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Fourier transform and uncertainity principle

  1. Apr 24, 2009 #1
    To find the frequency, Why do you need to consider the signal over long period of time?
    For example - if you look at a sine wave from 0-360 with two cycles, isn't it enough to get the frequency?
    I get the second part - you need a short time window to see sudden changes in frequency.
  2. jcsd
  3. Apr 24, 2009 #2


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    If you look at just one or a few wavelengths, you don't know if the signal has suddenly been cut off. The cut offs with their steep slopes correspond to very high frequencies.
  4. Apr 24, 2009 #3
    Thanks. I am sort of convinced but not entirely. So if the signal is taken over a long time period, even then the signal has to be cut off some where.
    I guess what you are saying is in case of long time, the cut off is so small it doesn't matter?

    Also, why does it take a lot of sines and cosines to make up a signal taken over small time interval, like a spike.
  5. Apr 25, 2009 #4


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    Well, long time is actually infinite - it's really in the definition of sine and cosine itself.

    It takes lots of sines and cosines to make a spike because a sine looks nothing like a spike, so you have to take lots of them, arrange them with the right phases to make sharp slopes, and cancel out the gentle slopes. The phase is very important, since you can also add sines of all frequencies and get white noise.

    It's somewhat annoying that we add up infinitely long signals to make a finite duration signal, especially if we would like to say something like "the frequency content changes with time", eg. in music. So there are variations of the pure Fourier transform eg. short-time Fourier transform, wavelet analysis etc. But in all such analyses, there is always some sort of time-frequency trade-off.
  6. Apr 25, 2009 #5
    But if you're only looking at the sine wave in a finite interval, how do you know that it's a sine wave outside said interval? It could drop off to zero, or it could turn into a polynomial, or it could become any one of an infinite number of functions.
  7. Apr 27, 2009 #6
    I am only interested in the interval (only in that particular window). A lot of real world signals are not uniform over a long period of time.
    Can you explain what you mean by -"it could turn in to a polynomial".
    Also, I don't understand why sine wave is the basic wave. what is a sine wave?
  8. Apr 27, 2009 #7


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    If you want a description of the sine wave use wikipedia. I'm not sure what answer you want because you started talking about sine waves so you seem to know what they are.

    Sinusoidal waves being the basic elements is something that can be taken from Fourier's series theory. Fourier conjectured that you could represent any periodic signal by using sine and cosine functions as an orthonormal basis, that is a summation of sines and cosines could fully reproduce any signal. He wasn't exactly correct, for example, discontinuous signals cannot be reproduced and take a look at the Gibb's ringing on any square wave. However, pretty much any real world signal is going to fall under the criterium for Fourier. The Fourier transform is a more generalized approach to the Fourier series.

    In the real world, yes you cannot do a true Fourier transform because it requires an infinite range of frequency/time samples. However, since we are generally only interested in a finite bandwidth or finite amount of time, what we do is use a window and sample the signal over the window. The requirements on the window are along the lines of what you want to know. For example, the bandwidth and the resolution in the frequency domain will put certain requirements on the size and sampling frequency of your window.
  9. Apr 27, 2009 #8

    Andy Resnick

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    In practical terms, you only need consider the signal for a timescale much longer than the period. Conversely, in order to construct a signal (from sines and cosines) that has a significant change within a duration Dt, one needs sines and cosines with periods much less than Dt.

    If your signal f(t) is 0 everywhere except for a 2-period duration of frequency 'v', the Fourier transform can be found like this:

    f(t) = Rect(2v)*Sin(v)
    F(q) = FourierTransform(f(t)) =Sinc(q*k)#D(q-k)+ Sinc(q*k)#D(q+k) where

    Rect(x) = 'rectangle function' = 0 everywhere except for an interval centered on x, where it is 1.
    Sinc(x) = Sin(x)/x
    D(x) = Delta function at x
    k is a scaling factor times v, k =2*pi/v, IIRC
    # is a convolution operation

    So, as the rectangle function gets narrower and narrower (shorter number of cycles), the Sinc function gets wider and wider; meaning there are more and more frequencies present in the original signal.
  10. Apr 27, 2009 #9
    born2bwire, the sine wave has a single frequency. So it can be called a fundamental wave. I googled after posting and found out. wiki doesn't have this definition.
    Andy, thanks for the explanation.
  11. Apr 27, 2009 #10


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    That doesn't mean that you can reduce all practical signals to a summation of sinusoidal waves. The reason for being able to do that is more complicated.
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