function
Find the transfer function, X 2(s )/ F( s ) for this system?
(a) Forces on M 1 due to only motion of M 1
(c) All Forces on M 1
(b) Forces on M 1 due to only motion of M 2
3/24/2015
Control Engineering
Control Engineering
Example: Transfer Function
(a) Forces on M 2 due to only motion of M 2
(b) Forces on M 2 due to only motion of M 1
Example: Transfer Function
From Force balance on M 1
( K + K ) X ( s) + ( f + 1
2
1
v1
f v 3 ) sX 1 ( s ) + M 1 s 2 X 1 ( s ) − F ( s ) − K 2 X 2 ( s ) − f v3 sX 2 ( s ) = 0
⇒ M 1 s 2 + ( f v1 + f v 3 ) s + ( K1 + K 2 ) X 1 ( s ) − [ f v 3 s + K 2 ] X 2 ( s ) = F ( s )
From Force balance on M 2
(K
Example: Transfer Function
Find the transfer function, X 2(s )/ F (s ) for this system?
a1 x + b1 y = c1 a2 x + b2 y = c2
x = −
b1
c1
b2
c2
a1
b1
a2
b2
,y=
+ K 3 ) X 2 ( s ) + ( f v2 + f v3 ) sX 2 ( s ) + M 2 s 2 X 2 (s ) − K 2 X 1 (s ) − f v3sX 1 (s ) = 0
⇒ − [ f v 3 s + K 2 ] X 1 ( s ) + M 2 s 2 + ( f v2 + f v3 )s + (K 2 + K 3 ) X 2 (s ) = 0
(c) All Forces on M 2
Control Engineering
2
a1
c1
a2
c2
a1
b1
a2
b2
Control Engineering
Exercise: Transfer Function
Find the transfer function X 2 (s )/ F( s ) for the system below:
Answer:
G(s) =
3s + 1 s s + 7 s + 5s + 1 3
2
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Control Engineering
Modeling Mechanical System (Rotational)
Control Engineering Find the transfer function
Control Engineering Find the transfer function
Problem: Systems with Gears θ2(s)/T1(s) for the system.
Control Engineering
Problem: Systems with Gears θ2(s)/T1(s) for the system.
Electromechanical System
Build a Electromechanical system and derive the relationship between input and output.
3/24/2015
Control Engineering
Modeling DC Motors
Derive the transfer function of an armature controlled D.C. motor as shown in figure below:
( s) K = E ( s ) s ( R + Ls )( I s + b ) + K θ i
Control Engineering
Control Engineering
Modeling DC Motors
Develop the block diagram of an armature controlled dc motor shown in Fig. below:
m
m
b
K m
Modeling DC Motors
The armature-controlled DC servomotor consists of the following components: 1. Armature: Current-carrying wire wrapped around a rotating member called the rotor. 2. Fixed field: Permanent magnets (often augmented as electromagnets) which create a fixed magnetic field that is perpendicular to the surface of the rotor.
Control Engineering
Modeling DC Motors
The job of the commutator is to reverse the direction of current so that the conductor experiences the same force while the rotor rotates. The magnetic field imposes a total force on the armature circuit of F = 2BLI a where B is the magnetic field strength, 2 L is the total length of the conductor that is perpendicular to the field, and I a is the armature current. Multiplying by the rotor’s radius r we obtain a torque
Tm ( t ) = rF = r 2BLI a (t ) = K m I a (t ) where
K m = r 2 BL
3/24/2015
Control Engineering
Modeling DC Motors
The output torque in the motor (d-c motor with a fixed field) is
Tm ( t ) = K m I a ( t )
Tm ( s ) = K m I a ( s )
where, T m is the output torque in the motor K m is the torque constant of the motor I a is the armature current
Control Engineering
Modeling DC Motors
Relationship between the motor speed and the rotational displacement,
( t ) =
ω
where, θ is
dθ ( t ) dt
the rotational displacement
Ω ( s ) = sθ ( s )
Control Engineering
Modeling DC Motors
The motor torque is equal to the torque delivered to the load, which can be expressed as
Tm ( t ) = TL ( t ) + Td ( t )
Tm ( s ) = TL ( s ) + Td ( s )
where, T L is the load torque T d is the disturbance torque
Control Engineering
Modeling DC Motors
The armature current I a generates a torque T m . However, any conductor travelling through a magnetic field will induce a voltage on the conductor. This voltage induces a current that opposes I a . Thus it is called the back electromotive force (back emf, V b ). It is proportional to the motor speed,
Vb ( t ) = Kb
dθ ( t ) dt
= Kbω ( t )
Vb ( s ) = K b Ω ( s )
K b is the voltage constant for motor
3/24/2015
Modeling Mechanical System (Rotational)
Control Engineering
Control Engineering
Modeling DC Motors
The load torque for rotating inertia is
T L ( t ) = J
d ω ( t ) dt
+ bω ( t ) T L ( s ) = JsΩ ( s ) + bΩ ( s )
where, ω is the angular speed J is the rotating inertia b is the rotational damping coefficient
Control Engineering
Modeling DC Motors
If the mechanical parameters of the rotor and any connected loads are known, then J m and D m can be obtained,
N 1 N 2
J m = J a + J L
2
N 1 N 2
Dm = Da + DL
2
Control Engineering
Modeling DC Motors
Applying KVL for armature circuit, Va ( t ) = Ra I a ( t ) + La
dI a ( t ) dt
+ Vb ( t ) Va ( s ) = Ra I a ( s ) + La sI a ( s ) + Vb ( s )
where, V a is the input voltage applied to the armature R a is the armature resistance La is the armature inductance
3/24/2015
Control Engineering
Modeling DC Motors
Control Engineering
Modeling DC Motors
A dynamometer test can then be applied to estimate the electrical parameters. This test relates torque and speed. Assume La is negligible, the following equation becomes Va ( t ) = Ra I a ( t ) + La
Va ( t ) =
Transfer function of the system neglecting any disturbances
( s) = V ( s ) s ( R θ a
K m
a
Ra Km
Modeling DC Motors
dt
+ Vb ( t )
Tm ( t ) + K bω ( t ) ⇒ Tm ( t ) = −
Kb K m Ra
Vb ( t ) = Kbω ( t ) Tm ( t ) = K m I a ( t )
(t) +
ω
K m Ra
Va ( t )
If the applied voltage is constant, then we have a linear relationship between torque and speed:
+ La s )( Js + b ) + Kb K m
Control Engineering
dI a ( t )
Tm ( t ) = −
Control Engineering
Kb Km Ra
(t ) +
ω
K m Ra
Va
Modeling DC Motors
Find the transfer function θL(s)/Ea(s) of an armature controlled D.C. motor as shown in figure below: Tm ( t ) = −
Kb K m Ra
(t ) +
ω
K m Ra
Va
Stall torque: Torque at which the motor is just unable to turn. No-load speed: Angular velocity at which motor runs without an imposed load. K V a Tstall = m V a ω no−load = Ra K b
3/24/2015
Control Engineering
Modeling DC Motors
Control Engineering
Modeling DC Motors
Km
Control Engineering θ m
Ea
Modeling DC Motors
( s) K = ( s ) s R ( J s + D ) + K m
a
= = =
( s) = θ ( s ) θ L
N 1
m
N 2
=
m
m
b
K m
Control Engineering
=
T stall
100
K b =
V a
=
ω no−load
s (12 s + 10 ) + ( 2 )( 5 )
0.417
1000
Answer: θ L
Ea
( s) 0.0417 = ( s ) s [ s + 1.667]
100 50
=2
Find the transfer function, G (s ) = θ L(s )/ Ea (s ), for the motor and load shown in Figure below. The torque-speed curve is given by T m = -8ω m + 200 when the input voltage is 100 volts.
5
= 0.1
=5
HomeWork: Modeling DC Motors
s ( J m s + Dm ) + K b K m Ra
100
500
V a
K m Ra
s [ s + 1.667 ]
=
Ra
G(s) =
1 / 20 s [ s + 15 / 2]
3/24/2015
Control Engineering
Biological System
Human
body composes of a number of dynamic systems. purpose is to find a dynamic relationship between body weight and calorie intake. Diet experts state that a moderately active person burns 20 calories/pound of body weight/day. Moreover, 4000 calories is roughly equivalent to 1 pound of body weight. Number of Calories taken, Ci = qi dt
Control Engineering Change
of body weight,
Our
Number
of Calories burned,
Biological System
dW =
Ci − C b
4000 4000dW = qi dt − 20Wdt 4000
dW dt
+ 20W = qi
Cb = 20Wdt
q i = rate of calorie intake (calories/day) t – days W = body weight (pounds)
Input, q i = rate of calorie intake (calories/day) Output, W = body weight (pounds)
Transfer
Function,
G ( s) =
W ( s) Qi ( s )
=
1 4000s + 20
