Challenge

Project - Speed Control of a Direct Current (DC) motor

Script Code

clc;
clear all;
% This file contains parameters for DC motor design
J= 0.01; %Inertia
B= 0.1; %friction
Km= 0.01; %back emf
Kt= 0.01; %torque
Ra= 1; %resistance
La= 0.5; %inductance
%% Design of PID Controller
A=[-B/J Kt/J; -Km/La -Ra/La];
B=[0; 1/La];

%% Placement of Poles
fn= 1;
wn= 2*pi*fn;
z= 0.5;
P1= -z*wn+wn*sqrt(z^2-1);
P2= -z*wn-wn*sqrt(z^2-1);
P= [P1; P2]
K= place(A,B,P)

k_dc= 1; %the DC gain of the system is 1
s= tf('s');
G1= k_dc*wn^2/(s^2+2*z*wn*s+wn^2);
pzmap(G1,'r')

Output

P =

  -3.1416 + 5.4414i
  -3.1416 - 5.4414i


K =

   38.3133   -2.8584

>> 

Summary

Poles are located on -2 and -3 which lie on the left-hand plane which indicates the system is stable.

Case-1

Output Scope

Summary

When Ki taken as 50 and step speed is taken as 10.

At 1.4 sec system gone to overshoot condition with 12.2 rpm of speed and 1.8 sec system gone to undershoot condition with 7.9 rpm of speed the reference signal simulates the actual signal at 4.5 sec and meet the Steady state condition.

Case-2

Output Scope

Summary

When Ki taken as 150 and step speed is taken as 10.

At 1.4 sec system gone to overshoot condition with 15.2 rpm of speed and at 1.8 sec system gone to undershoot condition with 6 rpm of speed again the system went for overshoot condition at 2.2 sec for 12.2 rpm and went for unndershoot condition at 2.5 sec for 8.1 rpm of speed, the reference signal simulates the actual signal at 5.8 sec and meet the Steady state condition.

Your portfolio, next

Ship work like this
in your own course.

Talk to a career counsellor

Free 20-minute call. We'll map the right course to where you want to work.

By submitting you agree to our Terms and Privacy Policy.