程序代写案例-DENM011

WORKED SOLUTIONS (2019)
Page 1 of 15


Module:

DEN408+DENM011 Title: Robotics
First
examiner:
Dr Ranjan Ve
pa 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.






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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,




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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.

















4
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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