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a) Design a lead compensation using a root locus that provides for an Mp < 10% and a rise time tr < 1 sec. b) Add an integral term to your controller so that there is no steady-state error in the presence of a constant disturbance, Td, and modify the compensation so that the specifications are still met. The code is: function prob_2_7a() % sys = tf(1,[1 0 4]); a = 2; b = 150; sysd = tf([1 a],[1 b]); figure(1) margin(sys) figure(2) OL = sys*sysd; margin(OL) figure(3) k = 0:10:1500; rlocus(OL,k),axis([-10,0,0,3]) w = 50; z = 1; k = rlocfind(OL,-z*w+w*sqrt(1-z*z)*i); figure(4) OL = k*OL; margin(OL) figure(5) sysCL = feedback(OL,1); y = step(sysCL,5); ymax = max(y); Mp = (ymax/y(end)-1)*100; step(sysCL,3),grid on title(['step with [Mp,k] = ',num2str([Mp,k])])

Solution for Take Home Exam 1  

, a) Design feedback assuming you have access to all the state elements. Place the closed- loop system poles at s = [-1 +- 1j, -0.5 +- 5j]....