Is there any posibility to apply Parseval identity(energy in time domain is equal to freq. domain) on One/Single-side Fourier transform?
In other word for "standard" FT the test Mathematica can be
nn = 250; T = 20.;
t = Range[0, T, T/nn];
f = Exp[-(t - 5)^2]*Sin[t - 5];
ts = t[[2]] - t[[1]];
ws = 2 π/ts/(nn);
F = RotateLeft[Abs[Fourier[f]], Floor[Length[f]/2]];
w = ws*Range[-Floor[nn/2] - 1, Floor[nn/2]];
q1 = Total[(w[[2]] - w[[1]])*Abs[F]^2];
q2 = Total[ts*Abs[f]^2];
Fp = F*Sqrt[(q2/q1)];
aries = FourierTransform[
Exp[-(τ - 5)^2]*Sin[τ - 5], τ, ω];
ListLinePlot[Partition[Riffle[t, f], 2], PlotRange -> All]
Show[ListPlot[Partition[Riffle[w, Fp], 2]],
Plot[Abs[aries], {ω, -7, 7}, PlotStyle -> Red],
PlotRange -> All]
One side FT is used for example damped oscilator where t=>0 where the solution can be Exp[-1 t]*Sin[2. t] . The script is
dt = 0.05;
t = Range[0, 10, dt];
fun = Exp[-1 t]*Sin[2. t];
vys = Assuming[ω > 0,
Integrate[
Exp[-tt]*Sin[2 tt]*Exp[I ω tt], {tt, 0, ∞}]];
r = RotateLeft[Im[Fourier[fun]], Floor[Length[Im[Fourier[fun]]]/2]];
r1 = RotateLeft[Fourier[fun], Floor[Length[Im[Fourier[fun]]]/2]];
an = Table[{ω, Im[vys]}, {ω, 0, 10, 0.1}];
dw = 2 π/dt/Length[t];
w = dw*Range[-Floor[Length[t]/2]+1, Floor[Length[t]/2]];
a = Total[t*fun^2]
b = Chop[Total[dw*r1*Conjugate[r1]]];
Show[ListPlot[Partition[Riffle[w, r], 2], PlotRange -> All],
ListLinePlot[Partition[Riffle[an[[All, 1]], an[[All, 2]]], 2],
PlotRange -> All], PlotRange -> All]
I have numerical data(in this case generated with function fun = Exp[-1 t]*Sin[2. t];). I like to evaluate integral((Integrate[Exp[-tt]*Sin[2 tt]*Exp[I ω tt], {tt, 0, ∞}]])) on this data. I am wondering if is possible to use fft(Fourier command) to evaluate this integral with numerical data. Note, there is analytical and numerical solution for comparision(This is test a script, in the final, there will be only numerical data). The graph shows numerical and analytical solution, the position/freq match is OK, but the amplitude don’t fit.
Thank you for your help and suggestions(I'm new in FT, so all answers will be gladly appreciate) and sorry for bad english.
vys2 = FourierTransform[Exp[-tt]*Sin[2 tt]*HeavisideTheta[tt], tt, \[Omega]]
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