advertisement

5. System Responses Standard forms of first and second-order systems with no zeros and unity gain are considered. 5.1 First-order systems DEU-MEE 5017 Advanced Automatic Control Assoc.Prof.Dr.Levent Malgaca, Fall 2015 5.1.1 Common step response characteristics 5.2 Second-order systems 5.2.1 Pole position for underdamped systems 5.3 Step response based second-order system specifications For a second-order system there is no general expression fof TR. TR can be derived emprically from the following figure. TR 0.8 2.5 n TP is found by differentiating the step response, and equating to zero.. Ts k k k n [ Feedback control systems book, page 130] where k is determinied by the defined percentage (% 2, % 5, %0.1). k is ofeten chosen as 4. t ss Tn (%0.13) Example k values for the defined percentage (depends on ). >> ksi=0.7;eb=0.02;k=-log(eb*sqrt(1-ksi^2)) k= 4.2487 >> ksi=0.7;eb=0.05;k=-log(eb*sqrt(1-ksi^2)) k= 3.3324 >> ksi=0.7;eb=0.001;k=-log(eb*sqrt(1-ksi^2)) k= 7.2444 Example 5.1 Design a second-order system under feedback control that a peak time 0.5 sec, and a % 5 overshoot. 5.4 Higher order systems Example 5.2 Example 5.3 Consider the transfer function in Example 5.2. Find the step response by Simulink and Laplace transform and compare the results. 5.4 Fundemental properties of linear systems The impulse and ramp responses can be found from the step response by using the derivative and integral properties. Example 5.4 For a transfer function given below, find the impulse and ramp responses form the step response and compare the results with “step” command in MATLAB. G (s ) The step response: The impulse response: 1 s 10 y step 0.1(1 e 10t ) y d y step (t ) e 10t dt t The ramp response: y ramp y step (t )dt 0.1(t 0.1(1 e 10t )) 0 MATLAB code: clc;clear; % Solution with step command ng=[1];dg=[1, 10]; p=roots(dg);p0=p(1);cksi_v2;t=0:dt:ts; g=tf(ng,dg); ys=step(g,t);plot(t,ys,'b') hold on g=tf([ng 0],dg); yi=step(g,t);plot(t,yi,'r') g=tf(ng,[dg 0]); yr=step(g,t);plot(t,yr,'g') % Analytical solution ysa=0.1*(1-exp(-10*t));plot(t,ysa,'ob') yia=exp(-10*t);plot(t,yia,'or') yra=0.1*(t+0.1*(1-exp(-10*t)));plot(t,yra,'og') Example 5.5 Consider the standart form of second-order transfer function (=0.15, n=2) given below. a) Find the impulse and ramp responses form the step response and compare the results with “step” command in MATLAB. b) Calculate the perfomance specifications. (TR,TP,TS, % OV, TD). Compare the results with those obtained with step command in MATLAB. G (s ) 4 s 2 0.6s 4 The step response: The impulse response: y step ? y d y step (t ) ? dt t The ramp response: y ramp y step (t )dt ? 0 5.5 Effect of zeros on the time-domain responses Example 5.6 5.6 Reduced order systems (Dominant poles):