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Unit step functions to find Laplace Transform

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    f(t) = cos (pi*t) if 1[itex]\leq[/itex] t <4 and 0 elsewhere

    using unit step functions to find Laplace Transform


    2. Relevant equations



    3. The attempt at a solution

    I came up with the unit step function f(t) = cos(pi*t) u(t-1) - cos(pi*t) u(t-4)
    in order to use the second shifting theorem f(t) must in the format of f(t-a) in this case a = 1 and 4.

    I add 1 and subtract 1 , add 4 and subtract 4 and I get

    f(t) = cos(pi(t-1+1) u(t-1) - cos(pi(t-4+4) u(t-4)
    = cos (pi(t-1)) u(t-1) +u(t-1) - cos(pi*(t-4) u(t-4)+u(t-4) << i think this step is not correct
    It should get -cos (pi(t-1)) u(t-1) - cos(pi*(t-4) u(t-4) but i dont know how to get there

    please help thank you
     
  2. jcsd
  3. Feb 5, 2013 #2

    jbunniii

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    It is not correct. Instead, let's go back to the previous step:
    $$f(t) = \cos(\pi(t-1+1)) u(t-1) - \cos(\pi(t-4+4)) u(t-4)$$
    I can write this as follows:
    $$f(t) = \cos(\pi(t-1)+\pi) u(t-1) - \cos(\pi(t-4)+4\pi) u(t-4)$$
    Now what can you say in general about ##\cos(x + \pi)##? What about ##\cos(x + 4\pi)##?
     
  4. Feb 5, 2013 #3

    Ray Vickson

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    Does the problem statement say you MUST use the unit step function? If not, just using the original form is as easy a way as any:
    [tex] \cal{L}(f)(s) = \int_0^4 e^{-st} \cos(\pi t) \, dt.[/tex]
     
  5. Feb 5, 2013 #4
    Thank you jbunniii I got it After that we use the trig identity of cos(x+y)=cos(x)-cos(y)-sin(x)sin(y)
    Thank you Ray Vickson for the tip
     
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