# Plot Fourier transform of $\sin (2 t)$

I'm trying to plot the Fourier transform of $\sin (2 t)$.

I tried using the FourierTransform Function (I'm expecting a peak at w = 2) but it gives me undefined at w = 2 because the DiracDelta function is not defined at DiracDelta[0]. Is there any way around this? I used

Plot[FourierTransform[Sin[2*t], t, w], {w, -3, 3}, PlotRange -> Full]


which has gaps at 2 and -2, but no height. How do I get a peak with some height there?

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You may find an answer to how to plot (approximations to the delta function) here: mathematica.stackexchange.com/q/3506/1783 – bill s Feb 16 '14 at 16:16

Since the transform has complex delta functions, you have to create a representation by hand.

FourierTransform[Sin[2*t], t, w]
(*
I Sqrt[\[Pi]/2] DiracDelta[-2 + w] - I Sqrt[\[Pi]/2] DiracDelta[2 + w]
*)


Here's one way that includes marking the axes by hand.

Plot[UnitBox[100 (2 - w)] - UnitBox[100 (2 + w)],
{w, -3, 3}, Exclusions -> {-2, 2}, PlotStyle -> Thick,
Ticks -> {Automatic, {{-1, Row[{-I Sqrt[Pi/2], Infinity}]}, {1,
Row[{I Sqrt[Pi/2], Infinity}]}}}]


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