Part 1: Short Questions
Question 1
You will be able to answer this question after you have studied Week 11.
Use the step response interactive resource to plot the step response of the following systems:
A first-order system with a gain of 4 and a time constant of 1 second;
b. Second-order system with a gain of 4, a natural frequency of 4 rad/s and a damping ratio of 2;
c. Second-order system with a gain of 4, a natural frequency of 4 rad/s and a damping ratio of 1;
d. Second-order system with gain of 4, a natural frequency of 4 rad/s and a damping ratio of 0.5;
A second-order system with a gain of 4, a natural frequency of 4 rad s−1 and a damping ratio of 0.5
Provide screenshots of four responses as your answer. Use the zoom feature to ensure the step response is shown from time of zero to a time when the response has settled to its final value.
3 marks out of 4
Question 2
You will be able to answer this question after you have studied Week 12
Identify whether each of the pole–zero diagrams shown in Figure 1 represent a stable system or an unstable system. Briefly explain your answers.
(a)
(Stable –all the poles are on the left of the imaginary axis in the s-plane)
(b)
(Unstable –Two poles are on the right of the imaginary axis in the s-plane)
(c)(Unstable –One pole is on the right of the imaginary axis in the s-plane)
(d)(Stable –all the poles are on the left of the imaginary axis in the s-plane)
4 marks out of 4
Question 3
You will be able to answer this question after you have studied Week 13.
A second-order process has the following properties:
k = 2
ζ = 0.4
ω = 3
What is the transfer function of this process?
Since the damping coefficient is less than 1, then the function takes the form of an underdamped second-order system in which the general equation is:
Y(t) = K (1 – cos ( + )); where Cos = ζ
Substituting the values of k, ζ, and ω in the equation;
Y(t) = 2 (1 – cos ( + )
Y(t) = 2 (1 – 1.091* cos ( + ))
Using the standard second order system transfer function (shown below) which is given in Chapter 4, Section 4.4, we have:
k = 2
2ζωn = 2.4
ωn2 = 9
0 marks out of 4
Question 4
You will be able to answer this question after you have studied Week 13. State whether each of the following two differential equations is linear or non-linear and, if linear, what is its order? Explain your answer.
a. 3y’’’ + 7y’’ + y’ + 4y = 3x + 1 —– Non-linear diff. equation (Variables in equation have powers more than 1). Order 3 (because highest power of y in the function is 3).
b. 3y’’’ + 7y’’ + y’ + 4y2 = 3x + 1—– Non-linear diff. equation (Variables in equation have powers more than 1). Order 3 (because highest power of y in the function is 3).
What are the orders of the following systems? Explain your answer.
c. Y(s)/X(s) = 3s/(5s4 + 3s3 + s2 + 2s +1) – Order 4 (highest power of s in the function is 4).
d. Y(s)/X(s) = (s2 +3s + 2)/(s3 + 2s2 + 3s + 5) – Order 3 (highest power of s in the function is 3).
4 marks out of 4
Question 5
You will be able to answer this question after you have studied Week 13.
A second-order system has the following open-loop transfer function:
G (s) = =
A PD controller is placed in the forward path of a closed-loop system. The derivative gain of the PD controller, K ’, is set to 0.4. This makes the transfer function of the PD controller:
C(s) = Kp (0.4s + 1)
Use the root-locus plotter to find the plot of the closed-loop poles by entering the combined transfer function C(s)G(s) without the proportional gain, so that the plot shows the position of the closed-loop poles as the proportional gain is varied from 0 to 100.
From the G(s) function;
G (s) =
Root loci poles = -1, -2; zeros = none
Combined transfer function C(s)G(s) = Kp (0.4s + 1) * = correct for combined transfer function.
Since C(s)G(s) + 1 = 0
+ =
Therefore;
Kp = – =
Determine the breakaway and the break-in points;
= 0 = –
Therefore;
= 0
= 0
Solving the equation;
S = -1.634; S = -3.366
The value of k for S = -1.634 will be;
K = – …substituting for k
K = – = – (-0.232044)/(0.866) = 0.268
The value of k for S = -3.366 will be;
K = – …substituting for k
K = – = – (3.231956)/(-0.866) = 3.732
You can not enter brackets in the RLP tool
The combined open-loop transfer function is:
1 mark
So, what is entered into the root-locus plotter is:
s+2.5/s^2+3s+2
1 mark out of 4
Question 6
You will be able to answer this question after you have studied Week 14.
A PID controller has the following transfer function and gains:
= Kp (1 + + sKD’)
Where Kp = 4, KI ’ = 0.25 and KD’ = 0.25 This controller introduces an open-loop pole at = 0 and two zeros. Calculate the position in the s-plane where you would find the zeros.
Substituting the values;
= 4 (1 + + 0.25s) = 4+ + s = = =
Since numerator of the function is a polynomial in s, the roots of the polynomial are zeros of the system.
Solving for the roots of the polynomial; Roots for the equation: …from analysis; s = -0.268; s = -3.732
Position in the S-plane where you would find the zeros:
s = -0.268; s = -3.732, s = 0.
4 marks out of 4
Question 7
You will be able to answer this question after you have studied Week 14.
A process has the following open-loop transfer function.
=
Assuming that this process is being controlled by a proportional controller with a gain of 10, find the steady-state error of the closed-loop system after a unit step input.
In finding the steady-state error; first re-write the denominator;
= =
You have forgotten the gain of 10
For X(s) is a unit step input; Y(s)/X(s) = G(s) and s = 0
Therefore, substituting for s = 0 in the equation;
yss = = 0.4
Since the proportional controller has a gain of 10; the steady state error will be;
Yss = 0.4*10 = 4
With a unit step input, we find the steady state response by setting s to zero.
Thus, the steady-state output is:
yss = 8/10 = 0.8
The steady state error would therefore be 1 – 0.8 = 0.2 or 20%
1 mark out of 4
Question 8
a. In Figure 6.3 in Book 2 (p. 78), there is a circuit that performs as a PID controller. The sub-circuit in the middle of the circuit takes the voltage input, Vin, and performs integration using an op-amp with a resistor, R3, and a capacitor, C1, to produce an output voltage which we will call V0. This then drives a current through R6 to produce an output labelled I for ‘integrator’. If the resistor R3 has a value of 1 kΩ, what should the value of the capacitor, C1, be to give a gain, V0 / Vin, of 5? (3 marks)
=
Substituting the provided values in the equation;
R3 = 1*103 Ω
= 5
Therefore; 5 = ;
C1 = 1/(5*) = 2 *10-4 F
The final op-amp should have a gain of –1. How is this achieved in the circuit? (1 mark)
The final op-amp showing gain of -1 is achieved by multiplying the error e(t) by the obtained gains for P, I, and D and the resulting outputs are summed by another inverting amplifier to get the final voltage output.
A gain of -1 is achieved by setting all the resistors, R5, R6 and R7 equal to Rf
3 marks out of 4
Question 9
You will be able to answer this question after you have studied Week 16. In the paper entitled ‘Soft computing approach to safe navigation of autonomous planetary rovers’ by Tunstel, Seraji and Howard (2000), the traction management system is described in Section 11.3.3 In it they describe how they trained a neural network to recognise soil types from images, and used that to estimate a traction coefficient which has a value of 0 to 1. This value is then fed into a fuzzy rule-based system to determine the speed of the vehicle. Use a diagram to show:
what shape membership functions are used;
Triangular shape membership functions used for low, medium, and high traction coefficients
How the membership functions are drawn on the traction coefficient axis?
Low membership
Medium membership
High membership
4 marks out of 4
Question 10
You will be able to answer this question after you have studied Week 17.
a. In the same paper described in Question 9, a neural network is used to determine the traction coefficient. From the information given in the paper, describe the architecture of the neural network, i.e. how many input neurons, hidden neurons and output neurons there are. (2 marks)
The architecture of the neural network consists of: four input neurons, two hidden neurons, and one output neuron.
The image in Figure 11.4b represents the network, the network has three fully connected layers i.e. each neuron output in one layer is connected to all the neuron inputs in the next layer
There are 40×40 = 1600 input neurons, 2 hidden neurons and one output neuron
b. When training a neural network, the data would usually be divided into three sets. What are these sets called and what is the purpose of each set? (2 marks)
Training set – It comprises randomly selected input and output pairs where the goal is to train the neural network to link the pairs, and errors are calculated to understand the areas for improvement.
Validation set – Consists of another set of randomly selected data but the goal is to introduce new data that the neural network has not seen before and evaluate how accurately it makes predictions based on the new data.
Test set – It consists of data used to test how well the already trained neural network performs.
3 marks out of 4
Part 11: In-depth Questions
Question 11 (20 marks)
You will be able to answer this question after you have studied Week 14.
This question asks you to analyse a second-order step response and design a controller using the root-locus method. You are advised to use no more than 500 words in your answer to this question. Figure 2 shows a unit step response of a second order process. It shows that there is a small overshoot and then the output settles at 1. At the bottom of the figure it shows the values of the pole location which are approximately at s = –1 + j and s = –1 – j.
a. Using the step response software, by trial and error, find the values for the damping ratio and the natural frequency that produces this response; (Hint: See Figure 2.5 on page 18 of Book 2). Make sure you include a screenshot of your step response. (3 marks)
Damping Ratio = 0.86
Natural frequency = 2.15
https://lpsa.swarthmore.edu/SecondOrder/SOI.html
Using the step response software
You should have used the OU step response SW and by trial and error found solutions to the
Poles at s = –1 + j and s = –1 – j.
0 marks out of 3
b. By substituting the values for the damping ratio, natural frequency and gain into the standard form of the second-order transfer function, the transfer function for this system is found to be:
G (s) =
Using this transfer function, show that the poles are at approximately s = –1 + j and s = –1 – j? (3 marks)
For the transfer function, pole are at locations of s where denominator = 0;
= 0
Applying quadratic equation formula;
S =
From the equation = 0; a = 1, b = 2, c = 2.
Substituting the values in the quadratic equation formula;
S = = =
Since j2 = -1; = = 2j
Therefore;
S = = -1 ± j
From the result, the two complex roots include;
S1 = -1 + j
S2 = -1 – j
3 marks out of 3
Figure 2 (a) Unit step response of a second-order process, (b) zoomed in on the overshot
It is decided that the process in Figure 2 will be controlled using a PD controller with a derivative gain, K ’ of 0.5, so that the transfer function of the PD controller is: C(s) = K (1 + 0.5s).
Using the combined transfer function of the forward path, C(s)G(s), use the root-locus software to plot the root locus with the value of K going from a minimum gain of 0 to a maximum of 50. Include a screenshot of your root-locus plot as your answer (3 marks)
From the G(s) function;
G (s) =
Root loci poles are s = –1 + j and s = –1 – j; zeros = none
Combined transfer function C(s)G(s) = Kp (1 + 0.5s) * =
Since C(s)G(s) + 1 = 0
+ = 0
Therefore;
Kp = –
Determine the breakaway and the break-in points;
= 0 = –
Therefore;
= 0
= 0
Solving the equation;
S = -0.586; S = -3.414
The value of k for S = -0.586 will be;
K = – …substituting for k
K = – – (1.1714)/(1.414) = -0.828
The value of k for S = -3.414 will be;
K = – …substituting for k
K = – – (6.8274)/(-1.414) = 4.828
You can not use brackets in the RLP and you do not enter Kp
The combined transfer function is:
The value entered into the root-locus plotter is 2+s/s^2+2s+2.The root locus looks like this:
0 marks out of 3
d. The root locus showed that at some value of gain the two closed-loop poles arrive at the real axis together. By zooming and shifting the root locus find the value of s at which point the two closed-loop poles meet on the real axis. Answers should be to 1 place after the decimal point. Remember to include a screenshot at an appropriate zoom level of your root-locus plot. (4 marks)
Not attempted 0 marks out of 4
e. The root-locus plot showed the position of the closed-loop poles as Kp was varied from 0 to 50. By reducing the value of the maximum gain, the root locus can be produced which stops when the two poles meet on the real axis and therefore the maximum gain is the value at this point where the two poles meet. Try lowering the maximum value of gain to find the value of the gain to 3 places after the decimal point where the two poles meet. Include a screenshot at an appropriate zoom level of your root-locus plot. (4 marks)
Not attempted 0 marks out of 4
f. Finally, we would like you to show that the value of gain found in part (e) and the value of s found in part (d) are correct by using the closed-loop transfer function to find the value of K where the two poles meet on the real axis. (3 marks)
Not attempted 0 marks out of 3
Q11 total 3 marks out of 20
Question 12 (20 marks)
You will be able to answer this question after you have studied Week 16. This question asks you to calculate the output of the fuzzy logic stable attitude controller described in the paper entitled ‘Soft computing approach to safe navigation of autonomous planetary rovers’ by Tunstel, Seraji and Howard (2000). You are advised to use no more than 500 words in your answer to this question.
The attitude controller in the above paper measured the pitch and roll of the rover. It then used a fuzzy logic system to determine the speed at which the rover should travel. It uses the membership functions shown in Figure 11.3 in the paper, together with 15 rules which are also summarised in Figure 11.3. However, there are no numbers on the axes and the membership functions for the output velocity are not shown. From other papers by the same authors it is possible to get a bit more information. Based on this, the three membership functions are shown below, this time with some numbers on the axes.
Figure 3: Roll
For roll the sets are:
a. NEGATIVE
b. ZERO
c. POSITIVE
Figure 4: Pitch
a. NEGATIVE HIGH
b. NEGATIVE LOW
c. ZERO
d. POSITIVE LOW
e. POSITIVE HIGH
Figure 5: Speed
For speed the sets are:
SLOW
MODERATE
FAST
Re-write the 15 rules in the form ‘If roll is NEGATIVE and pitch is NEGATIVE LOW then speed is MODERATE. (3 marks)
If roll is NEGATIVE and pitch is NEGATIVE HIGH then speed is SLOW
If roll is NEGATIVE and pitch is NEGATIVE LOW then speed is MODERATE
If roll is NEGATIVE and pitch is ZERO then speed is MODERATE
If roll is NEGATIVE and pitch is POSITIVE LOW then speed is MODERATE
If roll is NEGATIVE and pitch is POSITIVE HIGH then speed is SLOW
If roll is ZERO and pitch is NEGATIVE HIGH then speed is MODERATE
If roll is ZERO and pitch is NEGATIVE LOW then speed is MODERATE
If roll is ZERO and pitch is ZERO then speed is FAST
If roll is ZERO and pitch is POSITIVE LOW then speed is MODERATE
If roll is ZERO and pitch is POSITIVE HIGH then speed is MODERATE
If roll is POSITIVE and pitch is NEGATIVE HIGH then speed is SLOW
If roll is POSITIVE and pitch is NEGATIVE LOW then speed is MODERATE
If roll is POSITIVE and pitch is ZERO then speed is MODERATE
If roll is POSITIVE and pitch is POSITIVE LOW then speed is MODERATE
If roll is POSITIVE and pitch is POSITIVE HIGH then speed is SLOW
3 marks out of 3
b. With a pitch of –30 degrees and a roll of +10 degrees:
i. Using Figure 6, what are the membership values of pitch and roll? (4 marks)
Figure 6: A right-angled triangle. It shows that as you move along the horizontal to the right by 5, the vertical drops by 0.2.
From figure 6;
Membership value for pitch; At -30 degrees; 0.2 for what?
In taking the membership value for pitch at -30 degrees, it was realized that it is the same as +30 degrees and following the gradient of the triangle, it would be a negative value (-0.2). Since the membership value does not consider the sign, the absolute value of 0.2 was taken.
Membership value for roll; At +10 degrees; 1.0 – 2*0.2 = 0.6
In taking the membership value for roll at +10 degrees, it was realized it was two movements of five on the horizontal axis which led to a drop of 0.4 on the vertical axis leading to a value of 0.6.
i. With pitch at -30 degrees, the membership values are:
NH 0.2
NL 0.8
Z 0
PL 0
PH 0
With Roll at 10 degrees the membership values are:
N 0
Z 0.6
P 0.4
1 mark out of 4
ii. Using Table 2, write down the inferred membership values of speed. (4 marks)
Table 2: Minimum values of the two input conditions
Roll Pitch
NEGATIVE HIGH (0.2) NEGATIVE LOW (0.6) ZERO (1) POSITIVE LOW (0.6) POSITIVE HIGH (0.2)
NEGATIVE (0.6) 0.2 0.6 0.6 0.6 0.2
ZERO (1) 0.2 0.6 1 0.6 0.2
POSITIVE (0.6) 0.2 0.6 0.6 0.6 0.2
Roll Pitch
NEGATIVE HIGH NEGATIVE LOW ZERO POSITIVE LOW POSITIVE HIGH
NEGATIVE 0 0 0 0 0
ZERO 0.2 0.6 0 0 0
POSITIVE 0.2 0.4 0 0 0
In table 2, the negative and positive values for roll were taken as 0.6 (absolute values) and based on the outcome in part (i). Meanwhile, NEGATIVE HIGH and POSITIVE HIGH for pitch were taken as 0.2 and based on the outcome in part (i). Meanwhile, the NEGATIVE LOW and POSITIVE LOW for pitch were taken as 0.6 which is near the zero mark. The membership values of pitch and roll are maximum at ZERO with value of 1.0. The rule used for the minimum values of the two input conditions was that the minimum values of the roll and pitch at any particular point is taken as the inferred membership values for the speed. For instance, for roll NEGATIVE (0.6) and pitch NEGATIVE HIGH (0.2), the low of the two values (0.2) is taken as the inferred membership value of speed. Using this concept, the table 2 was completed as shown.
0 marks out of 4
iii. Using Table 3, write down the defuzzified value of speed when centre of area defuzzification is used. (5 marks)
Firstly, table 3 was completed based on the values in table 2 and the 15 rules in part a. From table 3, it was realized that the maximum values for;
Slow membership = 0.2
Moderate membership = 0.6
Fast membership = 1
Table 3: Maximum values for the membership functions of speed
Speed Rule 1 Rule 2 Rule 3 Rule 4 Rule 5 Rule 6 Rule 7 Rule 8 Rule 9 Rule 10 Rule 11 Rule 12 Rule 13 Rule 14 Rule 15
SLOW 0.2 0.2 0.2 0.2
MODERATE 0.6 0.6 0.6 0.2 0.6 0.6 0.2 0.6 0.6 0.6
FAST 1
Speed Rule 1 Rule 2 Rule 3 Rule 4 Rule 5 Rule 6 Rule 7 Rule 8 Rule 9 Rule 10 Rule 11 Rule 12 Rule 13 Rule 14 Rule 15 MAX
SLOW 0 0 0.2 0 0.2
MODERATE 0 0 0 0.2 0.6 0 0 0.4 0 0 0.6
FAST 0 0
Assuming a base equal to 100, the area of triangle for membership is height x 50;
Area for Slow = 50 x 0.2 = 10
Area for Moderate = 50 x 0.6 = 30
Area for fast = 50 x 1 = 50
Location for center of gravity, Dcog =
Substituting;
Location for center of gravity, Dcog =
Location for center of gravity, Dcog = 72.22
Therefore, the center of gravity is located at 72.22% on x-axis. This means that the defuzzified value of speed is 72.22%
To defuzzify the following equation can be used:
Where , and are the weights of each membership function. The weights are calculated by finding the area of each triangle, The membership function for speed shows three fully symmetrical triangles, however, their heights need altering to the membership values found in the previous table, 0.2, 0.6 and 0 respectively. The area is found by multiplying half the base by the height of the triangle, therefore, , and . Below is a diagram showing the membership values of speed reduced to points at the centre of area of the triangles:
Figure 11; a diagram showing the membership values at their centres of area.
The values of , and are the distances from the fulcrum, which is set at the 0 point, therefore, , and . The equation becomes:
Therefore, the defuzzified value of speed when centre of area defuzzification is used with a pitch of -30 degrees and a roll of +10 degrees is 0.4375 m s-1.
2 marks out of 5
C The attitude control is only one part of the speed control system. The other part is traction management, which takes an estimated value for the traction and also recommends a speed for the rover. How does the rover decide what speed to travel when faced with one speed from the attitude controller and one from the traction management system? (2 marks)
The rover uses neural network to decide on the weights of values from traction management system and the attitude control and based on patterns from training, selects the best option which is suited for the geological conditions.
In the paper it says that the minimum value of the two recommended speeds is taken
0 marks out of 2
d. Use Google Scholar to find the following paper. You do not have to download or read the paper. (2 marks)
What year was it published, where was the paper from and how many citations does it have? Provide a screenshot to support your answer.
Year published – 2001
Where was the paper from – California Institute of Technology, USA
How many Citations – 22
2 marks out of 2
Q12 total 8 marks out of 20
Q13 not attempted 0 marks out of 20
TMA 02 38%
