ch1_intro to robotic control system in emgineering

ch1_intro to robotic control system in emgineering

1. Linear Control of Manipulators
Robot Control System
2025
Robot Control System Linear Control of Manipulators 2025

2. Outline
1
Introduction
2
Feedback and Closed-Loop Control
3
Second-Order Linear Systems
4
Control Design
5
Control Law Partitioning
6
Trajectory Following
7
Disturbance Rejection
8
Modeling and Control of a single joint
9
Industrial Implementation
Robot Control System Linear Control of Manipulators 2025

3. Introduction to Linear Control
Linear control systems are widely used in industrial robotics
They are approximate methods since manipulator dynamics is
nonlinear
Key advantages:
Simple to implement
Well-established theoretical framework
Proven effectiveness in practice
Open-loop control:
No feedback
Sensitive to model errors and disturbances
Closed-loop control:
Uses sensor feedback
More robust to disturbances
Better performance
Robot Control System Linear Control of Manipulators 2025

4. Feedback Control Basics
Manipulator components:
Sensors at each joint (position/velocity)
Actuators for applying torque
Figure:
Typically, a trajectory can be generated by solving following
differential equation,
τ=M(θd)
¨
θd+V(θd,
˙
θd) +G(θd)
Control system computes appropriate actuator commands
Feedback used to compute servo error:
E=θd−θ,˙E=˙θd−˙θ
Robot Control System Linear Control of Manipulators 2025

5. Control System Design
Often the tracking of trajectory can be achieved by controllingEand˙E.
Manipulator problem is a MIMO problem
Construct a control system by treating each joint as a separate
system to be controlled
Design N-independent SISO system that is highly coupled
Figure:
Robot Control System Linear Control of Manipulators 2025

6. Spring-Mass System Example
Basic second-order system:
m¨x+b˙x+kx= 0
Figure:
Parameters:
m- mass,b- damping coefficient andk- spring stiffness
Characteristic equation obtained after taking Laplace Transform:
ms
2
+bs+k= 0
Robot Control System Linear Control of Manipulators 2025

7. System Response Types
Roots of the characteristic equation ares1ands2given by,
s1=
−b
2m
+
√
b
2
−4mk
2m
,s2=
−b
2m
−
√
b
2
−4mk
2m
Overdamped (b
2
>4mk)
Real, unequal roots
Sluggish, non-oscillatory response
Underdamped (b
2
<4mk)
Complex roots
Oscillatory response
Critically Damped (b
2
= 4mk)
Real, equal roots
Fastest non-oscillatory response
Robot Control System Linear Control of Manipulators 2025

8. System behavior can be modified using sensors, actuator and control
system
Figure:
Equation of motion is
m¨x+b˙x+kx=f
Force to be applied by actuator is
f=−kpx−kv˙x
Robot Control System Linear Control of Manipulators 2025

9. Control Law Design
Controller reads sensor input and computes actuator output
Figure:
Closed-loop dynamics:
m¨x+ (b+kv) ˙x+ (k+kp)x= 0
By setting controller gainskpandkv, we can get the desired behavior
of the system.
Gain selection for critical damping with desired stiffness is
b
′
= 2
√
k
′
Robot Control System Linear Control of Manipulators 2025

10. Control Law Design
Position Regulation Control
Example : If parameters of the system arem= 2,b= 2 andk= 2. For a
value ofkp= 16, the system becomes critically damped. find the closed
loop stiffness and the correspondingkv.
Solution :
Closed-loop stiffness =k+kp=2+16 = 18
Closed-loop friction coefficient = 2
√
mk= 2
√
2x18=12.
kv=12-2=10.
Robot Control System Linear Control of Manipulators 2025

11. Control Law Partitioning
Partition the controller into a model-based portion and a servo portion
System’s parameters appear only in model-based portion
Equation of motion of the given system is
m¨x+b˙x+kx=f
To design model-based control portion, choosefsuch that it reduces
the system so that it appears to be a unit mass
Second part of the control law makes use of feedback to modify the
system behavior
Robot Control System Linear Control of Manipulators 2025

12. The model-based portion of control law is of the formf=αf
′
+β,
whereαandβare functions or constants.
Iff
′
is taken as the new input, the system appears to be a unit mass
based on choice ofαandβ.
Control law becomes
m¨x+b˙x+kx=αf
′
+β
In order to make system appear as a unit mass from input, choose
α=mandβ=b˙x+kx.
With these assignments, system equation becomes ¨x=f
′
Control law to computef
′
can be given by,
f
′
=−kpx−kv˙x
Robot Control System Linear Control of Manipulators 2025

13. Overall law becomes
¨x+kv˙x+kpx= 0
Controller gains can be set up askv= 2
p
kp
Figure:
Robot Control System Linear Control of Manipulators 2025

14. Trajectory Following Control
Enhanced controller for trajectory tracking
e=xd−xis the tracking error
A servo control law for trajectory following is :
f
′
= ¨xd+kv˙e+kpe
Figure:
Error dynamics using control law design including model based
control part:
¨e+kv˙e+kpe= 0
Robot Control System Linear Control of Manipulators 2025

15. Disturbance Rejection
System with disturbance rejection capability has controller with an
additional input
¨e+kv˙e+kpe=fdist
Figure:
Iffdistis bounded bya, then the solution of differential equatione(t)
is also bounded
Iffdistis constant, steady state error may remain in the response.
Robot Control System Linear Control of Manipulators 2025

16. Iffdistis constant, steady state analysis is performed by analyzing
system at rest
Setting derivatives to zero, steady state equations are obtained as
kpe=fdistwhich is steady state error.
PID Control is used to eliminate steady state error
Updated control law is given by
f
′
= ¨xd+kv˙e+kpe+ki
Z
e dt
This results in error equation
¨e+kv˙e+kpe+ki
Z
e dt=fdist
Advantages of integral term:
Eliminates steady-state error
Improves disturbance rejection
Robot Control System Linear Control of Manipulators 2025

17. Modeling of DC Motor
A common actuator found in industrial robot is a DC Motor
It has two parts namely stator and rotor
The nonturning part of the motor (the stator) consists of a housing,
bearings, and either permanent magnets or electromagnets
Stator magnets establish a magnetic field across the turning part of
the motor (the rotor)
Rotor consists of a shaft and windings through which current moves
to power the motor
Current is conducted to the windings via brushes, which make contact
with the commutator
Robot Control System Linear Control of Manipulators 2025

18. The commutator is wired to the various windings (also called the
armature) in such a way that torque is always produced in the desired
direction.
Torque is generated by motor when current passes through the
winding, is given byF=qVXB, whereqis charge, moving with
velocity V through a magnetic field B, experiences a force F.
Torque developed by motor is proportional to the product of the
armature current and air gap flux (ϕ=Kfif)
τm=kmia (1)
When a motor is rotating, it acts as a generator, and a voltage
develops across the armature
The voltage generated for a given rotational velocity is
v=ke
˙θm (2)
Robot Control System Linear Control of Manipulators 2025

19. Motor-armature inductance
The major components of armature circuit are a voltage source,va,
the inductance of the armature windings, the resistance of the
armature windings,ra, and the generated back emf, v
The circuit is described by a first-order differential equation
laia+raia=va−ke
˙θm (3)
Figure:
Robot Control System Linear Control of Manipulators 2025

20. It is desirable to control the torque generated by the motor (rather
than the velocity) with electronic motor driver circuitry.
Drive circuits senses the current through the armature and
continuously adjusts the voltage sourcevaso that desired current
flows through the armature
A circuit called amplifier motor driver is used
Robot Control System Linear Control of Manipulators 2025

21. Effective Inertia
The torque applied to the rotor,τm, is given by (1) as a function of
the current flowing in the armature circuit
Figure:
to an inertial load
The gear ratioηcauses an increase in the torque seen at the load and
a reduction in the speed of the load, given by
τ=ητm (4)
Robot Control System Linear Control of Manipulators 2025

22. Also,
˙θ= 1/(η)˙θm (5)
Writing a torque balance equation for this system
τm=Im
¨θm+bm
˙θm+ (1/η)(I¨θ+b˙θ) (6)
whereImand I are the inertias of the motor rotor and of the load
bmandbare viscous friction coefficients for the rotor and load
bearings
Using above relation, motor torque is given by,
τm= (Im+ (1/η
2
))¨θm+ (bm+ (b/η
2
))˙θm (7)
It can also be given in terms of load variables by
τ= (I+η
2
Im)¨θ+ (b+η
2
bm)˙θ (8)
The termI+η
2
Imis called effective inertia and termb+η
2
bmis
called effective damping.
Robot Control System Linear Control of Manipulators 2025

23. Assumptions are made while modeling manipulators
The links are rigid bodies
Link inertia is constant
Nonlinear interacting terms such as the interactive and centrifugal
torques and gravity torque are ignored
Nonlinear effects such as coulomb friction, backlash and external
disturbances are ignored.
Joint actuators and their features
Electrical actuators require reduction gears of high ratio
It linearizes system dynamics and reduces coupling effects
useful for small to medium sized manipulators
Joint friction is increased along with elasticity and backlash
Robot Control System Linear Control of Manipulators 2025

24. Unmodeled Flexibility
the gearing, the shafts, the bearings, and the driven link are not
flexible
In reality, all of these elements have finite stiffness, and their
flexibility, if modeled, would increase the order of the system
If the system is sufficiently stiff, the natural frequencies of these
unmodeled resonances are very high and can be neglected compared
to the influence of the dominant second-order poles that are modeled
if the lowest structural resonance isωres, then we must limit our
closed-loop natural frequency according to the relationωn<= 0.5ωres
increasing gains leads to faster response and lower steady-state error,
but we now see that unmodeled structural resonances limit the
magnitude of gains
Typical industrial manipulators have structural resonances in the
range from 5 Hz to 25 Hz.
Robot Control System Linear Control of Manipulators 2025

25. Estimating resonant frequency
For the purposes of a rough estimate of the lowest resonant frequency
of beams and shafts,a lumped model of the mass is considered.
for estimating stiffness at the ends of beams and shafts; these lumped
models provide the effective mass or inertia needed for our estimation
of resonant frequency
Figure below shows the results of an energy analysis which suggests
that a beam of mass m be replaced by a point mass at the end of the
beam 0.23m and, likewise,that a distributed inertia of I be replaced by
a lumped 0.33 I at the end of the shaft.
Figure:
torsional resonance
Robot Control System Linear Control of Manipulators 2025

26. Control of a manipulator joint driven by a DC motor
Figure:
A single joint of a manipulator can be controlled with the partitioned
controller given by
α=Imax+η
2
Im
β=(b+η
2
bm)
˙
θ
τ
′
=¨θd+kv˙e+kpe
(9)
Robot Control System Linear Control of Manipulators 2025

27. The resulting closed loop dynamics is given by,
τdist= ¨e+kv˙e+kpe (10)
where gains are chosen as,
kp=ω
2
n=
1
4
ω
2
res (11)
kv= 2
p
kp=ωres (12)
Robot Control System Linear Control of Manipulators 2025

28. Industrial Robot Controller Architecture
Example: PUMA 560 Robot
Two-level hierarchy:
Master computer (LSI-11)
Joint microprocessors (6503)
Features:
Individual joint PID control
Position feedback from encoders
Current-controlled DC motors
Figure:
Robot Control System Linear Control of Manipulators 2025

29. Control System Implementation
Timing:
Master: 28ms cycle
Joint controllers: 0.875ms cycle
Functions:
Trajectory generation
Inverse kinematics
PID control computation
Motor current command
Figure:
Robot Control System Linear Control of Manipulators 2025

30. Conclusion
Linear control is effective for industrial robots
Key components:
Feedback control
PID control structure
Trajectory following
Disturbance rejection
Practical implementation considerations are crucial
Robot Control System Linear Control of Manipulators 2025