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ENS6160: Signal and System in Accelerometer

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Added on  2020-05-04

ENS6160: Signal and System in Accelerometer

   Added on 2020-05-04

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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 mass d2x(t)dtFrom newton’s law of motion:∑f=ma(i)md2z(t)dt=−cdz(t)dt−kz(t)+d2x(t)dt(ii)Rearranging equation (ii) gives: md2z(t)dt+cdz(t)dt+kz(t)=d2x(t)dt(iii)
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The accelerometer is a damped second order system. It can be expressed in homogenous second order differential equation as given in equation (iv) below.d2z(t)dt+2ζωndz(t)dt+ωn2kz(t)=0(iv)Where isωnnatural frequency of Piezoelectric accelerometer system and ζis damping ratio of the same system. At equilibrium point d2x(t)dt=0, thus, equation (iii) can be written 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 system ωn2=(mk)−1∴ωn=√km(vii)Damping ratio:2ζωn=(mc)−12ζ√km=(mc)−1Thus,ζ=c2√km(viii)Assumingz(t)=Aest,then:dz(t)dt=Asest∧d2z(t)dt=As2est(ix)Inserting equation (ix) in equation (v) becomes:Aest(s2+cms+km)=0Since Aest≠0,Then characteristic equation is therefore given in equation (x) belows2+cms+km=0(x)Eigenvalue are:s1=(−cm)+√(cm)2−4(km)2
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s2=(−cm)−√(cm)2−4(km)2From equation (x); i)if 0<ζ<1 the Piezoelectric accelerometer is underdamped: here, the system oscillates with a frequency equal to ωd=ωn√1−ζ2ii)if ζ=1the Piezoelectric accelerometer is critically damped: here,the Piezoelectric accelerometer slowly returns to equilibriumiii)if ζ1,the Piezoelectric accelerometer is over-damped: here, The Piezoelectric accelerometer returns to equilibrium fasterTaking m=4.3×106kg:k=508N/mFrom equation (vii), the natural frequency of accelerometer is: ωn=√km=√5084.3×10−6=10869.20121rad/sTaking ζ=1,then from equation (viii):ζ=c2√km⟹c=2ζ√km=2√4.3×106×508=0.09347513038kg/sSection 2: System analysis using MATLAB.For ζ=1Plotting the impulse-responses and step-responses of the piezoelectric accelerometer(for a duration 1 ms and time ‘step size’ of 10 ns) Figure 3 shows the step and impulse responses of the given system.
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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 stated system 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:
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