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Not strictly a physics question, anyone good at Fourier Transforms?

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data
    7NeLc.png


    2. Relevant equations

    7JlxA.png

    3. The attempt at a solution

    To be perfectly honest I have not attempted a solution thus far, my knowledge of the fourier transform is quite limited at the moment. I understand that I can use the above equation but what I want to know is if I can compute the transform for each step and then add them all together to create the final transform?

    TLDR: Can i add the transform of each step to form the final transform?
     
  2. jcsd
  3. Dec 7, 2011 #2
    yes
    when you integrate over the piecewise f that's pretty much what you're doing anyway

    [itex]F[\omega]=\int_{-\infty}^{\infty} f[t]e^{i \omega x}dt[/itex]

    just becomes;

    [itex]F[\omega]=A \int_{-2 \tau}^{- \tau} e^{i \omega x}dt + 2 A\int_{- \tau}^{\tau} e^{i \omega x}dt + A \int_{\tau}^{2\tau} e^{i \omega x}dt[/itex]

    which is what you'd get if you just added up each step

    at least that's how I'd see it, I haven't spent much time looking at fourier transforms myself
     
  4. Dec 7, 2011 #3

    LCKurtz

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    Just break the interval of integration [-2t,2t] up into three pieces where f(t) is nonzero.
     
  5. Dec 7, 2011 #4

    zcd

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    The fourier transform is linear, so you can treat the total signal as the sum of several step pulses and add their frequency responses after taking the FT.
     
  6. Dec 7, 2011 #5
    No, you should break it into 2 signals and add them together to get the signal you have. These two signals have a sinc function as their fourier transform.
     
  7. Dec 7, 2011 #6
    That's what I have written down on my page right now pretty much, except my i's are negative.

    I get the first integral in the second equation you've written to be = [itex]\frac{A}{i\omega}(e^{2i\omega\tau} - e^{i\omega\tau}[/itex]

    Do you think this is correct? I can't see a way to reduce that to a sine function yet but i'll work on the rest.
     
  8. Dec 7, 2011 #7
    Wait i'm confused, LCKurtz says break into 3 but Dickfore says break it into 2 signals?

    I've already broken it into three signals and i'm working on that, unless you can expand on your reason as to why that's a bad idea.
     
  9. Dec 7, 2011 #8
    Because 3 signals would be asymertic w.r.t. t = 0 and you will need to do a F.T. on 3 different things. With 2, they are both symmetric w.r.t. t = 0 and you apply the same F.T. twice.
     
  10. Dec 7, 2011 #9

    zcd

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    Again, FT is linear so you can break it however you want. The most obvious ways are to break it into 3 pulses -2τ to -τ (A), -τ to τ (2A), and τ to 2τ (A) or add 2 pulses -2τ to 2τ (A) to -τ to τ (A).
     
  11. Dec 7, 2011 #10

    LCKurtz

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    Both methods work. Dick is suggesting you can save a little work by using symmetry.
     
  12. Dec 7, 2011 #11
    Ah ok, so splitting it into 2 signals isn't immediately simpler as I have to find f(t) from -2tau to 0 but once I have this function and I have applied the F.T to it, it's symmetric and therefore simpler?

    hmm... I'm going to work on the more obvious method of 3 transforms as I don't know what f(t) would be from -2tau to 0.
     
  13. Dec 7, 2011 #12
    I would recommend (as this problem seems geared towards a beginner) just defining f(t) explicitly in terms of a mathematical expression (it will involve Heaviside step functions) and then simply plug it in the integral and do the integral. People who try to take conceptual shortcuts get themselves in trouble unless they really know what they are doing. I tell my students to get good at the rigorous, formulaic approach before they start taking conceptual shortcuts.
     
  14. Dec 7, 2011 #13
    The sign of the i doesn't really matter since everything is symmetric, but I did actually mean to put a minus sign there, it was a typo, so is the x in the exponent instead of the t..

    anyway, I just did the integration myself and got

    [itex]\frac{i A e^{-i \tau \omega }}{\omega }-\frac{i A e^{i \tau \omega }}{\omega }+\frac{i A e^{-2 i \tau \omega }}{\omega }-\frac{i A e^{2 i \tau \omega }}{\omega }[/itex]

    and if I'm not mistaken this should go into sins, and you'll have to add your sqrt 2 pi back in also and that should be you sorted
     
  15. Dec 7, 2011 #14
    I have a very similar result generic; except that my i's are on the denominator (which is just as mathematically sound as your result)

    The identity given to me for transforming exponentials into sine functions is:

    [itex]sin(A) = \frac{1}{2i}(e^{iA}-e^{-iA})[/itex]

    I'm just trying to work out how to apply that to what you and I have, I'll put my result below for comparison:

    [itex]F(\omega) = \frac{A}{i \omega sqrt{2\pi}}(e^{2 i \omega \tau}-e^{-2 i \omega \tau}) + \frac{A}{i \omega sqrt{2\pi}}(e^{i \omega \tau}-e^{- i \omega \tau})[\itex]

    It should just fall out into the answer provided, so I'm not too worried at this point.

    Thanks alot for all the help provided so far everyone, the discussion of different techniques has been quite enlightening.
     
  16. Dec 7, 2011 #15
    I can't seem to find the edit button so I've corrected my slashes in this post:

    [itex]F(\omega) = \frac{A}{i \omega \sqrt{2\pi}}(e^{2 i \omega \tau}-e^{-2 i \omega \tau}) + \frac{A}{i \omega \sqrt{2\pi}}(e^{i \omega \tau}-e^{- i \omega \tau})[/itex]
     
  17. Dec 7, 2011 #16
    if you use the fact that 1/i = -i you can find what I had
    [itex]\frac{A}{\sqrt{2 \pi }\omega }\left(i\text{ }e^{-i \tau \omega }-i e^{i \tau \omega }+i\text{ }e^{-2 i \tau \omega }-i\text{ }e^{2 i \tau \omega }\right)[/itex]

    then you use the fact that [itex]e^{-i \tau \omega }=cos(\tau \omega) - i sin(\tau \omega)[/itex]
    and then since cos is symmetric the cos parts in [itex]\frac{i A e^{-i \tau \omega }}{\omega }-\frac{i A e^{i \tau \omega }}{\omega }[/itex] go away leaving you with 2 sins, you then do the same for the e^2 parts and you should get the result you're looking for
     
  18. Dec 7, 2011 #17
    I used the identity:
    [itex]\frac{1}{i}(e^{i\omega\tau}-e^{-i\omega\tau}) = 2 sin(\omega \tau)[/itex]

    instead of the one you provided because it seemed to just cut out dealing with cosine terms altogether.

    this left me with:

    [itex]F(\omega) = \frac{2A\tau}{\sqrt{\omega\tau}} ( \frac{sin(2\omega\tau)}{2\omega\tau} + \frac{sin(\omega\tau)}{\omega\tau}) [/itex]

    When comparing this to the final solution it seems I have lost the 2 that should go before the [itex]sin(2\omega\tau)[/itex] term. I haven't provided my full working so no one will be able to see where I have gone wrong, but does this happen for your solution? I thought they were identical.
     
  19. Dec 7, 2011 #18
    yeah, it happened here too..
    could it just be a typo?
    I'm not sure where that extra 2 could come from
     
  20. Dec 7, 2011 #19
    I don't think it is, I haven't received any emails about it so I doubt so.

    My friend is working on the same piece at the moment so we'll see if he gets the same apparent error as us and then I'll start thinking it could be a typo.
     
  21. Dec 7, 2011 #20
    it seems the solution given is just confusing, as the 2s should cancel but they have both been written in. There needs to be that extra 2 there for the simplification to be valid, what we have written does not simplify to what we once had.

    If that makes any sense.
     
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