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Dtft of 1

  1. Jul 2, 2013 #1
    I'm stuck on calculating DTFT of 1.
    DTFT formula is:
    so dtft of 1 is:
    1) in case of w=1, sum becomes:
    doesn't it diverge?
    2) in case of w no 1, how the hell should that sum be calculated?
  2. jcsd
  3. Jul 2, 2013 #2
    [itex]\displaystyle \sum_{n=-\infty}^{\infty} e^{-jn \omega} = \sum_{n=0}^{\infty} e^{-jn \omega} + \sum_{n=-\infty}^{0} e^{-jn \omega} -1 = \sum_{n=0}^{\infty} e^{-jn \omega} + \sum_{n=0}^{\infty} e^{jn \omega} - 1[/itex].

    When the sum converges, each of the last two sums is a geometric series that you should be able to calculate. The intermediate steps may not be neat, but the final answer is very simple.

    However, there are values of [itex]\omega[/itex] for which the series is not convergent. You mention [itex]\omega=1[/itex]. I'm guessing you actually mean [itex]\omega=0[/itex] or [itex]e^{j \omega} = 1[/itex]. But this is not the only one!
    Last edited: Jul 2, 2013
  4. Jul 2, 2013 #3
    Thanks krome.
    yes I meant w = 0, so for w=0 the sum doesnt converge. but I found such formula for DFTF of 1:

    so for w=0 (and other 2πk numbers) I get 2π. did i get something wrong?
    also, for w not equal to 2πk, I get 0, so that sum must converge to 0.
    can you give any hint how should I calculate that sum? should I use eulers formula?
  5. Jul 2, 2013 #4
    You don't actually get [itex]2 \pi[/itex] at [itex]\omega = 0[/itex], you get infinity. You get [itex]2 \pi \delta ( \omega )[/itex] evaluated at [itex]\omega = 0[/itex], which is infinite or not well-defined at least.

    Yes, that is correct. You should get 0 when [itex]\omega \neq 2 \pi k[/itex] for any integer [itex]k[/itex].

    Do you know how to calculate the geometric series [itex]1+x+x^2+x^3+...[/itex]? If you do, then you should be able to calculate the two sums I wrote down. Except you run into a problem if [itex]x= 1[/itex]. That's precisely where you get infinity. Otherwise, you should get [itex]0[/itex].

    So for me, the logic goes as follows:
    (1) [itex]X( \omega ) = 0[/itex] except when [itex]\omega = 2 \pi k[/itex] for some integer [itex]k[/itex]. Therefore, [itex]\displaystyle X( \omega ) = \sum_{k= - \infty}^{\infty} c_k \delta ( \omega - 2 \pi k )[/itex] for some set of constants [itex]c_k[/itex].

    (2) To determine [itex]c_k[/itex], use the inverse DTFT: [itex]\displaystyle x(n) = \frac{1}{2 \pi} \int_{2 \pi k -\pi}^{2 \pi k + \pi} d \omega \, X( \omega ) e^{jn \omega}[/itex]. I've picked a particular region of integration of length [itex]2 \pi[/itex] here. [itex]X( \omega )[/itex] has periodicity [itex]2 \pi[/itex]. You are free to pick any full period over which to perform the integral. Plug in [itex]\displaystyle X( \omega ) = \sum_{m=- \infty}^{\infty} c_m \delta ( \omega - 2 \pi m )[/itex] and interchange the sum and the integral (check out Fubini's or Tonelli's theorems for more rigor regarding such interchanges; this is a mixed case between sums and integrals). You should get [itex]\displaystyle x(n) = \frac{c_k}{2 \pi}[/itex], which shows that [itex]c_k = 2 \pi[/itex] for all [itex]k[/itex] since [itex]x(n) = 1[/itex] for all [itex]n[/itex].
  6. Jul 2, 2013 #5


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    In fact, the series
    $$\sum_{n=-\infty}^{\infty} e^{-in\omega}$$
    diverges for all ##\omega##, because the terms do not converge to zero. Therefore the Fourier transform as defined by the sum does not exist. Your solution involving Dirac delta functions is a solution in the sense of distributions. It takes quite a lot more mathematical horsepower to make it rigorous. The nonrigorous method is to work in reverse: use the integral formula for the inverse Fourier transform to calculate the inverse DTFT of
    $$2\pi \sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$
    and show that the result is identically 1. Then in order to conclude that the DTFT of 1 is the indicated sum of Dirac delta functions, you need to employ the fact (if it is indeed a fact) that the DTFT and inverse DTFT are inverses of each other when working with distributions.
  7. Jul 3, 2013 #6
    I dont quite understand, my book said that [itex]\delta(n)[/itex] is 1 for n = 0 and 0 otherwise. now you're saying that it is infinity? is it dirac delta or kroneker delta? how should I distinguish? (on wikipedia, [itex]\delta[n][/itex] is kroneker and [itex]\delta(w)[/itex] is dirak, so I should distinguish by [] and (), but in my book both are (). My book is Schaum's Outlines of Digital Signal Processing
  8. Jul 3, 2013 #7


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    It is definitely the Dirac delta in this case. Any [] vs. () notation is nonstandard. It is very common to use ##\delta()## for either the Dirac or the Kronecker delta. If there is any chance of confusion, a careful author should indicate which one is intended.
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