COMPUTATIONAL METHODS IN ENGINEERING AND SCIENCE
EPMESC X, Aug. 21-23, 2006, Sanya, Hainan, China
©2006 Tsinghua University Press & Springer
Earthquake Response Analysis and Energy Calculation Based on Wavelet
Transform
Chen Wu *, Ruizhong Zhou
College of civil engineering and architecture, Fuzhou University, Fuzhou, Fujian, 350002 China
Email:Wuchen2001@21cn.com
Abstract In earthquake engineering, Fourier transform is a main way for spectrum analysis in the past. However,
signal is requested to be stationary and the result couldn’t offer time-frequency local character. Therefore, Fourier
transform isn’t suitable for dealing with seismic signal. Wavelet transform is a perfect way to analyze non-stationary
signal for its multi-resolution nature. Any seismic signal can be decomposed to some frequency channels by
multi-resolution analysis. Using this method, earthquake responses of MDOF system are deduced in this paper and
seismic energy & displacement energy in distinct frequency channel are discussed. With examples on MDOF system,
an argument that in elastic system total earthquake response can be obtained by adding all of the dynamic response
contributions in distinct frequency channel is demonstrated and the reconstructed signal after restraining high
frequency is applied to approximate model. With energy distribution, signal and response contributions of distinct
frequency channel are analyzed. At the end of this paper, some problems about wavelet transform are discussed.
Key words: Wavelet Transform, Multi-Resolution Analysis, Earthquake Responses, Energy Calculation
INTRODUCTION
In earthquake engineering, the most constructions have their natural modes and natural frequencies at elastic stage. For
seismic signal, frequency-domain analysis sometimes is more convenient than time-domain analysis [1]. In the 20th
century, Fourier transform is a main way for frequency spectrum analysis, but signal is requested to be stationary and
the result couldn’t offer time-frequency local characteristics. As we know, seismic signal is quite non-stationary and
time-frequency relationship just is the focus of people’s attention. Therefore, Fourier transform isn’t suitable for
dealing with seismic signal.
In recent years, wavelet transform has become a popular way to analyze non-stationary signal for its multi-resolution
nature. With multi-resolution analysis (MRA), various frequency components in a signal are decomposed into
different subspaces and dynamic response and energy distribution in distinct scale can be obtained. Hence wavelet
transform is an effective method to deal with seismic wave.
THEORETICAL ANALYSES
1. Multi-resolution analysis (MRA) MRA is the soul of wavelet transform. It is a spatial sequence {V j }
j∈Z
that
monotonicity, approximation, scaling, shifting and Riesz basis existence are satisfied in L2 ( R ) [2]. From a certain
MRA, a wavelet functionψ (t ) and its wavelet space {W j }
j∈Z
can be confirmed and L2 ( R) = ⊕ W j is held. As for
j∈Z
any f (t ) ∈ L ( R ) , there is
2
f (t ) =
∞
∞
∑ ∑d
j =−∞ k =−∞
ψ j ,k (t ), j , k ∈ Z
(1)
j ,k
where d j ,k =< f (t ),ψ j ,k (t ) > is a discrete wavelet transform for f (t ) , ψ j ,k (t ) = 2− j / 2ψ (2− j t − k ) is a binary wavelet
⎯ 1134 ⎯
function. Owing to the orthogonality of {ψ j ,k (t )}
j , k∈Z
,
{d }
j ,k
j , k∈Z
is orthogonal wavelet transformed and (1) is the
reconstruction for f (t ) .
For any j ∈ Z , W j is defined as the orthogonal complement space of V j in V j −1 ：
V j −1 = V j ⊕ W j & W j ⊥ W j ′ （ j ≠ j ′ ）
(2)
Figure 1: A tree corresponding to MRA
V j is the low-frequency component in V j −1 and it reflects signal’s “profile”. W j is the high-frequency component
and it reflects signal’s “detail”. As is shown in Figure1, a signal is decomposed step by step with MRA and the
previous low-frequency component is the only object for current decomposition. Where S is original signal, Aj and
Dj denote low-frequency signal and high-frequency signal in scale j respectively. The relation among them is
j
S = Aj + ∑ Dm [3], viz. any function f (t ) ∈ L2 ( R) can be reconstructed by low-frequency component whose
m =1
resolution is 2− j and high-frequency components whose resolution are 2− m (1 ≤ m < j ). Above idea just is famous
Mallat tower restructing algorithm：
j
f (t ) = f j (t ) + ∑∑ d m ,kψ m,k (t )
(3)
m =1 k∈Z
Table 1: Frequency distribution in MRA
Frequency channel
Frequency range
Frequency channel
Frequency range
A1
0-0.5
D1
0.5-1
A2
0-0.25
D2
0.25-0.5
A3
0-0.125
D3
0.125-0.25
A4
0-0.0625
D4
0.0625-0.125
In MRA, scale, resolution and frequency are interrelated. When scale increase, frequency resolution is falling and we
can get low-frequency characteristics. On the contrary, when scale is reduced, frequency resolution is rising and
high-frequency characteristics are got. If the highest frequency component is assumed as 1 and the maximum scale is
4, the frequency distribution can be seen as Table1.
2. Application of MRA in earthquake response analysis of MDOF system In the given references [4, 5], wavelet
transform is generally applied to signal-degree-of-freedom (SDOF) system. The feasibility of application in
multiple-degree-of freedom (MDOF) system is brought forward in reference [4]. In this paper, earthquake responses
of MDOF system with MRA are to be analyzed.
As for a dynamic system of N-DOFs linear construction, the differential equation of motion is expressed as:
[ M ]{ &&
x(t )} + [C ]{ x& (t )} + [ K ]{ x(t )} = −[ M ]{1} a&&g (t )
(4)
where [M], [C], [K] denote mass matrix, damping matrix and stiffness matrix respectively；{&x&} ,
acceleration, velocity and displacement of structural particle respectively;
{x&} , {x} are relative
T
{1} = {1, 1, ......, 1}N ×1 ; a&&g (t ) denotes
ground acceleration due to earthquake.
Using MRA, a seismic signal can be decomposed as:
a&&g (t ) = ∑ a&&gj (t )
(5)
j
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where a&&gj (t ) is signal component in scale j, which can be gained by Mallat algorithm. Then substitute (5) into (4),
we can get:
[ M ]{ &&
x(t )} + [C ]{ x& (t )} + [ K ]{ x(t )} = −∑ [ M ]{1}a&&gj (t )
(6)
j
If mode-superposition method is used, the displacement of particle i can be written:
N
N
k =1
k =1
xi (t ) = ∑ X ki qk (t ) = ∑ γ k X kiδ k (t )
(7)
where k is the number of vibration mode; qk (t ) is generalized coordinates; γ k is participating coefficient of mode;
X ki is amplitude; The displacement of equivalent SDOF system δ k (t ) can be obtained by Duhamel integral:
δ k (t ) =
1
ωd
t
∫ a&& (τ )e
0
−ξ k ωk ( t −τ )
g
sin ω d (t − τ )dτ
(8)
where the natural circular frequency for damping system ωd is to be defined by: ωd = ω 1 − ξ 2 .
Substitute (5) to (8), δ k (t ) can be given:
δ k (t ) =
1
ωd
∑ ∫ a&&
t
j
0
gj
(τ )e −ξk ωk ( t −τ ) sin ω d (t − τ ) dτ = ∑ δ jk (t )
(9)
j
Therefore, the displacement of particle i in a N-DOFs system is to be computed:
xi (t ) = ∑∑ γ k X kiδ jk (t ) = ∑ x ji (t )
j
k
(10)
j
It can be seen from the above equation that the total earthquake response can be obtained by adding all of the
dynamic response contributions in distinct frequency channel. If elastic time-history analysis is used to solve
equation (6), the same result can be drawn (example 1).
In addition, some superfrequency components in seismic signal are considered useless to dynamic response analysis
[6]. If these components are filtered, data size will be compressed, calculation will be simplified and signal noise
ratio will be raised. With wavelet analysis, a possibility of signal filtering is provided. When reconstructing a signal,
high-frequency coefficients in one scale or several scales are not inclusive so that the reconstructed signal doesn’t
contain these high-frequency components. Example 2 indicates that the reconstructed signal can be used in
approximate calculation and the precision is reliable.
3. Energy analyses The energy of original seismic signal is defined by [7]:
∞
E = ∫ a&&g (t ) 2 dt
(11)
−∞
With MRA, (5) is substituted into (11) as follow:
∞
E = ∫ [∑ a&&gj (t )]2 dt
−∞
j
∞
∞
= ∑ ∫ a&&gj (t ) 2 dt + 2 ∑ ∫ a&&gm (t )a&&gn (t )dt
j
−∞
m≠n
−∞
(12)
The orthogonality between two wavelet functions leads to the following relation between a&&gm (t ) and a&&gn (t ) :
∫
∞
−∞
a&&gm (t )a&&gn (t )dt = 0
( m ≠ n)
(13)
Hence,
∞
E = ∑ ∫ a&&gj (t ) 2 dt = ∑ E j
j
−∞
(14)
j
Similarly, the displacement energy for floor i can be written:
∞
∞
−∞
−∞
Q = ∫ xi (t ) 2 dt = ∫ [∑ x ji (t )]2 dt
j
∞
∞
−∞
−∞
= ∑ ∫ x ji (t ) 2 dt + 2 ∑ ∫ xmi (t ) xni (t )dt
j
m≠n
If (9), (10) and (13) are taken into account together, it’s easy to get:
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(15)
∫
∞
−∞
xmi (t ) xni (t )dt = 0
( m ≠ n)
(16)
Therefore, displacement energy can be obtained as follow:
∞
Q = ∑ ∫ x ji (t ) 2 dt = ∑ Q j
j
−∞
(17)
j
NUMERICAL EXAMPLES
1. Example 1: Earthquake response analysis with wavelet transform A ten-floor frame building is subjected to
70gal ElCentro wave. The duration is 10s and sampling period is 0.02s.The mass in every floor is 2.5×105kg.
Young’s modulus of concrete is 3.0×1010. Other parameters about each floor are shown in Table2. Now interlaminar
shear modulus is used, which is shown in Fig. 2.
Figure 2: Computational model
Table 2 Characteristic parameters of a building
Floor
F1
F2-F3
F4-F6
F7-F10
Height(m)
4.2
3.0
3.0
3.0
0.5×0.5
0.5×0.5
0.4×0.5
0.4×0.4
Column section
(m×m)
“db5” is regarded as the wavelet function by right of experience. N = 5 is chosen as the maximum scale because the
first natural frequency for this construction is 0.5438 Hz and Nyquist frequency of this signal is 25 Hz. As is shown
in Fig. 3(a), the original seismic wave is decomposed to six frequency channels with MRA (from low frequency to
high frequency).The center frequency of A5 is 0.3906 Hz, which is similar to the first natural frequency. By applying
six signal components to the building respectively, the top displacements due to wavelet components can be
calculated with elastic time-history analysis (Fig. 3(b)). Finally, all of the displacement contributions are superposed
and the total displacement is compared with the value due to original wave (Table 3).As illustrated in Table 3, two
results is quite alike. Furthermore, the top velocity and top acceleration are all calculated with wavelet transform and
results are same as the values due to original signal. Consequently an argument that in elastic system earthquake
responses can be obtained by adding all of dynamic response contributions of distinct frequency channel is
demonstrated.
In addition, as can be seen from Fig. 3, every picture describes the time information about one frequency channel so
that the relationship between time and frequency can be found, which could not be realized by Fourier transform.
2. Example 2: Application of MRA in approximate calculation In example 1, D1 and D2 are two highestfrequency components. In order to reduce data and develop computational efficiency, the two components are to be
filtered. In program, we can replace their wavelet coefficients with zero and then reconstruct signal with
“db5”wavelet function according to new wavelet coefficients. Top displacements result from original signal and
reconstructed signal are shown in Table4 respectively.
From Table 4, we can see that displacements due to two signals are fairly accordant and relative error is less than
±1% . It’s indicated that high-frequency components can be removed exactly with wavelet analysis and structural
dynamic properties still can be reflected after high-frequency components are removed. Therefore, this method is
effective and safe to dynamic approximate analysis.
⎯ 1137 ⎯
(a)
(b)
Figure 3: Wavelet components of distinct frequency channel and corresponding displacements
Table 3 Comparison on top displacements calculated by two ways (mm)
time
Wavelet
transform
Traditional
Method
t = 1s
t = 2s
t = 3s
t = 4s
t = 5s
t = 6s
t = 7s
t = 8s
t = 9s
t = 10s
3.3925
−2.435
20.9156
−10.8755
9.8214
−20.8445
23.8404
−31.0877
36.7573
−33.0123
3.3925
−2.435
20.9156
−10.8755
9.8214
−20.8445
23.8404
−31.0877
36.7573
−33.0123
Table 4 Comparison on top displacements with seismic signals of two kinds (mm)
time
t = 1s
t = 2s
t = 3s
t = 4s
t = 5s
Top displacement due to reconstructed signal
3.4101
−2.4585
20.9231
−10.9212
9.8963
Top displacement due to original signal
3.3925
−2.435
20.9156
−10.8755
9.8214
Relative error
0.519%
0.965%
0.0359%
0.420%
0.763%
time
t = 6s
t = 7s
t = 8s
t = 9s
t = 10s
Top displacement due to reconstructed signal
−20.9046
23.8444
−31.1283
36.842
−33.0105
Top displacement due to original signal
−20.8445
23.8404
−31.0877
36.7573
−33.0123
Relative error
0.288%
0.017%
0.131%
0.230%
−0.005%
3. Example 3: Energy analyses A three-floor frame building is subjected to 200gal ElCentro wave. The duration is
10s and sampling period is 0.02s.The mass in every floor: m1 = 2726kg, m2 = 2760kg, m3 = 2300kg, interlaminar
stiffness: k1 = 2.485×105N/m, k2 = 1.921×105N/m, k3 = 1.522×105N/m.
We still consider “db5” as the wavelet function. Because the first natural frequency for this construction is 0.6532Hz,
N = 4 is chosen as the maximum scale so that five wave bands are created. Then energy distributions of seismic
signal and displacement signal in distinct frequency channel can be calculated. (Fig. 4)
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Figure 4: Energy distributions of seismic signal and displacement signal in distinct frequency channel
Table 5 Displacement energies of distinct frequency channel and original signal
Floor
1
2
3
Channel A4
1903.20
7827.60
14247.00
Channel D4
147.25
186.23
255.66
Channel D3
5.38
3.50
2.55
Channel D2
0.07
0.08
0.10
Channel D1
0.0016
0.0027
0.0039
Sum of displacement energies
2056.91
8017.41
14505.31
Displacement energy due to original signal
2131.8
8477
15298
Relative error
3.5%
5.4%
5.2%
It can be seen from Fig. 4 that Elcentro wave has great components in the first three channels and the contributions
are decreasing in latter channels. Displacement energy of each floor is mainly focus on the first channel, so the first
channel (A4) has important influence on structural displacement and it is the key band for structural control. On the
contrary, high-frequency components have little effect on responses as reference [6] said.
In addition, from Table 5 we can see that the sum of displacement energy contributions is less than the energy due to
original signal. It is because the wavelet function is not a perfect filter and it isn’t infinite in length that energy in one
wave band is possible to leak to other channels after passing through the filter. The given materials show that
leakage rate is to decrease with increasing time-domain length of filter [8].
DISCUSSION ON WAVELET TRANSFORM
(1) Using wavelet transform, high resolutions can be obtained both in time-domain and in frequency-domain, which
is better than Fourier transform. However, wavelet transform essentially is Fourier transform with flexible
time-frequency window so that it doesn’t get rid of the limitation of Fourier transform thoroughly [9].
(2) The variety of wavelet basis make wavelet transform adaptable enough so that it is possible to construct or
choose optimal wavelet basis according to signal characteristics. However, once the basis is confirmed, it could not
be changed during decomposition and reconstruction. Hence it is possible for the basis to be the best in global but
the worst in local. Now it’s still a difficult problem to select an optimal wavelet basis [10]. Experience and
experiment are general way at present.
(3) Different wavelet basis has different properties and has different analytic ability to signal. In calculation,
different result is to be drawn if different basis is used and these results could not compare with each other.
(4) Wavelet basis is finite in length so that energy leaking is inevitable. However, if basis which is long enough in
time-domain length is selected, leakage rate will decrease [8].
⎯ 1139 ⎯
CONCLUSIONS
(1) MRA has favorable time-frequency localization. Using MRA, seismic signal can be decomposed to several
frequency channels and time information is to be gained. With this method, dynamic parameters in each channel can
be obtained and controlled conveniently. These parameters provide basic materials for structural control.
(2) For SDOF and MDOF elastic system, dynamic responses due to original seismic wave can be obtained by
superposing all of response contributions of each wave band.
(3) Using MRA, high frequency components in a signal can be restrained and signal’s dynamic properties still can be
represented by reconstructed signal. This method can be applied to approximate calculation.
(4) Theoretically, the sum of energy contributions of distinct channel equals to the energy due to original wave. In
calculation, finite length of wavelet basis and all kinds of energy loss will lead to energy leaking, but a small
quantity of error could not influence correctness in analysis and practicability in engineering.
(5) Wavelet transform is an effective way to analyze non-stationary signal, but it still is based on Fourier transform
so that further work should be made in fitting non-stationary signal. At present, it still is a puzzle to choose an
appropriate wavelet basis in application.
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Jiang JG, Zhou XH, Zou YH, Duan SW. Application of wavelet analysis to degree of single freedom dynamic
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