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What does a fourier transform do?

  1. May 29, 2014 #1

    joshmccraney

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    hey pf!

    physically, what does a fourier transform do? physically what comes out if i put velocity in?

    thanks!

    josh
     
  2. jcsd
  3. May 31, 2014 #2

    Simon Bridge

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    The fourier transform is not a physical transformation, it does not do anything physical.

    It is a mathematical tool which turns one function into another one.
    The interpretation depends on the form of the transform chosen and the context of the calculation.

    If you took the fourier transform of, say, the velocity-time function of some object, then you'd get the frequency domain form of that function. This is something that crops up in hydrodynamics, but you may be able to see the effect better transforming the v(t) function for SHM.
     
  4. May 31, 2014 #3

    joshmccraney

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    What do you mean By frequency domain with respect to the velocity? I tried reading online sources (and I'll continue to) but if you know perhaps you can save me some time?

    Thanks for your help!
     
  5. May 31, 2014 #4

    Simon Bridge

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    Well v(t) is the velocity in the time domain.

    Put v(t)=v.sin(wt) (SHM)
    then find V(w), it's forward fourier transform and see what you get.
     
  6. Jun 1, 2014 #5

    joshmccraney

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    ok, so we have $$F(x) = \int_{-\infty}^{\infty} e^{-i 2 \pi t x} \sin(\omega t) dt$$ there are probably easier ways to integrate, but if we integrate by parts twice, we arrive at $$F(x) = \left. \frac{\omega \cos(\omega t)+2 i x \pi \sin(\omega t)}{e^{-2 i x \pi t}(4 x^2 \pi^2-\omega^2)} \right|_{-\infty}^{\infty}$$ which implies
    $$F(x) = -\left. \frac{\omega \cos(\omega t)+2 i x \pi \sin(\omega t)}{e^{-2 i x \pi t}(4 x^2 \pi^2-\omega^2)} \right|_{-\infty}$$ which can be wrote as
    $$F(x) = e^{2 i x \pi t} \left. \frac{\omega \cos(\omega t)- 2 i x \pi \sin(\omega t)}{(4 x^2 \pi^2-\omega^2)} \right|_{\infty}$$ from here, if i use lopital's rule twice i will have a bunch of sine and cosine terms in the numerator multiplied by an increasing exponential. won't this diverge?
     
  7. Jun 1, 2014 #6

    Simon Bridge

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    Using your notation: $$F(x) = \int_{-\infty}^{\infty} e^{-i 2 \pi t x} \sin(\omega t)\; dt\\
    \qquad = \frac{\delta(x-\omega/2\pi)-\delta(x+\omega/2\pi)}{2i}$$

    Integrating exponentials is usualy easy - use Euler's formula to convert the sine, then you get a sum of integrals. Apply the integral definition of the Dirac delta function.

    Or you can just look it up:
    http://en.wikipedia.org/wiki/Fourier_transform#Distributions
     
    Last edited: Jun 1, 2014
  8. Jun 1, 2014 #7

    joshmccraney

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    is ##\delta (x)## the dirac delta function? can you explain how you arrived at this?

    thanks a ton for helping me!
     
  9. Jun 1, 2014 #8

    Matterwave

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    You can just express ##\sin(\omega t)## in terms of exponential functions:

    $$\sin(\omega t)=\frac{e^{i\omega t}-e^{-i\omega t}}{2i}$$

    Do you see how to arrive at his result now?
     
  10. Jun 1, 2014 #9

    joshmccraney

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    oh man, i didn't even see the bottom of your post! ok, ill give it a whirl.
     
  11. Jun 1, 2014 #10

    Simon Bridge

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    You were probably replying while I was editing my post to be a bit clearer ;)

    I had hoped that you would have gone from: ##v(t)=A\sin\omega_0 t## to using:
    $$V(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty v(t)e^{-i\omega t}\;dt\\
    \qquad = A\sqrt{2\pi}\frac{\delta(\omega-\omega_0)-\delta(\omega+\omega_0)}{2i}$$... or you may find it easier to think about ##v(t)=A\cos\omega t## so that ##V(\omega)## is real.

    So - back to post #1.
    A specific form of your question would be:
    "If ##v(t)## tells you how the velocity of an object undergoing SHM changes with time, what does ##V(\omega)## tell you?"

    Can you relate the values of V to the motion of the object?
    What happens if the motion is a bit more complicated?
     
  12. Jun 1, 2014 #11

    joshmccraney

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    ok simon, so i am on the same page with you now that the fourier transform over ##v_0 \sin (\omega t)## yields $$F(x) = v_0\frac{\delta(x-\omega/2\pi)-\delta(x+\omega/2\pi)}{2i}$$

    but what does this mean? having the dirac delta function without an integral is difficult for me to interpret. can you help?
     
    Last edited: Jun 2, 2014
  13. Jun 2, 2014 #12

    Simon Bridge

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    F is basically a distribution - sure.

    Notice that w/2pi is the frequency of the oscillations?
    Notice that F has dimensions of velocity?

    Explore what happens if f(t) is a sum of different-frequency sine waves? Maybe a product of them?

    Note: a mathematical operation on a function representing a physical process need not have a pat physical interpretation.

    This may help:
    https://www.physicsforums.com/showthread.php?t=217204
     
    Last edited: Jun 2, 2014
  14. Jun 2, 2014 #13

    joshmccraney

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    ok. so if i take ##v(t) = v_0 \sin(2\pi t / 2) \sin(2\pi t/3)## then we have a period of ##6## and a frequency of ##1/6##. i notice the ##5/6##, ##-5/6##, ##1/6##, ##-1/6## all show up in a sum of dirac delta functions (the sums and differences of the frequencies of the sine terms). whats happening here?
     
    Last edited: Jun 2, 2014
  15. Jun 2, 2014 #14

    Simon Bridge

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    As well as that - what happens to that constant out the front of the Dirac delta functions?
    What does it look like is happening? Looks like a discrete spectrum doesn't it?

    It is tricky to wrap you mind around what happens when you switch to the frequency domain.
    This is why I'm not just telling you.

    Sometime's it is clearer - like the Fourier transform of a position wavefunction is the corresponding momentum wavefunction. Wavefunctions are, similarly, best understood in integrals.
     
  16. Jun 2, 2014 #15

    joshmccraney

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    its looks like this constant is ##v_0 / 2i##

    yea, some kind of function that's identically zero except at the sums/differences of the sine waves' frequencies.

    what's the deal with these sums/differences? i would have thought, since fourier series decompose functions into fundamental frequencies of sine/cosine waves. perhaps because we have a product and not a sum?
     
  17. Jun 4, 2014 #16

    Simon Bridge

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    A standing wave is the sum of two travelling waves, same wavelength and amplitude, going in opposite directions.
    So Asin(wt) is made of two plane waves with amplitude A/2 headed in opposite directions.
    Alternatvely it is two phasors magnitude A/2 rotating in opposite directions.
    The Fourier transform reflects this, which is why it is so useful in signal processing.

    When you multiply two sine waves together, you can decompose that into a sum of sine waves.
    The argument of the sines involve the sum and difference of the frequencies.
    http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities$$A\sin(\omega_1 t) \sin(\omega_2 t) = \frac{A}{2}\cos (\omega_1-\omega_2)t-\frac{A}{2}\cos(\omega_1+\omega_2)t$$... so you'd expect the fourier transform to be non-zero at the sum and difference of the sine wave freqencies.
     
  18. Jun 5, 2014 #17

    joshmccraney

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    oh ok. from eulers identity. thanks!
     
  19. Jun 5, 2014 #18

    joshmccraney

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    you've been a big help!
     
  20. Jun 5, 2014 #19

    Simon Bridge

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    No worries - this is something that means different things in different contexts but you should have a better feel for what the transform is doing now. You'll get more when you study distributive integrals and measure theory.
     
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