5
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I want to create a piecewise function that repeats after t=T.

Clear[t, T, s];
T = 5;
s[t_] := \[Piecewise] {
   {t, 0 < t <= 1},
   {1, 1 < t <= 2},
   {-t + 3, 2 < t <= 3},
   {0, 3 < t <= T}
  }

To make it periodic, I add the following:

s[t_] := s[t - T] /; t > T
Plot[{s[t], Evaluate@D[s[t], t]}, {t, 0, 12}, Exclusions -> None]

It plots alright, but the derivative of s[t] stops at the end of T.

enter image description here

Question What is the correct way of defining and plotting the derivative and integral of a periodic waveform defined using the Piecewise function?

Thanks in advance for your help and suggestions.

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2
  • 1
    $\begingroup$ Plot[{s[t], D[s[x], x] /. x -> Mod[t, T]}, {t, 0, 12}]? $\endgroup$
    – kglr
    Jan 2, 2022 at 2:45
  • $\begingroup$ That's three good solutions on one page using Mod. kglr's solution (in the comments) is to find the derivative for the base case and plot it using Mod for different ranges. Nasser's solution is suitable for teaching canonical use of Piecewise. It defines a base function and then defines a periodic function based on it, perhaps using T as a parameter. Ted Ersek's solution shows the Gibb's phenomenon that alludes to MMa's use of Series being used in the background. My thanks to all the respondents. $\endgroup$
    – Syed
    Jan 2, 2022 at 6:39

2 Answers 2

4
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One way could be

Clear[t, T, s, s0];
T = 5;
s[t_] := Piecewise[{{s0[Mod[t, T]], t > T}, {s0[t], True}}]
s0[t_] := Piecewise[{{t, 0 < t <= 1}, {1, 1 < t <= 2}, {-t + 3, 
    2 < t <= 3}, {0, 3 < t <= T}}]

and now

Plot[s[t], {t, 0, 12}, Exclusions -> None]

Mathematica graphics

and

Plot[Evaluate[D[s[t], t]], {t, 0, 12}, Exclusions -> None]

Mathematica graphics

And

Plot[{s[t], Evaluate[D[s[t], t]]}, {t, 0, 12}, Exclusions -> None]

Mathematica graphics

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2
  • $\begingroup$ Thanks. Can you please elaborate why my method fails for the derivative but not for the plot? $\endgroup$
    – Syed
    Jan 1, 2022 at 21:14
  • 1
    $\begingroup$ @Syed I just noticed that when you had Clear[t,T,s]; T=5; s[t_]:=Piecewise[{{t,0<t<=1},{1,1<t<=2},{-t+3,2<t<=3},{0,3<t<=T}}]; s[t_] := s[t - T] /; t > T then D[s[t],t] gives zero for t>3. But changing it to s[t_]:=Piecewise[{{s0[Mod[t,T]],t>T},{s0[t],True}}]; s0[t_]:=Piecewise[{{t,0<t<=1},{1,1<t<=2},{-t+3,2<t<=3},{0,3<t<=T}}] now D[s[t], t] now gives correct derivatives mod T. $\endgroup$
    – Nasser
    Jan 2, 2022 at 4:27
2
$\begingroup$
Clear[s];

s[t_?(0<=#<=5&)]:=\[Piecewise]{
  {t,0<=t<=1},
  {1,1<t<=2},
  {-t+3,2<t<=3},
  {0,3<t<=5}
};

s[t_?(Element[#,Reals]&)]:=s[Mod[t,5]];

Edit by @Syed (to include plot)

Plot[{s[t], Evaluate[D[s[t], t]]}, {t, 0, 12}, Exclusions -> None]

enter image description here

**** Edit **** The derivative plotted above has errors.

Wolfram Tech Support said: Since there are conditions on the input, there is no symbolic derivative, so s'[real number] is computing an approximate derivative, which is of very low quality.

Another solution is given above that works much better.

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