Question

solving using MATLAB When b=4

R X *-9 40) 101G+ ba? exo tima G (S) 1 (s + 1)(s + 2)(s +b) + b=5 Where “b” in the forward transfer function is equal to the first non-zero digit of your ID Example: ID: 123456 b=6 ID: 123450 R and D are both unit-step input 1. Design a PI controller so that the system behaves with the fastest response with no overshoot a. Give the parameters of the controllers b. Provide the response of the system to a unit step reference input c. Analyze the stability of the system d. Sketch the response of the system if a disturbance input that occurs @t=5 sec. Evaluate the performance of the controller for disturbance rejection. e. Sketch the signal input to the Plant (Signal after the second comparator)

Answer 1

when b = 4,

the denominator of the plant is given by

(s+1)*(s+2)*(s+4) = s^3 + 7 s^2 + 14 s + 8

The denominator polynomial coefficients = [1 7 14 8]

The simulink model with PI controller is given below

The PI controller block is given below.

For faster response with no overshoot, you need to tune the controller by clicking on the tune button.

By tuning the controller for best response with no overshoot , the PI gains turn out to be

Kp = 7.2419 and

Ki = 5.5016

With the new designed controller gains, the PI block is given below.

Question b:

The unit step response is given below.

Question c:

MATLAB code to obtain the stability of the system

clc;
close all;
clear all;

% define the laplce variable s
s = tf('s');

% define the gains
Kp = 7.2419;
Ki = 5.5016;

% define the controller
Gc = Kp+Ki/s;

% define Gp
Gp = 1/((s+1)*(s+2)*(s+4));

% obtain the closed loop pole locations
figure;
pole(feedback(Gc*Gp,1))

The poles of the system are located at

ans =

-4.6330
-0.8822 + 1.0918i
-0.8822 - 1.0918i
-0.6027   

It is observed that the closed loop poles are located on the left side of jw axis. Therefore the system is going to be stable.

Question d: Disturbance rejection analysis

The simulink block diagram is given below.

The response is plotted below.

It is observed that PI control system has rejected the step disturbance and helped track the reference of unity.

Answer 2

when b = 4,

the denominator of the plant is given by

(s+1)*(s+2)*(s+4) = s^3 + 7 s^2 + 14 s + 8

The denominator polynomial coefficients = [1 7 14 8]

The simulink model with PI controller is given below

U Pl(s) $3-792-145+8 Step PID Controller Plant Scope

The PI controller block is given below.

Function Block Parameters: PID Controller X Controller: PI Form: Parallel Time domain: Continuous-time Discrete-time Main PID Advanced Data Types Controller parameters State Attributes O Proportional (P): 1 Compensator formula Integral (1): 2 P+12 s Tune... Initial conditions Source: internal Integrator: 0 External reset: none Ignore reset when linearizing OK Cancel Help Apply

For faster response with no overshoot, you need to tune the controller by clicking on the tune button.

PID Tuner (untitled1/PID Controller) - 3 Design mode: Extended Form: Parallel Type: PI Plot: Step Response: Reference tracking Show block response Hide parameters 1.5 Controller parameters Tuned Block 1 P 1 Amplitude 7.2419 5.5016 - 2 0.5 Block response Tuned response D N 0 0 2 4 6 12 14 16 18 8 10 Time (seconds) Performance and robustness Interactive tuning Bandwidth: > 0.712 rad/s 0.0983 0.983 9.83 Phase margin: 68 deg 0 45 90 Automatically update block parameters OK Cancel Apply Help

By tuning the controller for best response with no overshoot , the PI gains turn out to be

Kp = 7.2419 and

Ki = 5.5016

With the new designed controller gains, the PI block is given below.

Function Block Parameters: PID Controller Form: Parallel Controller: PI -Time domain: Continuous-time Discrete-time Data Types State Attributes Main PID Advanced Controller parameters - Compensator formula Proportional (P): 7.24187897709023 Integral (1): 5.50155434288195 P+12 S Tune... Initial conditions Source: internal Integrator: 0 External reset: none Ignore reset when linearizing OK Cancel Help Apply

Question b:

The unit step response is given below.

Scope ] X y 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 10 Time offset: 0

Question c:

MATLAB code to obtain the stability of the system

clc;
close all;
clear all;

% define the laplce variable s
s = tf('s');

% define the gains
Kp = 7.2419;
Ki = 5.5016;

% define the controller
Gc = Kp+Ki/s;

% define Gp
Gp = 1/((s+1)*(s+2)*(s+4));

% obtain the closed loop pole locations
figure;
pole(feedback(Gc*Gp,1))

The poles of the system are located at

ans =

-4.6330
-0.8822 + 1.0918i
-0.8822 - 1.0918i
-0.6027   

It is observed that the closed loop poles are located on the left side of jw axis. Therefore the system is going to be stable.

Question d: Disturbance rejection analysis

The simulink block diagram is given below.

Step disturbance у PICS) 1 $3-752-145+8 Plant Step reference PID Controller Scope

The response is plotted below.

Scope - 關 1.4 1.2 T- 0.8 0.6 14 0.2 口 口 5 10 15 Time offset: 0

It is observed that PI control system has rejected the step disturbance and helped track the reference of unity.