Essay About Pic And Dynamics Of Closed-Loop Systems

Eee 350 – Control System
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[pic 1]MARKSEEE350CONTROL SYSTEMS2016-2017ASSIGNMENT 1(b)NameIDMURNI NUR ATHIRAH BT ROSNAN125714AINA QISTINA BT MD TAHA125685Penalties for Plagiarism: Those who are caught to copy or allow others to copy part of or the entire work will have to share their marks for this assignment (i.e. if there are n reports that are similar, the marks will also be divided by n).Deadline: 31st October 2016 (4.30pm)Please submit to the administration office.Date of submission:31/10/2016_________________Lecturer:Dr Nur Syazreen AhmadLAB 2Aims of the experiment:To study the dynamics of closed-loop systems when subject to step input and step output disturbanceTo observe the dynamics of the systems when proportional and PI controls are includedTo design controllers to meet certain performance requirements3) Closed-Loop Response with P control3.1) Speed Control[pic 2]What is the (closed-loop) transfer function between [pic 3] and [pic 4] (in terms of[pic 5])? Denote this as [pic 6]Let [pic 7], calculate the steady-state gain of [pic 8].[pic 9]Let [pic 10] be a unit step input. Calculate the steady-state error for the closed-loop response when [pic 11].  [pic 12]The proportional control can be used to increase the speed of a first order system. Find the values of [pic 13] such that the time constant of the closed-loop system is less than 0.5s (show by calculation).[pic 14]Let [pic 15] be a proportional controller with a constant gain [pic 16], and [pic 17] as the closed-loop transfer function. Let [pic 18] be its output and plot the step responses of [pic 19], [pic 20] and [pic 21] in the same figure. (You have already obtained [pic 22]and [pic 23] from Lab 1).[pic 24][pic 25]Repeat the simulation with different values of gain (i.e. change [pic 26] from low to high values possible). For each [pic 27]that you have assigned, find the corresponding steady-state errors (ess) and time constants ([pic 28]). Make a conclusion on the results. You can do the iteration manually or perform a for loop, and your answer can be in table format or plots of ess vs K and [pic 29]vs [pic 30].[pic 31][pic 32]                    [pic 33][pic 34][pic 35][pic 36]                                            [pic 37][pic 38]       [pic 39]   [pic 40]From the previous step, verify your result in part (d) above. (i.e. the values of [pic 41] such that the time constant of the closed-loop system is at most 0.5s).From Previous Step:KESStc0.50.44440.55561.50.21050.26322.50.13790.17243.50.10260.12824.50.08160.10205.50.06780.08476.50.05790.07257.50.05060.06338.50.04490.05629.50.04040.0505From part (d), the result obtained is K > 0.6 when time constant is at most 0.5s. This is verified through the previous step as when K=0.5, the time constant is 0.5556s and as the K value increases, the time constant decreases.Plot (in the same figure) the step responses of the closed-loop system when [pic 42]and 10.[pic 43]   [pic 44][pic 45]For what values of K is the steady-state error zero? Explain your answer.[pic 46]But, T2(0) cannot equal to 1 because of loss of signals. However, when K approach infinity or a very large value, the ess   will approach 0.3.1) Steady-state error and steady-state trackingSuppose the closed-loop system is subject to an output disturbance, [pic 47] as shown in  Figure 3.[pic 48]What is the  transfer function from [pic 49]to [pic 50] (in terms of[pic 51])?[pic 52]What is the error due to [pic 53] (i.e. [pic 54]when [pic 55]and [pic 56])? Compute its corresponding steady-state value.[pic 57]Adjust the gain of the controller (calculate [pic 58]) such that the  steady-state error due to [pic 59] is at most 5%.[pic 60]Build the model as shown in Figure 3 in Simulink, set[pic 61], [pic 62] and the time frame from 0-10s, and export the input (time) and output to MATLAB. Show the response of the system when[pic 63]enters the system at [pic 64], and verify your calculation in (l).[pic 65]Show the response when [pic 66]enters the system at [pic 67]and [pic 68]enters the system at [pic 69].[pic 70]Write your observation and discussions.  A proportional control system is the system that the control effort is proportional to the error state and the K works as the gain of the system that can be control. 4) Closed-Loop Response with Integral Action and PI controlNote: For tasks (a)-(i), ignore the output disturbance, i.e. set [pic 71].[pic 72]                                                                                 (1)(Integral action)4.1) AnalysisWhat is the (closed-loop) transfer function between [pic 73] and  [pic 74] (in terms of [pic 75])?[pic 76]Let [pic 77]calculate the closed-loop transfer function and denote this as [pic 78][pic 79][pic 80]Let [pic 81] be a unit step input. Calculate the steady-state error for the closed-loop response. Compare with the results in the previous section (when only P control is used), and explain your answer.[pic 82]The steady-state error for P control 1/6 while PI control is 0. The integrator is added to the system as a compensator helps the system to achieve zero steady-state error.Based on [pic 83], and calculate the natural frequency, [pic 84]the damping ratio, [pic 85]and the corresponding peak overshoot [pic 86]and peak time, [pic 87]. Use the approximation[pic 88]     and   [pic 89][pic 90][pic 91][pic 92][pic 93][pic 94]Plot the step response of [pic 95], and verify your answers in part (d) above.[pic 96][pic 97]Plot the step responses of the system when [pic 98] in the same figure, and discuss your observations.[pic 99][pic 100][pic 101]4.2) DesignSuppose the controller (1) is used and the design specifications arethe peak overshoot is less than 10 %, andthe peak time is less than 2s.With what values of [pic 102]can both specifications be met? Explain your answer.[pic 103]                        [pic 104]KI = 2.From the result, the equations can be compared such that,[pic 105]Suppose both proportional and integral actions are included in the controller as follows:[pic 106]                                                                                 (2)(PI control)Find the closed-loop transfer function in terms of [pic 107]and [pic 108], and denote this as [pic 109][pic 110]With the specifications as in (g) above, design [pic 111]and [pic 112]to meet such requirements (you can use trial and error). Show the step response to verify your results.[pic 113][pic 114]Via Simulink, include a step output disturbance as in Figure 3. With the values of [pic 115]and [pic 116]obtained in (i) above, show the response when [pic 117]enters the system at [pic 118]and [pic 119]enters the system at [pic 120].[pic 121]Compare the response with your answer in Section 3.2 (n) (when only proportional controller is used).The steady-state error for PI system is zero while the P system is nonzero.5) Discussions and ConclusionsA zero steady-state error cannot be achieve with the P control system. Although the K value is put as a very large value, the steady-state error cannot be zero. However, when the integrator is added to the system, the zero steady state can be achieve. The time constant can be decreased as the K is increases.Evaluation Murni Nur Athirah Bt Rosnan:[pic 122]Evaluation Aina Qistina Binti Md.Taha: