Mechanical Engineering: Accelerometer System Analysis

This assignment delves into the analysis of an accelerometer system, a damped second-order system, through both mathematical and computational methods. The initial section focuses on deriving the governing equations of motion, including the derivation of the LCCDE and the expressions for damping ratio and natural frequency, culminating in determining the natural frequency value for the accelerometer. The subsequent section utilizes MATLAB to simulate the system's behavior, generating impulse and step response plots, along with frequency response analysis, to understand the system's characteristics under various damping conditions. Different damping ratios are implemented in MATLAB to analyze the system's step, impulse and frequency response. Finally, the analysis concludes with recommendations regarding the optimal damping ratio to achieve a critically damped response for efficient voltage output transmission and to avoid time wastage, and a discussion on the importance of the damping coefficient and its effect on the system's performance, providing a comprehensive understanding of the accelerometer's behavior.

ASSIGNMENT 1
2017
ASSIGNMENT
STUDENT NAME
UNIVERSITY AFFILIATION |
DATE OF SUBMISSION

ASSIGNMENT 2
ANALYSING A SIMPLE SYSTEM
Section 1: Mathematical Analysis of System (20 marks)
Note that the voltage output, V, is proportional to the change in the difference between x and y
due to piezoelectric effect.
(i) Free body diagram
(ii) The accelerometer is a damped second order system. It is expressed by the
homogenous second order DE for such a damped system as
Show that the motion of the accelerometer can be described by the LCCDE below,
For forced vibrations the equation of motion is given as follows,
newto n' s second law of motion∑ F=ma
Fapplied−Fdamping−Fspring=ma=m ¨z (t)
m ¨z ( t ) + Fdamping + Fspring=Fapplied
Applying the newton’s law to a free-body diagram of the mass,

ASSIGNMENT 3
m ¨x =∑ F=−Fb−Fk+ F
m ¨z =−b ˙z−kz + F
The equation is rearranged to standard to form a homogenous second order ODE,
m ¨z +b ˙z +kz =F
In standard form, it is expressed as,
¨z +
[ b
m ] ˙z +
[ k
m ] z= [ F
m ]
In the free body diagram, let b=C, and F/m = u
[ b
m ] ≅ 2 ζwn∧
[ k
m ] ≅ wn
2
¨z +2 ζwn ˙z +wn
2 z=u
For a free vibration system, u=0
¨z +2 ζwn ˙z +wn
2 z=0
(iii) The expression for the damping ration and natural frequency in terms of m, k, c
¿ [ k
m ] ≅ ωn
2 , ωn = √ k
m … natural frequency
¿ [ b
m ]≅ 2 ζwn , ζ= b
2 m wn
… damping ratio
(iv) Characteristic equation and eigen values for the system
¨z +2 ζwn ˙z +wn
2 z=0 … characteristic equation
Solution to the characteristic equation
z1 , z2=−c ± √c2−4 km
2 m
λ=−ζ ωn ± ωn √ ζ 2−1
(v) Natural response for system
(a) 0<ζ <1
(b) ζ =1
(c) ζ >1
Substituting in the equation above,
λ=−ζ ωn ± ωn √ ζ 2−1
For part a,
√ ζ2−1 , ζ <1 →underdamped

ASSIGNMENT 4
√ζ2−1 , ζ =1→ critically damped
√ζ2−1 , ζ >1 → overdamped
(vi) Determine the value of wn for the accelerometer
m=4.3 x 10−6 kg
k =508 N /m
ωn= √ k
m= √ 508
4.3 x 10−6 =1.0869 x 104
C=? ,ζ = C
2m wn
,C=2ζm wn
C=2 ζm wn =2 x 1 x 4.3 x 10−6 x 1.0869 x 104
C=0.0935
Section 2: System analysis using MATLAB (30 marks)
(vii) Impulse response and step response of the system
dr=1;
M = 4.3e-6; % units of kg

ASSIGNMENT 5
K = 508; % units of N/m
wn=sqrt(K/M); %natural frequency
B = 2*wn*dr*M; % units of N-s/m
num = 1;
den = [M B K];
sys = tf(num,den)
t=0:1e-9:1e-3;
step(sys,t)
grid on
figure (2)
impulse(sys,t)
grid on

ASSIGNMENT 6
(viii) Frequency response
%% frequency response
fw=0:100:20000;
figure(3)
freqs(num,den,fw)

ASSIGNMENT 7
(ix) Magnitude response at wn
wn =1/sqrt (2)
at f=-3db,
(x) Response of the system discussion
√(ζ^2-1) ,ζ<1 →underdamped
√(ζ^2-1) ,ζ=1 →critically damped
√(ζ^2-1) ,ζ>1 →overdamped
(xi) Repeat with the given damping ratios
Using ζ = 0.5,
dr=0.5;
M = 4.3e-6; % units of kg
K = 508; % units of N/m
wn=sqrt(K/M); %natural frequency
B = 2*wn*dr*M; % units of N-s/m
num = 1;
den = [M B K];
sys = tf(num,den)
t=0:1e-9:1e-3;
figure(1)
step(sys,t)
grid on
figure(2)
impulse(sys,t)
grid on
%% frequency response
fw=0:100:20000;
figure(3)
freqs(num,den,fw)
%% magnitude response

ASSIGNMENT 8

ASSIGNMENT 9
Using ζ = 2,
dr=2;
M = 4.3e-6; % units of kg
K = 508; % units of N/m
wn=sqrt(K/M); %natural frequency
B = 2*wn*dr*M; % units of N-s/m
num = 1;
den = [M B K];
sys = tf(num,den)
t=0:1e-9:1e-3;
figure(1)
step(sys,t)
grid on
figure(2)
impulse(sys,t)
grid on
%% frequency response
fw=0:100:20000;
figure(3)
freqs(num,den,fw)

ASSIGNMENT 10

ASSIGNMENT 11
(xii) Recommendation and conclusion
Vibration is oscillating motion of a particle or body about a fixed reference point. The
motion may be simple harmonic or complex. Setting the damping coefficient to 1. For
this value, the system no longer vibrates; Instead, the mass smoothly returns to its
equilibrium position x=0. If you need to design a system that returns to its
equilibrium position in the shortest possible time, then it is customary to select system
parameters such that a system of this kind is said to be critically damped. The
accelerometer requires the very lowest damping ratio to avoid a lot of time wastage
when transmitting a voltage output.