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Odd or even integration

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi,

    I'm having a problem comprehending the odd-even trigonometry properties when doing an integration and I hope someone here feel like explaining since I can't seem to find anything of this in my course literature.
    I suppose it's more or less of a integration problem.

    f(t) = (cost)(Θ(t + [itex]\frac{\pi}{2}[/itex]) - Θ(t - [itex]\frac{\pi}{2}[/itex]))


    2. Relevant equations
    3. The attempt at a solution

    When doing the integration of:
    [itex]\int (cost)(coswt - jsinwt) dt[/itex] from -pi/2 [itex]\rightarrow[/itex] +pi/2
    (The period is pi and the function is obviously even)

    Why do they then consider cost * -jsinwt to be negligible and equal to 0?

    http://www.wolframalpha.com/input/?i=integrate+cost*e^iwt+dt+from+-pi/2+to+pi/2
     
  2. jcsd
  3. Sep 6, 2014 #2

    PeroK

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    Are you sure this is correct?

    ##Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}) = Θ \pi##
     
  4. Sep 6, 2014 #3
    Yes. The function stretches from -pi/2 --> pi/2
    It's cut off
     
  5. Sep 6, 2014 #4
    1. ##\cos t(\cos\omega t - j\sin\omega t)## is neither even nor odd (w.r.t. ##t##), and its periodicity (as a function of ##t##) depends on ##\omega##.

    2. ##\cos\omega t-i\sin\omega t=e^{-i\omega t}##

    3. ##\cos t\cdot (-j)\sin\omega t## is an odd function (w.r.t. ##t##), so ##\int_{-a}^a \cos t\cdot (-j)\sin\omega t\ dt=0## for any real ##a##. In particular, ##\int_{-\pi/2}^{\pi/2} \cos t\cdot (-j)\sin\omega t\ dt=0##.

    I would assume, from context, that ##\Theta## is a function of ##t## here.
     
  6. Sep 6, 2014 #5
    Thanks. Seems logical :)
     
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