Plot a Dirac Delta Function?

Posted 4 months ago
943 Views
|
8 Replies
|
5 Total Likes
|
 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 Answer
8 Replies
Sort By:
Posted 4 months ago
 It's really a functional, not a function. There is no reason to expect one could "plot" it in the usual sense. Answer
Posted 4 months ago
 Delta Function is a generalized function, not a functional bud. But thanks anyways Answer
Posted 4 months ago
 Check the part about linear functionals. Answer
Posted 4 months ago
 You can approximate it. Plot[PDF[NormalDistribution[0, 10^-2], x], {x, -2, 2}, PlotRange -> {0, 1}]  Answer
Posted 4 months ago
 That’s it! Thanks! It’s undefined at zero so we gotta approximate, makes sense. Thanks!! Answer
Posted 4 months 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:  Answer
Posted 4 months 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] *) Answer
Posted 4 months 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! Answer