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ELEC3201W1 ROBOTIC SYSTEMS
Creation date：2024-05-18 17:58:30
Belonging category：电气工程electric engineering
ROBOTIC SYSTEMS

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ELEC3201W1

ROBOTIC SYSTEMS
DURATION 1440 MINS (24 Hours)
This paper contains 4 questions
Answer All questions
An outline marking scheme is shown in brackets to the right of each ques-
tion.
This examination contributes 100% of the marks for the module.
This version of the exam paper has been prepared with student revi-
sion for the kinematics topicin mind. Answers are given to the kine-
matics questions only
Please note the solutions are indicative of what would be expected
A foreign language dictionary is permitted ONLY IF it is a paper version
of a direct Word to Word translation dictionary AND it contains no notes,
additions or annotations.
16 page examination paper.
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Question 1.
(a) The Denavit Hartenberg (DH) Method is a commonly used approach for
obtaining the forward kinematic equations of robot manipulators. Which
statement(s) is/are correct?
(i) The homogenous transformation matrix derived using the DH Method
is unique
(ii) The DH Method uses left-hand coordinate frames
(iii) The DH Method is not applicable to manipulators with only prismatic
joints
(iv) In the DH Method, only two axes need to be chosen for every joint.
[2 marks]
(b) Give two reasons why it is beneficial to derive a robot’s forward kinematic
equations using the DH Method.
[2 marks]
(c) In the DH Method, the coordinate frame F0 is attached to the first joint.
What does the frame F1 represent? What do frames F2,F3, and so on,
represent (if present)?
[2 marks]
(d) Figure 1 depicts two joints of a robot manipulator. The coordinate frame
attached to the first joint is shown, along with three different possible
coordinate frames for the second joint. In the DH Method, which of
these coordinate frames are valid? Which is preferable?
[2 marks]
(e) Consider the diagram of the robot manipulator in Figure 2
(i) Determine the DH parameters of this manipulator, tabulate them,
and deduce three homogenous transformation matrices associated
with each joint. Marks will be awarded for the accuracy and clarity
of your working.
[6 marks]

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Joint 1 Joint 2
FIGURE 1: Robot joints with possible reference frames
Joint 1
Joint 2
Joint 3
FIGURE 2: Robotic manipulator with three joints, with joint variables θ1, d2 and θ3.

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(ii) Use these matrices to derive the homogenous transformation matrix
from base frame to end-effector. Marks will only be awarded for an
accurate expression and clear working.
[2 marks]
(iii) In the final homogenous transformation matrix, identify the rotation
matrix and the vector describing the (x, y, z) positions of the end
effector with respect to the base. Deduce the orientation of the end-
effector w.r.t. the base frame using the Euler angle convention.
[2 marks]
(iv) With this robot is it possible to independently control the (x, y) posi-
tions? Explain your answer.
[2 marks]
(f) The forward kinematics of an experimental robot are described by the
equation x1x2
x3
 =
l1 cos q1 + l2 cos q1 cos q2l1 sin q1 + l2 sin q1 cos q2
q3 − l2 sin q2

where qi are the joint variables, xi are the end-effector positions and li
the link lengths. Derive an equation describing the inverse kinematics of
the robot. Marks will be awarded for the accuracy and succinctness of
the final expression.
[5 marks]

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Indicative Solution for Question 1.
(a) The only statement which is correct is d) - only two axes (the z and the x axis need choosing, the y is
then determined since the coordinate frame is right-handed).
(b) Two reasons are:
• The DH method gives a systematic approach, which is useful for complex robotic systems
• It enables roboticists to communicate and exchange ideas in a common framework
(other answers also acceptable)
(c) The frame F1 represents the movement of link 1 with respect to the base frame. More generally, the
frame Fi represents the movement of link i with respect to link i− 1.
(d) In the Figure, the reference frame represented by the axes F1a = (x1a, y1a, z1a) and
F1a = (x1b, y1b, z1b) would both be equally valid choices of reference frame when applying the DH
method. Reference frame F1c = (x1c, y1c, z1c) does not represent a valid frame since, although x1c
and z1c are oriented correctly, y1c does not enable a right-handed-reference frame to be formed
(re-orienting this axis would correct this). In practice, one would probably choose reference frame
F1b since the x-axes are coincident and hence there would not be an extra term (d) representing the
offset between the two axes in the DH equations.
(e)
(i) The first part of the solution is to choose the z-axes and then one assigns reference frames to
each of these axes. Since F0 and FE have already been given, only two frames need to be
determined. Possible (but not unique) choices are shown below. Once the reference frames
have been chosen, the DH parameters can be deduced by inspection and tabulated
i θi di ai αi
1 θ∗1 −l1 l2 0
2 0 −d∗2 0 0
3 θ∗3 −l3 0 pi
These DH parameters can then be used to find the homogenous transformation matrices as
0
1T =

cos θ1 − sin θ1 0 l2 cos θ1
sin θ1 cos θ1 0 l2 sin θ1
0 0 1 −l1
0 0 0 1
 (1)
1
2T =

1 0 0 0
0 1 0 0
0 0 1 −d2
0 0 0 1
 (2)
2
ET =

cos θ3 − sin θ3 0 0
sin θ3 cos θ3 0 0
0 0 −1 −l3
0 0 0 1
 (3)
(different answers will result, depending on the choice of axes)
Copyright 2021 c© University of Southampton
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Joint 1
Joint 2
Joint 3
FIGURE 3: Robotic manipulator with three joints, with joint variables θ1, d2 and θ3.
(ii) The homogenous transformation matrix representing the end-effector position w.r.t base is
then given by
0
ET =
0
1 T
1
2T
2
ET
=

cos θ1 − sin θ1 0 l2 cos θ1
sin θ1 cos θ1 0 l2 sin θ1
0 0 1 −l1
0 0 0 1


1 0 0 0
0 1 0 0
0 0 1 −d2
0 0 0 1


cos θ3 − sin θ3 0 0
sin θ3 cos θ3 0 0
0 0 −1 −l3
0 0 0 1

=

cos(θ1 + θ3) − sin(θ1 + θ3) 0 l2 cos θ1
sin(θ1 + θ3) cos(θ1 + θ3) 0 l2 sin θ1
0 0 −1 −(l1 + l3 + d2)
0 0 0 1

(some trigonometric identities have been used)
(iii) The rotation matrix is easily identified as
0
ER =
cos(θ1 + θ3) − sin(θ1 + θ3) 0sin(θ1 + θ3) cos(θ1 + θ3) 0
0 0 −1

and the (x, y, z) positions as  l2 cos θ1l2 sin θ1
−(l1 + l3 + d2)

From the rotation matrix it is easy to see that the (yaw) orientation of the end effector will be
θ1 + θ3 radians around the z0 axis, and that the pitch of the manipulator will be pi radians around
the y0 axis. This agrees with what can be expected intuitively.
Copyright 2021 c© University of Southampton Page 6 of 16
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(iv) Independent control over the (x, y) positions is not possible with this robot since
px = l2 cos θ1 py = l2 sin θ1
Since, we can only control θ1 we can only control either px or py.
(f)
Note that the first two equations can be written as
px = (l1 + l2 cos q2) cos q1
py = (l1 + l2 cos q2) sin q1
Therefore
q1 = arctan(py/px)
This is the first of the IK equations. Note that the above two equations can also be squared and
added to get
p2x + p
2
y = (l1 + l2 cos q2)
2
which implies
q2 = arccos

√
p2x + p
2
y − l1
l2

which is the second of the IK equations. Finally, noting that
sin(arccos(x)) =
√
1− x2
we have, from the third element of the equation in the question
q3 = x3 + l2 sin q2
= x3 + l2
√√√√
1−
p2x + p
2
y + l
2
1 − 2l1
√
p2x + p
2
y
l22
= x3 +
√
l22 − l21 − p2x − p2y + 2l1
√
p2x + p
2
y
which is the third element of the IK equation.

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[25 marks]

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Question 2.
(a) Why would an engineer use the joint-space for trajectory planning? Ex-
plain your answer.
[2 marks]
(b) A robot has six joints and is required to move from a certain initial po-
sition to a final position, while passing through 5 intermediate points.
Assuming cubic splines are to be used to generate the trajectory, calcu-
late how many parameters will need to be calculated to enable this.
[1 mark]
(c) A robot is given a repetitive task to perform and a trajectory for a single
joint needs to be generated. The starting position of the joint is θ(0) =
45◦ and the starting velocity is θ˙(0) = 0. The end-point and velocity is
identical to that at the start. The robot must pass through a way point of
θ(tw) = 90
◦ with continuous velocity and acceleration of the joint. The
time to execute the task is 6 seconds with the way point to be reached
at 3 seconds.
(i) Set up a matrix equation which needs to be solved for the parame-
ters of the cubic functions. Marks will be awarded for accuracy and
clarity. You do not need to solve the equation.
[5 marks]
(ii) Could a “smaller” matrix equation be solved instead? Explain your
answer.
[1 marks]
(iii) Assume now that the robot joint should stop at the way-point and not
travel to the end-point. An engineer wants to check the maximum
joint velocity over this portion of the trajectory. Derive the time of
interest which s/he should check.
[4 marks]
(d) The forward kinematics of a robotic manipulator are given bypxpy
pz
 =
d1 cos θ2 cos θ3d1 cos θ2 sin θ3
−d1 sin θ2


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where the joint positions are d1, θ2 and θ3. Calculate the analytic Jaco-
bian of this manipulator and identify any limitations in mobility
[4 marks]
(e) A roboticist is developing a new mobile robot and needs to choose ac-
tuators and sensors for the system.
(i) Compare and contrast hydraulic and electric actuators. Give rea-
sons why each may be suitable for the system under consideration.
[4 marks]
(ii) The mobile robot features a manipulator with a gripper at the end.
The gripper needs to carry delicate objects. Explain the sensing
capabilities such a gripper would ideally be equipped with.
[4 marks]

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Indicative Solution for Question 2.
(a) Planning of trajectories in the task space involves choosing the end effector path directly in task
space and then determining the joint positions via inverse kinematics at each time-step. Planning of
trajectories in joint space involves choosing the start, end and possibly intermediate points in task
space. The joint positions corresponding to these points are then determined, but the trajectory of
the joints, from one point to another, is chosen in joint space. The advantages of task-space
trajectory planning is that more control over end-effector position is attained, but the drawback is
that the inverse kinematic equations have to be solved at each time instant to generate the
trajectories in joint space. Joint space removes these computational problems, but less control over
the end effector position is the drawback.
(Other answers possible)
(b) In total there are seven points, so 6 cubic functions are needed. Each cubic function has four
parameters to determine, so 24 parameters per joint and hence 6× 24 = 144 parameters in total.
(c)
(i) First assign the two cubic polynomials, the first for the portion of the trajectory between
t ∈ [0, tw] and the second for t ∈ [tw, tf ]:-
θ(t) = a10 + a11t+ a12t
2 + a13t
3
θ(t) = a20 + a21(t− tw) + a22(t− t2)2 + a23(t− tw)3
Joint velocity is then given by
θ˙(t) = a11 + 2a12t+ 3a13t
2
θ˙(t) = a21 + 2a22(t− t2) + 3a23(t− tw)2
and acceleration by
θ¨(t) = 2a12 + 6a13t
θ¨(t) = 2a22 + 6a23(t− tw)
Using the information about the start position and velocity - t = 0 - we have
45 = a10 (4)
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MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A PHIL2617 PHYS3116 MIE250 BCPM0008 ECN 620 QBUS6830 ECSE-415 ECON5102 MGMT20005 MAT 653 ECO 480 COMP 5/600 CSSE 5600 ME 571 7CCMMS61 CS 550 COMP2865 MATH3075 STATS 506 QTM 220 ISYS2120 FIT5106 STAT 153 MKTG3112 ECON2102 ELE2018 COMP3031 MSIN0010 SCC461 MATH237 CS246 CSC3065 CSE 101 ICTWEB411 COMP9313 ECOS 2025 FN620 MATH11111 MATH 3018 KE1068 FINA 6112 C++ MISM 6202 MSIN0105 CSC8631 MECO6935 COMP4121 ACCT2011 DPBS1150 EEL 3701C F19PL OPM 661 CSE 310 CS918 BIOL30001 ACCT3563 BANK3600 MET CS622 MATH367 ACCT5907 Continuing COMP5048 COMP61021 CS3334 GSOE9210 CSCI-570 COMP 5322 HW 2 ALY-6020 IEOR 4706 DATA 200 EBUS504 CGE12412 COMP4108 7SSMM800 ECMM171P MATE50002 CS225 RS 6003 BH2274 ELEC3111 MAT6669 ECON-UA 227 PHIL 1012 7SSMM603 TB028A P6 ECS708 CSC 427 CS6476 CSE 250A A1A2 ITS61504 EECS 376 model HW6 UV0010 ECON 6074 MG4F7 CMP-7030Y INFO3008 BFIP013 ECON 5068 MATH6002 CS3483 PSTAT 131 C31FM CSOR W4246 AAEC 5307 DATA ELEC362 COMSM0047 ECN6006 MA20224 BST351 MATH 323 MATH323 AA2021 COMP3811 COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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ACCOUNTING BUSS1040 CHEM 181 MAT136H1 GY 6183 COMP3170 AGRI90089 PHYSICS 1002 ECON10003 IEOR E4004 ACS61012 MATH 40550 MKTG2113 TELE4653 ENSC1004 ECE 232E MK726 MTRX2700 SHEET 3 CSCA08 COMP30027 Math 2XX3 statistic MMAN3200 BISM7213 ECMM109 ECON 2022 FNCE 20005 GR5067 CSCB09 ME 3017 CS 5100 PHYS 2903 COMS 4771 COMP222 COMP 1046 CSC148H1 ISYS2038 FINS2615 OT5206 COMP2048 A6 MECH5770M MA826/20 CMPT 726 STA457 INFS6071 INFS588 integral ECON 216 IRDR 0017 MA323 PA 15213 PSYC2001 ENV222 Statistics ACCT5930 ACCT12 LAND2121 ME 5405 COMM1180 TELE3113 A7 ECE 568 ACCT2019 INFS1200 ENGG1340 CS 1400 ASST3 FINM7409 SWEN30006 INFO2222 FIN20150 SOSS1000 CSSE7231 FINS5513 COMS3200 CECN 702 ECON4060H MN3515 HW2 PSYA02H3 ECON 2112 COMP90051 CS 1652 MA30059 AF1605 MKTG7501 ECO-6003B COM4521 CTM ECON3350 COMP 3804 ELEC9762 BISM 7255 ST22 MATH20014 MNE3122 COMP343 MATH5165 ELE7029 ENGR30002 SEEM3500 ECON7350 RISK5002 FINM7405 COMP90059 MOS 3320B STAT 431 BISM7221 MATH3014 MTH783P EBA 35302 ACS133 BUS2206 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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