Plotting Dirac Delta Function as colored arrows

Question

I want to produce graphs of Fourier transforms for lectures.

Using the answer from Calling Correct Function for Plotting DiracDelta I get a problem with the code mentioned below.

Definition of Mr. Fortuño

    ArrowsDeltaFunction[eqn_, x_] := 
  Module[{xsubs, listDeltas, coefDeltas, locationDeltas},
   xsubs = (x /. 
       Cases[eqn, DiracDelta[a__] :> Solve[a == 0, x], Infinity]) /. 
     x -> {};
   listDeltas = DiracDelta[x - x0] /. x0 -> xsubs;
   coefDeltas = Flatten[Coefficient[eqn, listDeltas]];
   locationDeltas = Flatten[xsubs];
   ar = Arrow[
     Table[{{locationDeltas[[i]], 0}, {locationDeltas[[i]], 
        coefDeltas[[i]]}}, {i, 1, Length[locationDeltas]}]];
   ds = Table[{EdgeForm[Opacity[0.8]], White, 
      Disk[{locationDeltas[[i]], 0}, 0.15]}, {i, 1, 
      Length[locationDeltas]}];
   ards = 
    Table[{Arrowheads[Medium], 
      Arrow[{{locationDeltas[[i]], 0}, {locationDeltas[[i]], 
         coefDeltas[[i]]}}], {EdgeForm[Opacity[0.8]], 
       White, {AspectRatio -> Automatic, 
        Disk[{locationDeltas[[i]], 0}, 0.05]}}}, {i, 1, 
      Length[locationDeltas]}]
   ];

and his example (modified with parameter lambda)

    Ft := (1 - I/2) Exp[-I t] + Cos[\[Lambda]1 t] + 
  2 Sin[\[Lambda]2 (t - 2)]

(*Fourier Transform*)
Fw := FourierTransform[Ft, t, w]
Fw
ReFw := ComplexExpand[Re[Fw]]
ImFw := ComplexExpand[Im[Fw]]

\[Lambda]1 = 3; \[Lambda]2 = 2;

Fw
ReFw := ComplexExpand[Re[Fw]]
ImFw := ComplexExpand[Im[Fw]]

(*Plot*)
Plot[{ReFw, ImFw}, {w, -5, 5}, PlotStyle -> {Thick, Red, Blue}, 
 PlotRange -> 3, AxesLabel -> {"w", "F(w)"}, 
 PlotLegends -> LineLegend[{Red, Blue}, {"Real part", "Imag part"}], 
 Epilog -> {Thick, Red, ArrowsDeltaFunction[ReFw, w], Blue, 
   ArrowsDeltaFunction[ImFw, w]}]

I get the correct picture

Using

Ft := A1 Sin[\[Omega]1 t + \[Phi]1] + A2 Sin[\[Omega]2 t + \[Phi]2];

with

A1 = 1; A2 = 1.25; \[Omega]1 = 4.6; \[Omega]2 = 4.3; \[Phi]1 = 0; \[Phi]2 = \[Pi]/2;

gives

    1.56664 DiracDelta[(4.3 + 0. I) - \[Omega]] + 
 I Sqrt[\[Pi]/2] DiracDelta[(4.6 + 0. I) - \[Omega]] + 
 1.56664 DiracDelta[(4.3 + 0. I) + \[Omega]] - 
 I Sqrt[\[Pi]/2] DiracDelta[(4.6 + 0. I) + \[Omega]]

and the picture is missing two arrows

Why is Mathematica substituting complex values with zero imaginary parts?

Why are the missing to arrows not plotted (the circles are there)?

Any help is appreciated

Please notice

Adding a comment to Francisco Rodríguez Fortuño and Ponkadoodles original answer was not possible, therefore I created this new question.

I have also visited the following links which did not solve my problem.

Is marked as having solutions Plot periodic function from Dirac delta function and links to the following two links that only offer manual solutions Plot Fourier transform of $\sin (2 t)$ Plot Dirac Delta function

Answer 1

Here is a code that plots your arrows correctly:

ca = Cases[
   1.56664 DiracDelta[(4.3 + 0. I) - x] + 
    I Sqrt[\[Pi]/2] DiracDelta[(4.6 + 0. I) - x] + 
    1.56664 DiracDelta[(4.3 + 0. I) + x] - 
    I Sqrt[\[Pi]/2] DiracDelta[(4.6 + 0. I) + x], 
   a_ DiracDelta[b_] :> {{a}, x /. Solve[b == 0]}];
di = Flatten[Tuples /@ ca, 1];
im = Select[di, Im[#[[1]]] != 0 &];
re = Select[di, Im[#[[1]]] == 0 &];
mm = Max[Abs[Join[re[[All, 1]], Im[im[[All, 1]]]]]] + 0.1;
Plot[{}, {x, -5, 5}, 
 Epilog -> {Red, Arrow[{{#[[2]], 0}, {#[[2]], #[[1]]}}] & /@ re, Blue,
    Arrow[{{#[[2]], 0}, {#[[2]], Im[#[[1]]]}}] & /@ im}, 
 PlotRange -> mm]
Clear[ca, di, im, re, mm]

He are more complex DiracDelta functions plotted:

-2 DiracDelta[x + 1] + 5 DiracDelta[(7 x + 3) (x - 4)] - 
 3 I DiracDelta[x - 3/2] + 2 I DiracDelta[x + 3]

Answer 2

In the first case, you use exact numbers for the frequencies, in the second case, machine numbers. Machine numbers only give exact results in special cases. This is the reason you get complex values from the FourierTransform. You can either rationalize your input or use Chop to get rid of the spurious imaginary part.

The missing arrows come from the fact, that you specify: DiracDelta[x - x0] /. x0 -> xsubs. This evaluates to DiracDelta[w +/- x0]. You compare this against: DiracDelta[x0 +/- w] what does not match in the case you have a -.

Another hint: if you rationalize your input take care of the fact that MMA simplifies e.g. DiracDelta[x-2/3] to 3 DiracDelta[-2+3x].