Laplace transform f ( t ) = f ( t + 2 π a ) = sin ⁡ a t | sin ⁡ a t |

Name: Laplace transform f ( t ) = f ( t + 2 π a ) = sin ⁡ a t | sin ⁡ a t |
Uploaded: 2022-02-23T17:22:57+00:00
Description: Laplace transform f ( t ) = f ( t + 2 π a ) = sin ⁡ a t | sin ⁡ a t |

Question

Laplace transform   f  (  t  )  =  f  (  t  +            2      π        a    )  =            sin      ⁡      a      t                      |            sin      ⁡      a      t              |

Adenomacu · Accepted Answer

Note that   T  =  2  π      /    a, and            sin      ⁡              a        t                            |            sin      ⁡              a        t                    |              =      {                                                +          1                          a                    t          ∈          (          0          ,          π          )                                      −          1                          a                    t          ∈          (          π          ,          2          π          )                        Then                              ∫          0                      2            π                          /                        a                          d        t                                      sin            ⁡                          a              t                                                          |                        sin            ⁡                          a              t                                      |                                                        e                      −            s            t                                              =                  ∫          0                      π                          /                        a                          d        t                          e                      −            s            t                          −                  ∫                      π                          /                        a                                2            π                          /                        a                          d        t                          e                      −            s            t                                                            =                              1            −                          e                              −                π                s                                  /                                a                                              s                −                                            e                              −                π                s                                  /                                a                                      −                          e                              −                2                π                s                                  /                                a                                              s                                                  =                              (            1            −                          e                              −                π                s                                  /                                a                                                    )              2                                s                    The full Laplace transform is then      ∫    0          ∞        d  t              sin      ⁡              a        t                            |            sin      ⁡              a        t                    |                    e          −      s      t        =            (      1      −              e                  −          π          s                      /                    a                            )        2                    s      (      1      −              e                  −          2          π          s                      /                    a                    )        =      1    s    tanh  ⁡                    π        s                    2        a

klastiesym · Accepted Answer

I did by let u = at, so the period T = 2pi

\begin{array}{l} f (t) = f (t + \frac{2 π}{a}) = \frac{\sin a t}{| \sin a t |} \\ L e t : f (u) = (u + 2 π) = \frac{\sin u}{| \sin u |} \\ F (s) = L [f (u)] = \frac{1}{1 - e^{- 2 π s}} \int_{o}^{2 π} e^{- s t} \frac{\sin a^{2} u}{| \sin a^{2} u |} d u \\ = \frac{1}{1 - e^{- 2 π s}} {\int_{o}^{π} e^{- s t} d u - \int_{π}^{2 π} e^{- s t} d u} \\ = \frac{e^{π s} - 1}{s (e^{π s} + 1)} \\ L [f (u)] = \frac{e^{π s} - 1}{s (e^{π s} + 1)} \\ => L [f (a t)] = \frac{1}{a} F (\frac{s}{a}) = \frac{e^{\frac{π s}{a}} - 1}{a \frac{s}{a} (e^{\frac{π s}{a}} + 1)} = \frac{e^{\frac{π s}{a}} - 1}{s (e^{\frac{π s}{a}} + 1)} = Y o u r s \end{array}

Laplace transform f(t)=f(t+(2pi)/a)=(sin at)/(|sin at|)

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