Question

control system

System Description: The figure 1 and 2 below show, respectively, components and block diagram of a motor and the measurements of velocity (via the tacho unit) and position (via the potentiometer). n represents the gearbox ratio between the rotating shaft and the output shaft. The left-hand side of the diagram represents the controller. A reference set point for the rotating shaft is entered in degrees and this is equivalent voltage. The error is calculated by subtracting the measured position nt gain, Kp. (or position controller) and converted to an from the desired position. This error is multiplied by a consta the resulting voltage used to control the motor. Input potentiomcter Reference input Output potentiometer Foedhack signal R. La Input device K Amplifier Emror measuring device Load Motor Gear train Figure 1: Components of a typical DC Motor position control system Set point Ria) (degrees) Set point R(a) (V Error, E(a) Output position Ya) (degreea) Output velocity a), (rad/s) Input conversion к Position gain+ control Kp Motor unit Kn 1 180 ts+1 voltage Tacho gain voltage control K Tacho unit voltage +leasuroment otentiomoter Figure 2: Block diagram of a typical DC Motor position control system The physical parameters of the system are: (Km) (Kp) (K.) (Ki) input conversion (K) (K) (n) (r.) 150 rad/s/V motor constant position gain control sensor gain 1 0.0278 degree/volts same as K. 0.02 V/rad/s 0.1 10 0.5s tacho unit gain tacho gain control gear box ratio time constant ii MATLAB Codes (Note: Show your codes to the instructor.) ii. Step response plot Note: Show the result to the instructor. iv. Response parameters Steady state value: Settling time Page Layout References Viai sert PART B Determine the step response of the system using MATLAB SIMULINK Hint Create SIMULINK model of the system; simulate and plot the response; report the system response parameters i. SIMULINK model Note: Show your model to the instructor. ii. Step response plot Note: Show the result to the instructor iv. Response parameters Steady state value Settling time Observation: Any difference between the two resalts? e6 of 6 rds 530 Enghsh (U.53

Answer 1

MATLAB code is given below in bold letters.

clc;
clear all;
close all;

% define the system parameters
Km = 150; % motor constant
Kp = 1; % position gain control
K0 = 0.0278; % sensor gain
K1 = K0; % input conversion
Kt = 0.02; % tacho unit gain
Kv = 0.1; % tacho gain control
n = 10; % gear box ratio
tau = 0.5; % time constant

% Now define the laplace variable
s = tf('s');

% the transfer function from R(s) to Y(s) is given as below
T = K1 * feedback(Kp*feedback(Km/(tau*s+1),Kv*Kt)*1/(n)*180/pi,K0);

% define time to plot the response
t = 0:0.001:0.5;

% now plot the unit step response of the system
figure;
step(T,t);grid on;
title('unit step response of the closed loop system');

RESPONSE IS PLOTTED BELOW.

unit step response of the closed loop system 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0,05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0,45 0.5 Time (seconds) Amplitude

Simulink model is given below to simulate.

untitled File Edit View Simulation Format Tools Help Normal 0.5 Km 1/n*180/pi tau.s+1 Unit Stenput conversion position control Scope motor unit Gear ratio gain Kv Kt tacho gain Tacho unit control KO sensor Ready ode45 100% Ci

The reponse of the simulation is plotted below.

Scope 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Time offset: 0