ANALYSIS
OF ERRORS IN THE CONVERSION
ACCELERATION
INTO DISPLACEMENT
Sangbo Han,
OF
Joog-Boong Lee
Division of Mechanical Engineering
Kyungnam University
449 Wallyoung-dong,
Masan, 63 I-70 1, Korea
ABSTRACT
1 INTRODUCTION
It is sometimes necessary to get the velocities and
displacements of the structure when the structural responses
are measured with accelerometers, There are two methods,
in general, to convert the acceleration signal into the
It turned out both the method
displacement signal.
produced a signiticant amount of errors depending on the
sampling resolution in time and frequency domain to
digitize the response signals. It is well known that to have
better resolution in time domain, one has to compromise
with the coarse resolution in frequency domain and visa
verse with fixed number of sampling. Therefore, with a
predetermined resolution in time and frequency domain,
converting high frequency signals in time domain and
converting low frequency signals in frequency domain will
produce biased errors. An effective way to convert the
acceleration signal into the displacement signal without
significant errors are studied here with the analysis on the
errors involved in the conversion process.
Accelerometers are the most frequently used transducers to
measure the vibration responses of the structures, and
information on the amplitudes, frequencies, and phase
differences of the measured accelerations are usually the
main objectives of the signal analysis involved in the
structural vibration test. Sometimes it is necessary to
retrieve the measured signals in the form of velocities and
displacements in cases such as active control of the structure.
While it is quite easy to extract the information on the
frequency components and corresponding amplitudes of the
velocities and displacements of the measured acceleration
signals, it is not an easy task to retrieve the time history of
the structural responses in the form of velocities and
displacements.
NOMENCLATURE
A, : Discrete Fourier coefficient of acceleration signal
D, : Discrete Fourier coefficient of displacement signal
V, : Discrete Fourier coefficient of velocity signal
X(f) : Fourier transform of displacement signal
a(t) : Time history of acceleration signal
0, : Time array digitized from a continuous acceleration
signal
E” : Relative error in evaluating the velocity from an
acceleration signal
Jo : Signal ti+equency
f* : Nyquist frequency of the measurement
There are two methods, in general, to convert the
acceleration signal into the displacement signal. One is
directly integrating the acceleration signal in time domain.
The other is dividing the Fourier transformed acceleration
signal by the scale factor of - w2 and taking the inverse
Fourier transform of it. It turned out both the method
produced a significant amount of errors depending on the
sampling resolution in time and frequency resolution to
digitize the response signals. It is well known that to have
better resolution in time domain, one has to compromise
with the coarse resolution in frequency domain and visa
verse with given number of sampling points. Therefore,
with a fixed resolution in time and frequency domain,
converting high frequency signals in time domain and
converting low frequency signals in frequency domain will
produce biased errors. The errors involved in the converting
process of the acceleration signal into displacement signal
are stated, and an effective way to convert the acceleration
signal into the displacement signal without significant errors
are studied here.
1408
,
2. STATEMENT
OF THE PROBLEM
We start our problem by defining the acceleration signal
measured with a digital signal analyzer with a fixed number
of digitization.
Therefore, the sampling resolution of the
measured signal in both time and frequency domain depends
on the record time T. The objective is to convert this
acceleration signal into displacement signal with the
assumption
that the acceleration signal is the best
representation of the structural response obtained by the
digital-analog converter of the analyzer.
domain method can be applied for both high and low
frequency signals. On the other hand, the frequency domain
method is not good for the signals with leakage. The direct
integration method can retrieve low frequency signals
whether they have leakage or not, but the method does not
work well for the signals with relatively high frequency
signals. Let’s examine the reason for the errors that are
involved in the converting procedure.
Considering following four different pure sinusoidal signals
can check the errors involved in the converting process.
The frequencies of the signals are 20 Hz, 20.3 Hz, 800 Hz,
800.3 Hz, respectively. It is assumed that all of the signals
were measured with a digital signal analyzer that has 2048
sampling points. Since the record time is fixed to be 1 sec.,
the time and frequency resolution of the signal is fixed,
which are 112037 second and I Hz, respectively.
Figure l(a) represents a time history of a single frequency
sinusoidal signal representing a theoretical displacement
signal. The frequency of the signal is 20 Hz and the total
record time T=lsec. and the signal is digitized with 2048
sampling points. Since the frequency of the signal is an
integer multiple of the frequency resolution, therefore, there
is no leakage in the measured signal [I].
Theoretically
generated acceleration signal of this sinusoidal signal is
converted into displacement signal by taking the Fourier
Transform of it and dividing the each frequency components
by the scale factor of --w* and taking the inverse Fourier
transform of it. We can see the original signal is nicely
converted into displacement without any signal distortion as
in Fig. I@).
1
-1 IY
0
0.4
02
0.4
Time
06
08
1
06
OS
1
1
0
(sec.)
Fig. I Tie histories of theoretical and converted
displacement from an acceleration signal
of 20 Hz using frequency domain method.
Now, we take exactly same procedure to convert
acceleration signal of frequency 20.3 Hz into displacement.
In this case, the converted displacement signal is totally
different from the original signal as in Fig. 2 (a) and (b).
1
Next, we convert both the 20 Hz and 20.3 Hz acceleration
signals into displacement signals by directly integrating
them in time domain. As we can see in Fig. 3, integrating
the signals in time domain nicely retrieve the corresponding
displacement signals. If the frequency of the acceleration
signal is increased to 800 Hz as in Fig. 4, the retrieved
signal fails to express rapidly changing peak values of the
signal.
From the results of converting single frequency acceleration
signal into displacement signal discussed above we can
draw following conclusion. When there is no leakage in the
signal, even though the condition can seldom be satisfied in
the real situations, time history of displacement signal of
given acceleration signal can be retrieved by using
frequency domain method.
In this case, the frequency
02
-IIY
0
’
’
y.u
0.2
0.4
06
0.8
1
0.2
0.4
Time
0.6
0.8
1
I
-1sol
0
(sec.)
Fig. 2 Time histories of theoretical and converted
displacement from an acceleration signal
of 20.3 Hz using tiequency domain method.
1409
and the inverse discrete Fourier transform is given by
1
N-l
a, = c A,eJ(2nkr’
A”
k=O
-1
0
0.2
0.4
02
0.4
06
0.6
0.6
06
E"
Time
(set
(2)
It is important to note that although the discrete Fourier
transform given in Eq. (I) does not provide enough
information to allow the continuous time series a(t) to be
obtained, it does allow all the discrete values of the series
{a,} to be retrieved exactly 121.
1
0
r=O,l,2;..,(N-I)
From the properties ofthe Fourier transform of the integrals,
the discrete Fourier transform of the velocity and the
displacement signals are given as
1
)
Fig. 3 Time histories of theoretical and converted
displacement Corn an acceleration signal
of 20 Hz using time domain method.
1
A,
j2nk
v, =-
I
D, zz...@~I2
k=O,l,2;..,(N-I)
A,
k=O,l,2;..,(&I)
(4)
Time histories of the velocities and the accelerations are
obtained by taking the inverse Fourier transform of the
coefticients in Eq. (3) and (4) as follows.
.,, 0 I
10 00s
. ’
0 01
0.025
0 0;
0.015
I-
v,
-
(b) Retrieved
displacement
=
N-l
If,,
eJ(2dd4r)
r=O,l,2;..,(N-I)
(5)
r=O,I,2;..,(N-I)
(6)
k=O
h’-,
d, = cDk
dc2nb’N)
k=O
-1
0
0 005
0 01
T!me
0015
(sec.)
0 02
0 025
The error involved in the transformation of the velocities
and displacements are from the scale factor of I / j2rrR and
-I l(2rrk)’
Fig. 4 Time histories of theoretical and converted
displacement from an acceleration signal
of 800 Hz using frequency domain method.
3. CONVERTING
DISPLACEMENT
ACCELERATION
JN FREQUENCY
INlW
DOMAIN
Suppose that the continuous time history of acceleration
response of the structure a(t) is not known and only equally
spaced samples are available. This acceleration signal is
represented by the discrete series {ur], r = 0, I, 2,..., (N-l),
where / = Y. At.
The discrete Fourier transform of the
series {a, ) is given by
-J(2*‘N)
k=0,1,2
/..., (N-I)
(1)
in Eq. (5) and (6).
Let’s explain the error with the results of the signal given in
Fig. 1. Theoretically, the Fourier transform of the single
frequency signal is delta function of S(f -fO)
where f0 is
the frequency of the signal. Therefore, theoretically, all the
other Fourier coefficients
are zero except at the
corresponding frequency value. But due to the digitization
error, each Fourier coefficients actually has some small
value, and when the Fourier coefficient of the frequency
component of the signal is divided by the correction factor,
there appears distortion in Fourier coefficients along the
tiequency axis as shown in Fig. 5. The amount of distortion
is much severer when there is leakage in the measured
signal as in Fig. 6, in which case the difference between the
maximum value of the Fourier coefficient and the minimum
value of the coefficients are relatively small so that the scale
factor plays significant roll in converting the acceleration
into displacement.
Suwose that the measured signal has
signal to noise ratio of 60 dB, which is common in practice,
then the discrete Fourier coefficients of the frequency
1410
component at high value of k would be divided by the
factor of -(2~k)~
and can be decreased by more than 60
dB. This will cause high frequency component appears less
than the low frequency noise components and the
displacement signal appears to have very big low frequency
components, which will distort the converted displacement
as shown in Fig. 2.
(a)
[;aTheoretical
0
200
loo
v(t) = x n(t)dt + v.
(7)
d(r) = 6 v(t)dt + do
(8)
The first source of
is due to the time
signal. Bias error
trapezoidal rule to
given as [3]
E =J&j(r)
2
I2
lOd0
0
200
600
(Hz)
600
1000
1o-5t
J
0
200
400
600
(b) Retrieved
600
,000
displacement
(9)
between the
(10)
And the Nyquist
where f0 is the signal frequency.
frequency of the measurement is determined by the
sampling resolution At as
fm=&
displacement
and initial
200
ii(t) = -(2&)3v(t)
‘loo
Frequency
-
For a pure sinusoidal signal, the relationship
acceleration and the velocity is given as
,000
Fig. 5 Absolute value of Fourier coefficients
of theoretical and converted displacement
signal of 20 Hz.
error occurred in the conversion process
resolution of the digitized acceleration
of the numerical quadrature using the
convert the acceleration into velocity is
O+r<At
(b) Retrieved
600
x
x
9
The velocity and displacement of the signal can be obtained
by directly integrating the acceleration signal in time
domain using the following definition.
velocity
-
600
s:
4. CONVERTING
ACCELERATION
SIGNAL INTO
DISPLACEMENT
SIGNAL IN TITHE DOMAIN
Here va and d, are the initial
displacement, respectively.
400
displacement
(11)
Therefore, the relative error in evaluating the velocity from
the acceleration of a pure sinusoidal signal is given as
(12)
400
Frequency
600
(Hz)
600
,000
Fig. 6 Absolute value of Fourier coefficients
of theoretical and converted displacement
signal of 20.3 Hz.
and the relationship between the frequency of the signal to
be analyzed within a certain amount of error and the
Nyquist frequency of the measurementis determined as
follows.
f.
= 3 3,?
=0.7287&f@
r
For example, if we want to evaluate the velocity by directly
integrating a sinusoidal acceleration signal within 5% of
error, the signal frequency should be lessthan 0.2685&
and within 1% of error, the signal frequency should be less
than O.l57Of,,
1411
The second source of error comes Tom the fact that there is
no information available on the initial conditions involved
with each integration scheme. Uncertain value of initial
velocity will produce a dc component during the successive
integration of the conversion process as shown in Figs. 7
and 8. One of the methods to eliminate the error due to the
uncertain initial value of the signal is filtering out the dc
component in every integration scheme or extrapolating the
acceleration signal to find out appropriate initial velocity.
1
E
E
0
-1
0
02
0.4
0.6
0.6
1
0.2
0.4
Time
0.6
0.6
1
2
E
E
o
I
-2L
0
(sec.)
Fig. 7 Time histories of theoretical and converted
displacement from an acceleration signal of
20 Hz with inaccurate initial condition.
1.5
-1
-
(a) Theoretical
5. CONVERSION
OF COMPLEX
As stated above, both the frequency domain method and
time domain method have certain amount of errors
depending on the frequency component of the signal. The
frequency domain method works well with the acceleration
signal measured without leakage. On the other hand, the
time domain method works well with the acceleration signal
whose frequency is well below the Nyquist bequency. But
these conditions are seldom satisfied in real situation. In
practice, structural response consists of both high and low
frequency component signals and leakage always happens in
To convert a complex signal with
the measurement.
minimum error in the frequency domain, there must be
some procedure to minimize
the leakage of the
measurement. The effect of the leakage on the conversion
error comes from the fact that noise components can
become significant after the Fourier coefficients are divided
by the scale factor of -CD’ Zero padding the noise signal
can reduce the effect of the conversion factor. Fourier
coefficients of a theoretical displacement signal with multifrequency component of 80.3 Hz, 400.3 Hz, 600.3 Hz and
those of converted displacement signal are shown in Fig. 9.
As expected, the low frequency noise components are
significant, and inverse Fourier transform of these
coefficients will produce a totally different displacement
signal as in Fig. 10. Zero padding the low frequency
components of the signal such as in Fig. I I retrieves the
displacement signal nicely except the starting point of the
signal as in Fig. 12. This may be due to the noise
components that are not zero padded.
(::w[
displacement
-
1
SIGNAL
(a) Theoretical
displacement
E
0.5
c
0.005
0.01
Time
(set
0.015
)
0.02
0 025
Fig. 8 Time histories of theoretical and converted
displacement Tom an acceleration signal of
800 Hz with inaccurate initial condition.
0
200
400
Frequency
600
(Hz)
600
t 000
Fig. 9 Absolute value of Fourier coefficients
of theoretical and converted displacement
signal with 3 frequency components.
1412
6. CONCLUSIONS
0.01
0 02
I-
_.__
0
0.01
0.03
(bj Retrieved
0 02
Time
0.04
disrhcement
0 03
(sec.)
0 04
0 OS
1
0 OS
Fig. 10 Time histories of theoretical and converted
displacement from an acceleration signal
without zero padded Fourier coefficients
When there is no leakage in the signal, even though the
condition can seldom be satisfied in the real situations, time
history of displacement signal of given acceleration signal
can be retrieved by using frequency domain method. In this
case, the frequency domain method can be applied for both
high and low frequency signals. The frequency domain
method is not good for the signals with leakage, and
unfortunately leakage always exists in the measured signals.
On the other hand, the direct integration method can retrieve
low frequency signals whether they have leakage or not, but
the method does not work well for the signals with
relatively high frequency components. Zero padding the
Fourier coefficients for the noise components using the
t?equency domain method appears to be the best in
converting a more complex acceleration signal into
displacement signal.
REFERENCES
[ 1] McConnell K. G., Vibration Testing, Theory and
Prucfice, John Wiley & Sons Tnc, New York, 1995.
[2] Newland D. E., An Introduction to Random Vibrations,
Spectral & Wmeler Analysis, 3rd. ed. John Wiley & Sons
Inc., New York, 1993.
[3] Burden R. L., Faires, J.D. and Reynolds A.C., Numerical
Analysis, Prindle, Weber & Schmidt, 1979.
Fig. 11 Time histories of theoretical and converted
displacement fi-om an acceleration signal
with zero padded Fourier coefficients.
2
E”
-2
0
-0
0.005
0 01
0 005
Time
0 01
(set
0.015
0.02
0 015
0.02
)
Fig. 12 Absolute value of Fourier coefficients
of theoretical and converted and zero
padded displacement signal with 3
frequency components
1413