Page 54 - Prosig Catalogue 2005

SOFTWARE PRODUCTS
FREQUENCY, HERTZ & ORDERS

any discussion on end effects. A section of the signal is shown in Figure
1 below.
Training & Support Figure 1 Time History of Two Sinewaves

Figure 6: H1(f) shown as modulus & phase
as shown below.
be seen through to the frequency response function. The DATS software If we FFT this composite signal then we get the modulus and phase plot
does, of course, provide a single step transfer function analysis. We have
deliberately used the long-hand form below to illustrate the steps in this
Condition Monitoring Figure 2 Standard FFT of Two Sinewaves
article.

As expected the amplitudes are 0.5 and 0.25 respectively – recall that
DATS for Windows gives half amplitudes. In the phase plot there is a 270°
phase change at 60 Hz and a 45° phase change (270° to 315° ) at 180 Hz.
Clearly the 45° phase change is as expected but why the 270° change?
Surely it should be 0° and 45° not 270° and 315° . The reason is because
Fourier analysis uses cosines and sines with the cosine, not the sine, for
the real part. A sine has a -90° or +270° shift relative to a cosine. In other
words the basis is a cosine wave not a sine!
Software All of the above is fairly basic signal analysis. Now suppose we have
a rotating shaft and we are measuring the vibrations of the shaft. The
nature of the rotating items is that the vibrations occur at multiples and
Figure 7: Complete DATS worksheet
at 3600rpm, which is 60 Hz, then we would expect to see responses at
It is necessary to understand that for the purposes of clarity in this article submultiples of the rotational speed. For example if the shaft is rotating
some important steps have been glossed over; windowing of the input multiples of this frequency. These multiples are the orders (or harmonics
for example. This is to assist in the basic understanding of the frequency in musical terms). First order is a frequency which is the same as the shaft
response function. rotational speed. In our example this is 60 Hz. The third order would be 3
* 60 = 180 Hz. The general relationship between the order, OR, the shaft
Frequency, Hertz & speed, R, in rpm, and the frequency, f, in Hz is
f = OR*(R/60)
Why use orders? The reason is of course that the order remains constant
Hardware Orders with shaft speed; first order is always at the shaft speed; second order
is always twice shaft speed and so on. This means we can step into the
The most common form of digitizing data is to use a regular time based
rotation and it is as if we were moving with the shaft. Instead of sampling
method. That is data is sampled at a constant rate specified as a number
at equal increments of time we sample at equal increments of rotation.
of samples/second. The Nyquist frequency, f , is defined such that f =
N
N
SampleRate/2. As discussed elsewhere Shannon’s Sampling Theorem
with the shaft rotational speed. Suppose we had a toothed wheel fixed to
tells us that if the signal we are sampling is band limited so that all the This is called synchronous sampling; we have synchronised our sampling
the shaft. Instead of a clock providing the command pulses to drive the
information is at frequencies less than f then we are alias free and have analogue to digital converter then pulses from each gear tooth will give us
N
a valid digitised signal. Furthermore the theorem assures us that we have equi angular or synchronous sampling at P samples/rev.
all the available information on the signal. We now have data which is sampled in units of a fraction of a rev rather
If we Fourier Analyze a signal, x(t), then we get its components expressed
than as a fraction of a second. If we Fourier transform this data we again
System Packages x(t). This is sometimes written in the form in increments of Orders not Hz. The result is that we now have a signal
at frequencies measured in Hz. This is completely reversible. That is if we
get a measurement as a function of a frequency type scale but now it is
have x(t) we can get X(f). Similarly if we have X(f) we can get back to
which gives the modulus and phase as a function of Orders rather than
as a function of Hz.
We do not have to sample synchronously to get orders because we can
As a simple example consider a test signal composed of two sinewaves.
use the relationship between frequency in Hz, f, order number, OR, and
The first one had an amplitude of 1.0, a frequency of 60Hz and a 0°
rotational speed, R. The procedure is then to FFT the time history, and by
phase. The second one had an amplitude of 0.5, a frequency of 180Hz
using the rotational speed to convert the frequency in Hz to a ‘frequency’
and a 45° phase. In this example we will sample at 2048 samples/second
in Orders. This is in principle fine for a constant speed but if the speed is
and acquire 2 seconds of data. The choice of 2048 samples/second is to
changing over the length of the FFT we have an incorrect result. Also it
54 ensure it exactly matches an FFT size. This is not necessary but it avoids
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