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Note: This questions is quite different from the ones referred to in the comments. Those deal with numerical questions, while this one is algebraic.

I have plots of the following type:

Plot[Cos[50 t] + Cos[51 t], {t, 0, 10}]

enter image description here

I would like to plot a envelope over this plot, i.e. another plot that joins all of maxima and minima of this plot respectively. Here is my attempt, but it's not exactly what I'd like:

Plot[{Cos[50 t] + Cos[51 t], Cos[t] + 1.5, -Cos[t] - 1.5}, {t, 0, 10}]

enter image description here

How can I generate the actual envelope?

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I'm guessing that's something like Cos[42x]+Cos[43x]? –  Mark McClure Jun 3 at 18:29
this is completely true but this is for the Fig.1, I want to access to sheath while I do not access to any formula for that. –  mostafa Jun 3 at 18:32
Well, you've got to give us some kind of input to start with. –  Mark McClure Jun 3 at 18:34
I plotted these ones with: Plot[{Cos[50 t] + Cos[51 t], Cos[t] + 1.5, -Cos[t] - 1.5}, {t, 0, 10}], I used of simulated functions 'Cos[t] + 1.5' and '-Cos[t] - 1.5' for the sheath. –  mostafa Jun 3 at 18:36
In that case, your sheath is given pretty much perfectly by $\pm 2\cos(t/2)$, but maybe that's not what you want? –  Mark McClure Jun 3 at 18:40

2 Answers 2

up vote 8 down vote accepted

Playing with the manipulate below might help. It's based on the the acoustics of beats.

  {Cos[a*t] + Cos[b*t], 2*Cos[(b - a) t/2], -2*Cos[(b - a) t/2]}, {t, 0, 10},
  PlotStyle -> {
   Directive[Black, Thick], 
   Directive[Black, Thick]}],
 {{a, 20}, 1, 50}, {{b, 21}, 1, 50}]

enter image description here

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Thank you so much. this is correct –  mostafa Jun 3 at 18:58

Don't mind me, I'm just having fun.

Grab the definition of HilbertTransform from this previous post, and then:

f[t_] := Cos[50 t] + Cos[51 t] + Sin[53 t] (* more sinusoids = more fun *)
g[t_] := Evaluate@HilbertTransform[f[τ], τ, t]
h[t_] := Abs[f[t] + I g[t]]
Plot[{f[t], h[t], -h[t]}, {t, 0, 10}, ImageSize -> Large, PlotPoints -> 100,
 PlotStyle -> {Automatic, Black, Black}]

enter image description here

You can see that the envelope has a nice analytical form:

Simplify[h[t], Assumptions -> t ∈ Reals]

$\left| e^{i t}-i e^{3 i t}+1\right|$

Further reading: analytic representation.

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This way though doesn't seem to catch when the envelope should go to zero. Shouldn't it e.g. vanish near 2? –  Ruslan Jun 4 at 8:34
Should it? It still looks like a sinusoid with small but nonzero amplitude there. (The envelope correctly goes to zero for the original function in the question, if that's what you're concerned about.) –  Rahul Narain Jun 4 at 8:55
Ah, OK then, you're right. –  Ruslan Jun 4 at 9:02

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