Question

Project 1 A Vibration Insulation Problem Passive isolation systems are sometimes used to insulate delicate equipment from unwanted vibrations. For example, in order to insulate electrical monitoring equipment from vibrations present in the floor of an industrial plant, the equipment may be placed on a platform supported by flexible mountings resting on the floor. A simple physical model for such a system is shown in Figure 4.P.1, where the mountings are modeled as an equivalent linear spring with spring constant k, the combined mass of the platform and equipment is m, and viscous damping with damping coefficient y is assumed. Assume also that only vertical motion occurs. In this project, we use this model to illustrate an important fundamental principle in the design of a passive isolation system. m Ty(t) Tult) FIGURE 4.P.1 Equipment supported by flexible mountings.

#5 is only I need in which we need to plot it on Matlab and I don't know how to plot it.

Answer 1

y(t) is plotted by taking Laplace Transform of differential equation and then plotting y(t) by taking inverse transform-

For step input u(t) and k=10

close all
clear all
syms s t
m =38; %mass of body
xi = 0.2; %damping ratio
k=10; %variable
T = 30; %Total time of simulation
i = 1;
wo = sqrt(k/m); %Natural frequency
G = (2*xi*wo*s+wo.^2)/(s^2+2*xi*wo*s+wo.^2);
g(t) = ilaplace(G/s,t);
t= 0:1:T;
u = ones(size(t)); %step input(all 1's for t>0)
for t= 0:1:T
y(i,1) =g(t);
i = i+1;
end
t = 0:1:T;
plot(t,u,t,y)
legend ('input','output');

1.6 input output 1.4 1.2. 0.8 0.6 0.4 4 0.2 0 0 5 10 15 20 25 30

As k will be increased, wo will increase and the response will become faster,

for k=100 (The value of k changed in above code)

1.6 input output 1.4 1.2 0.8 0.6 0.4 0.2 0 0 5 10 15 20 25 30

For

sinusoidal inputs-

Code gets modified slightly-

For k=10, input = 0.5 e^j20t (input frequency 20rad/sec)

close all
clear all
syms s t
m =38; %mass of body
xi = 0.2; %damping ratio
k=10; %variable
T = 30; %Total time of simulation
i = 1;
j = sqrt(-1);
wo = sqrt(k/m); %Natural frequency
G = (2*xi*wo*s+wo.^2)/(s^2+2*xi*wo*s+wo.^2);
g(t) = ilaplace(G/s,t);
t= 0:1:T;
for t= 0:1:T
u(i,1) = 0.5*exp(j*20*t); %Input
y(i,1) =g(t)*u(i,1);
i = i+1;
end
t = 0:1:T;
plot(t,u,t,abs(y))
legend ('input','output');

Output-

0.8 input output 0.6 0.4 0.2 -0.2 -0.4 -0.6 0 5 10 15 20 25 30

For k =100 (Response settles faster)

0.8 input output 0.6 0.4 0.2 V -0.2 -0.4 -0.6 0 5 10 15 20 25 30

For k =50, input frequency = 200rad/sec

Input amplitude =5

u(i,1) = 5*exp(j*200*t);

input output 6 2 -2 -4 -6 0 5 10 15 20 25 30