ENS6160: Signal and System in Accelerometer

Added on - 03 May 2020

  • 11

    Pages

  • 1811

    Words

  • 144

    Views

  • 0

    Downloads

Trusted by +2 million users,
1000+ happy students everyday
Showing pages 1 to 4 of 11 pages
ENS6160: SIGNAL AND SYSTEMGiven the accelerometer below:Figure 1: (a) Piezoelectric accelerometer attached to a large body. (b) Accelerometerrepresentation using mass, spring and damper.Section 1: Mathematical Analysis of SystemFrom figure 1 (b) the free body diagram is given in figure 2 below:Figure 2: FBDDeriving equation of motionUsing Hook’s law spring mass is given by:Spring force:Fs=kz(t)Damping Force,Fd=cdz(t)dtForce due to motion of the massd2x(t)dtFrom newton’s law of motion:f=ma(i)md2z(t)dt=cdz(t)dtkz(t)+d2x(t)dt(ii)Rearranging equation (ii) gives:md2z(t)dt+cdz(t)dt+kz(t)=d2x(t)dt(iii)
The accelerometer is a damped second order system. It can be expressed inhomogenous second order differential equation as given in equation (iv) below.d2z(t)dt+2ζωndz(t)dt+ωn2kz(t)=0(iv)Where isωnnatural frequency ofPiezoelectric accelerometersystem andζis dampingratio of the same system. At equilibrium pointd2x(t)dt=0,thus, equation (iii) can bewritten as: (iii) can be written as:d2z(t)dt+cdz(t)mdt+kz(t)m=0(v)From equation (iv) and (v):Natural frequency of the systemωn2=(mk)1ωn=km(vii)Damping ratio:2ζωn=(mc)12ζkm=(mc)1Thus,ζ=c2km(viii)Assumingz(t)=Aest,then:dz(t)dt=Asestd2z(t)dt=As2est(ix)Inserting equation (ix) in equation (v) becomes:Aest(s2+cms+km)=0SinceAest0,Then characteristic equation is therefore given in equation (x) belows2+cms+km=0(x)Eigenvalue are:s1=(cm)+(cm)24(km)2
s2=(cm)(cm)24(km)2From equation (x);i)if0<ζ<1thePiezoelectric accelerometeris underdamped: here, the systemoscillates with a frequency equal toωd=ωn1ζ2ii)ifζ=1thePiezoelectric accelerometeris critically damped: here,thePiezoelectric accelerometerslowly returns to equilibriumiii)ifζ1,thePiezoelectric accelerometeris over-damped: here, ThePiezoelectricaccelerometerreturns to equilibrium fasterTakingm=4.3×106kg:k=508N/mFrom equation (vii), the natural frequency of accelerometer is:ωn=km=5084.3×106=10869.20121rad/sTakingζ=1,then from equation (viii):ζ=c2kmc=2ζkm=24.3×106×508=0.09347513038kg/sSection 2: System analysis using MATLAB.Forζ=1Plotting the impulse-responses and step-responses of thepiezoelectric accelerometer(for a duration 1 ms and time ‘step size’ of 10 ns) Figure 3 shows the step and impulseresponses of the given system.
00.10.20.30.40.50.60.70.80.9110-30123410-5Impulse ResponseTime (seconds)Amplitude00.10.20.30.40.50.60.70.80.9110-300.5110-8Step ResponseTime (seconds)AmplitudeFigure 3:The impulse and step responses of the given systemFigure 4 given below is the plot of the magnitude and phase responses of the statedsystem over 0 Hz to 20,000 Hz with 100 Hz step size.102103104Frequency (rad/s)-150-100-500Phase(degrees)102103104Frequency (rad/s)2468Magnitude10-9Figure 4: The magnitude and phase response of the systemFrom the MATLAB plot and using data cursor:
desklib-logo
You’re reading a preview
Preview Documents

To View Complete Document

Click the button to download
Subscribe to our plans

Download This Document