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DTFT question regarding a pair

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    My book writes the following: using pair for the Discrete Time Fourier Transform:
    [itex]-a^{k}u[-k-1] <---(DTFT)---> \frac{1}{1-ae^{-iw}} for \left | a \right | > 1[/itex]

    2. Relevant equations

    Well, for the simple similar pair such as:

    [itex]a^{k}u[k] <---(DTFT)---> \frac{1}{1-ae^{-iw}} for \left | a \right | < 1[/itex]

    The calculation is pretty straightforward regarding the GP series.
    In the above however, I get lost where they get the result from.

    3. The attempt at a solution

    I've come to here, coming from DTFT definition and simplifying the unitstep boundaries to the sum of:
    [itex]\sum_{1}^{\infty}a^{-k}e^{iwk}[/itex]

    The question is, how do I get from here to the above?

    [itex]a^{k}u[k] <---(DTFT)---> \frac{1}{1-ae^{-iw}} for \left | a \right | < 1[/itex]

    I am thankful for any walk through you can give, since I spent way too much time on this problem. Is there something obvious I don't see about this one, or is the calculation of GP series I am mistaken about?

    Thanks in advance!
     
  2. jcsd
  3. Dec 5, 2013 #2
    Solved

    Finally I understood where I've done wrong. For anyone's interest, here it is:

    [itex]\sum_{1}^{\infty}a^{-k}e^{iw}=-\frac{a^{-1}e^{iw}}{1-a^{-1}e^{iw}}[/itex]

    From here one can solve the thing easily, which gives the correct condition that a>1 as well.

    The thread may be locked now.
     
  4. Dec 5, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    That is wrong unless you have written the sum on the left incorrectly.

    That is, of course, using the formula for the sum of a geometric series:
    [tex]a\sum_{n=0}^\infty r^n= \frac{a}{1- r}[/tex]

    [tex]\sum_{k=1}^\infty a^{-k}e^{ik\omega}= a^{-1}e^{ik\omega}\sum_{k=0}^\infty (a^{-1}e^{k\omega})^k= \frac{a^{-1}e^{i\omega}}{1- a^{-1}e^{i\omega}}[/tex]

    But without the "k" in the exponential, it would be just
    [tex]\sum_{k= 1}^\infty a^{-k}e^{i\omega}= a^{-1}e^{i\omega}\sum_{k=0}^\infty a^{-k}= \frac{a^{-1} e^{i\omega}}{1- a^{-1}}[/tex]
     
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