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Use Fourier analysis to construct an amplitude spectrum

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    A sensor yields a signal y(t) = |sin(120[itex]\pi[/itex]t)|
    a. Using Fourier analysis please construct an amplitude spectrum for this signal.

    2. Relevant equations
    A0 = [itex]\frac{1}{T}[/itex][itex]\int ^{-T/2}_{T/2}y(t) dt[/itex]
    An =[itex]\frac{2}{T}[/itex][itex]\int^{-T/2}_{T/2}y(t)cos\frac{2n\pi t}{T}dt[/itex]

    3. The attempt at a solution
    Because y(-t) = y(t), the function is even, and we can ignore Bn.

    My question is: what is the period? In prior examples, we used T = 2[itex]\pi[/itex].
    Therefore, will:
    A0 = [itex]\frac{1}{2\pi}[/itex] [itex]\int^{-\pi}_{\pi}[/itex] |sin(120[itex]\pi[/itex]t)| dt?
     
  2. jcsd
  3. Mar 12, 2013 #2

    LCKurtz

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    The period of ##\sin(bt)## is ##2\pi /b## so the period of your function without the absolute values would be$$
    T=\frac{2\pi}{120\pi}=\frac 1 {60}$$I would use that in half range formula. Note that by using the half range formula you can drop the absolute value.
     
    Last edited: Mar 12, 2013
  4. Mar 12, 2013 #3
    Thank you!

    By half range, do you mean integrating between 0 and 30?
     
  5. Mar 12, 2013 #4
    Also, would this mean that the function is now odd?
     
  6. Mar 12, 2013 #5

    LCKurtz

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    No. Half the period would be 1/120.

    No. You use the half range cosine formulas which gives the even extension of the function, which has the same effect as the absolute values. Look at the half range formulas in your text.
     
  7. Mar 12, 2013 #6
    Duh! Brain fart. Sorry, I don't have a text for this class, so I've been relying on google. Thanks again for your help!
     
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