Analysis of 1-DOF Torsional Mechanical System Control

This project report analyzes the dynamics and control of a 1-DOF torsional mechanical system, specifically the ECP M205. The study focuses on position and speed control using proportional (P) and proportional-derivative (PD) controllers. The introduction provides background on the torsional dynamic plant and its governing differential equation. The aim is to study the open-loop system and the response of the rotating disk to a rectangular pulse. The report details the system parameters, including mass and control effort. MATLAB code and plots are presented to illustrate the transient responses for various proportional gain (Kp) values. The analysis is extended to PD controllers, examining the effect of the derivative gain (Kd). Speed control is also investigated, with the derivation of the closed-loop transfer function and an examination of the impact of gain changes (Kp and Kd) on pole-zero locations using MATLAB. The conclusion summarizes the findings, highlighting the impact of parameter variations on system response and stability, as observed through MATLAB simulations. References to relevant literature are also provided.

Electronics
Dynamics and controls
Position and Speed Control of a 1 – DOF Torsional Mechanical System
Student Name –
Student ID -

Contents
Introduction...........................................................................................................................................2
Aim :......................................................................................................................................................3
Proportional Controller :.......................................................................................................................4
PD Controller.......................................................................................................................................10
Speed Control......................................................................................................................................15
Conclusion :.........................................................................................................................................23
References...........................................................................................................................................24

Introduction
The torsional Dynamic plant ECP M205 consists of an electromechanical plant. It has some
control hardware which can be controlled using a software interfacing. The system is capable
of supporting the specifications, generation of path and generating various plots. It can be
interfaced to a real world system. Its apparatus consists of many physical parts having bodies
with different connections. So, a study can be made for the linear drive’s flexibility, gears,
belts and also vibration effects can be studied for parts like robotic arms.
For the rotating disk, the differential equation used is as follows :
J d 2 θ / d t 2 + c d θ / d t = T ( t )
Here, θ = angle of rotation for the disk,
J = Moment of inertia for the disk
C = coefficient of viscous friction
T ( t ) = Torque produced by electric motor
Taking Laplace Transform:
Θ ( s ) / T ( s ) = 1 / J s 2 + s C = 1 / s ( J s + C )
Aim :
Open Loop 1 – DOF torsional system

The open loop transfer function is :
X ( s ) / Y ( s ) = k h w x 1 / s ( J s + C )
= 14.928 / s ( J s + C )
Since khw = 14.928
It includes open loop plant study and to study the response of a rotating disk to an externally
applied torque which has a rectangle shaped pulse.
Parameters of 1 DOF System :
For the ECPM205 system, the various parameters are as follows :
Mass of each cylinder = m = 500 g = 0.5 kg
Control effort = 0.25 V
Dwell time = 4000 ms
Number of repetitions = 1
We have to plot the angular velocity ( in degree per second ) versus the time.
A detailed study of the topic has been done [1] [2]. A study of torsional system dynamics for
low speed diesel engines on the basis of the instantaneous torque has been done [3].
Proportional Controller :
Transient Responses Plots for different values of Kp ( Proportional Gain ) for 1 co –
ordinate system :
Kp = 1
Matlab Code :

sys = tf ( [ 14.928 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;
step ( sys ) ;
subplot ( 2 , 1 , 2 ) ;
impulse ( sys ) ;
Figure 1
If Kp = 1, then step response is a straight line. Also, the impulse response takes a value of 15
V at steady state.
Kp = 0.005

Matlab Code :
sys = tf ( [ 14.928 * 0.005 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;
step ( sys ) ;
subplot ( 2 , 1 , 2 ) ;
impulse ( sys ) ;
Figure 2

If Kp = 0.005, then step response is a straight line. Also, the impulse response takes a value
of 0.075 V at steady state.
Kp = 0.01
Matlab Code :
sys = tf ( [ 14.928 * 0.01 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;
step ( sys ) ;
subplot ( 2 , 1 , 2 ) ;
impulse ( sys ) ;

Figure 3
If Kp = 0.01, then step response is a straight line. Also, the impulse response takes a value of
0.15 V at steady state.
Kp = 0.03
Matlab Code :
sys = tf ( [ 14.928 * 0.03 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;

step ( sys ) ;
subplot ( 2 , 1 , 2 ) ;
impulse ( sys ) ;
Figure 4
If Kp = 0.03, then step response is a straight line. Also, the impulse response takes a value of
0.45 V at steady state.
Kp = 0.07

Matlab Code :
sys = tf ( [ 14.928 * 0.07 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;
step ( sys ) ;
subplot ( 2 , 1 , 2 ) ;
impulse ( sys ) ;
Figure 5

If Kp = 0.07, then step response is a straight line. Also, the impulse response takes a value of
1.1 V at steady state.
PD Controller
The MATLAB plot command is used here.
Evaluation of parameters ( of the transient responses )
The effect of gain, Kp on the parameters
Transient Responses Plots for different values of Kd ( PD Controller ) for 1 co – ordinate
system :
The MATLAB plot command is used here.
Kd = 0.002
Matlab Code :
sys = tf ( [ 14.928 * 0.002 , 14.928 * 0.02 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;
step ( sys ) ;
subplot ( 2 , 1 , 2 ) ;
impulse ( sys ) ;

Figure 6
If Kd = 0.002, then step response is a straight line. Also, the impulse response takes a value
of 0.3 V at steady state.
Kd = 0.004
Matlab Code :
sys = tf ( [ 14.928 * 0.004 , 14.928 * 0.02 ] , [ 1 , 1 , 0 ] ) ;
subplot ( 2 , 1 , 1 ) ;