Materials for Ijima seminar: Chap. 5 June 22. 2020

5-1

5. Transforming seismic accelerograms into the Fourier finite series 5.1 Advantages of using Fourier series for computing the seismic responses of structures A trigonometric series uses the series of the functions of sine as well as cosine, and a Fourier series is the trigonometric series expressing an optional periodic function. Transforming a periodic function into the Fourier series is called the Fourier expansion or the Fourier transformation. The Fourier series of the periodic function x(t) with the period T is expressed as follows,

x(t) = akcos

2!kT

t + bksin2!kT

t"#$

%&'

k=0

(

) . (5.1.1)

The equation (5.1.1) is the infinite series, whereas computing the Fourier expansion really uses a finite series. Transforming a seismic record of acceleration into the Fourier series uses the number of the terms of the Fourier series decided from the number of data. Fig. 5.1 is the seismic accelerogram recorded at the Kobe Meteorological Agency in the Hyogo-ken Nanbu earthquake 1995. The number of the data of the accelerogram is 516 that the time interval between the data is 0.02 second. The Fourier series is derived from transforming the periodic function of the wave of Fig. 5.1 as a period. Fig. 5.2 is the Fourier spectrum showing the amplitude of the sinusoidal wave in the Fourier series obtained. The amplitude is a

k

2+ b

k

2 given by the coefficients in the equation (5.11), and the lateral axis is the frequency f = k /T . Transforming a physical phenomenon varying with time into the Fourier series

Time (s)

Acceleration (gal)

The max. of acc. -820.56gal

Fig. 5.1 Seismic wave of acceleration recorded at the Kobe Meteorological Agency in the Hyogo-ken Nanbu earthquake 1995.

Frequency (Hz)

(gal)

Fig. 5.2 Fourier spectrum of the seismic wave of acceleration in Fig. 5.1.

Fourier spectrum

ak

2+ b

k

2

5-2

means changing a way of observing the phenomenon from on a time axis to on a frequency axis, so that the phenomenon can be deeply understandable. For example, though Fig. 5.1 does not clearly show the approximate frequency of the principal wave in the earthquake, Fig. 5.2 clearly shows that the principal wave is large in the frequency area around 2 Hz. Another advantage of the Fourier transformation from seismic accelerograms is the followings. i) Computing inelastic seismic responses needs interpolating the time interval between recorded data of seismic accelerograms in order to keep the computational accuracy. The Fourier series changed from a seismic accelerogram is the continuous function, so that a time interval can be freely optional with the frequency characteristic unchangeable independently. For example, in the case of the linear interpolation between the recorded data, the less the interpolating interval is, the larger the components in high frequency area are. ii) Since the term of force in the equation of the motion becomes the functions of sine and cosine, the obtained solution is exact without using any assumption in the case of linear differential equations. 5.2 Transforming seismic accelerograms into the Fourier finite series When the seismic accelerogram is x(t) of the periodic function with the period T , the Fourier finite series is derived as follows. When the number of the data is N and when the time interval is !t , the period T is, T = N!t . (5.2.1) Further, giving each datum of the acceleration the numbering according to the order of time is, as follows, x

m= x(t

m),

m = 0,1,2,!!,N !1 , (5.2.2)

where, t

m= m!t . (5.2.3)

The number of the unknown coefficients ak, b

k are not infinite, but the number is decided

from the number N of the conditions shown by the equation (5.2.2). Determining the number of the terms as 0 to N / 2 and then substituting the equations (5.2.1) to (5.2.3) into the equation (5.1.1) gives,

xm= a

kcos

2!kmN

+ bksin2!kmN

"#$

%&'

k=0

N /2

( , (5.2.4)

where N is defined as an even number. The equation (5.2.4) includes k = 0 , so that the number of the unknown coefficients is N + 2 . However, the terms of k = 0 and k = N / 2 become as follows,

a0

cos2! • 0 •m

N+ b

0sin

2! • 0 •m

N= a

0 "

a0

2, (5.2.5)

aN /2cos

2! • N / 2( ) •m

N+ b

N /2sin2! • N / 2( ) •m

N= a

N /2cos

2! • N / 2( ) •m

N

! aN /2

2cos

2" • N / 2( ) •m

N, (5.2.6)

where dividing the coefficients a0

andaN /2

by 2 respectively is the reason why the expression determining the coefficients of cosine becomes consistent. The two equations show that the two in the unknown coefficients decrease, so that the number N of the unknown coefficients in the following equation can be decided.

5-3

xm=a0

2+ a

kcos

2!kmN

+ bksin2!kmN

"#$

%&'

k=1

N /2(1

) +aN /2

2cos

2! • (N / 2)•m

N (5.2.7)

In order to determine the coefficients, the equation that the order letter k in the equation (5.2.7) changes to l is defined as follows,

xm=a0

2+ a

lcos

2!lmN

+ blsin2!lmN

"#$

%&'

l=1

N /2(1

) +aN /2

2cos

2! • (N / 2)•m

N. (5.2.8)

Multiplying both sides of the equation (5.2.8) by cos 2!kmN

gives,

xmcos

2!km

N=a0

2cos

2!km

N+ cos

2!km

Nalcos

2!lm

Nl=1

N /2"1

#

+ cos2!km

Nblsin2!lm

Nl=1

N /2"1

# +aN /2

2cos

2!km

Ncos

2! • (N / 2)•m

N. (5.2.9)

Summing up respectively each side of the equations (5.2.9) changing from m = 0 to m = N !1 gives,

xmcos

2!km

Nm=0

N "1

# =a0

2cos

2!km

Nm=0

N "1

# + cos2!km

Nalcos

2!lm

Nl=1

N /2"1

#m=0

N "1

#

+ cos2!km

Nalsin2!lm

Nl=1

N /2"1

#m=0

N "1

# +aN /2

2cos

2!km

Ncos

2! • (N / 2)•m

Nm=0

N "1

# . (5.2.10)

The equation (5.2.10) indicates summing up on m after summing up on l . Since the result by changing the order from l to m of the sums is the same, Summing up on l after summing up on m gives,

xmcos

2!km

Nm=0

N "1

# =a0

2cos

2!km

Nm=0

N "1

# + al

cos2!lm

Ncos

2!km

Nm=0

N "1

#l=1

N /2"1

#

+ bl

sin2!lm

Ncos

2!km

Nm=0

N "1

#l=1

N /2"1

# +aN /2

2cos

2! • (N / 2)•m

Ncos

2!km

Nm=0

N "1

# . (5.2.11)

Changing the products of the trigonometric functions into the sum of them gives,

xmcos

2!km

Nm=0

N "1

# =a0

2cos

2!km

Nm=0

N "1

# +al

2cos

2! (l + k)m

Nm=0

N "1

# + cos2! (l " k)m

Nm=0

N "1

#$%&

'()l=1

N /2"1

#

+ bl

2sin2! (l + k)m

Nm=0

N "1

# + sin2! (l " k)m

Nm=0

N "1

#$%&

'()l=1

N /2"1

#

+ aN /22

cos2! • (N / 2 + k)•m

Nm=0

N "1

# + cos2! • (N / 2 " k)•m

Nm=0

N "1

#$%&

'()

. (5.2.12)

The right side of the equation (5.2.12) consists of the trigonometric finite series of all the terms. In the same way of deriving the equation (5.2.12), multiplying both sides of the equation

(5.2.8) by sin 2!kmN

, summing up respectively each side in the equations changing from m = 0

to m = N !1 , and then summing up on l after summing up on m gives,

xmsin2!km

Nm=0

N "1

# =a0

2sin2!km

Nm=0

N "1

# + al

cos2!lm

Nsin2!km

Nm=0

N "1

#l=1

N /2"1

#

5-4

+ bl

sin2!lm

Nsin2!km

Nm=0

N "1

#l=1

N /2"1

# +aN /2

2cos

2! • (N / 2)•m

Nsin2!km

Nm=0

N "1

# (5.2.13)

Changing the products of the trigonometric functions into the sum of them gives,

xmsin2!km

Nm=0

N "1

# =a0

2sin2!km

Nm=0

N "1

# +al

2sin2! (l + k)m

Nm=0

N "1

# " sin2! (l " k)m

Nm=0

N "1

#$%&

'()l=1

N /2"1

#

+ bl

2! cos

2" (l + k)m

N+

m=0

N !1

# cos2" (l ! k)m

Nm=0

N !1

#$%&

'()l=1

N /2!1

#

+ aN /22

sin2! • (N / 2 + k)•m

Nm=0

N "1

# " sin2! • (N / 2 " k)•m

Nm=0

N "1

#$%&

'()

. (5.2.14)

The unknown coefficients can be decided from concretely seeking the trigonometric finite series in the equations (5.2.12) and (5.2.14) separately in the cases of k = 0 , k = N / 2 , and k except them. The following formulas are useful for computing the trigonometric finite series. Deriving the equations (5.2.15) is shown in the following Supplements.

cosrxr=0

n!1

" = 2sinnx

2cos

(n !1)x

2sin

x

2 (5.2.15a)

sin rxr=0

n!1

" = 2sinnx

2sin(n !1)x

2sin

x

2 (5.2.15b)

i) k = 0 : The equations (5.2.12) and (5.2.14) change to,

xmcos

2! • 0 •m

Nm=0

N "1

# =a0

2cos

2! • 0 •m

Nm=0

N "1

# +al

2• 2 cos

2!lm

Nm=0

N "1

#l=1

N /2"1

#

+bl

2• 2 sin

2!lm

Nm=0

N "1

#l=1

N /2"1

# +aN /2

2• 2 cosm!

m=0

N "1

# , (5.2.16a)

0 = 0 +al

2• 0

l=1

N /2!1

" +bl

2• 0

l=1

N /2!1

" +aN /2

2• 0 . (5.2.16b)

Changing r to m as well as n to N , x to 2!l / N in the equation (5.2.15) concretely gives zeros to the second term, the third term and the fourth term in the right side of the equation (5.2.16a), as follows,

cos2!lmNm=0

N "1

# = 2sinN

2•2!lN

$%&

'()cos

N "12

•2!lN

$%&

'()sin

l!N

= 2sin l! cosN "12

•2!lN

$%&

'()sin

l!N

= 0 ,

(5.2.17a)

sin2!lmNm=0

N "1

# = 2sinN

2•2!lN

$%&

'()sin

N "12

•2!lN

$%&

'()sin

l!N

= 2sin l! sinN "12

•2!lN

$%&

'()sin

l!N

= 0 ,

(5.2.17b)

cosm!m=0

N "1

# = 2sinN!

2cos

(N "1)!

2sin

!

2= 0 . (5.2.17c)

This is because, since l is one or more in the equations of (5.2.17a) and (5.2.17b), sin l! = 0 and sin!l / N " 0 . Further, since N is even, sinN! / 2 = 0 . Therefore, the first term in the equation (5.2.16a) remains, so that a

0 becomes as follows,

xm

cos2! • 0 •m

Nm=0

N "1

# =a

0

2N $ a

0=

2

Nxm

cos2! • 0 •m

Nm=0

N "1

# . (5.2.18)

5-5

ii) k = 1 to N / 2 !1 : The concrete results of the equations of (5.2.12) and (5.2.14) in the case of l ! k are quite different from that in the case of l = k . (a) The case of l ! k : The trigonometric finite series in the two equations of (5.2.12) and (5.2.14) can be computed from using the equation (5.2.15). This case makes the five numbers of k , l + k , l ! k , N / 2 + k and N / 2 ! k the integer except zero. Therefore, like the equation (5.2.17), all the terms in the right sides in the two equations are zero. (b) The case of l = k : Since l ! k is zero, the trigonometric finite series of cosine is not zero in the two equations of (5.2.12) and (5.2.14) but the term is one, and then the sum is N . Namely, a

konly in the equation (5.2.12) has the nonzero series, as follows,

xm

cos2!km

Nm=0

N "1

# =N

2ak $ a

k=

2

Nxm

cos2!km

Nm=0

N "1

# . (5.2.19)

In the equation (5.2.14), bk has the nonzero series, and it is,

xm

sin2!km

Nm=0

N "1

# =N

2bk $ b

k=

2

Nxm

sin2!km

Nm=0

N "1

# . (5.2.20)

iii) k = N / 2 : In the equation (5.2.12), since the maximum of l is N / 2 !1 , l ! k consists in any l . Therefore, the finite series of the first to third term are zero, and the fourth term remains, as follows,

xm

cos2! (N / 2)m

Nm=0

N "1

# =aN /2

2cos

2! • 0 •m

N $

m=0

N "1

# aN /2

=2

Nxm

cos2! (N / 2)m

Nm=0

N "1

# . (5.2.21)

Both sides in the equation (5.2.14) become zero, The coefficients a

k, b

kin the equation (5.2.7) are expressed in one equation each by unifying

the equations of (5.2.18) to (5.2.21),

ak=

2

Nxm

cos2!km

Nm=0

N "1

# (k = 0,1,2,!,N

2) , (5.2.22a)

bk=

2

Nxm

sin2!km

Nm=0

N "1

# (k = 1,2,!,N

2"1) . (5.2.22b)

Supplements (1) Deriving the equation (5.2.15) The trigonometric finite series of the equation (5.2.15a) is defined as S ,

S = cosrxr=0

n!1

" = cos0 • x + cos x +!+ cos(n ! 2)x + cos(n !1)x (5.3.1)

Multiplying the equation (5.3.1) by sin(x / 2) and changing the products into the sums leave the two terms and delete the other terms, and then changing the sum of the two terms into the one term of the product gives,

S sinx

2= cos0sin

x

2+ cos x sin

x

2+!+ cos(n ! 2)x sin

x

2+ cos(n !1)x sin

x

2

= sin 0 +x

2

!"#

$%&' sin 0 '

x

2

!"#

$%&+ sin x +

x

2

!"#

$%&' sin x '

x

2

!"#

$%&+!

5-6

+ sin n ! 2( )x +x

2

"#$

%&'! sin n ! 2( )x !

x

2

"#$

%&'+ sin n !1( )x +

x

2

"#$

%&'! sin n !1( )x !

x

2

"#$

%&'

= sinx

2+ sin n !

1

2

"#$

%&'x = 2sin

nx

2cos

(n !1)x2

. (5.3.2)

The equation (5.2.15a) is derived from the equation (5.3.2), as follows,

cosrxr=0

n!1

" = 2sinnx

2cos

(n !1)x

2sin

x

2. (5.2.15a)

Like the equation (5.3.1), the trigonometric finite series of the equation (5.2.15b) is S ,

S = sin rxr=0

n!1

" = sin0 • x + sin x +!+ sin(n ! 2)x + sin(n !1)x . (5.3.3)

Multiplying the equation (5.3.3) by sin(x / 2) and computing the derivation like the equation (5.3.2) gives,

S sinx

2= sin0sin

x

2+ sin x sin

x

2+!+ sin(n ! 2)x sin

x

2+ sin(n !1)x sin

x

2

= cos 0 !x

2

"#$

%&'! cos 0 +

x

2

"#$

%&'+ cos x !

x

2

"#$

%&'! cos x +

x

2

"#$

%&'+!

+ cos n ! 2( )x !x

2

"#$

%&'! cos n ! 2( )x +

x

2

"#$

%&'+ cos n !1( )x !

x

2

"#$

%&'! cos n !1( )x +

x

2

"#$

%&'

= cosx

2! cos n !

1

2

"#$

%&'x = 2sin

nx

2sin(n !1)x2

. (5.3.4)

The equation (5.2.15b) is obtained from the equation (5.3.4), as follows,

sin rxr=0

n!1

" = 2sinnx

2sin(n !1)x

2sin

x

2. (5.2.15b)

(2) Nyquist frequency The maximum frequency in the Fourier finite series of a seismic accelerogram is called the Nyquist frequency. The Nyquist frequency in the equation (5.2.7) is the frequency of the maximum of k , and then k

max= N / 2 gives the Nyquist frequency f

max, as follows,

fmax

=N / 2

T=N / 2

N!t=1

2!t. (5.3.5)

Seismological observations in Japan are normally recording seismic accelerograms with the time interval of 0.01 second or 0.02 second, so that the Nyquist frequencies are 50Hz or 25Hz. When the seismic response analysis uses Fourier transformation, the response does not include the component in the frequency area more than 50Hz or 25Hz. (3) Resolving power of Fourier spectrum The frequency interval of the lateral axis in the Fourier spectrum of Fig. 5.2 is 0.097Hz. The frequency interval !f is called the resolving power in a Fourier spectrum. The time length of a seismic accelerogram as one period in the Fourier transformation decides the resolving power !f , as follows,

!f = fk+1 " fk =k +1

T"k

T=1

T. (5.3.6)