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DFT of sine function

  1. Sep 16, 2012 #1
    why does the DFT of sin(2∏nKo/N) give two value at k=K0 and k=-K0 though sin has only one frequency component and the DFT is complex , what does it mean? frequency cant have any phase the why it is complex?
    thanks
     
  2. jcsd
  3. Sep 17, 2012 #2

    rbj

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    it means that

    [tex] \sin \left( 2 \pi \frac{K_0}{N} n \right) = \frac{1}{2j}\left( e^{j 2 \pi \frac{K_0}{N} n} - e^{j 2 \pi \frac{-K_0}{N} n} \right) \ [/tex]


    can't quite decode that question.
     
  4. Sep 17, 2012 #3

    rbj

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    i was going to mention that, with the DFT, unless you apply the periodic nature of the results and move the component at [itex]N - K_0[/itex] to a bin you create at [itex] -K_0 [/itex], you won't see the component at [itex]-K_0[/itex]. [strike] the other thing to mention, is that you will see other components if your sine function does not have an exact integer number of cycles in the length [itex]N[/itex] data passed to the DFT. in other words [itex]N = k \times K_0[/itex] where [itex]k[/itex] is an integer. if you don't do that, you will see more than two components. [/strike] the other thing to mention, is that you will see other components if your sine function does not have an exact integer number of cycles in the length [itex]N[/itex] data passed to the DFT. in other words [itex]K_0[/itex] must an integer to see just two components at [itex]K_0[/itex] and at [itex]N - K_0[/itex].
     
    Last edited: Sep 17, 2012
  5. Sep 17, 2012 #4
    The bilateral Fourier transform applied to a sine function would result in a magnitude plot with two peaks like delta functions.

    The Fourier transform is essentially the inner product of a function with both sine and cosine functions for all frequencies. This includes negative frequencies.

    Cosine is unchanged when the sign of the argument is flipped, and sine is anti-symmetric about zero. It may be useful to know that the negative frequency components of a real function have conjugate symmetry; the signs of the cosine parts are the same for both positive and negative frequency components, where as the signs of the sine components are flipped (one is equal to negative one times the other).

    Nonetheless, over all real times and frequencies, the inner product of a sine wave with a +/-sine wave of the same frequency and the inner product of a cosine wave with a +/-cosine wave of the same frequency will be nonzero, where as the inner product of two sine waves of different frequencies (with any relative phase) will be equal to zero. This is why the magnitude plot would have two peaks; most of the inner products are zero, and only for the matching negative and positive frequencies would the inner product not be zero. For the DFT similar reasoning would be applicable.
     
    Last edited: Sep 17, 2012
  6. Sep 17, 2012 #5

    rbj

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    what's the inner product of a sine with a cosine of exactly the same frequency? don't they both have exactly the same negative and positive frequencies?
     
  7. Sep 17, 2012 #6
    Sine and cosine waves of the same (magnitude of) frequency are orthogonal; their inner products are zero. There is an easy way to show the following
    [itex]\int _{- \pi} ^{ \pi} cos(\theta)sin(\theta)d\theta = -\int _{0} ^{ \pi} cos(\theta)sin(\theta)d\theta + \int _{0} ^{ \pi} cos(\theta)sin(\theta)d\theta = 0[/itex]

    As such, it would take two sets of real components to express the result of a Fourier transform in general. This explains why each element of the Fourier transform has a cosine part, also known as a the real part, and a sine part, known as the imaginary part.
     
    Last edited: Sep 18, 2012
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