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Fourier transform of a modified impulse train

  1. Jul 26, 2007 #1
    I need to find the Fourier Transform (FT) of:

    [tex]x(t)=\sum^{\infty}_{n=-\infty}((-1)^{n}\delta(t-nT))[/tex]

    Not really sure how to solve this problem, so any help will be appreciated.

    Also, if you guys know a good reference for non-uniform sampling and reconstruction, please post it.
     
  2. jcsd
  3. Jul 26, 2007 #2

    Note: I think my answer below is correct, but I've never found a way to check my answers when finding the FT.... So, if someone with more experience could verify this, I would appreciate it!


    Notice first that all the impulses at odd values of [itex]n[/itex] are being subtracted, while all the impulses at even values of [itex]n[/itex] are being added. This suggests that we write [itex]x(t)[/itex] as follows:

    [tex]x(t)=\sum^{\infty}_{n=-\infty}\delta(t-2nT) - \sum^{\infty}_{n=-\infty}\delta(t-(2n+1)T) [/tex]

    [tex]=\sum^{\infty}_{n=-\infty}\delta(t-2nT) - \sum^{\infty}_{n=-\infty}\delta(t-2nT-T)[/tex]

    Now let

    [tex]\hat{x}(t) = \sum^{\infty}_{n=-\infty}\delta(t-2nT)[/tex]

    This is a standard impulse train with period 2T. From any table of basic Fourier Transforms:

    [tex]\hat{X}(j\omega) = \frac{2\pi}{2T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{2\pi n}{2T}\right) = \frac{\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{\pi n}{T}\right)[/tex]

    Now notice that [itex]x(t)[/itex], above, can be written as:

    [tex]x(t)=\hat{x}(t)-\hat{x}(t-T)[/itex]

    We already know the fourier transform for [itex]\hat{x}(t)[/itex]; now we just need to use the time shifting property to find the transform for [itex]\hat{x}(t-T)[/itex]:

    [tex]FT \{\hat{x}(t-T)\} = e^{-j\omega T}\hat{X}(j\omega)[/tex]

    Now, by linearity:

    [tex]X(j\omega)=\hat{X}(j\omega)-e^{-j\omega T}\hat{X}(j\omega)
    =\hat{X}(j\omega)[1-e^{-j\omega T}][/tex]

    And since we are dealing with an impulse train, the only values of [itex]\omega[/itex] we have to deal with are those at:

    [tex]\omega=\frac{\pi n}{T}[/tex] (since the impulse will be 0 everywhere else).

    From this we can write:

    [tex]e^{-j\omega T}=e^{-j\pi n }=(-1)^n[/tex]

    So our final answer will be:


    [tex]X(j\omega)=\frac{\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-\frac{\pi n}{T}\right)[1-(-1)^n][/tex]
     
    Last edited: Jul 26, 2007
  4. Jul 27, 2007 #3
    Thanks WolfOfTheSteps, I checked it out with a friend and it seems to be correct.

    Still, if anybody knows a good reference for non-uniform sampling and reconstruction, please post it.
     
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