FFT Use in DIAdem

7/21/2019 FFT Use in DIAdem

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FFT Use in NI DIAdem

Contents

What You Always Wanted to Know About FFT... ______________________________________ 2

FFT Basics _____________________________________________________________________ 2

A Simple Example _____________________________________________________________ 3

FFT under Scrutiny ____________________________________________________________ 4

FFT with Many Interpolation Points ________________________________________________ 4

An Exact Result _______________________________________________________________ 5

Transient Signals ______________________________________________________________ 5

Typical FFT Results ____________________________________________________________ 6

Square Wave ______________________________________________________________ 6Dirac Delta Function ________________________________________________________ 6

FFT Application _________________________________________________________________ 7

Using the Window Function _____________________________________________________ 7

Different Window Functions __________________________________________________ 7

Time Intervals ________________________________________________________________ 8

Averaging with Intervals _____________________________________________________ 8

Joint-Time-Frequency Analysis ________________________________________________ 9

Third/Octave Analysis _________________________________________________________ 10

FFT Functions _______________________________________________________________ 10

Why Autospectra? _________________________________________________________ 11

When does a PSD Make Sense? _____________________________________________ 11FFT with Two Time Signals _____________________________________________________ 12

Transfer Function _________________________________________________________ 12

Acquis it ion of Osci llation Sign als _________________________________________________ 15

General Information on the Acquisition of Measurement Data __________________________ 15

Precision of the AD convertor ________________________________________________ 15

Calibration of the Measurement Signals ________________________________________ 16

The Aliasing Effect ____________________________________________________________ 16

Lowpass Filter ____________________________________________________________ 17

Oversampling ____________________________________________________________ 17

Determining the Required Sampling Rate _________________________________________ 17

Determining Special Sampling Rates _____________________________________________ 17

Several Channels Simultaneously ________________________________________________ 18

Simple Multiplexer _________________________________________________________ 18

Scan Frequency and Sample Frequency _______________________________________ 18

Sample/hold _____________________________________________________________ 18

Several A/D Converters _____________________________________________________ 19

Comparing Results with FFT Analyzers ___________________________________________ 19

External Clocks in Rotation Speed Dependent Vibrations _____________________________ 19

Determining the Parameters for External Clocks _________________________________ 20

Special Topic: Acoust ic Measurement _____________________________________________ 21

Acoustic Signals _____________________________________________________________ 21


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To inspect high frequencies, the sampling rate must be appropriately fast. Fortechnical reasons the sampling rate must be at least 2.5 times higher than thehighest frequency in the signal that you want to analyze.

To resolve frequencies accurately the measurement must be accordingly long induration. The necessary measurement duration (in seconds) must be equal to or

exceed the inverse of the desired frequency resolution (in Herz).

A Simple ExampleThe following example analyzes aclean sine wave with an amplitude of5 Volts and a frequency of 30 Hz. Thesignal is acquired with a sampling rateof 1000 Hz over a period of 0.255seconds, acquiring in total 256 values.

The result should be an amplitudefrequency response with a peak of30 Hz and an amplitude of 5. Otherfrequencies should have amplitudes ofexact zeros.

The following figure showsthe actual result of thi ssimple FFT.

At first sight the calculation result doesnot meet the expectations. Instead ofone amplitude of 5 Volts at 30 Hz theresult is a number of amplitudes, ofwhich the greatest is 31.25 Hz and hasa value of approximately 4.15 Volts.

A more accurate look at the figurereveals two details that are responsiblefor the deviation from the expectedvalue.

The time signal (top) containsslightly more than seven oscillations(approximately 7.6). The FFT algorithmon the other hand assumes periodicsignals that can be repeated anynumber of times. This means the FFTrises at the end because it thinks thatafter the last value the signal startsover again at 0. The FFT does not

know that in actual fact the sine curvewas truncated in the middle of a wave.

The extract with a line display(bottom) shows that the actual

frequency of 30 Hz does not appear asa line in the result. Therefore the actualfrequency can only be inferred from itsnearest neighboring frequency lines.The frequency axis is divided intoequidistantly spaced samples. If onecounts the lines, the frequency 30 Hz

is at about sample 7.6.These two effects seem to be related.We will see later that the FFT returnsexactly the theoretically expected linewhen the measured sine curvecontains an exact integral number ofperiods. So, with some skill it ispossible to find a sampling rate thatcalculates the expected resultaccurately. In practice however, whenwaves are measured, it does not makesense to adjust the sampling rate tothe frequency of the already known

sine.

Incidentally: If the same calculation isexecuted with sine curves that haveslightly different frequencies or that do

Sine Signal (30 Hz) Acquired with 1000 Hz Sampling Rate

Real signal

with 30 Hz and 5V

Real result of the FFT

Extract with line display

Real signal

with 30 Hz and 5V

0 0.05 0.1 0.15 0.2 0.25

Time [s]

-5

0

5

Signal[V]

0 100 200 300 400

Frequency [Hz]

0

5

Amplitude[V]

0 10 20 30 40 50 60 70

Frequency [Hz]

0

1

2

3

4

5

Amplitude[V]


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The sine wave with the frequency30 Hz is only available for approxi-mately a fifth of the measurement time.The FFT now determines a highresolution but the exact frequency andamplitude are still not revealed.

Results cannot be improved by addingzeros. Adding short signals repeatedlyto one another does not help either.This does not improve the result.

Typical FFT ResultsSome FFT results for certain standardsignals characterize the behavior ofFFT functions. The results for sinecurves have been described in detail.

The following examples will showsome other typical results.

Square WaveSquare waves do not contain allfrequencies. They contain the mainfrequency and multiples (3, 5, 7, ...) ofthis frequency with a decreasingamplitude.

Theoretically this leads to an infiniteseries. Therefore aliasing effectsalways occur with square waves. Alater chapter will deal with this in moredetail. In the middle diagram the

effects are prevented through variousmeasures (higher sampling rate, filter,reduction).

Dirac Delta FunctionAn important signal for the FFT is atheoretically infinitesimally short pulse.

This ideal pulse has an evenlydistributed spectrum. In later exampleswe will see how a short impact (from a

hammer) triggers an oscillation thatcontains all frequencies.

A typical example for such a pulse isthe ringing of a tuning fork. The initialpulse contains all frequencies. After theinitial pulse the tuning fork continues tooscillate with its own particularresonant frequency.

Sine Signal (30 Hz) Acquired with 1000 Hz Sampling Rate

Extract with Line Display

0 0.25 0.5 0.75 1

Time [s]

-5

-2.5

0

2.5

5

Signal[V]

0 100 200 300 400

Frequency1 []

0

0.2

0.4

0.6

0.8

1

Ampl_Peak[]

15 20 25 30 35 40 45 50

Frequency1 []

0

0.2

0.4

0.6

0.8

1

Ampl_Peak[]

Rectangle Signal (30 Hz)

FFT with Aliasing Effect

FFT without Aliasing Effect

0 0.2 0.4 0.6 0.8

Time [s]

-5

-2.5

0

2.5

5

Oscillation

0 100 200 300 400 500

Frequency [Hz]

0

1

2

3

4

5

Amplitude

0 100 200 300 400 500

Frequency1 [Hz]

0

1

2

3

4

5

6

Amplitude1

Dirac Delta Function

FFT

0 0.025 0.05 0.075 0.1 0.125Time [s]

0

0.25

0.5

0.75

1

Crush

0 100 200 300 400 500

Frequency [Hz]

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

Amplitude


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hand side shows the result of a largeFFT with 16386 values. The frequencyresolution at 1.35 Hz is very high. Theright hand side shows the averagedresult of 8 FFT calculations with 2048

values each. The frequency resolutionis 10.7 Hz. The characteristics of thesetwo curves are roughly the same,however the amplitudes and the formof the peaks are very different as thedepicted detail shows.

Because the number of values in theFFT must always be a power of 2, apart of the data is not used. Apartitioning into several intervals leadsto more values being used becauseone can get closer to the number of19,000 values. It is also possible to

calculate overlapping intervals and tointegrate all values in the calculation;however some values are then usedtwice. The advantage of usingoverlapping intervals in windowfunctions such as Hanning is thatvibrations, which occur only briefly andare attenuated randomly, are includedin the neighboring interval evenstronger.

Appropriate application areas exist forall algorithms. For example, to inspecta sound wave to find out which of twosound sources that are close togetherholds the larger share of the totalsound, an evaluation with as manyvalues as possible is the right choice.In contrast, the above example showsone second of a human voice signal,which naturally does not contain anyexact constant frequencies. The peakof about 260 Hz is random and can beat a different position in the nextsecond.

If one looks at the time signal closely,

one can see that it changes radically inthe time period of just one second. Asound analysis can only either be anaverage of the total time whereby thefrequencies cannot be exactlydetermined, or the exact frequenciesare determined at a certain time pointof the measurement. In both cases anFFT with as many values as possibledoes not make sense. The rightapproach for this signal is therefore theaveraged FFT or a Third/Octaveanalysis.

Joint-Time-Frequency AnalysisA sound signal seems to containdifferent amplitudes and frequencies at

different times. To inspect this in detailthe FFT is executed with intervalswithout averaging, sometimes called aJoint-Time-Frequency Analysis. IF youdivide the above voice signal into 8

time segments and take the FFR ofeach, the results is 8 FFT curvesone for each time interval.

The calculation result shows that thereare substantial differences between theindividual FFT curves. The largestpeak at 260 Hz, for example, onlyoccurs briefly in interval 5 (in the timesignal after approximately 0.4seconds). Because the frequencyoccurs accurately only for a very shortperiod, the amplitude peak is high andvery narrow.

The signal can change considerablyfrom one tenth of a second to the next.So the question is whether it might notbe better to inspect shorter periods oftime. The following figure shows anevaluation with 37 intervals with 512values each.

This illustrates a fundamental problemof the FFT. The more accurate onewants to determine the time point atwhich a certain frequency occurs, themore inaccurate the frequencyresolution becomes. In this examplethe resolution is only 43 Hz. As alreadydescribed, more accurate frequencyresolutions require longer measure-ment intervals.

Sound Signal (22050 Hz)

260 Hz

260 Hz in the 5th interval

0 0.2 0.4 0.6 0.8

Time [s]

-40

-30

-20

-10

0

10

20

30

40

Soundpressure[-]

0 1000 2000 3000 4000 5000012345

678

2

4

6

8

0 500 1000 1500 2000 2500 3000

Frequency1 [Hz]

0

2

4

6

8

Amplitude1[-]


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For each application a compromisebetween exact time resolution andexact frequency resolution must befound.

Third/Octave AnalysisThe evaluation of sounds that can beheard by humans covers a very largefrequency range-- from about 20 Hz to20 kHz. The FFT provides equidistant(evenly distributed) frequency steps,which is impractical for such a largefrequency range. For sound data it isnot so much the exact signal frequencythat is important but the volume indifferent frequency ranges. The total

frequency domain is divided intostandardized ranges. The octaveanalysis divides the frequencies intooctaves (frequency is duplicated), thethird analysis divides the frequenciesinto thirds (three thirds are octave).The following example shows thethird/octave analysis together with theoriginal amplitude after the FFTcalculation.

The top axis system shows the total

frequency domain of the signal. Thelogarithmic division of the frequencyaxis highlights the importance of lowfrequencies. The dashed line shows

the octave analysis that summates allamplitudes that occur in the octave. 13amplitudes, whose mean frequenciesare at the frequency axis, represent thetotal frequency domain. The bottom

axis system shows a detail. Here themean frequencies of the third are atthe frequency axis. The meanfrequencies are standardized andenable the comparison of differentmeasurements.

Amplitudes (Peak) are summed byadding the squares and taking thesquare root of the result. Each of thethird/octave amplitudes shown abovecontains the sum of all the singleamplitudes in its band. If thefrequencies are very low, there are not

enough FFT values and some thirdsare not used.

If the FFT is executed in severalintervals, the results differ. However,the amplitudes of the individual thirdsand octaves remain approximately thesame, because the sum of amplitudesin a frequency band remains stableregardless of the number. In the lowerfrequency domain it can happenquicker that the FFT provides no oronly few amplitudes for the individualfrequency bands. Therefore the resultof the third/octave analysis improveswhen the calculation uses largeintervals.

In each octave band or third band thesum of the individual amplitudes isestablished. Therefore the amplitude ofthe band is always greater than thegreatest individual amplitude in thatband.

FFT FunctionsYou can use a number of different FFT

functions to evaluate the resultingamplitudes of the FFT calculation. Thefunctions are explained in the followingtext:

The FFT algorithm first determines twochannels which are named real part(R) and imaginary part (I) and whichare not relevant here. You candetermine the amplitudes and thephase shifting from the real part andthe imaginary part.

The amplitude, which is the most

important result of the simple FFT, iscalculated by the square root (R2+I

2).

This result is called the peakamplitude. All previous examplesshowed peak amplitudes.

37 FFTs over Every 512 Values260 Hz

700 Hz

Frequency Resolution: 43.16 Hz

01000

20003000

4000 50005

10

15

20

25

30

35

2.5

5

7.5

10

Octave

Third

Linear

Standardized Octave Center Frequency

Standardized Third Center Frequency

1 2 4 8 16 31.5 63 125 250 500 1000 2000 4000 8000

Frequency [Hz]

0.001

0.01

0.1

1

10

Ampl_Peak

50 63 80 100 125 160 200 250 315 400 500

Frequency [Hz]

0.001

0.01

0.1

1

10

Ampl_Peak


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Apart from the peak amplitude, theRMS amplitude is also relevant. TheRMS value is the quadratic integralmean value of an oscillation. TheRMS value can be calculated directly

from the time signal. A typicalapplication is the calculation of theRMS sound pressure from themeasured sound vibration. In absolutesine waves the RMS valuecorresponds to the amplitude dividedby the square root of two. This relationalso applies to the FFT resultsbecause the FFT divides time signalsinto sine amplitudes.

The square of the peak amplitude iscalled the Autospectrum and is alsorelevant in some calculations. The

same applies for the square of theRMS amplitude (RMS

2), which is also

called the Power Spectrum.

A further calculation method is the PSDamplitude (Power-Spectrum-Density),in which the power spectrum is dividedby the interval of the frequency lines.

Summary:

Peak amplitude: 2 +2

RMS amplitude:2 +2

2

Autospectrum: 2 +2

Power spectrum: 2 =2 +2

2

Power spectrumdensity: =

2 +2

2

Why Autospectra?A typical question is: Why do we needthese different FFT functions?

In practice the different functions allhave their relevance. It is important to

remember which function was used,because otherwise the data might beincorrectly interpreted.

The following example shows a typicalapplication with Autospectra.

The example analyzes a signal thatconsists of two sine frequencies. Theresults of three measurements withdifferent sampling rates andmeasurement periods are compared.

At the top one can see the signal withthe different measurement periods.The red measurement was executed

for an especially long period, and theblue measurement was executed withan especially high sampling rate.

In the middle one can see very clearlythat there are great differencesespecially next to and between the twosignal frequencies. The quality of theresults also depends on coincidence.The blue measurement is lucky andmeets the actual frequency quiteaccurately. This makes the read outmaximum amplitude more accurate.

In the lower part of the figure one cansee that always the same sum results,regardless of the different amplitudedevelopment. However, the summationmay only be executed with the twosquared FFT functions, the powerspectra, or the Autospectra, as shownhere.

With sound data the spectrum shows alarge number of peaks, but it is difficultto establish which frequency domainsoccupy the largest part in the totallevel.

In this case the cumulative curve offersa very good overview of the variousparts of the different ranges of aspectrum of the total level.

When does a PSD Make Sense?The PSD function determines thepower spectrum and divides theamplitudes by the interval of thefrequency lines. The objective of thiscalculation is an improvedcomparability of spectra with evenly

distributed frequency parts.PSD is not suitable for signals withdefined sine parts as was the case inthe previous example for Autospectra.

Signal Acquired with Different Sampling Rates and Measurement Times

Autospectra of the signals:

The extract shows the

relation between

number and amplitudeof the interpolation points.

Section

The number of Autospectra is the same.

The Autospectra and the RMS2

spectra

may be added up. The result is the

square of the sum spectrum.

With square root (end value) one receives

the sum of the amplitudes or the RMS values.

Sum

0 0.5 1 1.5 2 2.5

Time-10

-5

0

5

10

50 55 60 65 70

Frequency

0

20

40

60

80

100

120

0 200 400 600 800

Frequency

10-6

10-5

10-4

10-3

10-2

10-1

100

101

102

50 55 60 65 7010

-3

10-2

10-1

100

101

102


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hammer must correspond with the unitunder test in order to execute normalvibration measurements withoutdisturbances.

Stimulations with a hammer blow areeasy to execute. After the blow thoughthe decaying oscillation can only bemeasured for a short period of time.The time span should be sufficient forall relevant oscillations.

The following figure shows theindividual FFT analyses for the twosignals.

An ideal blow with a hammer would

have stimulated all frequenciescompletely evenly. In practice this doesnot succeed. However, the figureillustrates that all frequencies occur ina wide area and that the frequenciesare also contained in the wave signal.

The FFT of the oscillation shows atwhich frequencies the structureoscillates more and at which less. Thegreatest amplitude at approximately260 Hz is only that large because thestimulation there is so strong. Theresonance frequencies with a weakstimulation, for example 600 Hz, wouldbe considerably higher if thestimulation were ideal.

The transfer function calculatesamplitudes in proportion to therespective stimulation. Phase shiftingcan interpret results further.

For the results shown here it not yetclear whether the measured vibrationwas caused by the stimulation. If thestimulation is a hammer blow, it is easyto evaluate due to the time signals. Ifthe stimulation is a noise stimulation, itis not so easy. In this case thecoherent function helps.

The coherent function requires severaltime signals that are measuredconsecutively. A hammer blow triggersthe measurement (analog trigger). Ifthe stimulation is triggered by noise,

the measurement can start at any time.

If you compare the measurement, youreceive different results. If, for onefrequency, all amplitudes and phasesof the transfer function correspondexactly, the coherence is one. If theconcordance is very bad, thecoherence is approximately zero.Therefore the coherence functionshows very well which frequencydomains of the transfer function showusable results. It only makes sense to

evaluate frequencies that providereproducible transfer values.

Hammer Blow

Oscillation

0 0 .010 .02 0 .030 .04 0 .050 .06 0 .070 .08 0 .09 0 .1 0 .110 .12 0 .130 .14 0 .150 .16 0 .170 .18 0 .19 0 .2

Time [s]

0

0.01

0.02

0.03

0.04

0.05

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Time [s]

-1.5

-1

-0.5

0

0.5

1

1.5

FFT Hammer Blow

FFT Oscillation

0 500 1000 1500 2000

Frequency []

0

5E-05

0.0001

0.00015

0.0002

0 500 1000 1500 2000

Frequency []

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

Transfer Function

Phase

0 500 1000 1500 2000

Frequency []

0

500

1000

1500

2000

0 500 1000 1500 2000

Frequency []

-180

-90

0

90

180

Transfer Function: Mean Value of 7 Tests

Coherence

0 500 1000 1500 2000

Frequency []

0

500

1000

1500

2000

0 500 1000 1500 2000

Frequency []

00.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


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This figure shows the averaged resultsfrom seven tests. The coherenceshows very different qualities in thedifferent frequencies.

If the frequencies have high amplitudes

and thereby a strong transfer rate, thecoherence is usually very good.Frequencies with a low transfer ratevery often do not have a goodcoherence, because randomdisturbances prevail. If nothing is

transferred, the frequency and thephase cannot coincide.

If a bad coherence occurs at frequencypeaks, it might be that a vibration isbeing measured that was triggered by

disturbances and not by the hammerblow. Therefore vibrations should bemeasured in idle mode and not with arunning engine.


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Acquisition of Oscillation SignalsThe acquisition of oscillation signals determines the quality of the frequency analysis.

Important factors in the acquisi tion are the precision of the measurement equipment

(recorder, measurement amplifier, AD convertor), the sampling rate, and the duration ofthe measurement.

General Information on theAcquisit ion of MeasurementData

Precision of the AD convertorEvery AD convertor has a definedprecision which is determined by thevoltage range and the resolution. OftenAD convertors have a resolution of12 bit and a voltage range of -10 to+10V. If the resolution is 12 bit, thevoltage range is divided into 4096steps of each 0.0048828125V. Usuallythe range does not include the +10Vvalue.

Measured signals that do not use thecomplete voltage range are not asprecise. The following example showswhat results can look like if the voltage

range is not well used.

The vibration ranges fromapproximately -0.15 to +0.15 volt. Onlyabout 60 (approximately 1.5 percent)of the 4096 different voltages that theAD convertor can measure are used.

Warning:The steps always exist for A/Dconvertor data but are not always soobvious.

The voltage steps are especiallyconspicuous due to the high samplingrate. With a lower sampling rate (forexample 10 Hz) the steps would no

longer be visible but the inaccuracywould still exist. If the sampling rate islow, the AD convertor does not specifythe mean value of a specific period oftime, but picks out the current valuefrom a signal at a point of time. In thetransition range between two steps it isthus coincidence which step ismeasured, regardless of the precision

of the rest of the measurement chain.Further errors occur on the analog sidebetween the acquisition of the signaland the actual digitalization. Thus it isnot guaranteed that the AD convertoralways determines the "correct" step.

The measurement result providesfurther information:

In the transition range the valuesjump between two steps back andforth. Therefore it is possible to use thesmoothing function to calculate a curve

that seems to have a higher resolution.This only makes sense if themeasurement error is small enough.

The high sampling rate prevents ameaningful calculation of the signalincrease. In the top signal the increaseis either zero, one step up, and onestep down. To calculate increasesaccurately, the values must becompared in intervals that cover manysteps. Therefore the increase inmeasurement data converted by theAD convertor is only rarely possible.

The precision of the AD convertorcan be substantially improved if ameasurement amplifier or an ADconvertor with amplification is used.The signal can now be adjusted to theAD convertor and the measurementrange can be utilized better.

AD convertors with a 16 bitresolution can divide the measurementrange into 65,536 steps (16 timesmore than the 12 bit convertor).Special measurement devices oftencan measure substantially moreprecisely. The precision is thenspecified in decimal places and not inbit resolution. 12 bit resolutioncorresponds to 3 decimal places,

Sine Signal (0.5 Hz) Acquired with 10 kHz Sampling Rate

0 0.5 1 1.5 2

Time [s]

-0.1

-0.05

0

0.05

0.1

Sig

nal1[V]

0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6

Time [s]

0.08

0.09

0.1

0.11

0.12

0.13

Signal1[V]

Resolution of the AD Converter is 0.0048828125 Volts


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16 bit resolution corresponds to 4decimal places. Some devices canmeasure, for example, 8 decimalplaces precisely. To prevent disturbinginfluences from the computer, these

AD convertors are often located in aseparate chassis. Often a time periodis integrated and not justinstantaneously measured. Often theintegration lasts exactly 0.02 seconds(a 50 Hz net cycle) to average out netdisturbances. Higher sampling ratesare then no longer possible.

The technical data sheet ofmeasurement devices list the expectedmeasurement precision and thepossible errors that can occur duringmeasurements.

Calibration of the MeasurementSignalsDue to technical reasons, you cannotprevent the AD convertor from beingimprecise. Therefore it is important toestimate the possible errors and toensure that the precision meets therequirements.

First, you should specify the precisionyou need for your configuration. Thespecification determines how precisethe AD convertor must measure andwhich measurement amplification isnecessary.

After this, a test measurement ofknown values can determine whetherthe precision was really achieved.

The Aliasing EffectThe aliasing effect poses a big problemwhen acquiring oscillation signals. Thealiasing effect is sign of erroneousmeasurements due to too low

sampling rates in the digitization. Theeffect occurs when a vibration, whichhas a frequency greater than half ofthe sampling rate, is contained in thesignal. The following examplemeasures the 30 Hz vibrations fromprevious sections in this article with asampling rate of only 35 Hz. The redline in the top axis system shows thevibration which is simulated digitally bythe low sampling rate. The frequencyof the vibration is obviously incorrect.

The FFT of both measurementsillustrates that the actual frequency isshifted due to the incorrect samplingrate. If the sampling rate is 35 Hz, only

frequencies up to 17.5 Hz can bedisplayed correctly. The signalfrequencies above this limit do notsimply disappear, but emergesomewhere in the range between 0and 17.5 Hz. A simple rule is: Higherfrequencies are mirrored at the areaboundary.

The signal frequency (30 Hz) is12.5 Hz above the boundary of17.5 Hz and therefore appears 12.5 Hzbelow the boundary at 5 Hz. If thesignal frequency is exactly at the

sampling rate, the measurementvalues appear as with a constant(0 Hz). Even higher frequencies (35 Hzto 52.5 Hz) rise to 17.5 Hz an. Thisconvolution can be continued infinitely.If the recorder (for example, amicrophone) is capable to do so, thesignal can contain a vibration of3502 Hz which reappears at 2 Hz.

After digitization it is impossible todistinguish real frequencies from thewrong frequencies that were generatedby the aliasing effect. Therefore onemust ensure before digitization that nofrequencies above half the samplingrate are contained in the analog signal.

If an FFT determines large amplitudesbefore reaching half the sampling rate,signals must be just above thisfrequency. This means that you do notknow whether the measuredamplitudes are right or wrong. Even ifthe analog filters are very efficient thetop quarter of the FFT should be usedwith care.

There are several possibilities toprevent aliasing effects. These optionsdo not include digital filters becausedigital filters are not applied before the

Sine Signal (30 Hz) Acquired with 35 Hz Sampling RateOriginal

Sampling rate 35 HzHalf a sampling rate 17.5 Hz

0 0.1 0.2 0.3 0.4 0.5

Time [s]

-5

0

5

Signal[V]

0 10 20 30 40 50 60 70

Frequency [Hz]

0

1

2

3

4

5

Amplitude[V]


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digitization. By then the aliasing effectis irretrievably contained in the signal.

A popular disturbance frequency is thefrequency 50 Hz or 60 Hz of the powernetwork (wall socket and/or fluorescent

lights), which influences the measuredsignal in different ways. If the samplingrates are low (nearly always a fractionof 50 or 60), a random offset or a verylow beat frequency is measured in thesignal. If the sampling rate is 100 Hz, 2values per 50 Hz or 60 Hz vibration aremeasured. Depending on the start ofthe measurement, the values are atthe zero crossing and do not create adisturbance. However, if the values areat the antinode, a noisy effect is theresult. If one measures with 100 Hz or

120 Hz and observes differing degreesof noise at different starts of themeasurement, one should inspect thiswith a higher sampling rate. The powernetwork emerges as an FFT peak atexactly 50 Hz or 60 Hz.

Lowpass FilterTo eliminate higher frequencies directlyin the analog signal, lowpass filters areused. These filters are therefore calledanti-aliasing filters. Thereby all

frequency components above a certainlimit frequency are filtered out quitewell. This method can also causeproblems if different measurementtasks must be constantly solved withdifferent frequency ranges. Lowpassfilters which are freely programmableare quite expensive and not alwaysavailable. Therefore one has to resortto a number of fixed filters and therelated sampling rates.

It is also possible to use signalrecorders that have measurement

amplifiers with integrated lowpassfilters. The device or the respectivemanual usually indicate the frequencydomain to use for the calculation. Oftenseveral frequency domains can beselected. If a vibration recorder has, forexample, a frequency domain of 1 Hzto 1 KHz, higher frequencies cannot bemeasured. However, a sampling rateof 2 KHz does not incur an aliasingeffect.

Note finally that If the sampling rate ishigh enough, aliasing effects areprevented. It is worth scrutinizing thefrequency domains of recorders andfilters.

Warning:

There is a typical error that should beavoided. If the recorder providesfrequencies up to 1000 Hz, it is wrongto measure with a 1000 Hz sampling

rate. The rate must be at least2500 Hz.

OversamplingModern measurement devices, ADconvertors, and computers have somuch capacity that the optimalsampling rates and exact lowpassfilters can be disregarded. To measurea vibration at approximately 1 Hz witha transducer that has a frequencydomain of up to 100 Hz the sampling

rate can be 250 Hz. The signal doesnot contain aliasing effects anymorebut a lot of data.

A data reduction leads to the requiredresult. To do so, one can use a digitalfilter. However one can also use a FFTover all data to extract the requiredarea from the frequency domain (0100 Hz).

Determining theRequired Sampling Rate

The necessary sampling rate must be2.5 times the highest importantfrequency. The highest importantfrequency might be the monitoredvibration or a frequency in the signalthat might be causing disturbances. Ifnecessary, one should test whether aproblem has occurred with 50 Hzdisturbances (power network).

DeterminingSpecial Sampl ing Rates

A large number of all measurements iscarried out with integer sampling rates(100, 200, 500 1000, ...). As thenumber of values for a FFT must be apower of 2, the measurement durationis often 'warped'. The FFT frequenciesare then also at the 'warped' valuesbecause the frequency interval iscalculated with one divided by themeasurement duration.

For many evaluations it would bebetter to determine integer frequenciesthrough the FFT. A search function

such as the peak search in DIAdem,which can search for frequencies at thegreatest amplitude, then provides, forexample, 5 Hz and not 0.46875 Hz. Asampling rate with a power of 2


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Several A/D ConvertersA further possibility to guaranteecomplete simultaneity is the usage ofboards or devices with several A/Dconvertors.

Devices with one A/D convertor perchannel produce simultaneouslymeasured channels.

Some devices have two or four A/Dconvertors which measure severalchannels simultaneously. In practicethis is often sufficient.

Comparing Results withFFT Analyzers

To ensure that the measurement

results are correct and reproducible,one can compare different acquisitionsystems with each other. To do so, onedoes not use a random signal but, forexample, a sine generator that canprovide a signal which can be repeatedany number of times. The results ofcomputer-based measurementacquisition such as DIAdem and theresults of an FFT analyzer often differ.

The reasons for the deviationsbetween the various systems are notso much the different quality of A/D

convertors or analog technology, butsome of the already described details.Especially with clear sine signals ofsignal generators the A/D convertorsand filters do not have to be onehundred percent precise. It is moreimportant that the sampling rates andthe measurement duration in bothsystems are simultaneous. Additionallythe window function and the correctionparameters should be the same.

The internal operation mode in FFTanalyzers is not discernible at first

sight. To reproduce the results onemust read the documentation or studythe display in detail.

FFT analyzers often have integerfrequencies that were acquired with'warped' sampling rates. Analyzersalso often provide a number of even-numbered frequencies (for example,400 not 512 lines). This is not achievedwith a different FFT algorithm. Insteadthe top 112 calculated frequencies arenot displayed. This makes sense,

because the precision for thesefrequencies is low. It is also possiblethat the FFT analyzers only display thelower half of the calculated frequenciesbecause they operate internally with

twice the sampling rate. Theadvantage is that the analog lowpassfilters have a large scope.

As a rule, analyzer with multiplechannels measure inputs simul-

taneously without phase shifting.In comparison to normal computerdata acquisition technology, the FFTanalyzers comprise programmableanalog lowpass filters which usuallyprevent aliasing effects automatically. Ifone compares random distributedvibration signals and not pure sinesignals, one must prevent the aliasingeffects manually when one usesnormal computer data acquisitiontechnology.

External Clocks in RotationSpeed Dependent Vibrations

The FFT only provides precise resultswhen the time signal is monitored for alonger period of time. In this period oftime the signal must stay constant. Ifthe vibrations are rotation speeddependent, one must monitor a largernumber of revolutions to preciselyanalyze the frequencies andamplitudes that are produced byunbalance.

In this case the results are morecorrect if the data is measured over theangle of rotation and not over time. Forexample, if a quadrature encoder with32 pulses per revolution is mounted ona rotating system, the measurementcan run with an external clock. Now,precisely 32 values per revolution areacquired regardless of the rpm (revsper minute, revolutions per minute). Allrotation speed dependent vibrationsare contained in the input data for theFFT as integer numbers (for example16 in 512 values). As shown above,vibrations that are contained as integernumbers in the signal are representedexactly by the FFT. The external clockwith a pulse number of a power of 2can analyze rotation speed dependentvibrations precisely even if therotational speed changes.

In this case the sampling rate changesproportional to the rotational speed.After the FFT, one therefore receivesamplitudes over the order not over

time. In this case, the vibrations thatcorrespond with the rotational speedare at the first order; vibrations thatcorrespond with twice the rotational


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speed are at the second order and soon.

In machines, vibrations often occur thatcorrespond with exactly the rotationalspeed or a multiple of the rotational

speed. These vibrations are especiallystrong when they meet the resonancefrequencies of the machine.

Determining the Parametersfor External ClocksThe relationship between revolutions,resolutions and achieved order areasis analogous to the described relationbetween time-related data.

The order range stretches from 0 tohalf the number of measurementvalues per revolution.

The resolution of the order channelcorresponds to the reciprocal numberof the evaluated revolutions.

Example:

In a measurement with 64 measure-ment values per revolution and 1024measurement values, one receivesresults up to order 16 which are

divided into steps of 0.03125 orders.To evaluate higher orders, moremeasurement values per revolutionmust be measured. The number ofmeasured revolutions affects theresolution of the orders.

In machines and motors the first orderlies in the range of approximately3,000 rpm or 50 Hz. Higher orders arenecessary because many resonance,vibration, or noise problems haveconsiderably higher frequencies.

In this context the order analysis is anadditional topic. The order analysistests rotational speed dependentvibrations. The aim is to analyzeaccording to orders machine vibrationsin different states (to power up, to shutdown, different operational states).


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Special Topic: Acoustic MeasurementAcoustic measurements determine the volume and the frequencies contained in noise

signal. The noise range audible for humans is o f special interest. Acoust ic measure-

ments have some special features that can pose unusual problems compared to othertypical measurements.

The audible frequency range of sound lies between 16 and 20,000 Hz.The audible volume (acoustic pressure) lies between 10-4and 102Pascal (N/m2).To evaluate sound sources one needs a reproducible absolute precision.

The enormous measurement ranges inacoustic measurements require specialmeasures that are not necessary forother physical dimensions to thisextent. The frequency range requireshigh sampling rates and a large

number of data to receive a sufficientresolution in the lower frequencydomains. The measurement range withthe factor 10

6 between extremely loud

and barely audible cannot be handledwith normal A/D convertors. Thereforeone needs a special measurementamplifier that amplifies the microphonesignal so that it can resolve with anormal A/D convertor.

The high precision in the measurementof sound amplitudes is also unusual.Normally in vibrations, only the relative

differences of the amplitudes betweenthe various vibration signals and thedifferent frequencies are important. Inacoustic measurements slight changesto the volume are of crucial importance(for example when implementing noiseprotection measures).

Sophisticated measurement tasksrequire sophisticated measurementtechnology. In the field of acousticmeasurements several companiesspecializing in high-precision soundmeasurement technology haveestablished themselves.

Microphones with linear frequencyresponses at different volumes andclearly defined recording angle.

Measurement amplifiers with alarge measurement range (forexample, amplification 8x10 dB)

Measurement amplifiers withevaluation filters important for soundevaluation (for example, A-weighting)

Measurement amplifiers with direct

measurement of the volume in dB ordB(A)

Recorder with high playingprecision (even speed and volume)

Calibration devices to test themeasurement chainExact sound measurements can onlybe carried out with high-qualitymeasurement technology. Precisequantitative statements cannot be

derived for sound measurementsexecuted with normal microphonesand sound boards on the computer.The acoustic signal used in theexample above was measured withsound board and can only be used fora qualitative evaluation of thefrequencies but not for precisestatements on the volume.

Acoust ic SignalsUsually acoustic signals are measuredwith a microphone as acousticpressure in Pascal (N/m2). Theacoustic signal is an oscillation thatoverlays the normal air pressure.Therefore it does not make sense tospecify the acoustic pressure at aspecific time. The frequency and theoscillation amplitudes are of interest.

A relatively simple definition of theacoustic signal is possible without FFT.The calculation of the root-mean-square of the acoustic pressureprovides the first characteristic value of

the volume of the signal.First all individual acoustic pressuresare squared in order to determine theroot-mean-square. The square root isextracted from the mean value of allsquares (sum divided by the number).The abbreviation RMS stands for root-mean-square. In a pure sine vibrationthe rms value is equal to the amplitudeof the vibration divided by the squareroot of 2.

It is impractical to specify the acousticpressure in N/m

2because the acoustic

pressures occur in different dimensionsdepending on the situation. Thereforea logarithmic scale is used to specifythe acoustic pressure. To receive


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manageable dimensions the acousticpressure is referred to a base size.This base size is an acoustic pressureof 2*10

-5N/m

2for airborne noise which

is a barely audible acoustic signal. The

calculation is:

= 20logacoustic pressure

with L=Noise pressure level

The value defined as noise pressurelevel is specified in the unit dB(decibel).

Acoustic signals with an equal acousticpressure but different frequencies arereceived differently by the human ear.A sine tone of 200 Hz, for example, isnot considered as loud as a sine tone

of 2000 Hz with the same noise level.Several methods can adjust the rmsvalues to those heard by the humanear. The most common method is theA-weighting. The various frequencieshave correction values in dB, which areadded or subtracted from therespective rms values that aremeasured. For example, 10.9 dB aresubtracted from the sine at 200 Hz and

1.2 dB are added at the sine at2000 Hz. The noise pressure levelsthat are evaluated with the A-weightingare then specified in dB(A).

At 1000 Hz the correction value is

exactly zero. Therefore a signal with1000 Hz and an acoustic pressure(RMS) of N/m

2 is usually used for

calibration measurements. Theacoustic pressure corresponds with93.9794 dB or dB(A).

The signal can be filtered with acombination of highpass and lowpassfilters for the A-weighting. Such analogfilters are often integrated in themeasurement amplifiers. It is alsopossible to use digital filters aftermeasuring data.

A further option is to separate thesignal into individual frequencies withthe FFT. The frequencies can usesmoothing functions with therespective correction values. Theconversion is often executed after thirdand octave analyses because tablesfor the evaluation are standardized forthirds and octaves.

Documents

FFT Use in DIAdem