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Plot a Dirac Delta Function?

Luciano Pinheiro

Luciano Pinheiro, Texas Tech University

Posted 6 years ago

Good Evening All, I have stumbled across a Dirac Delta Function, when applying the Fourier Transform to a function. I am not quite sure how to plot it. Could anyone provide some guidance? Thanks! F[t_] := A*Sin[ t] g[k_] := FourierTransform[F[t], t, k] I A Sqrt[\[Pi]/2] DiracDelta[-1 + k] - I A Sqrt[\[Pi]/2] DiracDelta[1 + k

POSTED BY: Luciano Pinheiro

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Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

To be honest, I would be quite unhappy if a plot of `DiracDelta` would actually show anything! What you can do is making something visible by showing its convolution with some "sharp" function, e.g.: f[t_] := a Sin[t] g[\[FormalK]_] = FourierTransform[f[\[FormalT]], \[FormalT], \[FormalK]]; plotDelta[f_, k_] := Total@*ReIm@Convolve[f[\[FormalX]], HeavisideLambda[50 \[FormalX]], \[FormalX], k] Plot[Evaluate[plotDelta[g, k] /. a -> 1], {k, -2, 2}, PlotRange -> All] which gives:

To be honest, I would be quite unhappy if a plot of DiracDelta would actually show anything! What you can do is making something visible by showing its convolution with some "sharp" function, e.g.:

f[t_] := a Sin[t]
g[\[FormalK]_] = FourierTransform[f[\[FormalT]], \[FormalT], \[FormalK]];

plotDelta[f_, k_] := Total@*ReIm@Convolve[f[\[FormalX]], HeavisideLambda[50 \[FormalX]], \[FormalX], k]
Plot[Evaluate[plotDelta[g, k] /. a -> 1], {k, -2, 2}, PlotRange -> All]

which gives:

enter image description here

POSTED BY: Henrik Schachner

David Keith

Posted 6 years ago

The Dirac delta function is a Monster. It must be kept in a cage, called an integrand. Outside the cage, it makes no more sense than the Jabberwock. Inside the cage it may be tamed: Integrate[DiracDelta[x - a] f[x], {x, -Infinity, Infinity}, Assumptions -> Element[a, Reals]] (* f[a] *)

POSTED BY: David Keith

Luciano Pinheiro

Luciano Pinheiro, Texas Tech University

Posted 6 years ago

Love the analogy, David! And that's an elegant solution to my problem, based on the definition of the the Delta Func, because however the delta func is undefined when its argument is zero, the its integral is well defined by definition. thanks!