Control System Analysis - Assignment Example

The arm dynamics are represented by: Dynamic specification for the arm requires: (i) Percentage overshoot for a step input < 20% (ii) Rise time < 1.2 seconds (iii) Velocity constant, Kv  10 Part 1: 30% 1. Design a suitable phase-lead controller. You should explain in detail how your design has been developed. Your solution should include Bode plots constructed both manually and using an appropriate package such as MATLAB or Scilab. (20%) Steady-state error specification Putting Kv at 10 10 =270k K=10/270=0.037 The steady state error = % overshoot specification The percentage overshoot is related to the dumping ration by the equation from which we have graph figure 1 Figure 1 Phase margin (PM) is related to damping ration () by the equation From which we obtain the graph figure 2 Given that OS =20% from figure damping ration ()=0.45 and from figure 2 at ()=0.45 the phase margin is 460. Figure 2 The uncompensated phase margin from figure 3 is 1770 The maximum phase margin is calculated as Compensator magnitude 23dB From figure 4 at 23dB the frequency associated with this value is equal to and it represents the new 0-dB crossover frequency in the compensated system and from the figure 4 Figure 3. Bode plot of uncompensated system . Now equating to The parameters of the phase lead compensator are calculated on the basis of values obtained earlier =47 The transfer function of a phase-lead compensator is given as Thus Phase-lead compensator simulation results: Matlab Simulation clear all; wm = 0.217; % gain-crossover frequency alpha = 0.037; % phase-lead compensator parameter T = 1/wm/sqrt(alpha); % phase-lead compensator time constant K = 10; % DC compensator gain % Phase-lead compensator C(s) cnum = K*[T 1]; cden = [T*alpha 1]; % Open-loop sys G(s) gnum = [1]; gden = [1 1 0]; % Unity-Gain Feedback Loop H(s) hnum = [1]; hden = [1]; % Open-loop sys C(s)*G(s) numo = conv(cnum,gnum); deno = conv(cden,gden); % Closed-loop sys [gnumc,gdenc] = feedback(K*gnum,gden,hnum,hden,-1); [numc,denc] = feedback(numo,deno,hnum,hden,-1); bode(cnum,cden); Figure 4. Bode plot of phase-lead compensator . sys1 = tf(K*gnum,gden); sys2 = tf(numo,deno); [mag1,ph1,w]=bode(K*gnum,gden,logspace(-1,2,500)); [mag2,ph2,w]=bode(numo,deno,logspace(-1,2,500)); subplot(211); semilogx(w,20*log10(mag1),'r', w,20*log10(mag2),'b'); title('Bode Diagrams'); ylabel('Magnitude (dB)'); legend('uncompensated','compensated', -1); subplot(212); semilogx(w,ph1,'r', w,ph2,'b'); ylabel('Phase (deg)'); xlabel('Frequency (rad/sec)'); legend('uncompensated','compensated', -1); Figure 5. Bode plots of uncompensated and compensated systems. Figure 6. Response of uncompensated and compensated systems due to unit ramp input. 2 . Using the design rules in the Control Engineering Data Sheets, obtain the TF of the “equivalent” 2nd order TF (including controller). (5%) A higher order equation can be approximated to a second order equation by using the standard equation (2.3) From figure 3 we know PM=177 and =0.217 Thus resulting to Substituting in the equation (2.3) We have Which results to =0.187rad/s Now substituting for =0.187rad/s and in equation (2.1) We have The equivalent 2nd order TF is as shown in figure 8 where it can be seen PM changes from 1770 to 790 while changes from 0.217 to 0.371. 3 Compare the unit step responses of the equivalent 2nd order TF and the actual closed loop system, and comment on the accuracy of the equivalent. (5%) Figure 6. Step response of actual and second order equivalent. Question 2 For the digital design , after the system is designed with lead- compensator and the response was satisfactory, emulation digital design take place for discretezation the system using Bilinear at different sampling times in closed loop system Bilinear method of the compensated system are sensitive to the sampling time, at 0.5 s and identical to continuous with improvement in the transient characteristics and steady state error ess >> T=0.5;%sampling time >> Gz=c2d(H1,T,'zoh') Gz = 0.01588 z^3 + 0.01254 z^2 + 0.0001727 z + 2.007e-09 ----------------------------------------------------- z^4 - 1.047 z^3 + 0.047 z^2 - 2.567e-05 z + 6.305e-16 Sample time: 0.5 seconds Discrete-time transfer function. >> disp('The zero order Hold Method TF above \n')%step(Gz) The zero order Hold Method TF above \n >> z=tf('z',T) z = z Sample time: 0.5 seconds Discrete-time transfer function. >> O1=1; >> disp('Now using Digital Bilinear Method Approximation') Now using Digital Bilinear Method Approximation >> O=1 O = 1 >> Gz_lead=g_1*((2*(z-1))/(T*(z+1))-z1)/((2*(z-1))/(T*(z+1))-p1) Gz_lead = 0.03657 z^2 - 0.0008695 z - 0.03743 ----------------------------------- 0.9805 z^2 - 0.039 z - 1.02 Sample time: 0.5 seconds Discrete-time transfer function. >> GFz=Kc*Gz_lead*Gz; GTz = 2.148e-05 z^5 + 1.646e-05 z^4 - 2.216e-05 z^3 - 1.738e-05 z^2 - 2.392e -07 z - 2.781e-12 -------------------------------------------------------------------------------- 0.9805 z^6 - 1.066 z^5 - 0.9326 z^4 + 1.066 z^3 - 0.04793 z^2 + 2.594e -05 z - 2.781e-12 Sample time: 0.5 seconds Discrete-time transfer function. >> G_Fz=Kc*Gz_lead*Gz; >> GTz=feedback(G_Fz,1)%closed loop system GTz = 2.148e-05 z^5 + 1.646e-05 z^4 - 2.216e-05 z^3 - 1.738e-05 z^2 - 2.392e -07 z - 2.781e-12 -------------------------------------------------------------------------------- 0.9805 z^6 - 1.066 z^5 - 0.9326 z^4 + 1.066 z^3 - 0.04793 z^2 + 2.594e -05 z - 2.781e-12 Sample time: 0.5 seconds Discrete-time transfer function. >> step(GTz) Figure 8 >> T=2;%Two second sampling time >> Gz=c2d(H1,T,'zoh') Gz = 0.105 z^3 + 0.015 z^2 + 1.821e-08 z + 3.398e-23 --------------------------------------------------- z^4 - z^3 + 4.644e-06 z^2 - 4.345e-19 z + 4.232e-40 Sample time: 2 seconds Discrete-time transfer function. >> disp('The zero order Hold Method TF above \n')%step(Gz) The zero order Hold Method TF above \n >> z=tf('z',T) z = z Sample time: 2 seconds Discrete-time transfer function. >> O1=\n3; O1=\n3; | Error: Unexpected MATLAB operator. >> O1=1; >> disp('Now using Digital Bilinear Method Approximation') Now using Digital Bilinear Method Approximation >> O=1 O = 1 >> Gz_lead=g_1*((2*(z-1))/(T*(z+1))-z1)/((2*(z-1))/(T*(z+1))-p1) Gz_lead = 0.141 z^2 - 0.01391 z - 0.155 ----------------------------- 3.688 z^2 - 0.624 z - 4.312 Sample time: 2 seconds Discrete-time transfer function. >> GFz=Kc*Gz_lead*Gz; >> GTz=feedback(G_Fz,1)%closed loop sysyem GTz = 2.148e-05 z^5 + 1.646e-05 z^4 - 2.216e-05 z^3 - 1.738e-05 z^2 - 2.392e -07 z - 2.781e-12 -------------------------------------------------------------------------------- 0.9805 z^6 - 1.066 z^5 - 0.9326 z^4 + 1.066 z^3 - 0.04793 z^2 + 2.594e -05 z - 2.781e-12 Figure 9 Question 3 PID control PID has established itself as an important component in distributed control system and the PID controls are also embedded in many control systems meant for special purposes. More often the PID control incorporate logic, sequential functions, selectors as well as simple function blocks that are used in building of complex automation systems that find application in the production of energy, transportation and manufacturing industry.PID also in is important in some sophisticated control strategies including the model predictive controls that are organized in hierarchy. PID control is utilized at the lower level with the set points to controllers at the lower level being given by a multivariate controller. This makes PID to be high ranking amongst the other controllers and are found to be a vital component in a control engineer tool box. PID have managed to withstand many changes that have been witnessed in technology industry be it in mechanics and pneumatics or in microprocessors including electronic tubes, transistors and IC components. PID controllers have fully embraced micro-processor technology with almost all the currently being made incorporating this technology. With the inclusion of microprocessor technology PID is able to have other features including automatic tuning. Tuning PID allows the application of all general methods of control design in addition to there being other methods that are specifically made for application only in PID control and this are referred to as tuning methods. Notwithstanding he method that is utilized it is very important to put into consideration the key elements of control, uncertainty in the process, load disturbance sensor noise as well as reference signals. Ziegler and Nichols tuning methods are the most well known and they have had a lot influence on PID practice for 50 year. These methods have a basis in the characterizatrion of process dynamics by make use of a few parameters and simple equations representing the parameters of the controller. It is a wonder that there is overwhelming referencing of the methods this being attributable to their being able to give moderately good tuning that work in restricted situations. The explanation that can be given to this is that the methods are simple in addition to them being applicable in student exercises encountered in basic control courses. The step responses method is one of Ziegler and Nichols methods which is characterized by parameters a and L this being the only parameters. Wind up Even though most of the aspects associated with control system can be understood on the basis of linear theory, there are some non-linear aspects that need accounting for in almost all controllers. Wind up is a phenomenon that is brought about when integral action and saturations come into play. There are some limitation of some sought including a valve not being able to go beyond full opening and a motor not being able to go beyond a certain speed being some of typical examples. This means that having a system that have a wide spectrum of conditions in which it is supposed to work may result to control variable reaching the actuator limits. This results to breaking of the feedback loop resulting to the system operating as an open loop as a result of the actuator not going beyond its limit notwithstanding the process output. Suppose what is in use is an integrating action the there will continued integration of the error and as a result the integral term could end up being very large or in may wind up colloquially. This then calls for the error being assigned negatively for a considerably a long time before normalcy being achieved. Consequently any controller having integral action would give very high transients in case of actuator saturation. Reference K. H. Heong, G. Chong and L. Yun, (2005), PID Control System Analysis, Design, and Technology, IEEE Trans. On Control Systems Technology, vol 513, No. 4, pp. 559‐576. Read More