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Hello I would like to do the Fourier Transform of

H1'[t - L2[t]/c - L2primo[t]/c]

I know that its transform has to be:

i ω H1[ω]* Exp[- I ω (L2[t]/c + L2primo[t]/c)

I have considered L2[t] and L2primo[t] to be constant and not to be transformed...

In order to do that in mathematica I did this function:

Argomentof[e_] := Collect[e[[1, 2 ;;]], -1/c]
myFTd[f_, (q_: 1) expr_, ω_] := q*I* ω*f[ω]*Exp[I ω Argomentof[expr]];

if I do

myFTd[H1,L2primo[t]H1'[t - L2[t]/c - L2primo[t]/c] L2'[t])/c^2, ω]

I get:

(I Exp^(I ω L2primo[]) ω H1[ω])/c^2

that is wrong...anyone can help?

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    $\begingroup$ Why not use FourierTransform? For example: FourierTransform[H1'[t-a],t,s,FourierParameters->{0,-1}] $\endgroup$ – Carl Woll Sep 5 '18 at 23:43
  • $\begingroup$ cause I have a symbolic signals, I did not have defined the value of H1, so it does not compute anything..in addition when I have terms which multiply my signal to be transform i don't know how to deal with it! indeed that is what i get from your suggestion : I E^(-I a s) s FourierTransform[H1[t], t, s]..it did not compute H1[t] $\endgroup$ – Martina Sep 6 '18 at 7:21
  • $\begingroup$ But If I had another step like I define myFT for a symbolic element: myFT[f_[t_],[Omega]_]:=f[[Omega]] , and I do FourierTransform[H1[t], t, [Omega]] -> myFT[H1[t], [Omega]] I get this I E^(-I a s) s H1[s] ... thanks a lot for the idea! $\endgroup$ – Martina Sep 6 '18 at 7:45
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I think you can just use FourierTransform, and give an UpValue for H1. For example:

H1 /: FourierTransform[H1[t_], t_, s_] := OverHat[H1][s]

Then:

FourierTransform[H1'[t-a], t, s, FourierParameters->{0,-1}] //TeXForm

$i s e^{-i a s} \hat{\operatorname{H1}}(s)$

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