Question

1. [25%] Consider the closed-loop system shown where it is desired to stabilize the system with feedback where the control law is a form of a PID controller. Design using the Root Locus Method such that the: a. percent overshoot is less than 10% for a unit step b. settling time is less than 4 seconds, c. steady-state absolute error (not percent error) due to a unit ramp input (r=t) is less than 1. d. Note: The actuator u(t) saturates at +-5. e. You must do the first iteration “by hand sketch” and then you can continue future iterations by Matlab. i. Hint: You should consider steady-state requirement first and you will need two root locus plots. ii. Show the closed-loop simulation results assuming a unit step AND a unit ramp set-point for both the output c(t) and the controller u(t). Determine the integral square error (ISE) of the closed-loop system simulation subject to a unit step input. iii. Show the design process: (1) by hand sketching, and (2) verify using Matlab. Plant, G(s) R(s) U(s) 2 C(s) S2 + 5s + 4 Controller, Gc(s) Kas

Answer 1

As per the question, we need to design a PID controller in order to stabilize a system as shown in the figure below Plant, Gls) 2 as), K,+ S? + 5s + 4 Controller, Gels) Kas Such that the close-loop system hasa percent overshoot (Mp) of 10% for a unit step Setting time (ts) 4sec, steady-state error (ess for unit ramp input <1. Now, if we solve the above block diagram, the TF of the inner-loop will be T:(s) = G(s)H(s)/(1+G(s)H(s)] where G(s) =2/(s^2+55+4) and H(s) =Kds, Inserting the above result, wege T.(s) = (2.Kds)/(s^2+5+2Kd)s+4) ....-(1) Now consideringthe TF of the inner-loop, if we are moving on the outer-loop, the TF of the entire system with the PID controller having gains of Kp. Kd and Ki will be, T(s) = [2(s•Kp+Ki)*Kd)/(s^2+(5+2Kd+2KpKd)s+(4+2KiKd)] -...-(2) Now, as per the question, Mp=10% and ts=4sec As we know, Mp=exp(-15/11-5) and ts=4/(5 * Wn) Where is the damping catio and wn is is the natural frequency. Hence by computing the desired Mp and ts, we obtain, S = 0.6 andwn = 1.7 rad/sec. Now, from eq (2), we obtain (4+2 * Ki * Kd) = 1.7 Hence, Kis Kd= -0.56 -3)

And Kd+Kp.Kd =-2 ----(4) Now, if we utilize the third information, i.e, ess=1, then we will get S•R(3) ess =lims+01+G(5) H() which results Kd+Ki = 0.44... 15) Therefore, if we solve the Eq. (3), (4) and (5), we will obtain the following values of Ke Kd and ki. Such that KR = 2.57, Kd = -0.56, Ki = 1. The following matlab code can be utilized to realize the above system.

s=tf('s');
%% plant model
G = 2/(s^2+5*s+4);

%% PID parameter
Kp = 2.57;
Ki = 1;
Kd = -0.56;
C = pid(Kp,Ki,Kd);

%% Close loop system
G_CL = feedback(G*C,1);
%% compute root-locus
rlocus(G_CL)
%% compute step response
t=0:0.1:10;
step(G_CL,t)
%% compute ramp response
ramp=0.2*t;         % ramp input signal
[y,t]=lsim(G_CL,ramp,t); % compute ramp response
plot(t,y)

The root-locus plot of the above system is shown below.

Root Locus Imaginary Axis (seconds) 20 25 10 15 Real Axis (seconds)

The response for the step input and ramp input are shown below

Step Response Amplitude 10 Time (seconds)

Ramp Response NE 8 10 Time (sec.)