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{SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT 257 66 "Solving a 1-DOF System \+
Driven by Transient and Non-Periodic Forces" }}{PARA 258 "" 0 ""
{TEXT 256 29 "by Harald Kammerer, Germany, " }{TEXT -1 21 "maple@jadem
ountain.de" }}{PARA 259 "" 0 "" {TEXT -1 0 "" }}{PARA 259 "" 0 ""
{TEXT -1 106 "This worksheet compares the solutions of a single-degree
-of-freedom-system derived by different techniques" }}{PARA 259 "" 0 "
" {TEXT -1 50 "\nKEYWORDS: FFT, frequency domain, time domain, DOF" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "" {TEXT -1 181 "The au
thor expects that this worksheet will only be used for teaching and ed
ucational purposes and not for commercial profit without contacting th
e author for a licensed agreement. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}{PARA 0 "> " 0
"" {MPLTEXT 1 0 28 "with(plots):with(plottools):" }}}{EXCHG {PARA 3 "
" 0 "" {TEXT -1 12 "Introduction" }}{PARA 0 "" 0 "" {TEXT -1 99 "We co
nsider a simple mechanical system with one degree of freedom as shown \+
in the following figure." }}{PARA 260 "" 0 "" {METAFILE 317 213 213 1
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q:J:>@<:Uk:f:NZ;F:;jD:Dj@<:_cnZ:
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K=^:j=:E;gj^vx<^l?^:f??:>X:UK<:G;Gjysy=:JwK=>:;:C
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:f>JOvc?N_@^:f??:>X:UK<:GkAn:yayI:B:G;gjysy;j_P:m:cZ:>:^:n;j>JM;NYJ:YJ=vYxy:Z:n>n:;:C:UK=^:j=:E;MjJYjBXjJYJ:DJ;I:qB;_i:_r;AJd^:f?CJD:;b:UB:;JSN?an:^:aNLFt@>t>Z:f=<:iU:JGQJ<>:U;C:^:JroZ::UK;V:n=JELjZgt@nj>nt?f:NZ;J:^:
f?E:F;>Z>\\:>:G;gjy
sy;JtX:IJ:VuAjH`:K:oAG:;:C:US:GB:C:v:K:E;Iw;Iu;Iw;kRoA^[A>iANgA^:f??J<:C:>X>;n>n:yayI:JxkyC:B:ub;ju@jKEJeXJsYJX:UK;V:n=JELjQ]jfdJtXJQ\\JHjwwiyM:<:[V
::;JR^Z:jPn::f:NZ;:C:Uk<:M:;b:UB:N@ou;WC
<=JeDvoBFFYM<>:US:Gj@JSnTuF<=:W;JR>CCJ:f?Ej@JSNIEI<=:W;JNVBW:J
N`J;^::^hc>AAl@:VnCF:C::;SDOM<>:I;>:G;gjysy;JtJ_jgKjD:KZ:NYn:;J:J;
C:v:jL:OoMYjgDJ<>:Uk;:[Q>v:iAfOC:UK;^:J<:[AK:G
;Gjysy=:^YvYF;;jHJrwHCv:mNsYJX:UK;V:A?D:kKVf;:yW<=::Ff:vGU:JQdj:^::>??LA:>hDNZ:^::^Hk;U:joej::
:;SDO=;jMvYCB:f_;>:AB:>:1:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
71 "The system is loaded by a ground motion, described by the accelera
tion " }{XPPEDIT 18 0 "a[0](t);" "6#-&%\"aG6#\"\"!6#%\"tG" }{TEXT -1
92 ". In the following we consider different acceleration functions, d
escribed by the variables " }{XPPEDIT 18 0 "a[0](t)=spp0(t)" "6#/-&%\"
aG6#\"\"!6#%\"tG-%%spp0G6#F*" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "a[0]
(t)=sppa(t)" "6#/-&%\"aG6#\"\"!6#%\"tG-%%sppaG6#F*" }{TEXT -1 84 ". Th
e first one is a simple impulse motion, the second one is a series of \+
impulses. " }}{PARA 0 "" 0 "" {TEXT -1 58 "The motion of the mass m is
described by the displacement " }{XPPEDIT 18 0 "z(t)" "6#-%\"zG6#%\"t
G" }{TEXT -1 22 " and the acceleration " }{XPPEDIT 18 0 "a(t)" "6#-%\"
aG6#%\"tG" }{TEXT -1 68 ". The mass is supported on a combination of a
spring with stiffness " }{TEXT 259 1 "k" }{TEXT -1 42 " and a damper \+
with the damping resistance " }{TEXT 258 1 "d" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 56 "To solve the problem we use two different
usual methods:" }}{PARA 0 "" 0 "" {TEXT -1 50 "1. solve the equation \+
of motion in the time domain" }}{PARA 0 "" 0 "" {TEXT -1 171 "2. conve
rt the ground acceleration into the frequency domain, multiply the res
ult with the transfer function of the system and convert the solution \+
back to the time domain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 336 "The goal is to show the difference between the resu
lts of both methods and describe the reason for this difference. The b
ackground for the idea for this example is , that in many cases people
use the solution in the frequency domain to solve problems with trans
ient excitation with non-periodic excitations. This can result in fail
ures." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128
"To convert the function from the time domain to the frequency domain \+
and vice versa we use the Fast-Fourier-Transformation (see " }
{HYPERLNK 17 "FFT" 2 "FFT" "" }{TEXT -1 77 "). To use this function we
need a discrete number of points of the functions " }{XPPEDIT 18 0 "a
0(t)" "6#-%#a0G6#%\"tG" }{TEXT -1 22 ". This number must be " }
{XPPEDIT 18 0 "n = 2^p;" "6#/%\"nG)\"\"#%\"pG" }{TEXT -1 6 " with " }
{TEXT 260 1 "p" }{TEXT -1 75 " a non-negative integer. Additionally we
define here one more variable n2: " }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 25 "p:=8;\nn:=2**p;\nn2:=n/2+1;" }}}{EXCHG {PARA 3 "" 0 "
" {TEXT -1 32 "Description of The Ground Motion" }}{PARA 0 "" 0 ""
{TEXT -1 84 "For the first case, the ground motion is given by a singl
e impulse with the duration" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
8 "T1:=0.5;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "The ground acceler
ation is given by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "spp0(t
):=piecewise(t3*T1/2,0,1);" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 105 "Second we consider a series of impulses with the same du
ration. The time distance between the impulses is" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 10 "Tmax:=1.0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 54 "The ground motion is now described by the acceleration" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "sppa(t):=(1-cos(2*Pi*t)/abs(
cos(2*Pi*t)))/2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "In the follow
ing figures these functions are shown" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 75 "plot(spp0(t),t=0..5*Tmax,color=red,title=\"Excitation
by a single impulse\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81
"plot(sppa(t),t=0..5*Tmax,color=black,title=\"Excitation by a series o
f impulses\");" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 38 "Fourier Transfo
rm of the Ground Motion" }}{PARA 0 "" 0 "" {TEXT -1 283 "To use the Fa
st-Fourier-Transformation (FFT) we need numerical values of the accele
ration in discrete time steps. Further the FFT assumes that the functi
on which will be transformed is periodic. To use the FFT for solving n
on-periodic systems we assume that the function is periodic." }}{PARA
0 "" 0 "" {TEXT -1 177 "We will see now what happens with the single-i
mpulse ground motion after transforming into the frequency domain by u
se of the FFT. Therefore we calculate values of the function " }
{XPPEDIT 18 0 "spp0(t)" "6#-%%spp0G6#%\"tG" }{TEXT -1 5 " for " }
{TEXT 261 1 "n" }{TEXT -1 52 " discrete time steps inside the time int
erval [0,1)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "for i from \+
1 by 1 to n do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "ti[i]:=evalf((i-1
)/n*Tmax):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "x[i]:=evalf(subs(t=ti
[i],spp0(t))):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 90 "This values are stored in the array X. Ad
ditionally we define the array Y, which contains " }{TEXT 262 1 "n" }
{TEXT -1 41 " times the value 0, because the function " }{XPPEDIT 18
0 "spp0(t)" "6#-%%spp0G6#%\"tG" }{TEXT -1 20 " is a real function." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "X:=array([seq(x[i],i=1..n)]
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Y:=array([seq(0,i=1..
n)]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "The next figure shows th
e " }{TEXT 263 1 "n" }{TEXT -1 26 " discrete function points." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "display(curve([seq([ti[i],x
[i]],i=1..n)]),style=point,color=red,title=\"Discrete values of the ex
citation function\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "The Four
ier transform of the function " }{XPPEDIT 18 0 "spp0(t)" "6#-%%spp0G6#
%\"tG" }{TEXT -1 51 " is now calculated by use of the Maple-function F
FT" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "FFT(p,X,Y);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "The result is shown in the follow
ing figure. The black dots show the real part and red dots show the im
aginary part." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "display(cu
rve([seq([i,X[i]],i=1..n)],style=point),title=\"FFT-result of the exci
tation, real part\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "d
isplay(curve([seq([i,Y[i]],i=1..n)],color=red,style=point),title=\"FFT
-result of the excitation, imaginary part\");" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 110 "This result is not the final solution of the transfor
mation. To get this there are additional steps necessary." }}{PARA 0 "
" 0 "" {TEXT -1 203 "The FFT yields the Fourier transform for discrete
frequency values which are given by the time interval for which the f
unction is given. The distance of this discrete values for the frequen
cy is given by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "df:=1/Tma
x;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "The values which are yielde
d by the function FFT have to be divided by " }{TEXT 264 1 "n" }{TEXT
-1 28 ". So the constant parts are:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 14 "xs[1]:=X[1]/n;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 14 "ys[1]:=Y[1]/n;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "In the
considered cases of Fourier transforms of real functions, the constan
t part of the imaginary values is always 0." }}{PARA 0 "" 0 "" {TEXT
-1 106 "For i from 2 up to n2 the X[i]/n and Y[i]/n are the amplitudes
of the harmonic functions to the frequency " }{XPPEDIT 18 0 "(i-1)*df
" "6#*&,&%\"iG\"\"\"F&!\"\"F&%#dfGF&" }{TEXT -1 125 ". The values of X
[i]/n and Y[i]/n for i from n2+1 to n are the amplitudes for the negat
ive values for the frequency, that is " }{XPPEDIT 18 0 "-(i-1)*df" "6#
,$*&,&%\"iG\"\"\"F'!\"\"F'%#dfGF'F(" }{TEXT -1 140 ". But we know that
the real part, the X[i] belongs to the cosine function and the Y[i] \+
belongs to the sine function. So we use the relation" }}{PARA 0 "" 0 "
" {XPPEDIT 18 0 "X[i]/n*cos((i-1)*df)+X[n-i+1]/n*cos(-(i-1)*df) = 2*X[
i]/n*cos((i-1)*df);" "6#/,&*(&%\"XG6#%\"iG\"\"\"%\"nG!\"\"-%$cosG6#*&,
&F)F*F*F,F*%#dfGF*F*F**(&F'6#,(F+F*F)F,F*F*F*F+F,-F.6#,$*&,&F)F*F*F,F*
F2F*F,F*F***\"\"#F*&F'6#F)F*F+F,-F.6#*&,&F)F*F*F,F*F2F*F*" }{TEXT -1
4 " and" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Y[i]/n*sin((i-1)*df)+Y[n-i+1
]/n*sin(-(i-1)*df) = 2*Y[i]/n*sin((i-1)*df);" "6#/,&*(&%\"YG6#%\"iG\"
\"\"%\"nG!\"\"-%$sinG6#*&,&F)F*F*F,F*%#dfGF*F*F**(&F'6#,(F+F*F)F,F*F*F
*F+F,-F.6#,$*&,&F)F*F*F,F*F2F*F,F*F***\"\"#F*&F'6#F)F*F+F,-F.6#*&,&F)F
*F*F,F*F2F*F*" }{TEXT -1 17 ", so we get only " }{XPPEDIT 18 0 "n2=n/2
" "6#/%#n2G*&%\"nG\"\"\"\"\"#!\"\"" }{TEXT -1 24 " pairs of harmonic t
erms" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "for i from 2 by 1 t
o n2 do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "xs[i]:= 2*X[i]/n:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "ys[i]:= 2*Y[i]/n:" }}{PARA 0 "> "
0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "The d
iscrete values for the frequencies are" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 26 "for i from 1 by 1 to n2 do" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 15 "f[i]:=(i-1)*df:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "
od:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Now we have the excitation
transformed to series of harmonic terms:" }}{PARA 0 "" 0 "" {XPPEDIT
18 0 "Fourier(spp0(t)) = x[1]+sum(x[i]*cos((i-1)*df*2*Pi*t)-y[i]*sin((
i-1)*df*2*Pi*t),i = 2 .. infinity);" "6#/-%(FourierG6#-%%spp0G6#%\"tG,
&&%\"xG6#\"\"\"F/-%$sumG6$,&*&&F-6#%\"iGF/-%$cosG6#*,,&F7F/F/!\"\"F/%#
dfGF/\"\"#F/%#PiGF/F*F/F/F/*&&%\"yG6#F7F/-%$sinG6#*,,&F7F/F/F=F/F>F/F?
F/F@F/F*F/F/F=/F7;F?%)infinityGF/" }{TEXT -1 2 ". " }}{PARA 0 "" 0 ""
{TEXT -1 31 "The following figures show the " }{XPPEDIT 18 0 "n2" "6#%
#n2G" }{TEXT -1 94 " amplitudes of the amplitudes of the harmonic term
s of the Fourier transform of the excitation" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 106 "display(curve([seq([f[i],xs[i]],i=1..n2)],color
=blue,style=point),title=\"Amplitudes of the cosine terms\");" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "display(curve([seq([f[i],ys
[i]],i=1..n2)],color=red,style=point),title=\"Amplitudes of the sine t
erms\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "So the excitation is \+
described as Fourier series by:" }}{PARA 0 "" 0 "" {TEXT -1 20 "Sum of
cosine terms:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "xfc(t):= \+
sum(xs[ij]*cos(2*Pi*f[ij]*t),ij=1..n2):" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 21 "and sum of sine terms" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 48 "xfs(t):=-sum(ys[ij]*sin(2*Pi*f[ij]*t),ij=1..n2):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "The result is the sum of both part
s" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "xf(t):=xfc(t)+xfs(t):
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "The solution is shown in the \+
following figures" }{MPLTEXT 1 0 1 " " }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 80 "display(plot(xfc(t),t=0..Tmax,color=blue),title=\"Fou
rier series, blue: cosine\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 66 "display(plot(xfs(t),t=0..Tmax),title=\"Fourier series, red: sine
\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "display(plot(xf(t),
t=0..Tmax,color=black,thickness=2),title=\"Fourier series, black: cosi
ne+sine\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 293 "You see that this
series doesn't describe the actual function exactly. It is only an ap
proximation. The greater the number n the better the approximation. Le
t's now compare the result with the actual ground acceleration at the \+
same discrete time steps as used above to get this Fourier series." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "for i from 1 by 1 to n do"
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "xfi[i]:=evalf(subs(t=ti[i],xf(t))
):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 41 "The next figure show this discrete values" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 176 "display(plot(spp0(t),t=0..Tmax,col
or=black),curve([seq([ti[i],xfi[i]],i=1..n)],color=red,style=point), t
itle=\"black: actual excitation, red: Fourier series (discrete values)
\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 220 "We have seen that the Fo
urier transform in reality only approximates periodic functions. In th
e following figure we see that the calculated Fourier series is not re
ally an approximation for the single impulse excitation " }{XPPEDIT
18 0 "spp0(t)" "6#-%%spp0G6#%\"tG" }{TEXT -1 32 " but for the series o
f impulses " }{XPPEDIT 18 0 "sppa(t)" "6#-%%sppaG6#%\"tG" }{TEXT -1 1
"." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 148 "display(plot(spp0(t)
,t=0..5*Tmax,color=black, thickness=2),plot(xf(t),t=0..5*Tmax,color=re
d),title=\"black: actual excitation, red: Fourier series\");" }}}
{EXCHG {PARA 3 "" 0 "" {TEXT -1 36 "Description of the Mechanical Syst
em" }}{PARA 0 "" 0 "" {TEXT -1 68 "Here we consider a concrete system \+
with the following characteristic" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 12 "m:=2.*10**3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 12 "k:=3.*10**6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "d:=5.
*10**3;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "The natural frequency \+
and the damping ratio are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
22 "evalf(sqrt(k/m)/2/Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
25 "evalf(d/(2*m*sqrt(k/m)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "
The equation of motion for the considered system is" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 55 "bgl0:=m*(diff(s(t),t$2)+a0(t))+d*diff(s(t
),t)+k*s(t)=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "with the genera
l ground acceleration " }{XPPEDIT 18 0 "a0(t)" "6#-%#a0G6#%\"tG" }
{TEXT -1 56 ". Now we consider some different types for the solution.
" }}}{EXCHG {PARA 4 "" 0 "" {TEXT -1 77 "Solution with transient vibra
tion, excitation described by the Fourier series" }}{PARA 0 "" 0 ""
{TEXT -1 56 "First the constant part of the excitation is considered.
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sppi[1]:=xs[1]-ys[1]:" }}{PARA
0 "" 0 "" {TEXT -1 52 "This part is substituted into the equation of m
otion" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "bgl[1]:=subs(a0(t)=sppi[1]
,bgl0):" }}{PARA 0 "" 0 "" {TEXT -1 129 "The solution is reached by us
e of the Maple function dsolve. The motion should start from the rest,
so the initial conditions are" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "I
C:=s(0)=0,D(s)(0)=0:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "lsg[1]:=dso
lve(\{bgl[1],IC\},s(t)):" }}{PARA 0 "" 0 "" {TEXT -1 85 "And the rela
tive displacement of the system caused by this part of the excitation \+
is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "ski[1]:=value(rhs(lsg[1])):"
}}{PARA 0 "" 0 "" {TEXT -1 107 "In the same manner we get the displace
ment for every term of the Fourier series for the ground acceleration
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "for i from 2 by 1 to n2 do" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "sppi[i]:=xs[i]*cos(2*Pi*f[i]*t)-ys[
i]*sin(2*Pi*f[i]*t):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "bgl[i]:=sub
s(a0(t)=sppi[i],bgl0);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "lsg[i]:=d
solve(\{bgl[i],s(0)=0,D(s)(0)=0\},s(t)):" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 "ski[i]:=value(rhs(lsg[i])):" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 3 "od:" }}{PARA 0 "" 0 "" {TEXT -1 95 "The total relative
displacement between the body and the ground is finally the sum of al
l terms" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "sk(t):=evalf(sum(ski[ij]
,ij=1..n2)):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "The relative velo
city of the body is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "vk(t
):= diff(sk(t),t):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The relati
ve acceleration is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ak(t)
:= diff(sk(t),t$2):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The absol
ute acceleration is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "aak(
t):= diff(sk(t),t$2)+xf(t):" }}}{EXCHG {PARA 4 "" 0 "" {TEXT -1 89 "So
lution with transient vibration, excitation described by the actual gr
ound acceleration" }}{PARA 0 "" 0 "" {TEXT -1 130 "Now we consider the
case that the excitation is the single impulse ground acceleration. T
he equation of motion is the described by" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 50 "bgla:=convert(subs(a0(t)=spp0(t),bgl0),Heaviside):
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "And the solution is" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "lsga:=dsolve(\{bgla,s(0)=0,D
(s)(0)=0\},s(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ska(t)
:=value(rhs(lsga)):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "The relati
ve velocity of the body is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
24 "vka(t):= diff(ska(t),t):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "T
he relative acceleration is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
26 "aka(t):= diff(ska(t),t$2):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28
"The absolute acceleration is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 34 "aaka(t):=diff(ska(t),t$2)+spp0(t):" }}}{EXCHG {PARA 4 "" 0 ""
{TEXT -1 102 "Solution without transient vibration (steady state motio
n), excitation described by the Fourier series" }}{PARA 0 "" 0 ""
{TEXT -1 255 "the method to get the solution is nearly the same. But h
ere we don't need any initial conditions, only the integration constan
ts have to be set to zero. This yields the steady state motion. This i
s again first made for the constant part of the excitation." }}{PARA
0 "> " 0 "" {MPLTEXT 1 0 21 "sppi[1]:=xs[1]-ys[1]:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 52 "bgk[1]:=convert(subs(a0(t)=sppi[1],bgl0),Heaviside)
:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "sli[1]:=rhs(subs(\{_C1=0,_C2=0
\},dsolve(bgk[1],s(t)))):" }}{PARA 0 "" 0 "" {TEXT -1 75 "And then the
same way is used for all the other terms of the Fourier series" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "for i from 2 by 1 to n2 do" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "sppi[i]:=xs[i]*cos(2*Pi*f[i]*t)-ys[
i]*sin(2*Pi*f[i]*t):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "bgk[i]:=con
vert(subs(a0(t)=sppi[i],bgl0),Heaviside);" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 53 "sli[i]:=rhs(subs(\{_C1=0,_C2=0\},dsolve(bgk[i],s(t)))
):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 0 "" 0 "" {TEXT
-1 104 "The total relative displacement between the ground and the bod
y is the sum of all terms of the solution." }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 36 "sl(t):=evalf(sum(sli[ji],ji=1..n2)):" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 36 "The relative velocity of the body is" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "vl(t):= diff(sl(t),t):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The relative acceleration is" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "al(t):= diff(sl(t),t$2):" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The absolute acceleration is" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "aal(t):= diff(sl(t),t$2)+x
f(t):" }}}{EXCHG {PARA 4 "" 0 "" {TEXT -1 114 "Solution without transi
ent vibration (steady state motion), excitation described by the actua
l ground acceleration" }}{PARA 0 "" 0 "" {TEXT -1 45 "The solution of \+
the equation of motion is now" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 51 "sla(t):=rhs(subs(\{_C1=0,_C2=0\},dsolve(bgla,s(t)))):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "The relative velocity of the body \+
is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "vla(t):= diff(sla(t),
t):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The relative acceleration \+
is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ala(t):= diff(sla(t),
t$2):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The absolute acceleratio
n is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "aala(t):= diff(sla(
t),t$2)+spp0(t):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "At last we co
ntrol whether the result fulfill the equation of motion" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Ek(t):= (m*diff(ska(t),t$2)+d*diff
(ska(t),t)+k*ska(t))/m:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "
El(t):= (m*diff(sla(t),t$2)+d*diff(sla(t),t)+k*sla(t))/m:" }}}{EXCHG
{PARA 4 "" 0 "" {TEXT -1 16 "Plot the results" }}{PARA 0 "" 0 ""
{TEXT -1 79 "The results are not so easy to show in simple manner so w
e show them graphical." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Displac
ement" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "P[1]:=plot((sk(t))
,t=0..Tmax,color=black,title=\"Displacement with transient motion (Fou
rier)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "P[2]:=plot((sk
a(t)),t=0..Tmax,title=\"Displacement with transient motion (actual exc
itation)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "P[3]:=plot
((sk(t)),t=50*Tmax..51*Tmax,color=black,title=\"Displacement with tran
sient motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
103 "P[4]:=plot((ska(t)),t=50*Tmax..51*Tmax,title=\"Displacement with \+
transient motion (actual excitation)\"):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 94 "P[5]:=plot((sl(t)),t=0..Tmax,color=black,title=\"Disp
lacement, steady state motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 93 "P[6]:=plot((sla(t)),t=0..Tmax,title=\"Displacement
, steady state motion (actual excitation)\"):" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 8 "Velocity" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "
P[7]:=plot((vk(t)),t=0..Tmax,color=black,title=\"Velocity with transie
nt motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "
P[8]:=plot((vka(t)),t=0..Tmax,title=\"Velocity with transient motion (
actual excitation)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "
P[9]:=plot((vk(t)),t=50*Tmax..51*Tmax,color=black,title=\"Velocity wit
h transient motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 100 "P[10]:=plot((vka(t)),t=50*Tmax..51*Tmax,title=\"Velo
city with transient motion (actual excitation)\"):" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 91 "P[11]:=plot((vl(t)),t=0..Tmax,color=black,t
itle=\"Velocity, steady state motion (Fourier)\"):" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 90 "P[12]:=plot((vla(t)),t=0..Tmax,title=\"Velo
city, steady state motion (actual excitation)\"):" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 21 "Relative acceleration" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 107 "P[13]:=plot((ak(t)),t=0..Tmax,color=black,title=\"
Acceleration (relative) with transient motion (Fourier)\"):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "P[14]:=plot((aka(t)),t=0..T
max,title=\"Acceleration (relative) with transient motion (actual exci
tation)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "P[15]:=plot
((ak(t)),t=50*Tmax..51*Tmax,color=black,numpoints=500,title=\"Accelera
tion (relative) with transient motion (Fourier)\"):" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 115 "P[16]:=plot((aka(t)),t=50*Tmax..51*Tmax,
title=\"Acceleration (relative) with transient motion (actual excitati
on)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "P[17]:=plot((al
(t)),t=0..Tmax,color=black,title=\"Acceleration (relative), steady sta
te motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105
"P[18]:=plot((ala(t)),t=0..Tmax,title=\"Acceleration (relative), stead
y state motion (actual excitation)\"):" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 21 "Absolute acceleration" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 108 "P[19]:=plot((aak(t)),t=0..Tmax,color=black,title=\"A
cceleration (absolute) with transient motion (Fourier)\"):" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "P[20]:=plot((aaka(t)),t=0..Tmax,ti
tle=\"Acceleration (absolute) with transient motion (actual excitation
)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "P[21]:=plot((aak(
t)),t=50*Tmax..51*Tmax,color=black,numpoints=500,title=\"Acceleration \+
(absolute) with transient motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 116 "P[22]:=plot((aaka(t)),t=50*Tmax..51*Tmax,titl
e=\"Acceleration (absolute) with transient motion (actual excitation)
\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "P[23]:=plot((aal(t
)),t=0..Tmax,color=black,title=\"Acceleration (absolute), steady state
motion (Fourier)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "P
[24]:=plot((aala(t)),t=0..Tmax,title=\"Acceleration (absolute), steady
state motion (actual excitation)\"):" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 7 "Control" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "P[2
5]:=plot((Ek(t)),t=0..Tmax,color=black,title=\"Control (with transient
motion, actual axcitation)\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 88 "P[26]:=plot((El(t)),t=0..Tmax,title=\"Control (steady state mo
tion, actual excitation)\"):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "
Now the above defined plots are shown. First we consider the relative \+
displacement. This is the relative motion between ground and body." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "for i from 1 by 1 to 2 do d
isplay(P[i]) od;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 176 "Here we see \+
nearly no difference between the solution for the actual excitation an
d the approximation by the Fourier series. But now we consider the mot
ion after a longer time." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43
"for i from 3 by 1 to 4 do display(P[i]) od;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 484 "Of cause now there is a great difference, because the \+
actual excitation is a single impulse and yields a fading away vibrati
on after the impulse. In the considered time range the motion nearly v
anishes. The Fourier series describes a periodic excitation and the re
sult is a periodic motion. Only while the first period both ground mot
ions yield nearly the same result. When the number of Fourier terms in
creases, the difference between both motions during this first period \+
decreases." }}{PARA 0 "" 0 "" {TEXT -1 269 "Next we consider the stead
y state motion. Of course in the case of the single impulse it makes n
o sense to show this solution, because during the impulse the motion i
s a transient motion and the steady state is reached after longer time
, depending on the damping ratio." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 43 "for i from 5 by 1 to 6 do display(P[i]) od;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 323 "Now we see that in the case of th
e excitation by the Fourier series the solution is nearly the same as \+
the solution for the case of consider the solution with transient moti
on after a longer time. But the \"steady state motion\" in the case of
the solution with the actual ground motion is not the really steady s
tate motion." }}{PARA 0 "" 0 "" {TEXT -1 210 "So we see: The approxima
tion of the excitation by the Fourier series cannot describe the right
solution for a longer time period but only for the time where the exc
itation is approximated by the Fourier series." }}{PARA 0 "" 0 ""
{TEXT -1 169 "The same result yields the consideration of the relative
velocity, the relative acceleration and the absolute acceleration. Th
is plots are shown in the following figures" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 44 "for i from 7 by 1 to 24 do display(P[i]) od;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "At last the both control plots are
shown" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "for i from 25 by \+
1 to 26 do display(P[i]) od;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "B
oth plots show the negative excitation, so the control is all right."
}}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 40 "Solve the System in the Frequen
cy Domain" }}{PARA 0 "" 0 "" {TEXT -1 120 "First we deduce the general
solution for the problem. To do this we consider the general form of \+
the equation of motion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "
BGL:=M*(diff(Z(t),t$2)+A(t))+R*diff(Z(t),t)+K*Z(t)=0;" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 360 "To solve the problem in the frequency do
main the excitation must be converted to the frequency domain. So we h
ave to use the Fourier series to solve the problem. To get the general
solution we consider one harmonic part of the excitation. Further we \+
use the method of complex numbers to go on. Then one harmonic term of \+
the displacement of the ground motion is" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 34 "A(t):=diff(X0*exp(I*Omega*t),t$2);" }}}{EXCHG {PARA
0 "" 0 "" {TEXT -1 89 "Here we used the fact that the acceleration is
the second deviation of the displacement." }}{PARA 0 "" 0 "" {TEXT
-1 103 "We know that the displacement of the body has the same mathema
tical form like the excitation, so we set" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "Z(t):=Z0*exp(I*Omega*t);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 38 "Then we get for the equation of motion" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "BGL;" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 55 "This equation is solved with respect to Z0, this yields"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "LSG:=solve(BGL,Z0);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "The transfer function is defined \+
as the relation between the amplitude of the motion of the body and th
e amplitude of the excitation so we get the" }}{PARA 4 "" 0 "" {TEXT
-1 25 "Complex Transfer Function" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "V:=expand(LSG)/X0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 176 "Now we have to do some additional steps to separate the real a
nd the imaginary part of this function. This is here done without furt
her comment, because this is standard math. " }}{PARA 0 "" 0 "" {TEXT
-1 103 "(Because the solution has general manner it is here not easy t
o use the Maple functions \"Re\" and \"Im\".)" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 18 "Zaehler:=numer(V);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "Nenner:=denom(V);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 91 "Nenner2:=expand((tcoeff(Nenner,I)+coeff(Nenner,I)*I)*
(tcoeff(Nenner,I)-coeff(Nenner,I)*I));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 63 "Zaehler2:=expand(Zaehler*(tcoeff(Nenner,I)-coeff(Nenn
er,I)*I));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Now we have the ima
ginary part" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "Imaginaer:=s
implify(coeff(Zaehler2,I)/Nenner2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 42 "and the real part of the transfer function" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 45 "Real:=simplify(Zaehler2/Nenner2-Imaginaer*I)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "We substitute the concrete v
alues of our problem and get" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 55 "Imaginaerteil:=simplify(subs(\{M=m,K=k,R=d\},Imaginaer));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Realteil:=simplify(subs(\{M=
m,K=k,R=d\},Real));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "At last we
substitute the angular frequency " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG
" }{TEXT -1 18 " by the frequency " }{XPPEDIT 18 0 "f" "6#%\"fG" }
{TEXT -1 8 " and get" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "C:=
subs(Omega=f*2*Pi,Realteil);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 36 "S:=subs(Omega=f*2*Pi,Imaginaerteil);" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 123 "The real part and the imaginary part of the transfer fun
ction are shown in the following figures dependent on the frequency" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plot(C,f=0..10,title=\"Rea
l part of the transfer function\");" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 64 "plot(S,f=0..10,title=\"Imaginary part of the transfer
function\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 187 "The excitation \+
is with the Fourier transformation with use of the FFT function given \+
for discrete frequency values. So we calculate the transfer function f
or the same discrete frequencies" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 26 "for i from 1 by 1 to n2 do" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 29 "ci[i]:=evalf(subs(f=f[i],C));" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 29 "si[i]:=evalf(subs(f=f[i],S));" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "dis
play(curve([seq([f[i],ci[i]],i=1..n2)],style=point),title=\"Real part \+
of the transfer function, discrete values\");" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 121 "display(curve([seq([f[i],si[i]],i=1..n2)],style
=point),title=\"Imaginary part of the transfer function, discrete valu
es\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "We repeat the plot of t
he Fourier transform of the excitation" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 111 "display(curve([seq([f[i],X[i]],i=1..n2)],style=point
),title=\"Real part of Fourier transfom of the excitation\");" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "display(curve([seq([f[i],Y[
i]],i=1..n2)],style=point),title=\"Imaginary part of Fourier transfom \+
of the excitation\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 202 "We get \+
the Fourier transform of the system answer by multiply every term of t
he Fourier transform of the excitation with the corresponding term of \+
the transfer function. See that this terms are complex." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "for i from 1 by 1 to n2 do" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "USC[i]:=(ci[i]+I*si[i])*(X[i]+I*Y[i
]):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "UC[i]:= evalf(Re(USC[i])):"
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "US[i]:= evalf(Im(USC[i])):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 56 "We want to use the inverse Fast Fourier Transformation (
" }{HYPERLNK 17 "iFFT" 2 "iFFT" "" }{TEXT -1 201 "). So we have to sto
re the amplitudes for every harmonic part in arrays of that form like \+
the function FFT yields the result above. The members of the second p
art of the arrays (from n2+1 up to n) are" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 28 "for i from n2+1 by 1 to n do" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 19 "UC[i]:= UC[2*n2-i]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1
0 19 "US[i]:=-US[2*n2-i]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Now the two arrays which are need
ed by the function iFFT are defined." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 32 "UCa:=array([seq(UC[i],i=1..n)]):" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 32 "USa:=array([seq(US[i],i=1..n)]):" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "display(curve([seq([i,UCa[i]],i=1.
.n)]),title=\"Real part of the Fourier transform of the motion of the \+
system\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "display(curv
e([seq([i,USa[i]],i=1..n)]),title=\"Imaginary part of the Fourier tran
sform of the motion of the system\");" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 26 "Now iFFT yields the result" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 16 "iFFT(p,UCa,USa):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
75 "The following figures show the calculated relative acceleration of
the body" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "display(curve(
[seq([ti[i],USa[i]],i=1..n)]),title=\"Imaginary part of the relative a
cceleration\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "PF:=curv
e([seq([ti[i],UCa[i]],i=1..n)],color=green,title=\"Real part of the re
lative acceleration\"):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "
display(PF,title=\"Real part of the relative acceleration\");" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 161 "We see that the imaginary part of
the result nearly vanishes. This is right, because the excitation was
real, too. The small values result from numerical errors." }}{PARA 3
"" 0 "" {TEXT -1 27 "Comparison of the Solutions" }}{PARA 0 "" 0 ""
{TEXT -1 133 "We compare this result with the solutions above. The res
ult of the solution in the frequency domain is in the plots with green
color." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "display(P[13],PF
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "display(P[14],PF);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "display(P[17],PF);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "display(P[18],PF);" }}}
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }
{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG {PARA 0 "" 0 "" {TEXT
-1 376 "You see that the only plot with acceptable quality of the conf
ormity is the solution for the steady state motion with approximate th
e excitation by the Fourier series. The consideration in the frequency
domain yields results of bad quality in case of transient motion. (Re
member: We have seen above that the \"steady state motion\" in the las
t figure is not a real steady state)" }}}{EXCHG {PARA 261 "" 0 ""
{TEXT 265 11 "Disclaimer:" }{TEXT -1 240 " While every effort has been
made to validate the solutions in this worksheet, Waterloo Maple Inc.
and the contributors are not responsible for any errors contained and
are not liable for any damages resulting from the use of this materia
l." }}}}{MARK "264" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS
0 1 2 33 1 1 }