Lab # 5 Modeling of Translational Mechanical System
and to Calculate Its Various Responses
Objectives
To demonstrate the key components of a translational mechanical system, model the system
and finally to obtain various responses using computation and observe it on Rectilinear plant.
Equipment/Tools required:
Rectilinear Plant with controller interfaced with PC
Pre Lab :
Introduction:
The modelling elements of a translational mechanical system include mass, spring and
damper. Mass is an element which resists the motion due to inertia. According to Newton’s law,
Spring is an element which opposes motion due to elasticity. According to Hooke’s law,
Where K is Spring constant or stiffness of spring.
Damping is an element which opposes motion due to friction. If the friction is viscous
friction, then the frictional force is proportional to velocity, i.e.
Where B is damping coefficient.
These three modelling elements are symbolized in Figure 5.*.
(a) (b) (c)
Figure 5.* Modelling elements (a)Mass (b)Spring (b)Damper
Considering a single mass-spring-damper system (Figure 5.1) where an external force ‘f’
is applied on the mass ‘M’.
Figure 5.1 Single Mass-Spring-Damper system
The resistive forces aroused from mass, spring and damper are given be equations
mentioned above. According to D‘Alembert‘s principle, for any body, the algebraic sum of
externally applied forces and the forces opposing the motion in any given direction is zero.
Thus,
External force = Resistive forces
Taking Laplace Transformation on both sides and then re-arranging
Lab Tasks
Calculations:
1. Express the system shown in Figure 5.5 using modelling symbols. Your schematics
should look like as shown in Figure 5.6.
Figure 5.6 Mass-spring-damper system for the experiment
2. Make mathematical model of the system using laws of Electronics, Mechanics and
Physics (Single Mass-Spring-Damper system has been modelled as an example).
3. Calculate the transfer functions X1(s)/F(s), X2(s)/F(s) and X3(s)/F(s).
4. Calculate impulse, step, ramp, parabolic and sinusoidal responses by taking relevant
values of F(s).
5. Using Inverse Laplace, compute time domain equivalents of the responses of Step 4. You
can also use MATLAB.
6. Plot the time domain results obtained in Step 5.
SOLUTION:
In lab we only perform the task containing two mass in the system.
OBTAINED TRANSFER FUNCTION FOR X1(s)/F(s) :
X1(s)/f(s) = (𝑠2 + 𝑠 + 2)/(𝑠4 + 𝑠3 + 3𝑠2 + 𝑠 + 1)
MATLAB CODE:
clear all
close all
clc s =
tf('s');
TF=(s^2+s+2)/(s^4+s^3+3*s^2+s+1);
t=0:0.1:10
%impulse response
figure();
impulse(TF)
%step response
figure();
step(TF)
%ramp response
u=t; figure();
lsim(TF,u,t)
% parabola response
u=((t.^2)/2);
figure();
lsim(TF,u,t)
% sine response
u=sin(t);
figure();
lsim(TF,u,t)
OUTPUT FIGURES AND THEIR CHARACTERISTICS:
IMPULSE RESPONSE:
Figure 1 : impulse response of the transfer function defined in above matlab code .
Peak amplitude = 1.06
Peak time = 1.84 sec
Settling time = 27.4 sec
STEP RESPONSE:
Figure 2 : Step response of the transfer function defined in above matlab code .
Peak amplitude = 2.93 settling time = 25.6 sec
Overshoot(%) = 46.3 rise time = 1.69 sec Peak
time = 4.69 sec steady state final value = 2
RAMP RESPONSE:
Figure 3 : Ramp response of the transfer function defined in above matlab code .
Peak amplitude = 19.2
Peak time = 10 sec
PARABOLA RESPONSE :
Peak amplitude = 86.9 sec
Peak time = 10 sec
SINE SIGNAL RESPONSE:
Figure 4 : transfer function response by applying the sinusoidal signal.
Peak amplitude = -2.11
Peak time = 6.9 sec
OBTAINED TRANSFER FUNCTION FOR X2(s)/F(s) :
X2(s)/F(s) = 1/((𝑠2 + 1)(𝑠2 + 𝑠 + 2 + 1/(𝑠^2 +1))
MATLAB CODE :
clear all
close all
clc s =
tf('s');
TF=(1/((s^2+1)*(s^2+s+2+(1/s^2+1))));
t=0:0.1:10
%impulse response
figure();
impulse(TF)
%step response
figure();
step(TF)
%ramp response
u=t;
figure();
lsim(TF,u,t)
% parabola response
u=((t.^2)/2);
figure();
lsim(TF,u,t)
% sine response
u=sin(t);
figure();
lsim(TF,u,t)
OUTPUT FIGURES WITH THEIR CHARACRTERISTICS :
IMPULSE RESPONSE :
Figure 5 : system response when the impulse signal is applied to the system .
Peak amplitude = 5.94x10^26
Peak time = >900 sec
Settling time = undefined
STEP RESPONSE :
Figure 6 :step response of the above defined transfer function . the unit step signal is applied as an input
signal.
Peak amplitude = -3.74x10^26 Settling time = undefined
Overshoot(%) = NAN Rise time = N/A
Peak time = >900 sec Steady state final value = undefined
RAMP RESPONSE:
Figure 7 : system response when the Ramp signal is applied to it. graph shows both the input and the
system response .
Peak amplitude = -2.51
Peak time = 8.8 sec
PARABULA RESPONSE :
Figure 8 : system response when the input signal is parabula signal.
Peak amplitude = 3.97
Peak time = 6.5 sec
SINE SIGNAL RESPONSE :
Figure 9 :system respons shown in above graph when the input signal is of sinusoidal nature.graph
contains both the input and system response signal.
Peak amplitude = 4.37
Peak time = 10 sec
DESCRIPTION :
In this lab we solve the given mechanical systems to determine the
transfer function . First of all we separate all the mass system given in mechanical
system and make a fee body diagram for all the individual mass system. Then we
observe all the forces working on it, positive force if it moves the mass in
reference direction and negative if it works against the refrence direction. So all
the spring and damped force are negative forces for any mass system. Then we
find the relation between the reference direction to the applied force for the
entire system and which can be defined as transfer function .so by taking the ratio
of reference point to the applied force and then taking the laplace transform ot
this fraction we get the transfer function. After that we plot the transfer function
for the various defined input like its impulse response, step response , ramp
response and etc. we also learnt a new apparatus in this lab named as rectilinear
apparartus which is controlled by the software system interface software , using
this we upload the defined parameter to the controller and control the variation
in the system , we input the electro-mechanical system through this interface so
that the stepper motor execute the program connected to the controller
POST LAB :
Task 01
Figure 1: 3 DOF Plant
i) Consider k1 = 200, k2 = 400, k3 = 400, k4 = 0 and C1 = C2 = 0, c3=0.5. Find following
Transfer Functions;
ii) Find Step response;
OBTAINED TRANSERFER FUNCTIONS:
. Consider m1=m2=m3=1
𝑁1(𝑆)
= [0.25𝑠 4 + 0.25𝑠 3 + 600𝑠 2 + 400𝑠 + 160000]/[0.125𝑠 6 + 0.125𝑠 5 + 450𝑠 4
𝐷(𝑠)
+ 350𝑠 3 + 360000𝑠 2 + 160000𝑠 + 32000000]
𝑁2(𝑆) [200𝑠 2 + 200𝑠 + 160000]
=
𝐷(𝑠) 0.125𝑠 6 + 0.125𝑠 5 + 450𝑠 4 + 350𝑠 3 + 360000𝑠 2 + 160000𝑠 + 32000000
𝑁3(𝑆) 160000
=
𝐷(𝑠) 0.125𝑠 + 0.125𝑠 + 450𝑠 + 350𝑠 3 + 360000𝑠 2 + 160000𝑠 + 32000000
6 5 4
MATLAB CODE:
clear all
close all
clc
s = tf('s');
N1s = (0.25*s^4+0.25*s^3+600*s^2+400*s+160000);
N2s = (200*s^2+200*s+160000);
N3s = 160000;
Ds = 0.125*s^6+0.125*s^5+450*s^4+350*s^3+360000*s^2+160000*s+32000000;
% for N1(s)/D(s)
Tf1 = N1s/Ds;
figure();
step(Tf1)
% for N2(s)/D(s)
Tf2 = N2s/Ds;
figure();
step(Tf2)
% for N3(s)/D(s)
Tf3 = N3s/Ds;
figure();
step(Tf3)
OUTPUT FIGURES :
For N1(s)/D(s) :
Figure 10 : step response of the transfer function formmed by using the first mass system and appiled
force .
Peak amplitude = 0.0096 Settling time = 19.6 sec
Overshoot (%)= 92 Rise time = 0.113 sec
Peak time = 0.303 sec Steady state final value = 0.005
FOR N2(s)/D(s) :
Figure 11 : step response of the transfer function e=which containing the second mass system and
appiled force.
Peak amplitude = 0.00975 Settling time = 21.6 sec
Over shoot (%) = 95 Rise time = 0.0849 sec
Peak time = 0.287 sec Steady sate final value = 0.005
FOR N3(s)/D(s) :
Figure 12 : step response of the third transfer function contaning the third mass system and applied force
. graph show all the paramets.
Peak amplitude = 0.0102 Settling time = 19.1 sec
Over shoot (%) = 104 Rise rime = 0.0614 sec
Peak time = 0.961 sec Steady state final value = 0.005
Conclusion :
In this lab basically we learnt about the mechanical system . to solve the mechanical
system we first observe all the force on the mass or on the system , for this purpose we separated
all the mass system and observe as free body diagram.as the applied force moves the system in
the reference direction and all other resistive forces spring,damping and mass force acts along
opposite direction of the system.
So the first task in the lab is to derive the transfer function for the system given.as the
whole mechanical system contains two masses so we derive the transfer function for both the
masses with respect to the applied force F.then we plot this transfunction on matlab to observe
the different system responses such as impulse response,step reponse,ramp response and etc.
for all these inputs we observe the system behaviour whether it is stable or not for the specific
input signal. We also see some of the characteristics of the system in the MATLAB plot such as
peak aplitude, overshoot ,rise time etc. which defines the system output and which are required
by the user or customer.
