## Waterfall Analysis: Frequency Resolution and Smearing

When measuring noise and vibration in rotating machines, especially complex devices like automobile engines, it is very important to fully understand what is being measured and what analyses need to be performed on the resulting data. Even the most sophisticated measurement systems will not produce meaningful results if used incorrectly. The testing of noise & vibration during development can be important factors in a final design. Poor measurement or analysis can have huge consequences on the success of a new vehicle.

Are you using the optimal sample rate for your test? Will your rate of change of rotational speed lead to the most accurate results? Read on to find out more.

## Introduction

Waterfall analysis is a tool that is often used to evaluate the acoustic and/or vibratory performance of rotating machinery.  The process involves calculating spectra using fast Fourier transform (FFT) methods at incremental steps in rpm (revolutions per minute) as the rotational speed changes – either increasing (run up) or decreasing (coast down).  Because it takes time to sample enough data to fill a time block for a single FFT calculation (the actual calculation time is minimal), the rotational speed of the machine being tested will have changed from the beginning of the time block to the end of the time block. The changing of the rpm during the time required to capture each time block produces a phenomenon known as smearing of the data.  This article addresses the causes and effects of the smearing phenomenon to help the reader understand the considerations that must be addressed when acquiring and analyzing rotational data.

## Data Capture

The quality of the results obtained from a waterfall analysis is very dependent on the quality of the data acquisition.  To decide what parameters are important during the acquisition, such as…

• signal sampling frequency
• rpm sweep rate
• frequency resolution
• and so on…

you must pay attention to how you will analyze or process the data.

The first consideration is to decide what the rpm range is and what the highest order is that you want to analyze. These two values establish the frequency range (and by implication the sampling rate) required for the processing.  If the rpm sweeps from 600 rpm to 6000 rpm then the fundamental rotation frequency is 10 Hz (600 rpm/60 sec) to 100 Hz (6000 rpm/60 sec). It is assumed that you have a basic understanding of the workings of the test specimen being measured and are aware of the potential generators of vibration and/or sound and the likely rotation frequencies that the test piece will generate.

Example: A 6 cylinder internal combustion engine with accessories.

• The firing frequency will be 3 times the fundamental rotation frequency (1st order)
• Knowing that the valve train cam runs at ½ speed of the fundamental rotation speed, one expects ½ order and multiples of ½ order signal generation.
• With an alternator with a 6:1 pulley ratio, one would expect a 6th order source along with multiples of 6th order depending on how many poles the alternator has.
• An AC compressor might have a 2.5 pulley ratio and a 6 piston compressor, which might result in 2.5th order, 15th order plus multiples.

Using the highest order of interest being generated and the highest fundamental rotation frequency, you can calculate the highest frequency of interest. The sampling frequency is also based on this information.

If you have control over how fast the sweep up or coast down happens then it is always best to make this as long as reasonably possible.  However, varying the sweep time is not always an option, such as in a vehicle test when the specified operation is a wide open throttle operating mode. Figures 1 and 2 show the waterfall results with two different rates of increasing rpm.

Figure 1: Representing rpm (1st order rate of change) corresponding to 150 rpm/second)

Figure 2: Representing rpm (1st order rate of change) corresponding to 1200 rpm/second

It is clearly seen by observing the 3rd order, that when the sweep rate of the acquisition is more rapid the spectral energy is “smeared” over frequency.  The higher the order observed, the faster the rate of change of frequency/second.

## Frequency Domain Analysis

The next consideration is to decide on the optimal frequency resolution.  The process of calculating a spectrum using FFT techniques dictates the amount of time required to achieve a specific resolution – the actual capture time  is equal to the inverse of the desired analysis frequency resolution (frequency line spacing).

Example:

• If you need a frequency resolution of 1 Hz then the time required to capture enough data (time block) to achieve this resolution is 1 second.
• If you need a 2 Hz frequency resolution then you need only a 500 ms time block of data.

During the capture of this time block the rotational speed of the machine being measured (assuming a constantly sweeping speed) will be constantly changing.  If you capture the data from 600 rpm to 6600 rpm over a 10 second interval, then the rpm sweep rate (sometimes called slew rate) will be 6000 rpm/10 second interval or 600 rpm/second:

• The fundamental rotation frequency (1st order) will be changing at 10 Hz/second.
• The 3rd order will be changing at 30 Hz/second
• The 5th order be changing at 50 Hz/second

Assuming that you capture the data and analyze it to meet 1 Hz frequency resolution, the energy calculated for the 5th order frequency component of the signal will be present in 50 lines of the frequency spectrum.

For this reason, when you know that the rate of change of rpm is high, you must take care when selecting the frequency resolution for analysis .

Plotting a single spectrum (in this case the spectrum equivalent to 3000 rpm) and looking at the amplitude at the 5th order (nominally 250 Hz) it is obvious that the amplitudes of the peaks are different.  This is due to a combination of signal “smearing” (caused by the frequency of the signal changing during the capture of the time block) and “picket fence effect” or leakage. The width of these same peaks also becomes narrower (fewer frequency spectral lines) as the frequency resolution is decreased (1, 2, and 4 Hz).  The calculated RMS energy for each of these 3 analyses is shown in the plot legend.

Figure 3: Showing the order width for varying frequency resolution/frequency line spacing

## Summary

The discussion above shows the relationship between the test rpm sweep rate and the analysis frequency resolution/frequency line spacing; it becomes a trade-off between these 2 parameters.  The user typically wants the highest resolution (smallest frequency line spacing) but this requires a longer time block for calculating the FFT.  During the time required to acquire this time block, the rpm may change significantly, which results in spreading of the energy across multiple, adjacent frequency lines.

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#### John Mathey

Senior Technical Specialist at Prosig
John Mathey graduated with a MS degree from the University of Toledo in 1972. John has over 35 years of experience with instrumentation, measurement, and analysis. Twenty-five of those years were spent at Ford Motor Company solving and providing training for vehicle noise, vibration, and harshness (NVH) issues. He is now a technical specialist at Prosig USA, Inc. where he provides technical support to Prosig customers in the U.S.A.

## 2 thoughts on “Waterfall Analysis: Frequency Resolution and Smearing”

1. tatta seshu raghavacharyulu

Greetings Mr. John

Very nicely presented the water fall explanation.

For beginners please write some thing more clearly the approach of analysis to water fall compared to FFT signal analysis based on frequency.
Thanks
Tatta

Dear Mr John

Thank you for this clear explanation.

my question is: how to choose a proper sweep rate (i.e rpm/sec) for a rotor consists of a disc if I want to determine the resonance frequency by using a run-up method? and what should be speed step to waterfall the signal spectra? Also, the actual capture time (T-span)

These are the information of the system:

max speed = 3000 rpm (50 Hz)
# order = 5
frequency resolution = 0.5

expected first resonance frequency = 39 – 44 HZ

the minimum sweep rate of the rotor that I can control is 10 rpm/sec.

I look forward to hearing a response from you

Regards