# is there a way to avoid the Dirac Delta in the fourier transform?

I would like to compute the fourier transform of H1[t], as you can see the output of mathematica gives the the dirac delta:

n3p = {-1, 0, 0};

ep = {{1, 0, 0}, {0, -1, 0}, {0, 0, 0}};

h[t] = hp[t]*ep

hp[t] = h0*Cos[Wgw*t];

H1[t] = Integrate[h[t], {t, t - L3/c, t}].n3p.n3p

H1[ω] =
FourierTransform[H1[t], t, ω, FourierParameters -> {-1, 1}]


This is the output :

(I h0 DiracDelta[-Wgw + ω])/(2 Wgw) - ( I E^((I L3 Wgw)/c) h0 DiracDelta[-Wgw + ω])/(2 Wgw) - ( I h0 DiracDelta[Wgw + ω])/(2 Wgw) + ( I E^(-((I L3 Wgw)/c)) h0 DiracDelta[Wgw + ω])/(2 Wgw)

Is there a way to avoid the DiracDelta? And to keep it equal to 1?

Thank you!

• What do you mean by "keep it equal to 1"? – Henrik Schumacher Sep 25 '18 at 7:48
• I mean that I do not want to have the dirac delta in my FT output, there should be a link with the argument of your FT ..if the argument if >1 then you should not have the dirac delta in your output..but i am not able to figure out how to do that.. – Martina Sep 25 '18 at 7:55
• Are you saying that the result is incorrect? What result would you expect then? It seems to me that there's a mathematical misunderstanding here (i.e. the question isn't really Mathematica-related). – Szabolcs Sep 25 '18 at 8:35
• @Martina The Fourier transform of H1 is distribution and not a function. That's why it cannot be plotted. (People often say nonesense like the Dirac delta is a function that vanishes everywhere away from 0 but its integral over every interval containing 0 is 1. Which simply means: It is not a function; it is something different.) – Henrik Schumacher Sep 25 '18 at 8:46
• A Dirac $\delta$ cannot be plotted in the same sense as functions can, but it can be illustrated in various ways, e.g. by drawing a vertical line at the appropriate position. You could use the HalfLine graphics primitive for this. – Szabolcs Sep 25 '18 at 9:03

One can replace the Dirac delta function by one of its expressions in the form of a limit, and then take a finite value of the function, for example

n3p = {-1, 0, 0};

ep = {{1, 0, 0}, {0, -1, 0}, {0, 0, 0}};

h[t] = hp[t]*ep;

hp[t] = h0*Cos[Wgw*t];

H1[t] = Integrate[h[t], {t, t - L3/c, t}].n3p.n3p;
g[x_] := Exp[-(x/a)^2]/a/Sqrt[Pi]
f = FourierTransform[H1[t], t, \[Omega],
FourierParameters -> {-1, 1}] /. {DiracDelta -> g} // FullSimplify


Here we obtain

(E^(-((Wgw^2 + \[Omega]^2)/
a^2)) h0 (Sin[(L3 Wgw)/c - (2 I Wgw \[Omega])/a^2] +
I Sinh[(2 Wgw \[Omega])/a^2]))/(a Sqrt[\[Pi]] Wgw)


To display, we use this function

p[W_, h_, L_, c0_, a0_] :=
Plot[Re[f /. {Wgw -> W, h0 -> h, L3 -> L, c -> c0,
a -> a0}], {\[Omega], -3 W/2, 3*W/2}, PlotRange -> All,
AxesLabel -> {"\[Omega]", "H1(\[Omega])"}, WorkingPrecision -> 100]


Finally, we construct a distribution

p[1, 1, 1, 1, 1/100]


• Your math is not traditional math. Can you kindly give a reference to base your answer? TIA. – user64494 Sep 25 '18 at 17:17
• @user64494 Do you mean the representation of the Dirac delta function as a limit? – Alex Trounev Sep 25 '18 at 17:27
• It is not so simple and primitive: if you look in Wiki en.wikipedia.org/wiki/Dirac_delta_function as a first reading, you will understand that this limit is not a usual limit, but a limit in $\mathcal{D} '$ topology which is not implemented in Mathematica. – user64494 Sep 25 '18 at 17:38
• But this does not prevent us from using this representation to visualize the Fourier image containing the delta function. – Alex Trounev Sep 25 '18 at 17:49
• This is not a profanity, it's just an example. You can specify your example. Perhaps it will be better. – Alex Trounev Sep 25 '18 at 18:21