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Quick question on Fourier transform

  1. Sep 10, 2013 #1
    Hi all, as a physics student, I seldom use Fourier transform but from my understanding, given a periodic function you can decompose the function into sine function with different frequencies. Also, to get a ultra short pulse in time domain, this would require mixing many frequencies. I would therefore like to ask what happens if I pass my monochromatic light through a optical chopper. Is Fourier transform not applicable in this case since ultra short pulse generation does not require many frequencies?
     
  2. jcsd
  3. Sep 10, 2013 #2
    Ultra short pulses DO require many frequencies. Fourier transform theory applies
     
  4. Sep 10, 2013 #3

    boneh3ad

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    Of course when analyzing pulses, it may be advantageous to look at the wavelet transform instead depending on what you actually hope to get out of the analysis.
     
  5. Sep 10, 2013 #4

    marcusl

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    Fouier theory says that the spectrum Z of the product of two signals z(t) = x(t) y(t) is the convolution of their spectra, Z = X * Y. Your signal consists of a monochromatic wave times a periodic square wave, so its spectrum is the convolution of their spectra. You can easily figure it out from there.
     
  6. Sep 10, 2013 #5
    A single frequency like a pure sine wave is only a single frequency if it starts at the beginning of time and goes on forever. As soon as you start or stop it you introduce a step which is composed of higher frequencies. The sharper the step the higher are the included frequencies. So chopping introduces harmonics.
     
  7. Sep 10, 2013 #6
    I am aware that you need many frequencies to get short pulse. That's what Fourier transform tells us. In my example, I specifically said I only have single wavelength as source.

    So if I turn on and off my laser I introduce higher harmonics as well?

    Ain't this just the math? I am confused how higher frequencies are generated.
     
  8. Sep 11, 2013 #7

    marcusl

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    The rising and falling edges of the "chopper" are very short, and therefore contain high frequency components. An ideal square wave modulation has infinite slopes, thus having a spectrum that is non-zero to infinite frequencies.
     
  9. Sep 11, 2013 #8
    Don't really understand how this comes about. Is there any reference to explain in detail about this?
     
  10. Sep 11, 2013 #9
    That's an application of the uncertainty relation. When you chop down the beam you are making a measurement of the photon location. By the uncertainty principle the photon momentum becomes uncertain creating a spread on the wavelength. In other words, the uncertainty principle is the physical principle that expresses the spread in wavelength required by the mathematics of waves which can be studied and understood through Fourier transforms.
     
  11. Sep 11, 2013 #10

    marcusl

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    I second what dauto said. There is an uncertainty relation for Fourier transforms which is the analog of Heisenberg's Uncertainty Principle, and that applies to common classical signals. You can find some information on Wikipedia, and more in any text on Fourier transforms. I like Bracewell's text, which has an excellent chapter discussing relationships between the "two domains" (time and frequency).

    "Quote by dauto
    Ultra short pulses DO require many frequencies. Fourier transform theory applies

    I am aware that you need many frequencies to get short pulse. That's what Fourier transform tells us. In my example, I specifically said I only have single wavelength as source."

    Once you modulate the carrier (pure sine) with an envelope, you no longer have a pure sine of single wavelength. You have a time-varying envelope that may have a rich spectrum, as I indicated earlier.
     
  12. Sep 12, 2013 #11
    A square wave (& therefore loosly speaking a square pulse) is the sum of all the odd harmonics.

    Fourier decomposition of a (periodic) signal will give you a long string of harmonics along with their amplitudes that when added together will give you your original signal back. If the wave is square then you will only get odd harmonics. In practice you may well get low amplitude even harmonics as well.
     
  13. Sep 12, 2013 #12

    Dale

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    You might start here http://en.wikipedia.org/wiki/Gibbs_phenomenon

    A square wave has an infinite spread of frequencies. Gibbs ringing is what happens when you try to approximate a square wave with a finite spread of frequencies.
     
  14. Sep 13, 2013 #13

    Claude Bile

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    I think the key point here is that chopping/modulating a signal introduces extra frequency components.

    Claude.
     
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