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For complex periodic signal:

$-\frac{8 T \left(2 T \omega \sin (2 t \omega )-e^{-\frac{t}{T}}+\cos (2 t \omega )\right)}{\alpha ^2 \left(4 T^2 \omega ^2+1\right)}$

where $T,\alpha,\omega$ - parameters, $t$ - time

How to calculate the supremum in symbolic form?


This is what the Limit and MaxLimit commands show.

enter image description here


And here is the expression for signal itself

s=-((8 T (-E^(-(t/T)) + Cos[2 t \[Omega]] + 
    2 T \[Omega] Sin[2 t \[Omega]]))/(\[Alpha]^2 (1 + 
    4 T^2 \[Omega]^2)))
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  • $\begingroup$ What abouut -((8 T ( Cos[2 t \[Omega]] + 2 T \[Omega] Sin[2 t \[Omega]]))/(\[Alpha]^2 (1 + 4 T^2 \[Omega]^2)))? $\endgroup$ Commented May 20, 2021 at 8:36

2 Answers 2

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I assume that T,α,ω are positive. Apart from a constant factor, the main part of your expression is:

(-E^(-(t/T)) + Cos[2 t ω] + 2 T ω Sin[2 t ω])

The first term is negative, it goes to zero for large t, the 2 others are periodic. Therefore, the supremum is determined by the 2 periodic terms. The extrema are reached at t:

t0 = Solve[D[Cos[2 t \[Omega]] + 2 T \[Omega] Sin[2 t \[Omega]], t] == 0, t]; 

And the function values:

 Cos[2 t \[Omega]] + 2 T \[Omega] Sin[2 t \[Omega]] /. t0

enter image description here

The supremum is the second one of those values multiplied by the constant factor.

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  • $\begingroup$ Sqrt[(-((8 To)/(\[Alpha]^2 (1 + 4 To^2 \[Omega]^2))))^2 + (-(( 16 (To^2) \[Omega] )/(\[Alpha]^2 (1 + 4 To^2 \[Omega]^2))))^2 I got it like this $\endgroup$
    – ayr
    Commented May 20, 2021 at 8:48
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You might visualize the supremum (see @DanielHuber's answer, to fast for me ;-) ) by introducing dimensionsless time t/Tand parameter \[Omega]T=\[Omega] T

Manipulate[
 Plot[ {-((8   (-E^(-(\[Tau])) + Cos[2 \[Tau] \[Omega]T] + 
          2   \[Omega]T Sin[2 \[Tau] \[Omega]T]))/(  (1 + 
         4   \[Omega]T^2))),
   -((8   ( 
         Cos[2 \[Tau] \[Omega]T] + 
          2   \[Omega]T Sin[2 \[Tau] \[Omega]T]))/(  (1 + 
         4   \[Omega]T^2)))}, {\[Tau], 0, 5}, 
  AxesLabel -> {"t/T", "s[t]\[Alpha]^2/T"}], {{\[Omega]T, 1}, 0, 5, 
  Appearance -> "Labeled"}] 

enter image description here

The supremum function (original parameters) follows to

Simplify[-((8 (Cos[2 \[Tau] \[Omega]T] + 2   \[Omega]T Sin[2 \[Tau] \[Omega]T]))/(  (1 +4\[Omega]T^2))) T/\[Alpha]^2 /. {\[Omega]T -> \[Omega] T, \\[Tau] -> t/T }]   
(*-((8 T (Cos[2 t \[Omega]] +2 T \[Omega] Sin[2 t \[Omega]]))/(\[Alpha]^2 (1 + 4 T^2 \[Omega]^2)))*)
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  • $\begingroup$ Sqrt[(-((8 To)/(\[Alpha]^2 (1 + 4 To^2 \[Omega]^2))))^2 + (-(( 16 (To^2) \[Omega] )/(\[Alpha]^2 (1 + 4 To^2 \[Omega]^2))))^2 I got it like this $\endgroup$
    – ayr
    Commented May 20, 2021 at 8:48
  • $\begingroup$ The supremum function you are looking for should be -((8 T (Cos[2 t \[Omega]] +2 T \[Omega] Sin[2 t \[Omega]]))/(\[Alpha]^2 (1+4 T^2 \[Omega]^2))) I think. $\endgroup$ Commented May 20, 2021 at 8:53

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