SolutionofMeasurementsystemsandSignalProcessingPART1:MeasurementSystems1. a. Explain the hierarchy of standards. Explain the term standardb. Search the Internet to find an example of standard.(5p)SOLUTIONTheterm standardrefersto anobjcctor instrument,amethod ora procedure that providesavalue ofan acceprableaccuracyfor comparison.Aprimary standard defines the value ofthe unit to which it is associated.Secondarystandards,whilebasedon the primary standard,are more readily accessible and amenableforuse in a calibration,There is a hicrarchy of secondary standards; Atransfer standardmight be maintained by a national standards lab (such as NIST in the United States) tocalibrate industrial"laboratory standards".It is costly and time-consumingto certifyalaboratory standard, so theyare treated carefully and not used too regularly.A laboratonystandard would be maintained by a company tobe used to certify a more common in-housereference called the working standard.Aworking standard would be calibrated against thelaboratory standard.Theworking standard is used on a more regular basis to calibrateeveryday measurement devices or products being manufactured.Working standards are morethe normformostofus.Aworking standard is simplythe valueorinstrumentthatweassume is correct in checking the output operation of another instrument.Example: Agovernment lab maintains the primary standard for pressure.It calibrates a aninstrumentcalleda"deadweighttester"(see C9discussion)forhighpressure calibrations.Theseformitstransfer standardforhighpressure.Acompanythatmakespressuretransducers nceds an in-house standard to certify their products.They purchase twodeadweighttesters,They send onetesterto the national lab to be calibrated; thisbecomestheirlaboratory standard.On return, theyuse itto calibrate the other, this becomes theirworking standard.They test their manufactured transducers using the working standard-usually at one or two points over the transducer range to assure that it is working.Becausethe working standard is being used regularly,it can go out of calibration.Periodically,theycheck the working standard calibration against the laboratory standard.SeeASMEPTC19.2PressureMeasurementsforafurtherdiscussion.2. a. Why calibrate? What does calibrated mean?b.Drawanexampleofacalibrationcurve.(5p)
Solution of Measurement systems and Signal Processing PART 1: Measurement Systems 1. a. Explain the hierarchy of standards. Explain the term standard. b. Search the Internet to find an example of standard.(5p) 2. a. Why calibrate? What does calibrated mean? b. Draw an example of a calibration curve.(5p)
SOLUTION:The purpose ofa calibration is to evaluate and document the accuracy ofa measuring deviceA calibration should be performed whenever the accuracy level of a mcasured value must beascertained.An instrument that has been calibrated provides the engineer a basis for interpreting thedevice's output indication. It provides assurance in the measurement. Besides this purpose, acalibration assures the engineer that the device is working as expected.Aperiodic calibration of measuring instruments serves as a performance check on thoseinstruments and provides a level of confidence in their indicated values. Agood rule is tocalibrate measuring systems annually ormore often as needed.ISO 9000 certifications have strictrules on calibration results and thefrequencyofcalibration.3. Discuss interference in the test of Figure below.(5p)SOLUTION:In the example described byFigure 1.3,tests were run on different days on which the localbarometric pressure had changed,Between any two days of different barometric pressure,theboiling point measured would be different-this offset is due to the interference effect of thepressure.Considera test run over several days coincident with the motion of a major weatherfrontthrough the area.Clearly,this would impose a trend on the dataset.For example, themeasured boiling point may be seem as increasing from day to day.Byrunning overrandomdays separated bya sufficientperiod ofdays,so as nottoallowanyone atmospheric frontto impose atrend onthe data,the effects ofatmospheric pressurecanbe broken up into noise.The measured boiling point might then be high one test but then lowon the next, in effecl,making itlook likerandom data scater,ie.noise.4. What is hysteresis? Make a drawing. (2p)
3. Discuss interference in the test of Figure below.(5p) 4. What is hysteresis? Make a drawing. (2p)
SOLUTIONThe data are plotted below. The slope of a line passing through the data is 1.365 and the yintercept is 2.12. The data can be fit to the line y = 1.365x +2.12. Therefore, the staticsensitivity isK=1.365 for all x.P1.16108M6人4200246xM5. Find the input and output range of the calibration data and calculatethe corresponding span. (5p)SOLUTIONByinspection0.5≤x≤100cm0.4≤y≤253.2VTheinput range(x)is from0.5to100 cm.Theoutputrange (y)isfrom0.4to253.2V.Thecorrespondingspansaregivenbyr= 99.5cmro=252.8V6.For the calibration data of Table 1.5, determine the static sensitivity ofthe system at(a) X = 5;(b) X = 10; and(c) X = 20.(5p)
5. Find the input and output range of the calibration data and calculate the corresponding span. (5p) 6. For the calibration data of Table 1.5, determine the static sensitivity of the system at(a) X = 5;(b) X = 10; and(c) X = 20.(5p)
SOLUTION:The data reveal a linear relation on a log-log plot suggesting y=bxThat is:logy=log (bx)=logb+mlogxorY=B+mXFrom the plot,B=O, so thatb =I, and m =L.2.Thus, wefind from the calibration therelationshipy=xi2BecauseK=[dy/dxk=1.2x02,weobtainP1.11x [cm]K[V/cm]51.661000101.90100202.18WIA101We should expect that errors would-0.1propagate with the same sensitivity as the10.110100data, Hencefor y-f(x),as scnsitivityincreases, the influence of the errors onyX[cm]due to errors in x between would increase.7. Research on Internet the following test standards Write a shortofthe(200-word)reportthat describes theintent andanoverviewstandards
7. Research on Internet the following test standards Write a short (200-word) report that describes the intent and an overview of the standards
SOLUTIONa)thermostatSensor/transducer:bimetallicthermometerOutput: displacement of thermometer tipController: mercury contact switch (open:furnace off; closed:furnace on)b)speedoneterMethod 1:Sensor: usually a mechanically coupled cabkeTransducer: typically a dc generator that is turned by the cable producing an electricalsignalOutput: typically a pointer/scale (note: often a galvanometer is used to convert theelectrical signal in a mechanical rotation of the pointer)Method 2:Sensor; A magnet altached to the rotating shafiTransducer: A Hall Effect device that is stationary but detects each sensor passage bycreating voltage pulseSignal Conditioning: A pulse counting circuit; maybe also digital-analog converter (ifanalog readout is used)Output: An analog or digital readout calibrated to convert pulses per minute to kph ormph.Portable CDStereo Playerc)Sensor: laser with optical reader (reflected light signal differentiates between a""and"0")Transducer: digital register (stores digital information for signal conditioning)Signal conditioning: digital-to-analog converter and amplifier (converts digital numbersto voltages and amplifies the voltage)Ouput: headset/speaker (note:the headese/speaker isa second transducer inthis systemconverting an electrical signal back to a mechanical displacement)d)anti-lockbruking syslemSensor: brake activation switch senses brakes 'on'; encoder counts wheels revolutions perunit timeSignal conditioning: timing circuitOutput: a feedback signal that pulsates brake action overriding the driver's constant pedalpressuree)audio speakerSensor: coil (to which the input terminals are connected)Transducer: coil-magnet-speaker cone that acts as a miniature electrical de motorresponding to changes in current applied.Output: speakerconedisplacement8. List the important characteristics of input andoutputsignalssanddefine each.(5p)
8. List the important characteristics of input and output signals and define each.(5p)
SOLUTION:1Magnitude-generally refers to the maximum value ofa signal2.Range-differenceberweenmaximumandminimumvaluesofasignal3.Amplitude-indicative of signal fluctuations relativeto the mean4.Frequency-describesthetimevariationofasignal5.Dynamic-signal istime varying6. Static -signal does not change over the time period of interest7.Deterministic -signal can be described by an equation(other than a Fourier series or integral approximation)8.Non-deterministic-describes a signal which has no discernible pattern of repetitionandcannot be described bya simple equation.9. Determine the average and rms values for the functiony(t) = 25 + 10sin 6t over the time periods(a) 0 to 0.1 s,(b) 0.4 to 0.5 s,(c) 0 to 1/3 s,and(d) 0 to 20 s.Comment on the nature and meaning of the results interms of analysis of dynamic signals.(5p)SOLUTION:For the function y()dand-[oaYrmThus in general.(30+2cos6元l)d=12-5ael1-4[0(5-)+(in6m-sin6m)6.2-4,LandX(30+2cos6m）dt90i+inc12元6元
9. Determine the average and rms values for the functiony(t) = 25 + 10 sin 6t over the time periods(a) 0 to 0.1 s,(b) 0.4 to 0.5 s,(c) 0 to 1/3 s, and(d) 0 to 20 s.Comment on the nature and meaning of the results in terms of analysis of dynamic signals.(5p)
The resultingvalues areJm=31.02a)J=31.01b)J=28.99Jmm=29.00c)y=30Jm=30.03d) j=30mm=30.0310. Determine the value of the spring constant that would result in aspring-mass system thatwould execute one complete cycle of oscillation every 2.7 s, for a massof o.5 kg.Whatnatural frequency does this system exhibit inradians/second?(5p)SOLUTION:Since区0-Vm(asshowninassociationwithequation2.7)andT-2元_1=2.7sof0=2.33rad/sThenaturalfrequency isthenfound as=2.33rad/sAndk0=2.33=V0.5kgk=2.71N/m(kg/sec)11.A spring withk=5000N/cm supports a mass of 1kg.Determine thenatural frequency of this system in radians/second and hertz. (5p)
10. Determine the value of the spring constant that would result in a spring-mass system that would execute one complete cycle of oscillation every 2.7 s, for a mass of 0.5 kg. Whatnatural frequency does this system exhibit in radians/second?(5p) 11. A spring with k = 5000 N/cm supports a mass of 1 kg. Determine the natural frequency of this system in radians/second and hertz. (5p)
SOLUTION:The natural frequency may be determined,N100cm5000Kcmm=707.1rad/s01kgmand0f==112.5Hz2元12. For the following sine and cosine functions determine the period, thefrequency in hertz,and the circular frequency in radians/second. (Note: trepresents time in seconds).a. sin 10t/5b. 8 cos 8tc. sin 5n t for n=1 to(5p)SOLUTION:a)@=2元/5rad/sJ=0.2HzT=5sb)=20rad/sf = 3.18 HzT =0.31sc)0=3n元 rad/sf= 3n/2 HzT = 2/(3n)sn=1 to 8o13. Express the functiony(t) = 4 sin 2 t + 15 cos 2 tin terms of (a) acosine term only and (b) a sine term only .(5p)
12. For the following sine and cosine functions determine the period, the frequency in hertz,and the circular frequency in radians/second. (Note: t represents time in seconds).a. sin 10t/5b. ∞(5p) n=1 to ∞ cosine term only and (b) a sine term only .(5p)
SOLUTION:FromEquations2.10and2.11)μ=tan"By=Ccos(ot-a)A9"=tan"4y=Csin(or+$)BandwithAcosot+Bsinat=A+Bcos(al-)Acosot+Bsincot=A+Bsin(ot+)we findC=A+B=V15°+4=15.52§= tan"4= 0.26 rad15g"= tan~15-=1.31 rad4andy=15.52cos(2元t-0.26)Answer(a)y=15.52sin(2元l+1.31)Answer(b)Part 1: Data Acquisition14.Determine the alias frequency that results from sampling fl atsample rate fs:a. fl = 60 Hz, fs = 90 Hz c. fl = 10 Hz, fs = 6 Hzb. fl =1.2 kHz; fs = 2 kHz d. fl = 16 Hz, fs = 8 Hz(5p)
Part 1: Data Acquisition 14. Determine the alias frequency that results from sampling f1 at sample rate fs:a. f1 = 60 Hz; fs = 90 Hz c. f1 = 10 Hz; fs = 6 Hzb. f1 = 1.2 kHz; fs = 2 kHz d. f1 = 16 Hz; fs = 8 Hz(5p)
SOLUTIONThe Nyquist or folding frequcncy is defined as fn = f/2. All frequencies in the sampledsignal that are above fn will be folded back to lower frequencies below fn,a process depictedbythe folding diagram (Figure 7.3).(a) fj = 60 Hzf = 90 Hzfn = fj/2= 45 Hz.The ratio, t/fy = 1.33, that is f = L.33fn. Referring to The folding diagram, a frequency of1.33fy will be folded back to a frequency of 0.67fn. So f, = 60 Hz but in the sampled signal itbehaves as fa=0.67fn=30 Hz and out-of-phase with the original signal.f, = 2000 Hz(b) f = 1200 Hzfn= fJ2= 1000HzThe ratio, t'ty =1.2, that is f= 1.2f,Referring to Thefolding diagram, a frequencyof 1.2fnwill be folded back to a frequcncy of 0.8fn. So fi =1200 Hz but in the sampled signal itbehaves as f, = 0.8fy = 960 Hz and out-of-pbase with the original signal.(c) fi = 10 Hz1=6Hzfx= f/2=3Hz.The ratio, t/f = 3.3, that is f = 3.3f. Referring to The folding diagram, a frequency of 3.3fnwill be folded back to a frequcncy of o.7fn. So fi -1o Hz but in the sampled signal itbehaves as f, = 0.7fv = 7 Hz and in-phase with the original signal.(d) fi = 16 Hz=8HzIN=1/2=4HzThe ratio, f/fy = 4, that is f= 4fn. Refering to The folding diagram (and projccting itsbehavior beyond the values sbown), a frequency of 4fe will be tolded back to a frequency of0. So fi = 16 Hz but in the sampled signal it behaves as fa = 0 Hz. It will be seen as a constant(de) signal15. Convert the analogue signal E(t) = (4 + 2 sin 4 t + 3 sin 16t)Vinto a discrete timesignal using a sample rate of 32 Hz. Build the discretetime signal and its amplitude and phase spectra. Then try at fs = 16 Hzand at fs = 40 Hz. Discuss results.(10p)
into a discrete timesignal using a sample rate of 32 Hz. Build the discrete time signal and its amplitude and phase spectra. Then try at fs = 16 Hz and at fs = 40 Hz. Discuss results.(10p)