Question

PROBLEM 4 A unity feedback closed loop control system is displayed in Figure 4 (a) Assume that the controller is given by G (s)-2. Based on the lsim function of MATLAB, calculate and obtain the graph of the response for 0,(1)-a. Here a ; 0.5%, Find the height error after 10 seconds, (b) In order to reduce the steady-state error, substitute G. (s) with the following controller: K2 This is a Proportional-Integral (PI) controller. Repeat part (a) in the presence of PI controller, and juxtapose the steady state error of the previous controller with the Pl controller. An+ Desired- attitude Kn Actual attitude -10 G,(G) (2 +3.5s +6) Figure 4: A unity feedback closed loop control system

Answer 1

with proportional controller:

matlab code:

s1 =

20 (s+5)
-------------------------
s (s+10) (s^2 + 3.5s + 6)

Continuous-time zero/pole/gain model.

sys =

20 (s+5)
------------------------------------------
(s+10.14) (s+2.239) (s^2 + 1.122s + 4.405)

Continuous-time zero/pole/gain model.

response:

Linear Simulation Results 14 12 10 Input: In(1) Time (seconds): 10.9 Amplitude: 5.45 System: sys Time (seconds): 10.9 Amplitude: 5.16 10 15 20 25 Time (seconds)
with proportional + integral controller:

matlab code:

t=0:0.001:25;
k=2*(-10)*(-1);
z=[-0.5 -5];
p=[0 0 -10 -1.7500+1.7139*i -1.7500-1.7139*i ];
s1=zpk(z,p,k) %open loop transfer function
sys=feedback(s1,1) % close loop transfer function
ip=0.5*t; % ramp input with 0.5 as amplitude.
lsim(sys,ip,t)

output:

s1 =

20 (s+0.5) (s+5)
---------------------------
s^2 (s+10) (s^2 + 3.5s + 6)

Continuous-time zero/pole/gain model.

sys =


20 (s+5) (s+0.5)

------------------------------------------------------
  
(s+10.13) (s+1.904) (s+0.7302) (s^2 + 0.7337s + 3.549)
  
  

Continuous-time zero/pole/gain model.

response:

Linear Simulation Results 14 12 10 System: sys Time (seconds): 10.9 Amplitude:5.44 10 15 20 25 Time (seconds)