Answers
UNIVERSITY OF SOUTHAMPTON ELEC3201W1
SEMESTER 1 EXAMINATION 2020 - 2021
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.
Copyright 2021 c© University of Southampton Page 1 of 16
An
sw
er
s
ELEC3201W1
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]
Copyright 2021 c© University of Southampton Page 2 of 16
An
sw
er
s
ELEC3201W1
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.
Copyright 2021 c© University of Southampton
TURN OVER
Page 3 of 16
An
sw
er
s
ELEC3201W1
(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]
Copyright 2021 c© University of Southampton Page 4 of 16
An
sw
er
s
ELEC3201W1
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
TURN OVER
Page 5 of 16
An
sw
er
s
ELEC3201W1
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
An
sw
er
s
ELEC3201W1
(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.
Copyright 2021 c© University of Southampton
TURN OVER
Page 7 of 16
An
sw
er
s
ELEC3201W1
[25 marks]
Copyright 2021 c© University of Southampton Page 8 of 16
An
sw
er
s
ELEC3201W1
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

Copyright 2021 c© University of Southampton
TURN OVER
Page 9 of 16
An
sw
er
s
ELEC3201W1
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]
Copyright 2021 c© University of Southampton Page 10 of 16
An
sw
er
s
ELEC3201W1
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)
0 = a11 (5)
Similarly using information about the finish position and velocity - t = tf - we have
45 = a20 + a21(tf − tw) + a22(tf − t2)2 + a23(tf − tw)3 (6)
0 = a21 + 2a22(tf − t2) + 3a23(tf − tw)2 (7)
Next, at t = tw we know that θ(tf ) = 90; thus we have
90 = a10 + a11tw + a12t
2
w + a13t
3
w (8)
90 = a20 (9)
Finally, because velocity and acceleration are continuous at the way point we have
a11 + 2a12tw + 3a13t
2
w = a21 (10)
2a12 + 6a13tw = 2a22 (11)
Copyright 2021 c© University of Southampton
TURN OVER
Page 11 of 16
An
sw
er
s
ELEC3201W1
Noting that tf = 6 and tw = 3 we can stack equations (4)-(11) in matrix-vector form as
45
0
45
0
90
90
0
0

=

1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 0 0 1 3 9 27
0 0 0 0 0 1 6 27
1 3 9 27 0 0 0 0
1 0 0 0 0 0 0 0
0 1 6 27 0 −1 0 0
0 0 2 18 0 0 −2 0


a10
a11
a12
a13
a20
a21
a22
a23

(12)
(Other matrices possible, depending on the order of “stacking”)
(ii) Yes, a small matrix equation could be solved. From the working above, it is clear that we have
expressions for a10, a11 and a20 directly. Therefore we could set up an expression in the
unknowns a12, a13, a21, a22 and a23 - this would lead to a matrix-vector equation involving a 5× 5
matrix.
(iii) We effectively ignore the second portion of the trajectory and just consider the part between
θ(0) = 45 and θ(tw) = 90. At both points the velocities are zero. From the expressions above we
have
θ˙(t) = a11 + 2a12t+ 3a13t
2
which is actually
θ˙(t) = 2a12t+ 3a13t
2
since a11 = 0. Now the maximum/minimum of this function occurs when
2a12 + 6a13t = 0
Also from our expression for velocity, we have that, at t = tw:
0 = 2a12(tw) + 3a13t
2
w = tw(2a12 + 3a13tw)
So since tw 6= 0, we have
a12 = −3
2
a13tw
Using this in our expression for max/min velocity we have
t = − a12
3a13
=
1
2
tw
So we need to check the velocity at the half-way time i.e 1.5 seconds.
(d) The Jacobian matrix is easily calculated from basic differentiation
J(q) =
cos θ2 cos θ3 −d1 sin θ2 cos θ3 −d1 cos θ2 sin θ3cos θ2 sin θ3 −d1 sin θ2 sin θ3 d1 cos θ2 cos θ3
− sin θ2 −d1 cos θ2 0

Copyright 2021 c© University of Southampton Page 12 of 16
An
sw
er
s
ELEC3201W1
To determine constraints on the velocity we need to calculate |J(q)| and determine it is zero.
|J(q)| = cos θ2 cos θ3
∣∣∣∣ −d1 sin θ2 sin θ3 d1 cos θ2 cos θ3−d1 cos θ2 0
∣∣∣∣
+ d1 sin θ2 cos θ3
∣∣∣∣ cos θ2 sin θ3 d1 cos θ2 cos θ3− sin θ2 0
∣∣∣∣
− d1 cos θ2 sin θ3
∣∣∣∣ cos θ2 sin θ3 −d1 sin θ2 sin θ3− sin θ2 −d1 cos θ2
∣∣∣∣
= cos θ2 cos θ3(d
2
1 cos
2 θ1 cos θ3)
+ d1 sin θ2 cos θ3(d1 cos θ2 cos θ3 sin θ2)
− d1 cos θ2 sin θ3(−d1 cos2 θ2 sin θ3 − d1 sin2 θ2 sin θ3)
= d21 cos
3 θ2 + cos
2 θ3 + d
2
1 sin
2 θ2 cos
2 θ2 cos θ2 + d
2
1(cos
3 θ2 sin
2 θ3 + cos θ2 sin
2 θ2 sin
2 θ3)
= d21 cos
3 θ3(sin
2 θ3 + cos
2 θ3) + d
2
1 cos θ2 sin
2(θ2(sin
2 θ3 + cos
2 θ3)
= d21 cos θ2(sin
2 θ2 + cos
2 θ2)
= d21 cos θ2
So singularity when d1 = 0 and θ2 = pi, 3pi/2
(e)
(i) Students would be expected to provide advantages of electric actuators (convenience, precise
control, ease of maintenance etc), disadvantages of electric actuators (poor force/torque levels,
size/power issues etc) and then also discuss the advantages of hydraulic actuators (high levels
of force production, maintenance of force under “zero” power, good size/force ratio) and their
disadvantages (difficulty in control, maintenance issues, cleanliness issues etc).
Somewhat open-ended question with different answers acceptable.
(ii) Students would be expected to suggest that the manipulator behaviour would be improved with
sensors which can detect object proximity, the force exerted on an object, and whether or not
an object is slipping or deforming. This information can then be used to adjust gripper force or
orientation. The concept of soft robotics could also be discussed.
Somewhat open-ended question with different answers acceptable.
Copyright 2021 c© University of Southampton
TURN OVER
Page 13 of 16
An
sw
er
s
ELEC3201W1
[25 marks]
Copyright 2021 c© University of Southampton Page 14 of 16
An
sw
er
s
ELEC3201W1
Question 3.
A camera is taking images of a working area in a robotic system as shown
in Figure 4.
(a) Due to technical difficulties, the images taken by the camera have low
contrast with low visibility. Propose and describe techniques to in-
crease the visibility of images.
[4 marks]
(b) Also, due to low light and radiation, images are contaminated with some
noise. Propose and describe filters to remove or to reduce the noise.
[5 marks]
(c) In the manufacturing process, there are three kinds of objects with three
different shapes appearing in front of the camera in the working area.
The colour and intensity of the objects are distinctively different from the
background.
(i) Advance an algorithm which can recognise these three objects.
[11 marks]
(ii) Explain if it would be possible to use deep learning methods to
classify these three objects.
[5 marks]
[25 marks]
FIGURE 4: Robotic system with camera
Copyright 2021 c© University of Southampton
TURN OVER
Page 15 of 16
An
sw
er
s
ELEC3201W1
Question 4.
(a) Consider a swarm of ground robots operating in the New Forest, per-
forming a monitoring task - the robots have to approach a number of
pre-specified GPS marked trees and take pictures of the tree barks for
further analysis. For the task, the robots of the swarm have to traverse
large distances in an energy efficient manner. Describe the three main
approaches that you would employ for your ground robots to traverse
the forest, along with their individual merits and demerits.
[10 marks]
(b) In tuning the parameters for your robot controllers for the above task, you
decide to use an evolutionary algorithm. Describe the genetic operators
employed in your algorithm and provide an example of how they work.
[5 marks]
(c) Besides the above mentioned forest monitoring application, briefly ex-
plain two other application scenarios for robot swarms.
[2 marks]
(d) Explain with justification, which of the following is/are NOT a self-organizing
system.
(i) A Swiss watch comprising over 1000 components.
(ii) An ant colony comprising over 100,000 ants.
(iii) A flock of over a 100 starlings.
(iv) The development of a multicellular organism with over 30 trillion
cells.
[8 marks]
[25 marks]
Copyright 2021 c© University of Southampton
END OF PAPER
Page 16 of 16