WORKED SOLUTIONS (2019)

Page 1 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

1a The reference frames, the first being fixed to the base while the other three are fixed to the

links and are chosen according to the DH convention. The link parameters are

summarized in the table.

Table Link parameters for the first three links of the PUMA 560 manipulator

Link no. ai, i di i

Link 1 0 90° h 1

Link 2 a1 0 0 2

Link 3 a2 0 0 3

3 3

1b. The homogeneous transformation for the first link is,

−

−

=

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

1 0 0 0

1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

0 0 cos sin

0 0 sin cos

11

11

1 ,0 h

θθ

θθ

T , that is,

−

=

−

−

=

1 0 0 0

0 1 0

0 cos 0 sin

0 sin 0 cos

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

1 0 0 0

1 0 0

0 0 cos sin

0 0 sin cos

11

11

11

11

1 ,0 hh

θθ

θθ

θθ

θθ

T

.

The homogeneous transformation for the second link is,

−

=

1 0 0 0

0 1 0 0

0 0 1 0

0 0 1

1 0 0 0

0 1 0 0

0 0 cos sin

0 0 sin cos 1

22

22

2 ,1

a

θθ

θθ

T ,

−

=

1 0 0 0

0 1 0 0

sin 0 cos sin

cos 0 sin cos

2122

2122

2 ,1

θθθ

θθθ

a

a

T .

The homogeneous transformation for the third link is,

−

=

1 0 0 0

0 1 0 0

0 0 1 0

0 0 1

1 0 0 0

0 1 0 0

0 0 cos sin

0 0 sin cos 1

33

33

3 ,2

a

θθ

θθ

T ,

−

=

1 0 0 0

0 1 0 0

sin 0 cos sin

cos 0 sin cos

3233

3233

3 ,2

θθθ

θθθ

a

a

T .

4

4

4

1c. The co-ordinates of a point in a frame located at the end effector may be expressed in

terms of the base coordinates as,

3 3

0,1 1,2 1,3 0,31 1 1

= =

0p p pT T T T

where pi is the position vector of the point in the ith frame, i= 0, 1, 2, 3. Hence,

0,3 0,1 1,2 1,3=T T T T .

4

WORKED SOLUTIONS (2019)

Page 2 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

1c

WORKED SOLUTIONS (2019)

Page 3 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

2a)

5

2a) Projecting the end effector position on the x, y, and z axes, where the x axis is aligned with

the x2 axis, y is aligned with the y2 axis and z with the z2 axis, the position of joint 2

relative to the end-effector axis is,

( )[ ]Te slslclcl 02332223322,22 ++−== rr ,

Projecting the position of joint 3 in the x, y, and z axes, the position of joint 2 relative to

the joint 3 is,

( )[ ]Tslcl 022223,21 −== rr .

Hence, the vector positions of joint 2, 2r relative to the end-effector and 1r relative to

joint 3, ( )[ ]Tslslclcl 023322233222 ++−=r , ( )[ ]Tslcl 022221 −=r ,

[ ]T1000 =z ,

where the quantities ic , is , ikc and iks are defined as iic θcos = , iis θsin = , ( iikc θ= cos

the ( )000 ,, zyx frame.

The top view of the first two links in the robot manipulator with the local joint coordinates

is shown below. The first two links are equivalent to a two-link manipulator.

2+1

WORKED SOLUTIONS (2019)

Page 4 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

2b) As far as first joint, it is prismatic with displacement d1, assumed positive in the positive z

direction.

The linear velocity of the prismatic joint is:

10

1

d

z

y

x

z=

.

Assembling the preceding results, for the joints 1 it follows that,

11

0

dd J

z

0

v

=

=

ω

,

=

0z

0J .

The unit vector in the positive z direction is:

[ ]T1 0 00 =z .

5

1

1

2c) Now consider the second and third joints. To determine the relevant components of the

screw vectors, consider the second joint by itself.

The second and third joints are both revolute joints rotating about parallel axes with axes

passing through the joint axes and are in the plane of the paper but in the positive z

direction. The first joint is prismatic while the next two joints are revolute joints rotating

about parallel axes with axes passing through the joint axes and are in the plane of the

paper but in the positive z direction.

Hence the angular velocity vectors of each of the revolute joints are,

i

i

θ

ω

ω

ω

0

3

2

1

z=

, i =2 and 3, where

=

1

0

0

0z .

The velocity vector of the end effector relative to centre of rotation of joint 2 while

assuming that joint 3 is fixed, is given in the ( )000 ,, zyx frame by:

220

2

rz ×=

θ

z

y

x

where 2r is the vector distance of the location of joint 2 relative to the end effector

location in the ( )000 ,, zyx frame.

2

3

WORKED SOLUTIONS (2019)

Page 5 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

2c) Considering the third joint only, the velocity vector of a fixed point at joint 3 relative to

the end effector, while assuming that the joint 2 is fixed is given in the ( )000 ,, zyx frame

by:,

( )1220

2

rrz −×=

θ

z

y

x

where 1r is the vector distance of joint 2 relative to the joint 3 in the ( )000 ,, zyx frame.

Assembling the preceding results, it follows that,

( )

=

−××

=

3

2

1

3

2

1

120200

003

2

1

0

θ

θ

θ

θω

ω

ω

dd

z

y

x

J

rrzrzz

zz

and that,

( )

−××

=

120200

00

0

rrzrzz

zzJ .

3

2c) Thus the 46 × Jacobian matrix for the transformation from the end-effector to the base co-

ordinates is of the form,

( )

−××

=

120200

00

0

rrzrzz

zzJ ,

=

1

0

0

0z ,

−

−=

0

22

22

1 sl

cl

r and

+

−−

−=

0

23322

23322

2 slsl

clcl

r .

2

WORKED SOLUTIONS (2019)

Page 6 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

3a Let q1= 1θ , angle of rotation of the first link w.r.t. the local horizontal, positive counter

clockwise and 2q = 2θ , angle of rotation of the second link w.r.t. the first, positive

counter clockwise.

The height of the c.g. of the first link from the axis of the first revolute joint is,

111 sinθcgLY = . For the second link it is, ( )122112 sinsin θθθ ++= cgLLY and for the tip

mass it is, ( )12211 sinsin θθθ ++= LLYtip .

Increase in the potential energy of the body is, [ ] ( )[ ]

[ ] ( )[ ]122112111

122112111

sinsinsin

sinsinsin

θθθθ

θθθθ

++++

+++=

LLgMLgM

LLgmLgmV cgcg

.

Hence, ( ) ( ) ( )212222112111211 sinsin θθθ ++++++= LMLmgLMLMLmLmgV cgcg

which may be written as, ( )2122111 sinsin θθθ +Γ+Γ= ggV

where, ( )1211121111 LMLMLmLm cg +++=Γ , ( )222222 LMLm cg +=Γ

3

3a The horizontal position of the c.g. of the first & second link and the tip mass, positive

east, are, 111 cosθcgLX = , ( )122112 coscos θθθ ++= cgLLX ,

( )12211 coscos θθθ ++= LLX tip , 111 cosθLX j =

Horizontal velocities of the c.g.’s of the masses, 1111 sinθθ cgLX −= , 1111 sinθθ LX j −= , ( ) ( )121221112 sinsin θθθθθθ ++−−= cgLLX , ( ) ( )12122111 sinsin θθθθθθ ++−−= LLX tip .

Vertical velocity of the c.g.’s of the masses, 1111 cosθθ cgLY = , 1111 cosθθ LY j = , ( ) ( )121221112 coscos θθθθθθ +++= cgLLY , ( ) ( )12122111 coscos θθθθθθ +++= LLYtip

( ) ( ) ( ) ( )2222222221211212111 21212121 tiptipjj YXMYXmYXMYXmT +++++++= .

3

3a ( ) ( )( ) ( ) ( )( )21212211121212211122 coscossinsin θθθθθθθθθθθθ +++++++=+ LLLLYX tiptip

( ) ( ) ( ) ( )( )

( ) ( ) ( )( )12112112121

12

2

12

22

12

2

21

2

1

22

1

2

1

22

coscossinsin2

cossincossin

θθθθθθθθθ

θθθθθθθθθ

+++++

++++++=+

LL

LLYX tiptip

( ) ( ) 21212121222212122 cos2 θθθθθθθ ++++=+ LLLLYX tiptip ; 21212121 θ LYX jj =+

( ) ( ) 2121212122221212222 cos2 θθθθθθθ ++++=+ cgcg LLLLYX ; 21212121 θ cgLYX =+ .

Substituting & Simplifying,

( ) ( )

+++

++++=

2

21

2

2

2

1

2

12

2

21

2

2

2

1

2

12

2

1

2

11

2

1

2

111 2

1

2

1

2

1

2

1 θθθθθθθθ LLMLLmLMLmT cgcg

( ) ( )( )211212222 cos θθθθ +++ LLMLm cg

3

WORKED SOLUTIONS (2019)

Page 7 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

3a

The kinetic energy of rotation of the rods is, ( )221222212112 2

1

2

1 θθθ ++= kmkmT .

The total kinetic energy is, =+= 21 TTT

( ) ( )( ) ( )( )( )22122222222212121221211 2121 θθθ ++++++++ LMkLmLMMmkLm cgcg( ) ( )( )211212222 cos θθθθ +++ LLMLm cg .

1

3a Hence the total kinetic energy may be expressed as,

( ) ( )211221221222111 cos2121 θθθθθθθ ++++= IIIT ,

where ( ) ( ) 21222112121111 LMmLMkLmI cgcg ++++= ,

( ) 1221222221 LLLMLmI cg Γ=+= , ( ) 2222222222 LMkLmI cgcg ++= ,

3

3b Hence the Lagrangian may be defined as, L T V= − . The Euler Lagrange equations are,

d

dt

L

q

L

q

Q

i i

i

∂

∂

∂

∂

− = where, iiq θ= ; Qi are the generalised forces other than those

accounted for by the potential energy function and are equal to the torques applied by the

joint servo motors, iT .

( ) ( ) 221212122111

1

cos2

θθθθθθ

θ∂

∂

++++= IIIT

( )21222121

2

cos

θθθθ

θ∂

∂

++= IIT ; 0

1

=

θ∂

∂ T

; ( )( )211221

2

sin

θθθθ

θ∂

∂

+−= IT ;

( ) ( )2122111

1

coscos

θθθ∂

∂

+Γ+Γ= gg

q

V

; ( )2122

2

cos

θθ∂

∂

+Γ= g

q

V

.

3

( ) ( ) 221212122111

1

cos2

θθθθθθ

θ∂

∂

++++= IIIL ,

( )21222121

2

cos

θθθθ

θ∂

∂

++= IIL , 1

1

Γ−= g

q

L

∂

∂

, ( )( )2112212

2

sin

θθθθ∂

∂

+−Γ−= Ig

q

L

,

With ( ) ( )21221111 coscos θθθ +Γ+Γ=Γ , ( )21222 cos θθ +Γ=Γ .

3

WORKED SOLUTIONS (2019)

Page 8 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

3c Hence the two Euler-Lagrange equations of motion are,

( )2 1 2111 21 2 22 21 2

21 2 221 2 22 1 2 1

1 1

2 2

2

cos cos

sin

cos

I I I I

I

I I

T

g

T

θ θ θθθ θ θ

θ θ θ θ

− + + +

+

+

Γ

+ = Γ

.

3

3c

Let

+

=

21

1

2

1

θθ

θ

ω

ω

Hence the two Euler-Lagrange equations of motion are,

2 2

1 1 111 21 2 22 21 2 1 2

21 2 221 2 22 2 2 21

cos cos

sin

cos

TI I I I

I g

I I T

ωθ θ ω ωθ

θ ω ω

Γ+ + −

+ + = Γ

.

2

3c The equations may also be written in the state space and in matrix form in terms of 1ω

and 2ω , and in terms of 1θ and 2θ , where,

11 ωθ = ; 122 ωωθ −= .

The matrix equations of motion in terms of 1ω and 2ω , are:

2 2

1 1 111 21 2 22 21 2 1 2

21 2 221 2 22 2 2 21

cos cos

sin

cos

TI I I I

I g

I I T

ωθ θ ω ωθ

θ ω ω

Γ+ + −

+ + = Γ

,

where

( ) ( )21221111 coscos θθθ +Γ+Γ=Γ ; ( )21222 cos θθ +Γ=Γ .

1

WORKED SOLUTIONS (2019)

Page 9 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

Q4a

The strategy of Computed torque control involves feeding back, to each of the joint

servos, a signal that cancels all of the effects of gravity, friction, the manipulator inertia

torques as well as the Coriolis and centrifugal torques. All of these forces are computed

on the basis of Euler-Lagrange dynamic model. All these effects are treated as

disturbances that must be cancelled at each of the joints. Additional feedbacks are then

provided to put in place the desired inertia torques, viscous friction torques and stiffness

torques which are selected so the error dynamics behaves in a desirable and prescribed

manner.

4+3

4b)

Thus the feedback law may be assumed to be a computed torque controller. In the general

case the computed torque controller inputs are,

( )qqhD ,ˆˆ

2

1

2

1

2

1 +

≡

=

v

v

T

T

T

T

c

c

.

In the case of the inverted pendulum on a cart,

2

2

M m mLc x FmL s

mLc mL mgLs

θ

θ τ

+

− =

.

Hence traction force and joint torque are,

2

1

2

2

M m mLc vF mL s

vmLc mL mgLs

θ

τ

+

= −

,

where the auxiliary control inputs are defined as,

1 2 2

2

2 2des des desn n n n

desdes des

x x xxv x

v

ω ω ω ω

θ θθ θ θ

= + + − −

.

Thus the equations of motion reduce to,

2

1

2

2

vxd

vdt θ

=

.

3

3

4

WORKED SOLUTIONS (2019)

Page 10 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

4c Let [ ] TTdes des desx θ=q be the demanded angular position commands to the joint servo

motors .

The tracking error is given by,

1

2

des

des act

des

xe x

e θθ

= − = −

q q

and satisfies the equation,

=

+

+

0

0

2

2

12

2

1

2

1

e

e

e

e

e

e

nn ωω

and is exponentially stable, i.e. poles are strictly negative real parts.

Robot arms are designed so they don’t oscillate and consequently an optimum choice of

the damping ratio ζ is unity or very slightly greater than unity. Hence the poles

(eigenvalues) are chosen to be strictly negative.

2

3

1

2

WORKED SOLUTIONS (2019)

Page 11 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

5a For the measurement of the acceleration a sensor based on the piezo-electric effect is often

employed. A piezoelectric crystal or ceramic is employed and it can be modelled as a mass

supported on a spring and damper. The inertial force acting on the piezoelectric material

generates a stress across two faces on the material which in turn generates an output

charge.

3

5a Consider a schematic diagram of an accelerometer as shown in the figure above. The

direction of sensitivity is shown on the right of the figure The ground motion is given as

( )ty . The equation of motion in terms of ( )tx ,

( ) ( ) mgyxcyxkxm −−−−−= .

Let the relative displacement of the mass, relative to the base be, ( ) ( ) ( )tytxtxr −= , Then

it follows that,

mgymkxxcxm rrr −−=++ .

As the stiffness constant is usually very high, neglecting the relative inertial term in

comparison with the spring force,

( ) ( )GymgymmgymxckxF rr −−=+−=−−=+= ,

Where F is the force acting on the piezoelectric material surface, gG −= is the

component of the acceleration due to gravity in the direction of measurement. Thus,

GAvpiezo −∝ where A is the ground acceleration given by, yA = . This is the principle of

acceleration measurement.

2

2

1

WORKED SOLUTIONS (2019)

Page 12 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

5b A DC servo motor includes a feedback loop around a conventional DC motor, which is driven by

a voltage. The motor output shaft of a conventional DC motor, rotates at constant speed

depending on the applied input voltage.

Typically for the position control of a DC motor it is essential that the measured position must be

the primary feedback as it is compared with the demanded position. A voltage proportional to the

error between the demanded and actual positions is used to drive the motor. In addition a signal

proportional to the rate of position is also added to the feedback. The closed loop system is of the

form shown below.

+ K G(D)+yi

_

e

H(D)

y

In the figure, ( ) 2 21 1H D K d dt K D= + ≡ + .

Unlike servo motor, stepper motors are versatile, long-lasting, simple motors that can be

used in many applications. Stepper motors are used without feedback in many

applications. This is because unless a step is missed, a stepper motor steps a known angle

each time it is moved. Thus, its angular position is always known and no feedback is nec-

essary.

3

3

1

3

5c One popular method of implement a control system is to use pulses are illustrated in

figure 5c and is known as Pulse Width Modulation (PWM) control. The cycle time, Tc , is

maintained a constant and the pulse on-time is T+ .

1

The ratio of the pulse on-time of the pulse cycle time defines the duty cycle and the

average D. C. value of the drive signal is: d cV V T T+= . Referring to the figure, the error

(e) input is defined by the voltage V. The gain K is effectively varied by changing the

pulse width or on-time. The control input to the motor is dV .

t

O

V

Tc

T+ T-

Figure 5c Pulse Width Modulation

1

2

WORKED SOLUTIONS (2019)

Page 13 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

The principal advantage of PWM control over the traditional method described earlier is

the simplicity of the drive electronics and the ease of interfacing to a micro-controller

circuit. PWM is easy to implement with microprocessors.

Commonly, to control the speed of a DC motor, the applied voltage is changed; as the

voltage increases or decreases, the speed of the motor is changed accordingly. However,

when the speed is to be controlled by a microprocessor, changing the voltages to

servomotors requires many bits of information. Alternatively, using PWM, many voltage

levels can be achieved with one single-level input voltage, namely, the high voltage (say,

5 volts) and one output bit. To do this, the voltage on the output port of the processor is

turned on and off repeatedly, many times a second, so that by varying the length of time

that the voltage is on or off, the average effective voltage will vary.

A major disadvantage of the PWM method is the high levels of noise generated by the

drive electronics which makes it unreliable for precision applications.

3

WORKED SOLUTIONS (2019)

Page 14 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

6a

The primary factors that must be considered in selecting a sensor for a robotic application apart

from its weight and size are,

Resolution and dynamic range: The resolution may be defined as the smallest increment of the

measurand that can be detected with certainty by the instrument. The dynamic range is the

minimum reading of the instrument output corresponding to the minimum input which is

necessary to cause a measurable change in the instrument from zero input to the maximum reading

or output corresponding to the maximum allowable input to the instrument. Thus the range

represents to which the sensor may be employed in practice. It is also stated as a ratio of the

maximum to the minimum reading that may be obtained from a sensor while the difference

between the maximum and minimum is specified as the span.

Consistency or Repeatability: When a measurement of a quantity made by an observer is

identical or sufficiently close to another measurement of the same quantity, repeated under

identical experimental conditions, methods and apparatus, it is said to be repeatable or

reproducible or consistent.

Accuracy: Accuracy is a term used to qualify the error as being within certain limits with a

specified probability. Accuracy is the degree of correctness with which a method of measurement

yields the true value of a quantity, usually expressed in terms of the relative magnitude of the

error.

Predictability: Predictability is the property of being able to estimate the static and dynamic

errors in the mesurand, to the same level of accuracy as in a measurement of it, with a specified

sensor, without actually performing the measurement.

Sensitivity: Typically, the sensitivity is the ratio of the sensor output to the corresponding

incremental change in the magnitude of the measurand.

4

2

2

2

2

6b Consider the equivalent circuit of a linear potentiometer.

x

vin

vout

com

Resistance

= R/unit length

Moving

wiper

i

i1

iload

Rload

L

Equivalent circiuit of a linear potentiometer

The total current is related to the load current by the relation:

loadiii += 1 .

The voltage drop aross the load is equal to the voltage drop across the wiper and the com

port. Thus,

( ) loadload RixLRi ×=−×1 .

Hence,

3

WORKED SOLUTIONS (2019)

Page 15 of 15

Module:

DEN408+DENM011 Title: Robotics

First

examiner:

Dr Ranjan Vepa Second

examiner:

Dr M. Hasan Shaheed

Qu. No. Worked Solution Marks

( )xLR

R

ii loadload

−

×=1 and ( )

−

+×=

xLR

R

ii loadload 1 .

It follows that,

( )

( )

+−

−

×=

load

load RxLR

xLRii and ( )( )

+−

×−

×=

load

load

load RxLR

RxLR

iv .

The total circuit rsistance with the wiper in position is,

( )

( )

( )

( )( )

( ) load

load

load

load

load

Total RxLR

LRxLxRR

RxLR

RxLR

xR

RRxL

xRR

+−

+−×

=

+−

×−

+=

+

−

+=

111

and ( )( )( ) load

load

Total RxLR

LRxLxRR

iRiv

+−

+−×

×=×= .

Thus, ( )( )( )

( )

( )( )

+−

×−

=

+−×

×−

=

LRxLxR

RxL

LRxLxRR

RxLR

v

v

load

load

load

loadload

.

When loadR is very large relative to R, i.e. ∞→loadR ,

( )

( )( )

( ) ( )

L

xL

LR

RxL

LRxLxR

RxL

v

v

load

load

load

loadload −

=

×−

≈

+−

×−

=

Hence, the output voltage is a linear function of the position of the wiper from one end.

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6c The need for tactile sensing: The sense of touch, one of 5 human senses (vision, hearing, smell,

taste and touch), provides particular information that cannot be perceived by any of the other

senses. Touch may interpret object texture, hardness and temperature as well as the mechanical

state of an object such as vibration or movement. Tactile sensors are devices that measure the

parameters of contact between the sensory surface and an object. Thus tactile sensing involves

detection and measurement of the spatial distribution of forces perpendicular to a predetermined

sensory area, and the subsequent interpretation of the spatial force distribution.

The application in Robotics: A tactile sensors are used to control a manipulator for grasping an

unknown object based on tactile feedback.

The tactile sensor placed on each jaw of the manipulator's gripper and serves as force distribution

sensor, which is then used to control the applied force/pressure applied by the gripper on the

object by force-feedback control.

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