ANALYSIS OF ERRORS IN THE CONVERSION OF ACCELERATION INTO DISPLACEMENT

Transcription

1 ANALYSIS OF ERRORS IN THE CONVERSION OF ACCELERATION INTO DISPLACEMENT Sangbo Han, Joog-Boong Lee Division of Mechanical Engineering Kyungnam University 449 Wallyoung-dong, Masan, 63 I-70, Korea ABSTRACT 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 displacement signal. It turned out both the method 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. 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 INTRODUCTION 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. 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. 408,

2 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. 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, 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 sec., the time and frequency resolution of the signal is fixed, which are 2037 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@). 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). 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 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 IY OS Fig. I Tie histories of theoretical and converted of 20 Hz using frequency domain method. -sol -IIY y.u I Fig. 2 Time histories of theoretical and converted of 20.3 Hz using tiequency domain method.

3 E" Time (set ) Fig. 3 Time histories of theoretical and converted displacement Corn an acceleration signal of 20 Hz using time domain method. and the inverse discrete Fourier transform N-l a, = c A, ej(2nkr k=o is given by A r=o,l,2;..,(n-i) (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 2. From the properties ofthe Fourier transform of the integrals, the discrete Fourier transform of the velocity and the displacement signals are given as v, =- j2nk A, k=o,l,2;..,(n-i) I D, A, k=o,l,2;..,(&i) (4).,, 0 I 0 00s ; I- - (b) Retrieved displacement T!me (sec.) Fig. 4 Time histories of theoretical and converted of 800 Hz using frequency domain method. 3. CONVERTING ACCELERATION INlW DISPLACEMENT JN FREQUENCY 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,,2 /..., (N-I) () 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. N-l v, = If,, ej(2dd4r) r=o,l,2;..,(n-i) (5) k=o h -, d, = cdk dc2nb N) r=o,i,2;..,(n-i) (6) k=o The error involved in the transformation of the velocities and displacements are from the scale factor of I / j2rrr and -I l(2rrk) in Eq. (5) and (6). Let s explain the error with the results of the signal given in Fig.. 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 40

4 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 CONVERTING ACCELERATION SIGNAL INTO DISPLACEMENT SIGNAL IN TITHE DOMAIN The velocity and displacement of the signal can be obtained by directly integrating the acceleration signal in time domain using the following definition. v(t) = x n(t)dt + v. (7) [;a- -(a) Theoretical s : x 9 loo lod0 displacement ,000 - (b) Retrieved displacement loo Frequency (Hz) Fig. 5 Absolute value of Fourier coefficients of theoretical and converted displacement signal of 20 Hz. d(r) = 6 v(t)dt + do (8) Here va and d, are the initial velocity and initial displacement, respectively. The first source of error occurred in the conversion process is due to the time resolution of the digitized acceleration signal. Bias error of the numerical quadrature using the trapezoidal rule to convert the acceleration into velocity is given as [3] o-5t J ,000 - (b) Retrieved displacement E =J&j(r) 2 O+r<At I2 For a pure sinusoidal signal, the relationship acceleration and the velocity is given as (9) between the ii(t) = -(2&)3v(t) (0) where f0 is the signal frequency. And the Nyquist frequency of the measurement is determined by the sampling resolution At as fm=& () ,000 Frequency (Hz) 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 measurement is determined as follows. Therefore, the relative error in evaluating the velocity from the acceleration of a pure sinusoidal signal is given as (2) 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 less than & and within % of error, the signal frequency should be less than O.l57Of,, 4

5 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. E E o -2L I Fig. 7 Time histories of theoretical and converted of 20 Hz with inaccurate initial condition. 5. CONVERSION OF COMPLEX SIGNAL 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 the measurement. To convert a complex signal with 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, Hz, 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. 0. 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. 2. This may be due to the noise components that are not zero padded (a) Theoretical displacement (::w[ - (a) Theoretical displacement E 0.5 c Time (set ) t 000 Frequency (Hz) Fig. 8 Time histories of theoretical and converted displacement Tom an acceleration signal of 800 Hz with inaccurate initial condition. Fig. 9 Absolute value of Fourier coefficients of theoretical and converted displacement signal with 3 frequency components. 42

6 6. CONCLUSIONS _ OS I- (bj Retrieved disrhcement OS Fig. 0 Time histories of theoretical and converted 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 [ ] McConnell K. G., Vibration Testing, Theory and Prucfice, John Wiley & Sons Tnc, New York, 995. [2] Newland D. E., An Introduction to Random Vibrations, Spectral & Wmeler Analysis, 3rd. ed. John Wiley & Sons Inc., New York, 993. [3] Burden R. L., Faires, J.D. and Reynolds A.C., Numerical Analysis, Prindle, Weber & Schmidt, 979. Fig. Time histories of theoretical and converted displacement fi-om an acceleration signal with zero padded Fourier coefficients. 2 E Time (set ) Fig. 2 Absolute value of Fourier coefficients of theoretical and converted and zero padded displacement signal with 3 frequency components 43