Data Processing Assignment

Project 1
Data processing and analysis:
Note; ωn = √ 1
LC
While; ζ = 1
2 R √ C
L
Then, ω 2−ω 1
ωn =2ζ
¨V 0 + 2ζωn ˙v0+ v0ωn2 = 2 ζωn ˙vn
¨V 0 = 1
δ t2 [v0i+2 -2v0i+1 + v0i]
= 1
δt [v0i+1 - v0i]
δt = 1
Fs
Hence, ωn = 2 πf n
Therefore; ωn *δt = 2 πfn
fn
v0i+2 = v0i+1(2-2ζ* 2 πfn
fn ) – vo0(1-2ζ* 2 πfn
fn + ( 2 πfn
fn )2 + 2ζ( 2 πfn
fn )(Vni+1 – Vni)
where; i= 0→ N −1
a) Use Fourier Transforms to estimate the natural frequency of the system
and its damping factor.
Mx + cx + kx = fsinωt………………………………(1)
From the calculation below
1

The value of ζ = 0.79995
While ω = 0.873
Lets take the passanger vehicle has a weight of; f = 10000 N
While x(0) = 0.064 and x(0) = 0
The above equation is similar to the unforced system except that it has a forcing function.
To solve this equation we have to reduce it into first order differential equation
Let
x = v ………………..(2)
but
v = [(f/m)sinωt – (c/m)v – (k/m)x] ……….(3)
the second order differential equation has been reduced to the first order differential equation
therefore;
x = y(1)
x = v = y(2)
equation (2) and (3) then reduced to
y(1) = y(2)
y(2) = [(f/m)sin(ωt) + (-c/m)*y(2) – (k/m)*y(1)]
to calculate for the value of c
we compaire equation (1) with the following generalized equation
x + 2ζωn + ωn2 = fsinωt
equating the coefficient
c/m = 2ζωn
ωn2 = k/m
from the calculation below
ωd = 0.873 rad/sec
2

ωn = 0.874 rad/sec
ζ = 0.03328
damped time period Td = 2π/ωd = 7.2 sec
for 5 time cycle the interval should be 36 sec, since the plot should indicate both the transient
and steady state responses, the time interval will be increased
function yp = forced(t,y)
yp = (y(2);(((f/m)*sin(ωn*t))-((c/m*y(2))-(k/m)*y(1))));
The following code represent the main code
span = (0 5);
y0=(0.064;0);
(t,y)=ode45(forced,tspan,y0);
Plot(t,y(:,1));
grid on
xlabel(time)
ylabel(Displacement)
title(Displacement vs time)
hold on
tspan represent the time interval and y0 represent the initial condition for y(1) and y(2), which in
turn represent the displacement x and the first derivative f x
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A0 = 2
N ∑
n=1
N
X ncos ¿ 2 π
N
=
2
N [0.064∗cos 2 π
8 +0.052∗cos 2 π∗2
8 +0.042∗cos 2 π∗3
8 + 0.034∗cos 2 π∗4
8 +0.028∗cos 2 π∗5
8 + 0.022∗cos 2 π∗6
8
= 0.0000961941
B0 = 2
N ∑
n=1
N
X ncos ¿ 2 π
N
=
2
N [0.064∗sin 2 π
8 +0.052∗sin 2 π∗2
8 + 0.042∗sin 2 π∗3
8 + 0.034∗sin 2 π∗4
8 +0.028∗sin 2 π∗5
8 +0.022∗sin 2 π∗6
8 +
= 0.01909
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Amplitude = √0.00009619412+ 0.019092
= 0.01909
Phase of harmonic = tan-1 0.01909
0.0000961941 = 89.710
Period per circle = 0.9 seconds
Total period = 0.9 * 8 = 7.2 seconds
Minimum frequency = N
0.9 = 1
0.9 = 1.111 Hz
Maximum frequency = 9
0.9 = 10 Hz
Frequency at the Centre = 5
0.9 = 5.556 Hz
Hence, natural frequency = ωn = 2 π
T = 2∗π
7.2 = 0.873 rad/sec
While, damping factor = ζ = f max−f min
2∗f centre =10−1.111
2∗5.556
= 0.79995
b) Obtain the same parameters using time-domain analysis.
close all;
clear all;
clc;
ωd = 2П/T =0.873 rad/sec
ωn = ωd ¿ √ ¿)
ζ = 1/sqrt(1+( 2 π
δ )
2
¿
δ=¿ = 1/nln(x(t)/x(t+nT))
Filename= project 2
X(t) = 0.064 m
X(t + nT) = 0.012 m
5

δ=¿1/8ln(0.064/0.012)
= 0.2092
end.
Damping factor = ζ =
1
√1+ ( 2 π
0.2092 )2
= 0.03328
Natural frequency = ωn = 0.873
√1+0.033282
= 0.874 rad/ sec
c) Comment on discrepancies
The percentage difference of the natural frequency = 0.874−0.873
0.873 ∗100=0.115 %
This is a clear indication that the natural frequencies on the two cases are almost similar or
almost equal
The percentage difference on the damping factor = 0.79995−0.03328
0.79995 ∗100=95.84 %
The damping factors used in the two cases are totally difference
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