Frequency Fundamentals

Frequency Fundamentals

In Machinery Vibrations

Ali M. Al-Shurafa, Vibrations EngineerSaudi Electricity Company- Ghazlan Power PlantSaudi Arabia

1. Definition of FrequencyPeople are familiar with frequency more when it is mentioned in the context of sound. You can

recognize high frequency sounds and describe them as “acute”. Sounds with low frequencies are saidto be “dense”. Actually, people recognize individuals from their voices, which is simply the frequencyof their voices when they talk. What more exciting is normal people perform “fault diagnosis” on otherpeople. When you are sick, your friends know you are so from the “frequency of your voice”. Invibration, the concept is similar!

Frequency is a measure of the occurrence rate of a periodic event. Vibration signals havefrequency because they are periodic. The event, in the vibration signals, is the movement of the bodyback and forth. So, basically, frequency is the repeatability of an event per unit time.

For a plot of vibration in time domain (time waveform), x-axis represents the time elapsing andy-axis reflects the value (amplitude) of the vibration. One cannot the extract the frequency of thesignal, from this plot, directly. But for a simple (single) waveform, the frequency can be calculated.

1.1 From Time to Frequency DomainWith the aid of Fast Fourier Transformation, FFT, same information of the previous plot is

transferred to another domain, frequency domain. Hence, the x-axis in the new plot is frequencyinstead of time. This plot is called frequency spectrum or simply spectrum.

The typical units of time, in the time waveform plots, is millisecond, msec. The frequency inthe spectrum has many units like Hz and cpm. More details will be covered in the coming sections.

The complex wave, in the figure above, is composed of three simple sine waves. Each sinewave has its own frequency, while overall wave cannot be defined completely by any one of these (orother) frequencies. FFT decomposes the complex wave to its simple sine wave components that havedistinct frequencies.

1.2 Sinusoidal WavesComplex waves (with multiple different sine waves) can have a frequency but usually

frequency is referred to harmonic (or simple sinusoidal) waves. A sinusoidal wave has a single

frequency. With only one frequency and one amplitude, the sinusoidal wave is completely defined,unlike the complex waveform signals.

The relation between the period and frequency for a sinusoidal wave is:

F= 1/ p

where F= wave frequency, Hzp= complete sinusoidal wave period, second.

This relation holds only for sinusoidal of filtered signals. Remember, there is no singlefrequency for unfiltered signals.

1.2 Example 1Find the frequency, in Hz, of the sine wave plotted below. Time unit is msec.

Solution: First find the period of a complete cycle and in this case p = 6.5 msec.Convert the time units into second. Then

F(Hz)= 1/p =1/(6.5/1000) =153.8 Hz.

2. Frequency UnitsFrequency can be expressed independently from the machine speed. This frequency can be

described as absolute and the units in this case are Hz, cps, rad/sec and cpm. If the vibration frequencyis compared to the speed of the machine, the vibration frequency is expressed in orders.

2.1 Hz or cpsMr. Hertz, French Scientist, defined this unit. It is the basic unit for frequency and is defined as

Hz = 1/secIt is used in general vibration. Hz is not used directly in the trigonometric functions with the

standard form, see next paragraph. If one considers a cycle to be unitless, then for simplicity he cansay 1/sec= cycle/sec. Hence,

cps = HzWhere cps stands for cycle per second.

2.2 rad/secThis unit is more found in the scientific literature and typically used in the trigonometry

equations. For instance,

D(t) = d sin(F*t)

where D = running displacement amplitude, mt = the time (independent variable), secd = maximum displacement amplitude in the signal, mF= signal frequency, rad/sec

Remember, all the trigonometric functions are evaluated at an angle (which is the argument betweenthe parentheses). The unit of angles is radians. So the units inside the parentheses must be rad. Tocompensate the seconds coming from the time ,t , the correct units of F must be rad/sec.

[rad] = angle θ = (F*t) = [rad/sec*sec] = [rad]From basic mathematics,

2 *pi*rad = 360 deg or rad = 57.3 deg

That means if the frequency is 1 rad/sec, each second the object will rotate 57.3 deg or 16% ofa complete cycle. Refer to the figure below.

2.3 cpmThis is the commonly used unit in the machinery vibration because of its analogy with rpm,

speed of the machine. cpm is cycle per minute. The relation between cpm and cps is the conversionfactor from minute to second. Explicitly,

cpm = 60 * cps = 60 * Hz

2.4 OrderWhen the vibration frequency is considered with respect to the rotational speed (running

frequency) of the machine, frequency is expressed in a dimensionless ratio called order. Usually orderis expressed in this format: 1X, 7X, 0.48X etc.

Frequency order, n X= Vibration Frequency / Machine Speed

Note that both vibration frequency and machine speed must be consistent (of same units). Theuse of the order as the unit of frequency is extremely important in the fault diagnosis.Order is the slope of the vibration lines in the Cascade Plots. Harmonics are the integer multiples of aparticular frequency, usually 1X. Harmonics and orders are similar but not exactly the same.

2.4 Conversion Table

Hz cps cpm rad/sec orderHz 1 1 60 2pi -cps 1 1 60 2pi -cpm 1//60 1//60 1 2pi / 60 -

rad/sec 1/2pi 1/2pi 60/2pi 1 - order - - - - 1

Table is used to convert the units in the left-side column to the topside row by multiplying theoriginal quantity by the value in the table.One cannot find the order of the vibration unless the object has a rotational speed and the speed isknown.

2.5 ExamplesExample 3 : A 4-pole motor has a vibration frequency of 3550 cpm. Calculate the vibrationfrequency in the following units: a) Hz b) rad/s c) cps d)order.

Solution:a) F(Hz) = F(cpm)/60 = 3550/60 = 59.2 Hzb) F(rad/sec) = F(cpm)*2pi/60 = 371.7 rad/secc) F(cps) = F(Hz) = 59.2 cpsd) A 4-pole motor has a synchronous speed of 1800 rpm. Then,

Frequency Order = 3550/1800 ≈ 1.97 X

Example 4: Calculate the frequency of the first example in the following units: a) cps b) cpm c) rad/sd)orderSolution:

a) F(cps) = F(Hz) = 153.8 Hzb) F(cpm) = F(cps) *60 = 9228 cpmc) F(rad/sec) = F(cps)*2pi = 966.3 rad/secd) Order cannot be found without knowing the speed of the machine.

3. Natural FrequencyVibrations as a response could have a frequency of either the excitation frequency (fault force

frequency) or a frequency of the structure (natural frequency). Natural frequency is a physicalcharacteristic of any structure, regardless of the forces applied on the structure. It is function ofstiffness available in the structure and mass distribution. In vibration analysis, natural frequency isimportant for many reasons. One of them is to avoid the problem of resonance, when both excitationfrequency and natural frequency coincide.

There are several methods to find the natural frequency. Each method is appropriate fordifferent applications. These methods include:

1. Analytical solution to differential equations,2. Numerical solution to differential equations (e.g. finite element and finite deference

methods)3. Physical testing (e.g. hummer test).

First method is rarely implemented in field. It requires mathematical modeling of the machineand advanced mathematical procedures to have the solution (natural frequency of the machine).

The second method is performed with the aid of a computer and special software like ANSYS.The structure is modeled in the software as a series of small elements comprising the actual structure.The software is fed with data about the structure (mass distribution, stiffness properties, boundaryconditions, forces etc.). The computer will solve the physical equations for the small elements andlater will combine these solution “somehow” and end up with the solution of the structure as a whole.

The third method is applied in the field occasionally. An input (force) is applied on the system(machine) and the response (vibration) is measured and analyzed to find the natural frequency.Impulse (or impact) force is imposed on the machine usually by a hammer while a data collector ismounted on the machine to measure the response.

In all the three methods, vibration analyst must know the basic equation of the naturalfrequency. For the simplest case, the system can be modeled as a spring-mass system with mass, m,and stiffness, k.

Mk=ω where ω in rad/sec, k in N/m and M in kg

The physical meaning of this concept is illustrated by the following examples.

1. When a machine is loose, its stiffness decreases and hence (from the equation) itnatural frequency decreases.

2. The natural frequency of an accelerometer is much higher compared to the naturalfrequency of velocity transducer. One reason for this is the mass of theaccelerometer is much less compared to that of a velocity transducer.

3. Long pipes are provided with supports to reduce the resonance vibration. Thesupports act as a stiffener. Consequently, stiffness increases and natural frequency sodoes.

When a resonance problem occurs, the solution usually is detuning the excitation and naturalfrequencies away enough from each other.

4. Fault FrequenciesFaults generate forces with certain frequencies. Frequencies of many common faults can be

predicted by calculations. For simple analysis there are charts showing fault frequencies and somecomments. Many of these charts should be considered as guidelines because they were developedbased on experience in machines covered by the person who developed these charts. The actualfrequencies depend heavily on the machine design and that is why a single fault can have differentfrequencies for different designs. Or that is why there are several different charts in their contents.Next is presented a summary chart with same basic faults. Then, rolling element-bearing frequenciesare given as an example of cases with different calculations based on design information available.Refer to fault diagnosis articles for more details.

4.1 Summary Table

Serial No. Fault Frequency1 Imbalance 1X2 Misalignment 1X, 2X, 3X3 Gear Mish Frequency

(GMF= rpm*No of teeth) 1GM, 2GM, 3GM

4 Vane Pass Frequency(VPF= rpm *No. of vanes) 1VPF

5 Motor electrical Problems 2Line Frequency (2LF)6 Rolling Element Bearings FTF ,BPFI, BPFI, BSF

4.2 Rolling Element Bearing FrequenciesThere is more than one set of equations to calculate the frequencies associated with rolling

element-bearing problems. Based on the available bearing information and design, the calculationaccuracy improves. These abbreviations are used with all the sets.

where N = Shaft Speed in Hzn = Number of ElementsD = Pitch Diameter of Rolling Elementsd = Rolling Element Diameterβ = Contact Anglefi =Inner Race Rotational Frequencyfo= outer Race Rotational FrequencyB= Diameter of Rolling ElementsP= Pitch Diameter of Rolling Elements

Formula Set 1If no information about the bearing is available, good approximations of bearing tone

frequencies for most common bearings are as follows:

Outer Race Fault (Order) = n * RPM * 0.4

Inner Race Fault (Order) = n * RPM * 0.6

Fundamental Train Frequency (Order) = 0.4 * RPM

Formula Set 2If contact angle is available, use this set.

Formula Set 3

If the contact angle is unknown or equal to zero, use the following formulas.

Formula set 4There is also a set of general formulas for cases where motions of both races are possible.

FTF = (fi/2)(1-B/P cosβ ) + (fo/2)(1+B/P cosβ )BPFI = (N/2) fo-fi (1+B/P cos β)

BPFI = (N/2) fo-fi (1-B/P cos β)

BSF = (P/2B) fo-fi [1-(B/P)2 cos2 β ]

5. Frequency Ranges

5.1 For Fault DetectionAs stated earlier vibration faults are associated with particular frequencies. In practice, these

frequencies may appear in addition to their harmonics. Sometimes, the fault frequency itself is notimportant as its harmonics. For example, it is normal to see an amplitude at the gear mish frequencybut you should not have harmonics of the gear mish frequency. For this reason, the frequency spanneeded to investigate the machine is wider than the frequency of the fault itself. The table shownunderneath summarizes and exemplifies typical spans for typical problems.

5.2 For Measure SelectionVibration amplitudes (displacement, velocity and acceleration) are referred to as measures.

Because they are related to each other through differentiation or integration, a plot of a particularmeasure can be re-plotted with another measure. For example, the figures below are for the same dataacquired but the amplitude measure is changed from displacement to velocity and to acceleration.

Notice that amplitudes at higher frequencies are magnified. This is the effect of frequency!

An amplitude in a spectrum will be clearly identified only if the spectrum is plotted in thecorrect measure. If the spectrum measure is not selected correctly, important amplitudes might beunderestimated. Use the figures below as guidelines to select the proper measure for the spectrum.

The Figures below illustrate the significant effect of the frequency on the amplitude of a signalwhen it is integrated or differentiated. In all the plots, w (rad/sec) is the frequency as the runningvariable.

Original signal is collected from an accelerometer and filtered to a single frequency of 10rad/sec. Notice that amplitude of the signal is 3 (say in/sec2 peak ).

When this signal is integrated to convert it from a acceleration to velocity, the amplitude isshortened.Recall: velocity = v = )*10cos(*10/3)10sin(3( tdttadt −== ∫∫

Similarly, the velocity is integrated in order to find the vibration as a displacement measure.Signal amplitude reduces even further.

5.3 For Vibration Pick-ups SelectionEach type of pick-ups works best in certain frequency range. Use the coming figures to select

the proper type. There are many considerations to be taken when selecting pick-up type, vibrationfrequency expected to be monitored is a major consideration.

6. Frequency Spectrum

6.1 ResolutionResolution is the ability of data collector to detect two amplitudes, close to each other in the

frequency spectrum, as distinct amplitudes (at different frequencies). Higher resolution gives precisevalues of frequencies. The importance of frequency resolution is appreciated when two signals, veryclose to each other, are collected and interpreted as single signal, by mistake. The resolution must behigh enough to distinguish the two signals and to give accurate frequencies. Resolution and frequencyhave same units. 150 cpm is typical resolution. The equation to calculate the resolution is

Min. Resolution = WF * Fmax/(2*Lines)

Where WF = the window factor, UnitlessFmax = maximum frequency in the frequency spectrum, HzLine = Total number of lines in the frequency spectrum, unitless.

Example 5: A 2-pole motor, with 100-rpm slip, came from maintenance after rewinding. Afterinstallation, motor had 2X vibrations shown by a poor resolution spectrum. Two possible causes forthis symptom could be: misalignment and electrical problem with 2LF. Another vibration reading witha correct resolution is required. Calculate the frequency resolution required to detect the faultfrequency.

Solution:Motor speed is n = 3600-100=3500 rpm2LF= 2*60 =120 Hz =120*60 = 7200 cpm2X frequency (due to misalignment) = 7000cpm.The analyzer must be able to pick up signal with frequency difference less than 200 cpm or 3.3

Hz. A good resolution can be half the frequency difference or less, say 50 cpm.

6.2 Frequency FiltersIn vibration analysis, filters are used to allow signals of certain frequencies and forbid signals

with other frequencies.Filters that allow signals only with low frequencies are called Low Pass Filters, while filters

that allow signals only with high frequencies are called High Pass Filters. Examples are given in thefigures below. When both low and high pass filters are combined, they generate a Band Pass Filter,which allow only signals with a specific range of frequency to pass.

Data collectors are not capable of collecting or processing signals with all frequencies. Sofilters are always used. Examples are Anti-aliasing Filter, Cut-off Frequency Filter and MaxFrequency Filter.

Default setting of common data collectors demonstratesCut-off frequency of 300 cpmMax. Frequency of 60000 cpm

This is like a bandwidth filter (300-60000cpm). This range is quite reasonable for routinecondition evaluation and basic diagnosis. Usually frequencies below or above these limits aregenerated from noise (not machine related).

6.3 Full Spectrum and Negative Frequency

Full spectrum is generated by collecting data from two probes (horizontal and vertical) in thesame time. The x-axis and y-axis are frequency and amplitude respectively. In the conventional (half)spectrum frequency starts from zero and increases. In the full spectrum the frequency can increase inboth positive and negative directions. But the meaning of negative frequency is that the rotor vibrates

at that frequency in the opposite direction of shaft spinning. So, the positive frequency componentsindicate forward precession while the negative components indicate backward (or reverse) precession.It is not possible to generate half spectrum from full spectrum and the opposite is not possible also.Most malfunctions show forward precession. Rubbing in a fluid film bearing will have negativefrequency.

7. ReferencesTodd Reeves, Basic Signal Processing for Vibration Data Collection.Bob Cecil, Vibration Primer.M960 help system (from SKF).Paul Goldman ,Application of full spectrum to rotating machinery diagnostics. Orbit.Wolfson Maintenance.Charles Jackson, The Practical Vibration Primer.