\u00a0
\nInstrumentation and Control\/Assignment 2: Control<\/p>\n
Intended Learning Outcomes assessed by this work
\n1. Understand the basic underpinning theory of linear continuous-time dynamical systems and be
\nable to simulate their behaviour;
\n2. Manipulate and solve differential equations using Laplace transforms and interpret the system
\npoles and zeros.
\nHand-in date: 18 April 2016
\nNote that the Hand-in date is a \u2018hard\u2019 deadline, in the sense that if the coursework is handed in late
\nthen a zero mark is awarded. You are encouraged to submit this coursework well in advance of the
\ndeadline date.
\nMarks allocated
\nQuestion 1: 10 marks
\nQuestion 2: 20 marks
\nQuestion 3: 20 marks
\nQuestion 4: 20 marks
\nQuestion 5: 20 marks
\nPresentation: 10 marks
\nCoursework Submission
\nThe completed coursework should be submitted, over the counter, to the Assignment Office before
\n16.00 hours on the deadline date. The coursework must include a printed bar-coded coversheet. The
\nbar-coded coversheets can be generated from NOVA, online assessments system, which can be
\naccessed by all students from https:\/\/webapp.coventry.ac.uk\/Nova\/NovaMain.aspx.
\nControl Coursework J. E. Trollope
\nPage 2 of 7
\nPlease note that the work submitted should be your individual effort. The penalties for plagiarism
\nare severe. Evidence of plagiarism will be dealt with according to the rules and regulations outlined
\nin the Student Handbook (Faculty of Engineering and Computing).
\nAssessment Criteria
\nThere are 100 marks in total for this assignment and a breakdown of these marks is provided in the
\nassignment details. Of these marks, 90 marks are for accuracy in the mathematics, clear logical
\narguments, calculations and use of software packages (if appropriate). The remaining 10 marks are
\nfor the general presentation.
\nMarks may be deducted for:
\n1. writing mathematics without suitable explanatory text (as a general rule, all mathematics
\nshould be contained in proper, full sentences);
\n2. mathematical errors (including errors of notation);
\n3. poor proof-reading (for example, poor spelling, poor grammar, etc.);
\n4. poor readability, for example, confusing explanations, etc.;
\n5. not suitably commenting any computer software package commands.
\nItems 1 and 2 will normally lead to deductions from the marks for accuracy, whereas items 3-4 relate
\nto the presentation mark. (Suitable allowance will be made for students with a recognised disability
\nsuch as dyslexia.)
\nTypically, to obtain a first class mark (= 70%), the report should have few mathematical errors and all
\nwork should be suitably explained. Similarly, to obtain a pass mark (= 40%), the report should
\ntypically show evidence that the basic concepts are understood, but several errors and\/ or lack of
\nproper explanations may be present. It should be understood that these are only typical cases and
\nthat the marking scheme will determine the final mark obtained.
\nControl Coursework J. E. Trollope
\nPage 3 of 7
\nQuestion 1
\na) Suppose ( ) and ( ) are the Laplace transforms of the reference ( ) and system output ( )
\nrespectively. If the blocks 1, 2, , and in Figure 1 are transfer functions in the Laplace variable
\n , e.g. 1( ), find the closed loop transfer function of the form ( )
\n ( )
\n.
\n(10 marks)
\n 1
\nR(s) Y(s)<\/p>\n
+
\n\u2013
\n 2<\/p>\n
\n+
\n\u2013
\nFigure 1: Linear time-invariant system
\nControl Coursework J. E. Trollope
\nPage 4 of 7
\nQuestion 2
\na) For the closed loop system configuration in Figure 2, obtain an expression for the overall closed
\nloop transfer function
\n(4marks)
\nb) Hence or otherwise write down the closed loop characteristic equation
\n(2 marks)
\nc) Use the Routh-Hurwitz stability criterion to find the values of for which the system is stable
\n(6 marks)
\nd) Sketch the root locus for this system and indicate the main points
\n(4 marks)
\ne) Use MATLAB to confirm your findings in part d)
\n(4 marks)
\n\u2013
\n( + 1)
\n( \u2013 1)( + 2)<\/p>\n
Figure 2: Closed-loop transfer-function
\nControl Coursework J. E. Trollope
\nPage 5 of 7
\nQuestion 3
\nA system is configured as a negative unity feedback control system having a transfer function
\n ( ) =
\n1
\n( + 1)( + 4)
\nwhich is cascaded with a proportional controller having a gain .
\na) With the gain initially set to unity, determine the closed loop transfer function and calculate the
\nsystem poles.
\n(3 marks)
\nb) With the gain set to unity apply a unit step at time = 0 (assuming zero initial conditions). Use
\nMATLAB to obtain the step response. Comment on the form of the response i.e. describe the key
\nfeatures (Hint: relate this to the inverse Laplace transform of ( ).
\n.(6 marks)
\nc) In an attempt to reduce the steady state error, the gain is increased to 10. With this value of
\nrepeat steps a) and b). Check your results using MATLAB.
\n(6 marks)
\nd) The proportional controller is now replaced by a proportional plus integral controller. You are now
\nrequired to produce a closed-loop system response which is critically damped (i.e. fastest response
\nwithout overshoot). Determine the controller gains required. Implement the controller in MATLAB
\nand demonstrate that the response is as required.
\n(5 marks)
\nControl Coursework J. E. Trollope
\nPage 6 of 7
\nQuestion 4
\nThe system given by the transfer function in Question 3 is now configured in state space form
\n = +
\n = <\/p>\n
where is the system matrix, is the input vector, is the output vector, and are the system
\ninput and output, respectively, and is the system state vector.
\n(a) By defining 1 = and 2 = write down the corresponding phase variable canonical form.
\n(4 marks)
\n(b) The system is now configured for state variable feedback in order to control the dynamic
\nbehaviour of the system, i.e. =
\n where <\/p>\n
is a feedback vector containing controller gains.
\nMaking use of modal control the slow eigenvalue is to be relocated such that its response is now 10
\ntimes faster. Determine the feedback vector to achieve this response, and demonstrate the
\nperformance using MATLAB.
\n(6 marks)
\n(c) Explain what is meant by observability and controllability. Show that the system is both
\nobservable and controllable using the modal matrix method.
\n(3 marks)
\n(d) Write down the Kalman observability and controllability test matrices, hence or otherwise
\nconfirm your results in (c).
\n(3 marks)
\n(e) Work back from the state space representation obtained in (a) to show how a transfer function
\nrepresentation can be obtained. Hence or otherwise write down the open-loop characteristic
\nequation and show that the eigenvalues of the system matrix are the same as the system poles.
\n(4 marks)
\nControl Coursework J. E. Trollope
\nPage 7 of 7
\nQuestion 5
\nIn the UK, research is being undertaking on the feasibility of railway carriages coupling and
\nuncoupling whilst on the move. The new railway concept is to increase the railway network capacity,
\nconnectivity and to reduce the cost of travel and carbon emissions.
\nAs an engineer your task is to design a control system for the future frontal structure of the railway
\ncarriage. It is envisaged this new design will be transformed from today\u2019s passive design to an active
\ndesign, with a stiffness controller to deal with all possible scenarios. The premise of the future
\nfrontal structure is to absorb the collision energy between the coupling trains in an effective manner
\nto minimise the jerk (change in acceleration) experienced by the occupants on-board the carriages.
\nThe control system must be designed for the following two scenarios:
\n1) A train consisting of three carriages coupling onto a train consisting of 1 carriage with a
\nspeed difference of 5kmh (each carriages mass is 43 tonnes)
\n2) A train consisting of five carriages coupling onto a train consisting of 1 carriage with a speed
\ndifferent of 7khm (each carriage mass is 43 tonnes)
\nIn both cases the front of the larger train couples up to the rear of the single carriage.
\nYou may wish to consider the following points:
\n The energy resulting from the coupling action can be determined by using the conservation
\nof momentum and energy equations
\n Based on the passive design case, the coupling energy is distributed between the carriages
\nbased on the mass ratios i.e. three carriages mass versus one carriage mass (3:1) as in
\nscenario one
\no Is this desirable No. Thus, the need the active design of the frontal structure
\n The use of a mass spring damper model for the frontal structure of the railway carriage
\no This is used to capture the displacement (y-axis) versus time (x-axis) of the coupling
\nmechanism and further possible simulation models e.g. force versus deformation
\n(relating to work done\/delta energy)
\n The use of feedback control to achieve desired outcome
\n List any assumptions made
\n The use of MATLAB to simulate the design
\nThis open ended question is designed for you to demonstrate your understanding of the control
\nlectures by applying your knowledge to a realistic control system design problem. We are not
\nexpecting completed designs \u2013 rather a good understanding of the control problem will achieve high
\nmarks.
\n(20 marks)<\/p>\n","protected":false},"excerpt":{"rendered":"
\u00a0 Instrumentation and Control\/Assignment 2: Control Intended Learning Outcomes assessed by this work 1. Understand the basic underpinning theory of linear continuous-time dynamical systems and be able to simulate their behaviour; 2. Manipulate and solve differential equations using Laplace transforms Read More …<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-93823","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/93823","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=93823"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/93823\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=93823"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=93823"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=93823"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}