Systems Modelling and Analysis Assignment 3

Systems Modelling and Analysis Assignment 3
Due: Monday 26/10/2020 by 11:00:00 pm. To be submitted individually using the online multiple choice/llin-the-blank system on Canvas.
Input Format:
Do not add + in front of positive numbers, e.g. 0:1 should be used not +0:1. If the answer is negative then
use – in front, e.g., -0:1 is ne.
Add a zero before decimals, e.g., 0:1 should be used not :1
Do not use scientic notation, e.g., 12400 should be used not 1:24e4.
Do not add units, e.g., 0:1 should be used not 0:1m.
If you do not follow these input format the answers will be marked wrong.
Assignment Background
This assignment is a continuation of Assignment 1 and 2, where you have modelled the DC-motor driven
positioning system. No material from Assignment 1 or 2 is required to complete this assignment.
Recall that V (s) = L fv(t)g is the voltage supplied to the DC motor and Y (s) = L fy(t)g is the dartboard’s
vertical position. The transfer function GOL(s) = Y V ((ss)) is exactly
GOL(s) = 50:9090s2 + 18512:3966s + 64793:3884
s6 + 492:3962s5 + 48199:8236s4 + 285994:0771s3 + 3665435:2617s2 + 11197107:4380s
Please run the code in MATLAB:
num = [50.9090 18512.3966 64793.3884];
den = [1 492.3962 48199.8236 285994.0771 3665435.2617 11197107.4380 0];
G_OL = tf(num,den);
Assume zero initial conditions when required..
Part 1: Dartboard Positioning System, Sinusoidal Inputs
Q1-2. Using MATLAB, plot the Bode diagram of GOL(s) (with a grid). Observe the behaviour at low
frequencies and at high frequencies (for both magnitude and phase).
(Q1) Is the gain smaller at low or high frequencies? [Canvas Input: Select Low or High]
(Q2) Is the phase lag smaller at low or high frequencies? [Canvas Input: Select Low or High]
Q3-4. Apply sinusoids of varying amplitude (v = Asin(t)) and observe the steady-state behaviour of y(t)
(i.e. yss(t)). For each question below, you are given a specic value A. Use this value of A to answer the
question.
(Q3) What is the absolute value of the phase lag between yss(t) and v(t)? (The answer should be
in degrees and in the range [0o; 360o))?
(Q4) What is the average value of yss(t)?
[Canvas Input: Two signed numbers – 5% tolerance allowed]
Q5-6. Now apply sinusoids of varying frequency (v = 100 sin(!t)) and observe the steady state behaviour
of y(t) (i.e. yss(t)). For each question below, you are given a specic value !, use this value of ! to answer
the question.
(Q5) What is the absolute value of the phase lag between yss(t) and v(t)? (The answer should be
in degrees and in the range [0o; 360o))?
(Q6) For yss(t); what is the maximum deviation from its average value?
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[Canvas Input: Two signed numbers – 5% tolerance allowed]
Q7. Amplify the voltage supplied to the motor by a constant A, so that v(t) = A sin(t) is supplied to the
motor.
Find the largest integer A such that jy(t)j ≤ 0:25[m]. [Canvas Input: One integer]
Hint: It is sucient to use a trial-and-error approach.
Q8. Using MATLAB, produce the Nyquist plot of GOL(s). Which of the following methods could we
use to modify GOL(s) so that its Nyquist plot intersects with the real axis at -1? [Canvas Input:
A multiple choice question]
Note: There may be multiple methods, choose all that apply.
Part 2: Dartboard Positioning System, Closed-Loop Control and
Discrete Systems
To simplify the problem now approximate the open-loop system with a lower order system as follows:

GOL(s) = s(s+3)(s+1+7 0:25(:s5+4) j)(s+1-7:5j). Part 1 has shown us that tuning parameters A and w for input v =

Asin(wt) to achieve a desired behaviour of y(t) is not ecient, so we turn again to closed-loop control. A
position sensor is purchased to measure y(t) accurately. First, use a controller C(s) = K such that the
voltage supplied to the motor is v = K(r – y), where r is a reference signal.
Q9-12. Produce the Bode Diagram for the closed-loop system with K = 50. On the Canvas quiz, you will
be given dierent values of !.
(Q9-10) Find the values of gain [dB] for the given values of !. [Canvas Input: two signed numbers – 5%
tolerance allowed]
(Q11-12) Find the values of phase [deg] for the given values of !. (The answer should be in degrees and in
the range [0o; 360o)) [Canvas Input: two signed numbers – 5% tolerance allowed]
Q13-15. For K = 70, plot r(t) = 2 sin (!t) and the corresponding y(t) for ! = 0:593 [rad/sec], ! = 5:93
[rad/sec], ! = 59:3 [rad/sec] (two graphs on the same axes for each frequency). Capture enough time to
show the steady-state response of y(t).
Find the value of y(t) – r(t) when t is 100 s for ! = 0:593; 5:93; 59:3 [rad/sec]. [Canvas Input: Three
signed numbers – 5% tolerance allowed]
For K = 12 and the reference signal r(t) = 0:1 sin(t), we will nd the magnitude and the phase-shift of yss(t)
in three ways:
Q16-17. Analytically, by substituting s = j · 1 into the transfer function of the closed-loop system. For
yss(t) = M sin (t + φ), write M2 and tanφ as irreducible fractions. This means there is a unique pair
of positive integers a1 and b1 such that M2 = ab11 , and a1 and b1 have no common divisors except one.
Similarly, tan(φ) = ab22 . Find a1 and a2. [Canvas Input: Two positive integer – exact solution required]
Q18-19. Graphically, by reading values from the response plot y(t) after it reaches a steady-state. For the
response plot y(t), let the rst peak in y(t) after t = 50 sec be at time ty. Let t1 be the closest r(t) peak
preceding time ty and t2 be the closest r(t) peak succeeding ty.
(Q18) What is y(ty)? [Canvas Input: One signed numbers -5% tolerance allowed]
(Q19) Select the equation that approximates the phase [deg] of yss(t) from Figure 1. [LMS Input:
One integer]
Q20-21. Graphically, by reading values from the Bode diagram of the closed-loop system. Using the Bode
diagram, nd the magnitude [dB] and phase [deg] of yss(t). (The answer of the phase should be in
degrees and in the range [0o; 360o))[LMS Input: Two signed numbers accurate to 3 signicant gures]
Q22-23. We wish to avoid the dartboard going outside the boundary at a steady-state mode (if there is
a nite number of exits – it is acceptable). Assume that we would like to operate at high frequencies (i.e.
2
! > 1 rad/s). If K = 225 and the total width of the safe zone is 1000 [mm], which frequencies ! should not
be used where the reference signal is r(t) = 0:55 sin (!t)[m]?
Using the Bode diagram, nd the range ! 2 [!min; !max] [rad/sec] such that the reference signal
is not suitable. [Canvas Input: Two signed numbers – 5% tolerance allowed]
Set K = 150 from now on forming a closed loop system GCL(j; !). We would like to address the long
phase-lag produced in this system when r(t) = 0:1sin(t).To improve performance, we will use an additional
controller with transfer function C1(s) = a0:0001 bs+1 s+1 outside the feedback loop such that
V (s) = K (C1(s)R(s) – Y (s))
(R(s) is the Laplace transform of r(t)). We would like the output y(t) to exactly match the reference signal
r(t) = 0:1sin(t) and can do so by tuning the values of a and b in the controller.
Q24-25. Find jGCL(j!)j [unitless] and GCL(j!) [deg] for ! = 1 rad/sec for the original closed
loop system GCL(j!). (The answer of the phase should be in degrees and in the range [0o; 360o)) [Canvas
Input: Two signed numbers – 5% tolerance allowed]
Q26-27. Find a and b such that the output for the new system matches the reference signal
(yss(t) = r(t) = 0:1 sin(t)). [Canvas Input: Two signed numbers – 5% tolerance allowed]
Q28-31. Reect on the usage of closed-loop control for the positioning system. What are the benets which
can be achieved by using it, compared to open-loop control?
Select true or false for the following statements
The closed loop control system:
(Q28) does not require a sensor. [Canvas Input: Select True or False]
(Q29) can allow the poles of the system to be moved to the LHP and make the system exponentially stable.
[Canvas Input: Select True or False]
(Q30) can allow better tracking of a reference signal. [Canvas Input: Select True or False]
(Q31) results in a system that is more robust to errors due to the presence of feedback. [Canvas Input:
Select True or False]
After analysing the system in the continuous time, now investigate how the system behaves in the discrete
time.
Q32. Start with converting the continuous-time system GOL(s) into discrete-time system GOL(z) using
the bilinear transformation (Tustin’s Method) with the sampling time T = 1s. Let the open-loop transfer
function be GOL(z) = a0zz44++ba11zz33++ba2z2z 2+2+b3az3+z+b4a4 . What is the value of b2? [Canvas Input: one signed
numbers – 5% tolerance allowed]
Q33. Then calculate the discrete-time frequency wd [rad/s] corresponding to the continuous-time operating
frequency wa = 1 [rad/s]. What is the value of wd? [Canvas Input: one signed numbers – 5% tolerance
allowed]
1. t
y – t1
2. t1 – ty
3. t
y – t2
4. t2 – ty
5. t
y –
12
(t1 + t2)
6. 1
2(t1 + t2) – ty
7. (ty – t1) 180 π
8. (t1 – ty) 180 π
9. (ty – t2) 180 π
10. (t2 – ty) 180 π
11. (ty – t1) 180 π
12. (t1 – ty) 180 π
13. (ty – t2) 180 π
14. (t2 – ty) 180 π
15. ty – 12(t1 + t2) 180 π
16. 1 2(t1 + t2) – ty 180 π
17. ty – 1 2(t1 + t2) 180 π
18. 1 2(t1 + t2) – ty 180 π
19. 360 (ty – t1)
20. 360 (t1 – ty)
21. 360 (ty – t2)
22. 360 (t2 – ty)
23. 360 ty – 12(t1 + t2)
24. 360 12(t1 + t2) – ty
Figure 1: Selection options for Q19
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Part 3: Filtering Signals
You will now be investigating the eect of dierent transfer functions on multi-tone audio les by analysing
their frequency content and the frequency response of the transfer functions.
Note that in all of Part 3 you may ignore the inuence of the transient response. That is, you only need
consider the steady-state response given by the frequency response G(j!).
Q34. The magnitude of the Discrete Fourier Transform of two signals are given in Figure 2. Signal 2 is
Signal 1 after being passed through some transfer function. Based on the gures, what type of transfer
function was used? [Canvas Input: Multiple Choice]
Figure 2: Signal Before and After Passing Through Transfer Function
a. Low Pass ( a
s + a
)
b. High Pass ( s
s + a
)
c. Band Pass ( s
(s + a)(s + b))
d. Band Reject ( s2 + b2
s2 + as + b2 )
In the Canvas quiz introduction, you are given 10 audio les (.wav). Each audio le is a selection of multiple
pure tones, 5 seconds long. Download the audio les. Use the function audioread() to load these audio
les into your MATLAB workspace.
Example (‘tones1.wav’):
>> [signal1, Fs1] = audioread(‘tones1.wav’);
signal1 will be a vector of numbers representing the audio signal ‘tones.wav’. Fs1 will be the sampling
frequency, in Hz.
The magnitude of the Discrete Fourier Transform for ‘tones1.wav’ is given in Figure 3, for your reference.
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Figure 3: Frequency Content of tones1.wav
Q35-36. Load ‘tones2.wav’ and, using MATLAB, determine the frequency content of the signal. What
are the two frequencies (Hz) present in the signal with the largest magnitudes? [Canvas Input:
Two signed numbers – 5% tolerance allowed]
‘tones[1-5].wav’ are 5 dierent multi-tone recorded audio signals. These signals are used as the inputs to
some transfer functions and recorded again to produce 5 new audio signals: ‘output[a-e].wav’. Some noise is
introduced into the audio signals during the recording process.
Q37. Match the original audio les (‘tones[1-5].wav’) to the corresponding output audio les
(‘output[a-e].wav’). [Canvas Input: Matching]
You are also given a data matrix ‘tf_data.mat’. Load the le into your MATLAB workspace (>> load(‘tf_data.mat’)).
This matrix represents the frequency response, G(j!), of 16 dierent transfer functons. Each original audio
le was used as the input to one of these transfer functions to produce the output les. We will now identify
which transfer functions were used.
The matrix is 16×25001, each row (1-16) representing a dierent transfer function. The following 16 Figures
are the Bode Diagrams for each transfer function in the same order as is presented in the matrix, for your
reference.
The matrix gives 25001 complex values for each transfer function. Each complex value is the transfer function
evaluated at a dierent frequency. These frequencies (in Hz) are given in ‘freq.mat’.
Hint: You may nd the MATLAB command frd() useful to load the data into a frequency response data
model. In this way you can more easily plot the Bode plots using the bode() command. For example:
>> G = frd(resp,omega);
>> bode(G);
Be careful with units, omega should be in rad/s.
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Figure 4: TF 1 Figure 5: TF 2
Figure 6: TF 3 Figure 7: TF 4
Figure 8: TF 5 Figure 9: TF 6
6
Figure 10: TF 7 Figure 11: TF 8
Figure 12: TF 9 Figure 13: TF 10
Figure 14: TF 11 Figure 15: TF 12
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Figure 16: TF 13 Figure 17: TF 14
Figure 18: TF 15 Figure 19: TF 16
Q38. Use these complex values to determine the eect the transfer functions will have on the frequencies present in the original signals. Match the original audio les to the corresponding transfer
functions. [Canvas Input: Matching]
Note: ‘TF 1’ is the transfer function described by the rst row of the data matrix and the rst Bode plot
given. TF 2-16 follow in the same fashion.
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