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Arguments of the FT and DTFT

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  1. Nov 22, 2016 #1
    Could someone explain the intuition behind the variables of the FT and DTFT? Do I understand it correctly ?

    For FT being [itex]X(f)[/itex], I understand that [itex]f[/itex] is a possible argument the frequency, as in number of cycles per second.

    FT can be alternatively parameterized by [itex]\omega = 2 \pi f [/itex] which specified the number of cycles in radians, which results in also appending a division by [itex]2 \pi [/itex] to the transform.

    For DTFT, I am told that we use the "digital frequency" [itex]\Omega[/itex] which ranges from 0 to [itex]2\pi[/itex] (or from[itex]-\pi [/itex] to [itex]\pi[/itex]). I vaguely understand that this is because of the periodicity, since [itex]X(\Omega)[/itex] is a periodic expansion of the FT.

    Is that all there is? If so, why do the fourier pairs differ in the case of the FT and the DTFT?
    (For example, see http://www.mechmat.ethz.ch/Lectures/tables.pdf [Broken]

    I have also come accross the DTFT in the form of [itex]X(e^{j\omega})[/itex]. What's that about?
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Nov 23, 2016 #2

    Svein

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    In short: FT is a mathematical operation, DTFT is an algorithm for computers to create something that approximates an FT.
     
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